NCERT Solutions for Class 12 Maths Chapter 9 Differential Equations – PDF Download

Q: For each of the differential equations in Question 11 to 12, find a particular solution satisfying the given condition: 47. (x3+x2+x+1)dydx=2x2+x

The given D.E is (x3+x2+x+1)dydx=2x2+x ⇒dy=(2x2+xx3+x2+x+1)dx⇒dy=2x2+xx2(x+1)+(x+1)dx=2x2+x(x+1)(x2+1)dx Integrating both sides we get, ∫dy=∫2x2+x(x+1)(x2+1)dx Let, 2x2+x(x+1)(x2+1)=Ax+1+Bx+cx2+1 ⇒2x2+2=A(x2+1)+(Bx+c)(x+1)=Ax2+A+Bx2+Bx+Cx+C=(A+B)x2+(B+C)x2+(A+C) Comparing the co-efficient we get, A+B=2−−−−−−(1)B+C=1−−−−−−(2)A+C=0−−−−−−(3) Subtracting equation (1) – (2), we get A+B−(B+C)=2−1→A−C=1 But from equation (3) A=−C so, we get, A(−C)−C=1→−2C=1→C=−12&A=−(−12)=12 And putting value of A in equation (1), 12+B=2⇒B=2−12=4−12=32 Putting value of A,B and C in 2x2+x(x+1)(x2+1)=12x+1+32x−12x2+1=12(x+1)+32(xx2+1)−12(1x2+1) Hence, the integration becomes ∫dy=∫12(x+1)dx+∫34(2xx2+1)dx−∫12(1x2+1)dx⇒y=12log(x+1)+34log(x2+1)−12tan−1x1+c Given, At x=0,y=1 Then, 1=12log1+34log1−12tan−1(0)+C ⇒1=0+0−0+C{?log1=0tan−10−0}⇒c=1 ∴ The required particular solution is: y=12log(x+1)+34log(x2+1)−12tan−1x+1=14[2log(x+1)+3log(x2+1)]−12tan−1x+1=14[log(x+1)2+log(x2+1)3]−12tan−1x+1=14[log(x+1)2(x2+1)3]−12tan−1x+1

Q: 41. (ex+e−x)dy−(ex−e−x)dx=0

Given, (ex+e−x)dy− (ex−e−x)dx=0 ⇒ (ex+e−x)dy= (ex−e−x)dx⇒dy=ex−e−xex+e−xdx Integrating both sides ⇒∫dy=ex−e−xex+e−xdx {? ∫f| (x)f (x)dx=log|x|} ⇒y=log|ex+e−x|+c is the required general solution.

Updated on Aug 6, 2025 12:15 IST

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Differential Equation Class 12 covers basic concepts of the chapter, general and particular solutions of a differential equation, methods to solve a first-order, first-degree differential equation, formation of differential equations, and applications of differential equations in different areas. It is an important chapter as an in-depth study of differential equations is extremely significant in all modern scientific investigations.
Differential Equations Class 12 NCERT Solutions is an accurate and reliable study material for Class 12 students preparing for the CBSE Board exam and entrance tests like JEE Mains. The student can also download the Differential Equations Class 12 PDF from this page to get well-structured solutions of all the NCERT textbook questions of this chapter. To get access to the NCERT solutions of all the Maths, Physics, and Chemistry of Class 11 and Class 12, check - NCERT Solutions Class 11 and 12.

Table of content

Quick Summary of Differential Equations – Class 12 Maths
Class 12 Math Chapter 9 Differential Equations NCERT Solution PDF
Class 12 Math Chapter 9 Differential Equations: Key Topics, Weightage
Important Formulas of Differential Equation Class 12
Class 12 Differential Equations Exercise-wise Solution
Differential Equations Exercise 9.1 Solutions

Quick Summary of Differential Equations – Class 12 Maths

Here is an overview of the Class 12 Differential Equations:

The chapter states that a differential equation is an equation involving derivatives of the dependent variable with respect to the independent variable.
A polynomial equation in its derivatives is the degree of a differential equation.
The order of the highest order derivative occurring in the differential equation is the order of the differential equation.
Homogeneous differential equation - a differential equation that can be expressed as math ml code are homogenous function of degree zero.

To read the chapter-wise important topics and get free PDFs, check Class 12 Maths NCERT Solutions.

Class 12 Math Chapter 9 Differential Equations NCERT Solution PDF

To get access to the step-by-step solutions of all the NCERT questions of the Differential Equations Class 12, download the PDF from the link given here. The solutions are ideal for CBSE Board exam preparation and other competitive exams like JEE Mains.

Class 12 Math Differential Equations NCERT Solutions: Free PDF Download

Class 12 Math Chapter 9 Differential Equations: Key Topics, Weightage

Differential Equation Class 12 is an important chapter as the concepts are used in all modern scientific investigations. Students should master the concepts to score high in the CBSE Board exam and in other competitive exams. Find below the topics covered in this chapter:

Exercise	Topics Covered
9.1	Introduction
9.2	Basic Concepts
9.3	General and Particular Solutions of a Differential Equation
9.4	Methods of Solving First Order, First Degree Differential Equations

Differential Equation Class 12 Weightage in JEE Mains

Exam	Number of Questions	Weightage
JEE Main	1-2 questions	3.3% to 7%

Important Formulas of Differential Equation Class 12

Differential Equations Important Formulae for CBSE and Competitive Exams

Students can check the important formulae for CBSE and other engineering entrance exams such as JEE Main, CUSAT CAT, and more... below;

General Solution:
$y = \int f(x)dx + C$
Variable Separation:
$\int \frac{1}{g(y)}dy = \int f(x)dx$
Linear Differential Equations:
- General form: $\frac{dy}{dx} + Py = Q$
- Solution form: $y \cdot e^{\int P d x} = \int Q \cdot e^{\int P d x} d x + C$
Homogeneous Equations:
- Substitution: $y = vx$ or $x = vy$
- Simplified separable form: $\frac{dy}{dx} = f\left(\frac{y}{x}\right)$
Exact Differential Equations:
- General form: $M(x, y)dx + N(x, y)dy = 0$
- Solution: $\int Mdx + \int \left(N - \frac{\partial}{\partial y} \left( \int Mdx \right)\right)dy = C$
Exponential Growth/Decay Differential Equations: $\frac{d y}{d t} = k y Solution: y = y_{0} e^{k t}$

Class 12 Differential Equations Exercise-wise Solution

Class 12 Differential Equations is an important chapter, which consists of essential concepts of Calculus. Various exercises of Differential Equations deal with key concepts such as order and degree of a differential equation, formation of differential equations by eliminating arbitrary constants, and methods to solve them, including variable separation, homogeneous equations, and linear differential equations. Shiksha provides exercise-wise solutions with step-by-step explanations, Students can take the help of solutions to master these concepts. Students can check the Exercise-wise Chapter 9 Differential Equation Class 12 math solutions here.

Differential Equations Exercise 9.1 Solutions

Class 12 Differential Equations Exercise 9.1 focuses on Fundamental concepts such as understanding the order and degree of a differential equation, identifying differential equations from given relationships, and forming differential equations by eliminating arbitrary constants from given functions. Differential Equation Exercise 9.1 Solutions consists of 12 Questions. Students can check the complete Exercise 9.1 Solutions below;

Class 12 Chapter 9 Differential Equations Exercise 9.1 NCERT Solution

Determine order and degree (if defined) of differential equations given in Questions 1 to 10:

Q1. $\frac{d^{4} y}{d x^{4}} + s i n (y') = 0$

A.1. The highest order derivation present in the differential equation (D.E.) is $\frac{d^{4} y}{d x^{4}}$ , so its order is 4.

As, the given D.E.is not a polynomial equation in its derivative ,its degree is not defined.

Q2. y' + 5y = 0

A.2. The highest order derivation present in the D.E. is y , so its order is 1.

As the given D.E. is a polynomial equation in its derivative its degree is 1.

Q3.

{(\frac{d s}{d t})}^{4} + 3 s \frac{d^{2} s}{d t^{2}} = 0

A.3. The highest order derivation present in the D.E. is $\frac{d^{2} s}{d t^{2}}$ so its order is 2 .

As the given D.E. is a polynomial equation in its derivative its degree is 1.

Q4.

\frac{d^{2} y}{{(d x^{2})}^{2}} + c o s (\frac{d y}{d x}) = 0

A.4.

\frac{d^{4} y}{d x^{4}}

As the given D.E. is not a polynomial equation in its derivative, its degree is not defined.

Commonly asked questions

For each of the differential equations in Question 11 to 12, find a particular solution satisfying the given condition:

47. $(x^{3} + x^{2} + x + 1) \frac{d y}{d x} = 2 x^{2} + x$

The given D.E is $(x^{3} + x^{2} + x + 1) \frac{d y}{d x} = 2 x^{2} + x$

$\begin{array}{l} \Rightarrow d y = (\frac{2 x^{2} + x}{x^{3} + x^{2} + x + 1}) d x \\ \Rightarrow d y = \frac{2 x^{2} + x}{x^{2} (x + 1) + (x + 1)} d x = \frac{2 x^{2} + x}{(x + 1) (x^{2} + 1)} d x \end{array}$

Integrating both sides we get,

$\int d y = \int \frac{2 x^{2} + x}{(x + 1) (x^{2} + 1)} d x$

Let, $\frac{2 x^{2} + x}{(x + 1) (x^{2} + 1)} = \frac{A}{x + 1} + \frac{B x + c}{x^{2} + 1}$

$\begin{array}{l} \Rightarrow 2 x^{2} + 2 = A (x^{2} + 1) + (B x + c) (x + 1) \\ = A x^{2} + A + B x^{2} + B x + C x + C \\ = (A + B) x^{2} + (B + C) x^{2} + (A + C) \end{array}$

Comparing the co-efficient we get,

$\begin{array}{l} A + B = 2 - - - - - - (1) \\ B + C = 1 - - - - - - (2) \\ A + C = 0 - - - - - - (3) \end{array}$

Subtracting equation (1) – (2), we get

$\begin{array}{l} A + B - (B + C) = 2 - 1 \\ \to A - C = 1 \end{array}$

But from equation (3) $A = - C$ so, we get,

$\begin{array}{l} A (- C) - C = 1 \\ \to - 2 C = 1 \\ \to C = - \frac{1}{2} \\ & A = - (- \frac{1}{2}) = \frac{1}{2} \end{array}$

And putting value of A in equation (1),

$\begin{array}{l} \frac{1}{2} + B = 2 \\ \Rightarrow B = 2 - \frac{1}{2} = \frac{4 - 1}{2} = \frac{3}{2} \end{array}$

Putting value of A,B and C in

$\begin{array}{l} \frac{2 x^{2} + x}{(x + 1) (x^{2} + 1)} = \frac{\frac{1}{2}}{x + 1} + \frac{\frac{3}{2} x - \frac{1}{2}}{x^{2} + 1} \\ = \frac{1}{2 (x + 1)} + \frac{3}{2} (\frac{x}{x^{2} + 1}) - \frac{1}{2} (\frac{1}{x^{2} + 1}) \end{array}$

Hence, the integration becomes

$\begin{array}{l} \int d y = \int \frac{1}{2 (x + 1)} d x + \int \frac{3}{4} (\frac{2 x}{x^{2} + 1}) d x - \int \frac{1}{2} (\frac{1}{x^{2} + 1}) d x \\ \Rightarrow y = \frac{1}{2} l o g (x + 1) + \frac{3}{4} l o g (x^{2} + 1) - \frac{1}{2} t a n^{- 1} \frac{x}{1} + c \end{array}$

Given, At $x = 0, y = 1$

Then, $1 = \frac{1}{2} l o g 1 + \frac{3}{4} l o g 1 - \frac{1}{2} t a n^{- 1} (0) + C$

$\begin{array}{l} \Rightarrow 1 = 0 + 0 - 0 + C {\begin{cases} ? l o g 1 = 0 \\ t a n^{- 1} 0 - 0 \end{cases}} \\ \Rightarrow c = 1 \end{array}$

$∴$ The required particular solution is:

$\begin{array}{l} y = \frac{1}{2} l o g (x + 1) + \frac{3}{4} l o g (x^{2} + 1) - \frac{1}{2} t a n^{- 1} x + 1 \\ = \frac{1}{4} [2 l o g (x + 1) + 3 l o g (x^{2} + 1)] - \frac{1}{2} t a n^{- 1} x + 1 \\ = \frac{1}{4} [l o g {(x + 1)}^{2} + l o g {(x^{2} + 1)}^{3}] - \frac{1}{2} t a n^{- 1} x + 1 \\ = \frac{1}{4} [l o g {(x + 1)}^{2} {(x^{2} + 1)}^{3}] - \frac{1}{2} t a n^{- 1} x + 1 \end{array}$

41. $(e^{x} + e^{- x}) d y - (e^{x} - e^{- x}) d x = 0$

Given, $(e^{x} + e^{- x}) d y - (e^{x} - e^{- x}) d x = 0$

$\begin{array}{l} \Rightarrow (e^{x} + e^{- x}) d y = (e^{x} - e^{- x}) d x \\ \Rightarrow d y = \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} d x \end{array}$

Integrating both sides

$\Rightarrow \int d y = \frac{e^{x} - e^{- x}}{e^{x} + e^{- x}} d x {? \int \frac{f^{|} (x)}{f (x)} d x = l o g | x |}$

$\Rightarrow y = l o g | e^{x} + e^{- x} | + c$ is the required general solution.

58. In a culture the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present.

Let ‘x’ be the number of bacteria present in instantaneous time t.

Then, $\frac{d x}{d t} \propto x$

$\Rightarrow \frac{d x}{d t} = k x, w h e r e, k =$ constant of proportionality.

$\Rightarrow \frac{d x}{x} = k d t$

Integrating both sides,

$\begin{array}{l} \int \frac{d x}{x} = \int k d t \\ \Rightarrow l o g x = k t + c \end{array}$

Given, at $t = 0, x = x_{0} (s a y) t h e n,$

$l o g x_{0} = c (I n i t i a l, x_{0} = 100000)$

So, the differential equation is

$\begin{array}{l} l o g x = k t + l o g x_{0} \\ \Rightarrow l o g x - l o g x_{0} = k t \\ \Rightarrow l o g \frac{x}{x_{0}} = k t \end{array}$

As the bacteria number increased by 10% in 2 hours.

The number of bacteria increased in 2hours $= 10 % \times 100000 = 10000$

Hence, at t=2,

$x = 100000 + 10000 = 110000$

So, $\begin{array}{l} l o g \frac{110000}{100000} = 2 k \\ \Rightarrow k = \frac{1}{2} l o g (\frac{11}{10}) \end{array}$

Hence, $l o g \frac{x}{x_{0}} = [\frac{1}{2} l o g \frac{11}{10}] \times t$

$w h e n, x = 200000,$ then we get,

$l o g \frac{200000}{100000} = \frac{1}{2} l o g \frac{11}{10} \times t$

$\begin{array}{l} \Rightarrow 2 l o g 2 = l o g (\frac{11}{10}) \times t \\ \Rightarrow t = \frac{2 l o g 2}{l o g (\frac{11}{10})} h o u r s \end{array}$

Determine order and degree (if defined) of differential equations given in Questions 1 to 10:

1. $\frac{d^{4} y}{d x^{4}} + s i n (y') = 0$

The highest order derivation present in the differential equation (D.E.) is $\frac{d^{4} y}{d x^{4}}$ , so its order is 4.

As, the given D.E.is not a polynomial equation in its derivative, its degree is not defined.

22. Kindly Consider the following

Kindly go through the solution

37. $\frac{d y}{d x} = \frac{1 - c o s x}{1 + c o s x}$

The given D.E is

$\frac{d y}{d x} = \frac{1 - c o s x}{1 + c o s x}$

By separable of variable,

$\begin{array}{l} \Rightarrow d y = \frac{1 - c o s x}{1 + c o s x} d x {\begin{cases} c o s 2 x = 1 - 2 s i n^{2} x \\ = 2 s i n^{2} x = 1 - c o s 2 x \\ = 2 s i n^{2} \frac{x}{2} = 1 - c o s x \\ c o s 2 x = 2 c o s^{2} x - 1 \end{cases}} \\ \Rightarrow d y = \frac{2 s i n^{2} \frac{x}{2}}{2 c o s^{2} \frac{x}{2}} d x \\ \Rightarrow d y = t a n^{2} \frac{x}{2} d x \end{array}$

Integrating both sides,

$\begin{array}{l} \int d y = \int t a n^{2} \frac{x}{2} d x {s e c^{2} x = 1 + t a n^{2}} \\ \Rightarrow y = \int (s e c^{2} \frac{x}{2} - 1) d x \end{array}$

$\Rightarrow y = \frac{t a n \frac{x}{2}}{\frac{1}{2}} - x + c$ c = constant

$\Rightarrow y = 2 t a n \frac{x}{2} - x + c$ is the general solution.

39. $\frac{d y}{d x} + y = 1 (y \neq 1)$

Given, $\frac{d y}{d x} + y = 1$

$\Rightarrow \frac{d y}{d x} = 1 - y = - (y - 1)$

By separable of variable,

$\frac{d y}{(y - 1)} = - d x$

Integrating both sides,

$\begin{array}{l} \int \frac{d y}{(y - 1)} = - \int d x \\ \Rightarrow l o g | y - 1 | = - x + c \\ \Rightarrow | y - 1 | = e^{- x + c} \\ \Rightarrow y - 1 = \pm e^{- x} . e^{c} \end{array}$

$\Rightarrow y = 1 + A c$ where $A = \pm e^{c}$

Is the general solution.

42. $\frac{d y}{d x} = (1 + x^{2}) (1 + y^{2})$

Given, $\frac{d y}{d x} = (1 + x^{2}) (1 + y^{2})$

$\Rightarrow \frac{d y}{(1 + y^{2})} = (1 + x^{2}) d x$

Integrating both sides

$\begin{array}{l} \int \frac{d y}{(1 + y^{2})} d y = \int (x^{2} + 1) d x \\ \Rightarrow t a n^{- 1} \frac{y}{1} = \frac{x^{3}}{3} + x + c \end{array}$

$\Rightarrow t a n^{- 1} y = \frac{x^{3}}{3} + x + c$ is the general solution.

65. Kindly Consider the following

Kindly go through the solution

Integrating both sides,

18. Kindly Consider the following

Given, $y = x s i n x$

So, $y^{|} = x \frac{d}{d x} s i n x + s i n x \frac{d x}{d x} = x c o s x + s i n x$

Now, L.H.S of the given D.E $= x y^{|}$

$= x (x c o s x + s i n x)$

$= x^{2} c o s x + x s i n x$

Kindly Consider the following

25. $\frac{x}{a} + \frac{y}{b} = 1$

Given: Equation of the family of curves $\frac{x}{a} + \frac{y}{b} = 1 . . . . . . . . . . (i)$

Differentiating both sides of the given equation with respect to x, we get:

$\begin{array}{l} \frac{1}{a} + \frac{1}{b} \frac{d y}{d x} = 0 \\ \Rightarrow \frac{1}{a} + \frac{1}{b} y^{'} = 0 \end{array}$

Again, differentiating both sides with respect to x, we get:

$\begin{array}{l} 0 + \frac{1}{b} y^{"} = 0 \\ \Rightarrow \frac{1}{b} y^{"} = 0 \\ \Rightarrow y^{"} = 0 \end{array}$

Hence, the required differential equation of the given curve is $y^{"} = 0$

21. x + y = tan^-1 y : y²y' + y²+ 1 = 0

Given, $x + y = t a n^{- 1} y$

Differentiate with ‘x’ we get

$\begin{array}{l} 1 + \frac{d y}{d x} = \frac{1}{1 + y^{2}} \frac{d y}{d x} \\ = 1 + y^{|} = \frac{1}{1 + y^{2}} y^{|} \\ = (1 + y^{|}) (1 + y^{2}) = y^{|} \\ = 1 + y^{2} y^{|} + y^{|} + y^{2} = y^{|} \\ = y^{2} y^{|} + y^{2} + 1 = 0 \end{array}$

$∴$ The given $f x^{n}$ is a solution of the given D.E

40. $s e c^{2} x t a n y d x + s e c^{2} y t a n x d y = 0$

Given, $s e c^{2} x t a n y d x + s e c^{2} y t a n x d y = 0$

Dividing throughout by ‘ $t a n x t a n y$ ’ we get,

$\begin{array}{l} \frac{s e c^{2} x t a n y}{t a n x t a n y} d x + \frac{s e c^{2} y t a n x}{t a n x t a n y} d y = 0 \\ \Rightarrow \frac{s e c^{2} x}{t a n x} d x + \frac{s e c^{2} y}{t a n y} d y = 0 \end{array}$

Integrating both sides we get,

$\begin{array}{l} \int \frac{s e c^{2} x}{t a n x} d x + \int \frac{s e c^{2} y}{t a n y} d y = l o g c \\ \Rightarrow l o g | t a n x | + l o g | t a n y | = l o g c {\int \frac{f^{|} (x)}{f (x)} d x - l o g | f (x) |} \\ \Rightarrow l o g | (t a n x + t a n y) | = l o g c \end{array}$

$\Rightarrow t a n x t a n y = \pm c$ is the required general solution.

43. $y l o g y d x - x d y = 0$

Given, $y l o g y d x - x d y = 0$

$\begin{array}{l} \Rightarrow y l o g y d x = x d y \\ \Rightarrow \frac{d y}{y l o g y} = \frac{d x}{x} \end{array}$

Integration both sides,

$\int \frac{d y}{y l o g y} = \int \frac{d x}{x}$

Put log $y = t \Rightarrow \frac{1}{y} = \frac{d t}{d y} \Rightarrow \frac{d y}{y} = d t$

Hence, $\int \frac{d t}{t} = \int \frac{d x}{x}$

$\begin{array}{l} \Rightarrow l o g | t | = l o g | x | + l o g | c | = l o g | x c | \\ \Rightarrow t = \pm x c \end{array}$

$\Rightarrow l o g y = a x$ where $a = \pm c$

$\Rightarrow y = e^{a x}$ is the general solution.

44. $x^{5} \frac{d y}{d x} = - y^{5}$

Given, $x^{5} \frac{d y}{d x} = - y^{5}$

$\Rightarrow \frac{d y}{y^{5}} = - \frac{d x}{x^{5}}$

Integrating both sides

$\begin{array}{l} \int \frac{d y}{y^{5}} = - \int \frac{d x}{x^{5}} \\ \Rightarrow \int y^{- 5} d y = - \int x^{- 5} d x \end{array}$

$\begin{array}{l} \Rightarrow \frac{y^{- 5 + 1}}{(- 5 + 1)} = - \frac{x^{- 5 + 1}}{(- 5 + 1)} + c \\ \Rightarrow \frac{1}{- 4 y^{4}} = \frac{1}{4 x^{4}} + c \end{array}$

$\Rightarrow \frac{1}{y^{4}} = \frac{1}{x^{4}} + 4 c$ is the general solution.

45. $\frac{d y}{d x} = s i n^{- 1} x$

Given, $\frac{d y}{d x} = s i n^{- 1} x$

$\Rightarrow d y = s i n^{- 1} x d x$

Integrating

46. $e^{x} t a n y d x + (1 - e^{x}) s e c^{2} y d y = 0$

Given, $e^{x} t a n y d x + (1 - e^{x}) s e c^{2} y d y = 0$

Dividing throughout by $(1 - e^{x}) t a n y$ we get,

$\begin{array}{l} \frac{e^{x} t a n y}{(1 - e^{x}) t a n y} d x + \frac{(1 - e^{x}) s e c^{2} y}{(1 - e^{x}) t a n y} d y = 0 \\ = \frac{e^{x}}{1 - e^{x}} d x + \frac{s e c^{2} y}{t a n y} d y = 0 \end{array}$

Integrating both sides

$\begin{array}{l} = \int \frac{- e^{x}}{1 - e^{x}} d x + \int \frac{s e c^{2} y}{t a n y} d y = c l o g c \\ = - l o g | 1 - e^{x} | + l o g | t a n y | = c l o g c \\ = l o g \frac{t a n y}{1 - e^{x}} = l o g c \\ = \frac{t a n y}{1 - e^{x}} = c \end{array}$

$= t a n y = (1 - e^{x}) c$ is the general solution.

53 Find the equation of the curve passing through the point (0,-2) given that at any point (x,y) on the curve the product of the slope of its tangent and y-coordinate of the point is equal to the x-coordinate of the point.

The slope of the tangent to then curve is $\frac{d y}{d x}$

$\begin{array}{l} \frac{d y}{d x} . y = x \\ \Rightarrow y . d y = x d x \end{array}$

So,

Integrating both sides,

$\begin{array}{l} \int y . d y = \int x d x \\ \Rightarrow \int \frac{y^{2}}{2} = \frac{x^{2}}{2} + c \\ \Rightarrow y^{2} = x^{2} + A, W h e r e, A = 2 c \end{array}$

As the curve passes through (0, -2) we have,

$\begin{array}{l} {(- 2)}^{2} = 0^{2} + A \\ \Rightarrow A = 4 \end{array}$

$∴$ The equation of the curve is

$y^{2} = x^{2} + 4$

56. In a bank, principal increases continuously at the rate of r% per year. Find the value of r if Rs 100 doubles itself in 10 years (loge 2 = 0.6931).

Let P, r and t be the principal rate and time respectively.

Then, increase in principal $\frac{d P}{d t} = P \times r %$

$\begin{array}{l} \Rightarrow \frac{d P}{d t} = P . \frac{r}{100} \\ \Rightarrow \frac{d P}{P} = \frac{r}{100} d t \end{array}$

Integrating both sides,

$\begin{array}{l} \int \frac{d P}{P} = \int \frac{r}{100} d t \\ \Rightarrow l o g P = \frac{r t}{100} + c \\ \Rightarrow P = e^{\frac{r t}{100} + c} \end{array}$

Given at t=0,P=100

So, $100 = e^{\frac{r \times 0}{100} + c}$

$\begin{array}{l} \Rightarrow 100 = e^{0} \times e^{c} \\ \Rightarrow e^{c} = 100 (? e^{0} = 1) \end{array}$

And at $t = 10, P = 2 \times 100 = 200$

So, $200 = e^{\frac{r 10}{100} + c}$

$\begin{array}{l} \Rightarrow 200 = e^{\frac{r}{10}} . e^{e} \\ \Rightarrow e^{\frac{r}{10}} = \frac{200}{100} = 2 \end{array}$

$\begin{array}{l} \Rightarrow \frac{r}{10} = l o g 2 \\ \Rightarrow \frac{r}{10} = 0.6931 \\ \Rightarrow r = 6.931 \end{array}$

Hence, the rate is 6.931%

57. In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years

(e^0.5 = 1.645).

Let P and t the principal and time respectively.

Then, increase in principal $\frac{d P}{d t} = P \times 5 %$

$\Rightarrow \frac{d P}{P} = \frac{5}{100} d t$

Integrating both sides,

$\begin{array}{l} \Rightarrow \int \frac{d P}{P} = \int \frac{1}{20} d t \\ \Rightarrow l o g P = \frac{t}{20} + c \\ \Rightarrow P = e^{\frac{t}{20} + c} \end{array}$

At, t=0, P=1000

So, $\begin{array}{l} 1000 = e^{\frac{0}{20} + c} \\ \Rightarrow e^{c} = 1000 \end{array}$

And at t=10,

$\begin{array}{l} P = e^{\frac{10}{100} + c} = e^{0.5} . e^{c} \\ \Rightarrow P = 1.648 \times 1000 = 1648 \end{array}$

P = ?1648

62. $(x - y) d y - (x + y) d x = 0$

The Given D.E. is $(x - y) d y - (x + y) d x = 0$

$\begin{array}{l} \Rightarrow (x - y) d y = (x + y) d x \\ \Rightarrow \frac{d y}{d x} = \frac{x + y}{x - y} = \frac{x (1 + \frac{y}{x})}{x (1 - \frac{y}{x})} = \frac{1 + \frac{y}{x}}{1 - \frac{y}{x}} = f (\frac{y}{x}) \end{array}$

Hence, the given D.E. is homogenous.

Let, $y = v x = \frac{y}{x} = v, s o, \frac{d y}{d x} = v + x \frac{d v}{d x}$ in the D.E

Then, $\begin{array}{l} v + x \frac{d v}{d x} = \frac{1 + v}{1 - v} \\ \Rightarrow x \frac{d v}{d x} = \frac{1 + v}{1 - v} - v = \frac{1 + v - 1 + v^{2}}{1 - v} = \frac{1 + v^{2}}{1 - v} \\ \Rightarrow [\frac{1 - v}{1 + v^{2}}] d v = \frac{d x}{x} \end{array}$

Integrating both sides,

$\begin{array}{l} \int \frac{1 - v}{1 + v^{2}} d v = \int \frac{d x}{x} \\ \Rightarrow \int \frac{1}{1 + v^{2}} d v - \frac{1}{2} \int \frac{2 v}{1 + v^{2}} d v = l o g x + c \\ \Rightarrow t a n^{- 1} v - \frac{1}{2} l o g (1 + v^{2}) = l o g x + c \end{array}$

Putting back $v = \frac{y}{x}, w e g e t,$

$\begin{array}{l} = t a n^{- 1} \frac{y}{x} - \frac{1}{2} l o g | 1 + \frac{y^{2}}{x^{2}} | = l o g x + c \\ = t a n^{- 1} \frac{y}{x} - \frac{1}{2} l o g | \frac{x^{2} + y^{2}}{x^{2}} | = l o g x + c \\ = t a n^{- 1} \frac{y}{x} - \frac{1}{2} [l o g (x^{2} + y^{2}) - l o g x^{2}] = l o g x + c \end{array}$

$\begin{array}{l} = t a n^{- 1} \frac{y}{x} - \frac{1}{2} l o g (x^{2} + y^{2}) + \frac{1}{2} l o g x^{2} = l o g x + c \\ = t a n^{- 1} \frac{y}{x} - \frac{1}{2} l o g (x^{2} + y^{2}) + l o g {(x^{2})}^{\frac{1}{2}} = l o g x + c \\ = t a n^{- 1} \frac{y}{x} - \frac{1}{2} l o g (x^{2} + y^{2}) + l o g x = l o g + c \\ = t a n^{- 1} \frac{y}{x} = \frac{1}{2} l o g (x^{2} + y^{2}) + c \end{array}$

38. Kindly Consider the following

Kindly go through the solution

54. At any point (x, y) of a curve, the slope of the tangent is twice the slope of the line segment joining the point of contact to the point (–4, –3). Find the equation of the curve given that it passes through (–2, 1).

The slope of tangent is $\frac{d y}{d x}$ and slope of line joining line (-4,-3) and point say P(x,y)

$\frac{y - (- 3)}{x - (- 4)} = \frac{y + 3}{x + 4}$

So, $\frac{d y}{d x} = 2 (\frac{y + 3}{x + 4})$

$\Rightarrow \frac{d y}{y + 3} = \frac{2}{x + 4} d x$

Integrating both sides,

$\begin{array}{l} \int \frac{d y}{y + 3} = \int \frac{2}{x + 4} d x \\ \Rightarrow l o g | y + 3 | = 2 l o g | x + 4 | + l o g | c | \\ \Rightarrow l o g | y + 3 | = l o g {(x + 4)}^{2} + l o g | c | \\ \Rightarrow l o g | y + 3 | = l o g | c {(x + 4)}^{2} | \\ \Rightarrow y + 3 = c_{1} {(x + 4)}^{2}, w h e r e, c_{1} = \pm c \end{array}$

Since, the curve passes through (-2,1) we get,

$\begin{array}{l} y = 1, a t, x = - 2 \\ \Rightarrow 1 + 3 = c {(- 2 + 4)}^{2} \\ \Rightarrow 4 = c \times 4 \\ \Rightarrow c = 1 \end{array}$

$∴$ The equation of the curve is $y + 3 = {(x + 4)}^{2}$

55. The volume of the spherical balloon being inflated changes at a constant rate. If initially its radius is 3 units and after 3 seconds it is 6 units. Find the radius of balloon after t seconds.

Let ‘r’ and U be the radius and volume of the spherical balloon.

Then, $\frac{d U}{d t} = k,$ k = constant

$\begin{array}{l} \frac{d}{d t} (\frac{4}{3} π r^{3}) = k \\ \Rightarrow 4 π r^{2} \frac{d r}{d t} = k \\ \Rightarrow 4 π r^{2} d r = k d t \end{array}$

Integrating both sides,

$\begin{array}{l} \int 4 π r^{2} d r = \int k d t \\ \Rightarrow \frac{4}{3} π r^{3} = k t + c \end{array}$

Given at t = 0, r = 3

So, 4π(3)³ = c

C = 36π

And, at t=3, r=6

So, $\frac{4}{3} π {(6)}^{3} = 3 k + 36 π (c = 36 π)$

$\begin{array}{l} \Rightarrow 288 π - 36 π = 3 k \\ \Rightarrow k = \frac{252 π}{3} = 84 π \end{array}$

Hence, putting value of c and k in,

$\frac{4}{3} π r^{3} = k t + c$ , we get,

$\begin{array}{l} \frac{4}{3} π r^{3} = 84 π . t + 36 π \\ \Rightarrow r^{3} = \frac{3}{4 π} (84 π . t + 36 π) \\ \Rightarrow r^{3} = 63 t + 27 \\ \Rightarrow r = {[63 t + 27]}^{\frac{1}{3}} \end{array}$

59. The general solution of the differential equation $\frac{d y}{d x} = e^{x + y}$ is:

(A) e^x + e^-y = C

(B) e^x + e^y = C

(C) e^-x + e^y = C

(D) e^x + e^-y = C

The given D.E. is

$\begin{array}{l} \frac{d y}{d x} = e^{x + y} \\ \Rightarrow \frac{d y}{d x} = e^{x} . e^{y} \\ \Rightarrow \frac{d y}{e^{y}} = e^{x} d x \end{array}$

Integrating both sides,

$\begin{array}{l} \int \frac{d y}{e^{y}} = \int e^{x} d x \\ \Rightarrow \frac{e^{- y}}{- 1} = e^{x} + c_{1} \\ \Rightarrow e^{- y} = - e^{x} - c_{1} \\ \Rightarrow e^{- y} + e^{x} = c, w h e r e, c = - c_{1} \end{array}$

$∴$ Option (A) is correct.

60. $(x^{2} + x y) d y = (x^{2} + y^{2}) d x$

The given D.E. is

$\begin{array}{l} \frac{d y}{d x} = \frac{x^{2} + y^{2}}{x^{2} + x + y} = \frac{x^{2} (1 + \frac{y^{2}}{x^{2}})}{x^{2} (1 + \frac{y}{x})} = F (x, y) \\ T h e n, F (λ x, λ y) = \frac{λ^{2} x^{2}}{λ^{2} x^{2}} [\frac{1 + \frac{λ^{2} y^{2}}{λ^{2} x^{2}}}{1 + \frac{λ y}{λ x}}] = λ^{0} F (x, y) \\ = λ^{2} F (x, y) \end{array}$

Hence, $F (x, y)$ is a homogenous $f x^{n}$ of degree 2.

To solve it, let

$\begin{array}{l} y = v x, s o t h a t, \frac{d y}{d x} = v . \frac{d x}{d x} + x \frac{d v}{d x} \\ v = \frac{y}{x} \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x} \end{array}$

The D.E. now becomes,

$\begin{array}{l} v + x \frac{d v}{d x} = \frac{1 + v^{2}}{1 + v} \\ \Rightarrow x \frac{d v}{d x} = \frac{1 + v^{2}}{1 + v} - v = \frac{1 + v^{2} - v - v^{2}}{1 + v} = \frac{1 - v}{1 + v} \\ \Rightarrow (\frac{1 + v}{1 - v}) d v = \frac{d x}{x} \end{array}$

Integrating both sides,

$\begin{array}{l} \int \frac{1 + v}{1 - v} d v = \int \frac{d x}{x} \\ \Rightarrow \int \frac{1 + v}{1 - v} d v = l o g x + c \\ \Rightarrow \int \frac{2 - (1 - v)}{1 - v} d v = l o g x + c \\ \Rightarrow \int \frac{2}{1 - v} d v - \int 1 d v = l o g x + c \\ \Rightarrow \frac{2 l o g | 1 - v |}{- 1} - v = l o g x + c \\ \Rightarrow l o g {(1 - v)}^{2} + v = - l o g x - c \end{array}$

Put $v = \frac{y}{x},$

$\begin{array}{l} l o g {(1 - \frac{y}{x})}^{2} + \frac{y}{2} = - l o g x + c \\ \Rightarrow l o g {(\frac{x - y}{x})}^{2} + l o g x = - \frac{y}{x} - c \end{array}$

$\begin{array}{l} \Rightarrow l o g [{(\frac{x - y}{x})}^{2} \times x] = - \frac{y}{x} - c \\ \Rightarrow \frac{x - y}{x} = e^{- \frac{y}{x} - c} = c_{1} e^{- \frac{y}{x} - c}, w h e r e, c_{1} e \end{array}$

$\Rightarrow {(x - y)}^{2} = c_{1} . x e^{- \frac{y}{x}}$ is the required solution of the D.E.

61. Kindly Consider the following

$y' = \frac{x + y}{x}$

The given D.E. is

$\begin{array}{l} y^{1} = \frac{x + y}{x} \\ = \frac{d y}{d x} = 1 + \frac{y}{x} = f (\frac{y}{x}) \end{array}$

Hence, the given D.E is homogenous.

Let, $y = v x \to \frac{y}{x} = v, s o t h a t, \frac{d y}{d x} = v + \frac{x d v}{d x}$

So, the D.E. becomes

$\begin{array}{l} v + \frac{x d v}{d x} = 1 + v \\ \Rightarrow \frac{x d v}{d x} = 1 \\ \Rightarrow d v = \frac{d x}{x} \end{array}$

Integrating both sides,

$\begin{array}{l} \int d v = \int \frac{d x}{x} \\ v = l o g | x | + c \end{array}$

Putting $v = \frac{y}{x}$ back we get,

$\frac{y}{x} = l o g | x | + c$

63. $(x^{2} - y^{2}) d x + 2 x y d y = 0$

The Given D.E. is

$\begin{array}{l} (x^{2} - y^{2}) d x + 2 x y d y = 0 \\ \Rightarrow 2 \times y d y = - (x^{2} - y^{2}) d x \\ = \frac{d y}{d x} = - \frac{(x^{2} - y^{2})}{2 x y} = \frac{(y^{2} - x^{2})}{2 x y} \\ = \frac{1}{2} \frac{(\frac{y^{2} - x^{2}}{x^{2}})}{(\frac{x y}{x^{2}})} \\ = \frac{1}{2} \frac{[{(\frac{y}{x})}^{2} - 1]}{(\frac{y}{x})} = f (\frac{y}{x}) \end{array}$

Hence, the given D.E. is homogenous.

Let, $y = v x \Rightarrow \frac{y}{x} = v, s o t h a t, \frac{d y}{d x} = v + x \frac{d v}{d x}$ in the D.E

$\begin{array}{l} \Rightarrow v + x \frac{d v}{d x} = \frac{1}{2} [\frac{v^{2} - 1}{v}] \\ \Rightarrow x \frac{d v}{d x} = \frac{v^{2} - 1}{2 v} - v = \frac{v^{2} - 1 - 2 v^{2}}{2 v} = \frac{- 1 - v^{2}}{2 v} \\ \Rightarrow (\frac{2 v}{1 + v^{2}}) d v = - \frac{d x}{x} \end{array}$

Integrating both sides we get,

Putting back $v = \frac{y}{x}$ we get,

$\begin{array}{l} \Rightarrow x [\frac{y^{2}}{x^{2}} + 1] = c \\ \Rightarrow \frac{y^{2} + x^{2}}{x} = c \end{array}$

$\Rightarrow y^{2} + x^{2} = x c$ is the required solution.

64. $x^{2} \frac{d y}{d x} = x^{2} - 2 y^{2} + x y$

The given D.E. is

$\begin{array}{l} x^{2} \frac{d y}{d x} = x^{2} - 2 y^{2} + + x y \\ \Rightarrow \frac{d y}{d x} = \frac{x^{2} - 2 y^{2} + + x y}{x^{2}} = 1 - \frac{2 y^{2}}{x^{2}} + \frac{y}{x} \\ \Rightarrow \frac{d y}{d x} = 1 - 2 {(\frac{y}{x})}^{2} + \frac{y}{x} = f (\frac{y}{x}) \end{array}$

Hence, the D.E. is homogenous $f x^{n}$

Let, $y = v x . = \frac{y}{x} = v$ so that, $\frac{d y}{d x} = v + x \frac{d v}{d x}$ is the D.E.

Thus, $v + x \frac{d v}{d x} = 1 - 2 v^{2} + v$

$\begin{array}{l} \Rightarrow x \frac{d v}{d x} = 1 - 2 v^{2} \\ \Rightarrow \frac{d v}{1 - 2 v^{2}} = \frac{d x}{x} \end{array}$

Integrating both sides,

66. ${x c o s (\frac{y}{x}) + y s i n (\frac{y}{x})} y d x = {y s i n (\frac{y}{x}) + x c o s (\frac{y}{x})} x d y$

The given D.E is

. $\begin{array}{l} {x c o s (\frac{y}{x}) + y s i n (\frac{y}{x})} y d x = {y s i n (\frac{y}{x}) - x c o s (\frac{y}{x})} x d y \\ \Rightarrow \frac{d y}{d x} = \frac{{x c o s \frac{y}{x} + y s i n \frac{y}{x}} y}{{y s i n \frac{y}{x} - x c o s \frac{y}{x}} x} = \frac{x y c o s \frac{y}{x} + y^{2} s i n \frac{y}{x}}{x y s i n \frac{y}{x} - x^{2} c o s \frac{y}{x}} \end{array}$

$\Rightarrow \frac{\frac{y}{x} c o s \frac{y}{x} + {(\frac{y}{x})}^{2} - s i n \frac{y}{x}}{(\frac{y}{x}) s i n \frac{y}{x} - c o s \frac{y}{x}}$ {Dividing numerator and denominator by $x^{2}$ }

$= f (\frac{y}{x})$

Hence, the given D.E is homogenous.

Let, $y = v x = \frac{y}{x} = v$ so that $\frac{d y}{d x} = v + x \frac{d v}{d x}$ in the D.E.

Then, $v + x \frac{d v}{d x} = \frac{v c o s v + v^{2} s i n v}{v s i n v - c o s v}$

$\begin{array}{l} \Rightarrow x \frac{d v}{d x} = \frac{v c o s v + v^{2} s i n v}{v s i n v - c o s v} - v = \frac{v c o s v + v^{2} s i n v - v^{2} s i n v + v c o s v}{v s i n v - c o s v} \\ \Rightarrow \frac{v c o s v}{v s i n v - c o s v} \\ \Rightarrow (\frac{v s i n v - c o s v}{v c o s v}) d v = \frac{2 d x}{x} \end{array}$

Integrating both sides,

$\begin{array}{l} \int \frac{v s i n v - c o s v}{v c o s v} d v = 2 \int \frac{d x}{x} \\ \Rightarrow \int t a n v d v - \int \frac{1}{v} d v = 2 l o g | x | + l o g | c | \\ \Rightarrow l o g | s e c v | - l o g | v | = l o g x^{2} + l o g c \\ \Rightarrow l o g | \frac{s e c v}{v} | = l o g c x^{2} \\ \Rightarrow \frac{s e c v}{v} = c x^{2} \\ \Rightarrow s e c v = c x^{2} v \end{array}$

Putting back $\Rightarrow v = \frac{y}{x} = c x^{2} \frac{y}{x} = c x y$

$\begin{array}{l} s e c \frac{y}{x} = c x^{2} \frac{y}{x} = c x y \\ \Rightarrow \frac{1}{c o s \frac{y}{x}} = c x y \\ \Rightarrow \frac{1}{c} = x y c o s \frac{y}{x} \end{array}$

$\Rightarrow x y c o s \frac{y}{x} = c_{1}$ where $c_{1} = \frac{1}{c}$

67. $x \frac{d y}{d x} - y = x s i n (\frac{y}{x}) = 0$

The given D.E. is $\frac{x d y}{d x} - y + x s i n (\frac{y}{x}) = 0$

$\begin{array}{l} \Rightarrow \frac{x d y}{d x} = y - x s i n (\frac{y}{x}) \\ \Rightarrow \frac{d y}{d x} = \frac{y - x s i n (\frac{y}{x})}{x} = \frac{y}{x} - s i n \frac{y}{x} = f (\frac{y}{x}) \end{array}$

Hence, the given D.E. is homogenous.

Let, $y = v x = \frac{y}{x} = v$ so that $\frac{d y}{d x} = v + x \frac{d v}{d x}$ in the D.E.

Then, $v + x \frac{d v}{d x} = v - s i n v$

$\begin{array}{l} \Rightarrow x \frac{d v}{d x} = - s i n v \\ \Rightarrow \frac{d v}{s i n v} = - \frac{d x}{x} \\ \Rightarrow c o s e c v d v = - \frac{d x}{x} \end{array}$

Integrating both sides we get,

$\begin{array}{l} \int c o s e c v d v = \int - \frac{d x}{x} \\ \Rightarrow l o g | c o s e c v - c o t v | = - l o g x + l o g c \\ \Rightarrow l o g | c o s e c v - c o t v | = l o g \frac{c}{x} \\ \Rightarrow c o s e c v - c o t v = \frac{c}{x} \end{array}$

Putting back $v = \frac{y}{x}$ we get,

$\begin{array}{l} c o s e c \frac{y}{x} - c o t \frac{y}{x} = \frac{c}{x} \\ \Rightarrow \frac{1}{s i n \frac{y}{x}} - \frac{c o s \frac{y}{x}}{s i n \frac{y}{x}} = \frac{c}{x} \end{array}$

$\Rightarrow x [1 - c o s \frac{y}{x}] = c s i n (\frac{y}{x})$ is the solution of the D.E.

68. $y d x = x l o g (\frac{y}{x}) d y - 2 x d y = 0$

The given D.E is

$\begin{array}{l} y d x + x l o g (\frac{y}{x}) . d y - 2 x d y = 0 \\ \Rightarrow y d x = [2 x d y - x l o g (\frac{y}{x}) d y] \\ \Rightarrow y d x = [2 x - x l o g (\frac{y}{x})] d y \\ \Rightarrow \frac{d y}{d x} = \frac{y}{2 x - x l o g (\frac{y}{x})} = \frac{y}{x (2 - l o g \frac{y}{x})} \\ = \frac{\frac{y}{x}}{2 - l o g \frac{y}{x}} = f (\frac{y}{x}) \end{array}$

Hence, the given D.E is homogenous.

Let, $y = v x = \frac{y}{x} = v$ so that $\frac{d y}{d x} = v + x \frac{d v}{d x}$ in the D.E.

Then, $v + x \frac{d v}{d x} = \frac{v}{2 - l o g v}$

$\begin{array}{l} \Rightarrow x \frac{d v}{d x} = \frac{v}{2 - l o g v} - v = \frac{v - 2 v + v l o g v}{2 - l o g v} = \frac{v l o g v - v}{2 - l o g v} \\ \Rightarrow \frac{2 l o g v}{v [l o g v - 1]} d v = \frac{d x}{x} \\ \Rightarrow \frac{1 + 1 - l o g v}{v [l o g v - 1]} d v = \frac{d x}{x} \\ \Rightarrow \frac{1 - [l o g v - 1]}{v [l o g v - 1]} d v = \frac{d x}{x} \end{array}$

$\Rightarrow [\frac{1}{v (l o g v - 1)} - \frac{1}{v}] d v = \frac{d x}{x}$

Integrating both sides we get,

$\begin{array}{l} \int [\frac{1}{v (l o g v - 1)} - \frac{1}{v}] d v = \int \frac{d x}{x} \\ \int \frac{d v}{v (l o g v - 1)} - l o g v = l o g x + l o g c \end{array}$

Let, $l o g v - 1 = t, s o, \frac{d}{d v} (l o g v - 1) = \frac{d t}{d v}$

$\begin{array}{l} \Rightarrow \frac{1}{v} = \frac{d t}{d v} \\ \Rightarrow \frac{d v}{v} = d t \\ \Rightarrow \int \frac{d t}{t} - l o g v = l o g x + l o g c \\ \Rightarrow l o g t - l o g v = l o g x + l o g c \\ \Rightarrow l o g | l o g v - 1 | - l o g v = l o g c x \end{array}$

Putting back $v = \frac{y}{x}$ we get,

$\begin{array}{l} l o g | l o g (\frac{y}{x}) - 1 | - l o g \frac{y}{x} = l o g \frac{c}{x} \\ = l o g [\frac{l o g (\frac{y}{x}) - 1}{\frac{y}{x}}] = l o g \frac{c}{x} \\ = \frac{y}{x} [l o g (\frac{y}{x}) - 1] = c x \end{array}$

$= l o g (\frac{y}{x}) - 1 = c y$ is the required solution.

69. $(1 + e^{\frac{x}{y}}) d x + e^{\frac{x}{y}} (1 - \frac{x}{y}) d y = 0$

The given D.E. is

$\begin{array}{l} (1 + e^{\frac{x}{y}}) d x + e^{\frac{x}{y}} (1 - \frac{x}{y}) d y = 0 \\ \Rightarrow (1 + e^{\frac{x}{y}}) d x = - e^{\frac{x}{y}} (1 - \frac{x}{y}) d y \\ \Rightarrow \frac{d x}{d y} = - \frac{e^{\frac{x}{y}} (1 - \frac{x}{y})}{1 + e^{\frac{x}{y}}} = f (\frac{x}{y}) \end{array}$

Hence, the given D.E. is homogenous.

Let, $x = y v = \frac{y}{x}$ so that $\frac{d y}{d x} = v + y \frac{d x}{d y}$ in the D.E.

Then, $v + y \frac{d v}{d y} = \frac{- e^{v} (1 - v)}{1 + e^{v}}$

$\begin{array}{l} \Rightarrow y \frac{d v}{d y} = \frac{v e^{v} - e^{v}}{1 + e^{v}} - v = \frac{v e^{v} - e^{v} - v - v e^{v}}{1 + e^{v}} \\ \Rightarrow y \frac{d v}{d y} = - \frac{(e^{v} + v)}{1 + e^{v}} \\ \Rightarrow (\frac{1 + e^{v}}{e^{v} + v}) d v = - \frac{d y}{y} \end{array}$

Integrating both sides we get,

$l o g | e^{v} + v | = - l o g | y | + l o g | c |$

Putting back $v = \frac{x}{y}$ we get,

$\begin{array}{l} l o g | e^{\frac{x}{y}} + \frac{x}{y} | = l o g | \frac{c}{y} | \\ = e^{\frac{x}{y}} + \frac{x}{y} = \frac{c}{y} \end{array}$

$= x + y e^{\frac{x}{y}} = c$ is the general solution.

For each of the differential equations in Questions from 11 to 15, find the particular solution satisfying the given condition

70. (x + y) dy + (x – y) dx = 0; y = 1 when x = 1

The given D.E. is

$\begin{array}{l} (x + y) d y + (x - y) d x = 0 \\ = (x + y) d y = - (x - y) d x \\ = \frac{d y}{d x} = \frac{y - x}{x + y} = \frac{\frac{y - x}{x}}{\frac{x + y}{y}} = \frac{\frac{y}{x} - 1}{1 + \frac{y}{x}} = f (\frac{y}{x}) \end{array}$

i.e, homogenous

Let, $y = v x = v = \frac{y}{x}$ so that $\frac{d y}{d x} = v + y \frac{d y}{d x}$ in the D.E.

Then, $v + x \frac{d v}{d x} = \frac{v - 1}{v + 1}$

$\begin{array}{l} = x \frac{d v}{d x} = \frac{v - 1}{v + 1} - v = \frac{v - 1 - v^{2} - v}{v + 1} = - \frac{(v^{2} + 1)}{v + 1} \\ = [\frac{v + 1}{v^{2} + 1}] d v = - \frac{d x}{x} \end{array}$

Integrating both sides,

$\begin{array}{l} \int \frac{v + 1}{v^{2} + 1} d v = \int - \frac{d x}{x} \\ = \frac{1}{2} \int \frac{2 v}{v^{2} + 1} d v + \int \frac{1}{v^{2} + 1} d v = - l o g x + c \\ = \frac{l o g | v^{2} + 1 |}{2} + t a n^{- 1} v = - l o g x + c \end{array}$

Putting back $v = \frac{y}{x}$ we get,

$\begin{array}{l} = \frac{1}{2} l o g | \frac{y^{2}}{x^{2}} + 1 | + t a n^{- 1} \frac{y}{x} = - l o g x + c \\ = \frac{1}{2} [l o g (y^{2} + x^{2}) - l o g x^{2}] + t a n^{- 1} \frac{y}{x} + l o g x = c \\ = \frac{1}{2} l o g (y^{2} + x^{2}) - \frac{1}{2} l o g x^{2} + l o g x + t a n^{- 1} \frac{y}{x} = c \\ = \frac{1}{2} l o g (y^{2} + x^{2}) - l o g x + l o g x + t a n^{- 1} \frac{y}{x} = c \\ = \frac{1}{2} l o g (y^{2} + x^{2}) + t a n^{- 1} \frac{y}{x} = c \end{array}$

Given, $y = 1, w h e n, x = 1$

So, $= \frac{1}{2} l o g (1^{2} + 1^{2}) + t a n^{- 1} \frac{1}{1} = c$

$= \frac{1}{2} l o g 2 + \frac{π}{4} = c$

Hence, the particular solution is

$\frac{1}{2} l o g (1^{2} + 1^{2}) + t a n^{- 1} \frac{y}{x} = \frac{1}{2} l o g 2 + \frac{π}{4}$

6. (y′′′) + (y′′) + (y′)⁴ + y⁵ =0

The highest order derivative present in the D.E. is $y^{| | |}$ so its order is 3.

As the given D.E. is a polynomial equation in its derivation, its degree is 2.

The degree of the differential equation ${(\frac{d^{2} y}{d x^{2}})}^{3} + {(\frac{d y}{d x})}^{2} + s i n (\frac{d y}{d x}) + 1 = 0$ is:

(A) 3

(B) 2

(C) 1

(D) Not defined

In the given D.E,

$s i n \frac{d y}{d x}$ is a trigonometric function of derivative $\frac{d y}{d x}$ . So it is not a polynomial equation so its derivative is not defined.

Hence, Degree of the given D.E. is not defined.

$∴$ Option (D) is correct.

12. The order of the differential equation $2 x^{2} \frac{d^{2} y}{d x^{2}} - 3 \frac{d y}{d x} + y = 0$ is:

(A) 2

(B) 1

(C) 0

(D) Not defined

The highest order derivative present in the given D.E. is $\frac{d^{2} y}{d x^{2}}$ and its order is 2.

$∴$ Option (A) is correct.

Kindly Consider the following

2. y' + 5y = 0

The highest order derivation present in the D.E. is y, so its order is 1.

As the given D.E. is a polynomial equation in its derivative its degree is 1.

3. ${(\frac{d s}{d t})}^{4} + 3 s \frac{d^{2} s}{d t^{2}} = 0$

The highest order derivation present in the D.E. is $\frac{d^{2} s}{d t^{2}}$ so its order is 2 .

As the given D.E. is a polynomial equation in its derivative its degree is 1.

4. $\frac{d^{2} y}{{(d x^{2})}^{2}} + c o s (\frac{d y}{d x}) = 0$

$\frac{d^{4} y}{d x^{4}}$ As the given D.E. is not a polynomial equation in its derivative, its degree is not defined.

$\frac{d^{2} y}{d x^{2}} = c o s 3 x + s i n 3 x$

$\frac{d^{4} y}{d x^{4}}$ As the given D.E. is a polynomial equation in its derivative, its degree is 1.

7. y''' + 2y'' + y' = 0

The highest order present in the D.E. is $y^{| | |}$ so its order is 3.

As the given D.E. is a polynomial equation in its derivative, its degree is 1.

8. y′ + y = e^x

The given order derivative present in the D.E. is $y^{|}$ so its order is 1.

As the given D.E. is a polynomial equation in its derivative, its degree is 1.

9. y'′ + (y')² + 2y = 0

The highest order derivative present in the D.E. is $y^{| |}$ so its order is 2.

As the given D.E. is a polynomial equation in its derivative, its degree is 1.

10. y'′ + 2y' + sin y = 0

The highest order derivative present in the D.E. is $y^{| |}$ so its order is 2.

As the given D.E. is polynomial equation in its derivative, its degree is 1.

13. y = e^x + 1 : y″ – y′ = 0

Given $f x^{n}$ is $y = e^{x} + 1$

Differentiating with $x$ we get,

$y^{|} = \frac{d y}{d x} = e^{x}$

Again,

$y^{| |} = \frac{d^{2} y}{d x^{2}} = e^{x}$

Substituting value of $y^{| |}$ and $y^{|}$ in the given D.E. we get

$L . H . S . = y^{| |} - y^{|} = e^{x} - e^{x} = 0 = R . H . S$

$∴$ The given $f x^{n}$ is a solution of the given D.E.

14. y = x² + 2x + C : y′ – 2x – 2 = 0

Given, $f x^{n}$ is $y = x^{2} + 2 x + c$

So, $y^{|} = 2 x + 2$

Substituting value of $y^{|}$ in the given D.E. we get,

$L . H . S . = y^{|} - 2 x - 2 = 2 x + 2 - 2 x - 2 = 0 = R . H . S$

$∴$ The given $f x^{n}$ is a solution of the given D.E.

15. y = cos x + C : y′ + sin x = 0

Given, $f x^{n}$ is $y = c o s x + c$

So, $y^{|} = - s i n x$

Putting the value of $y^{|}$ in the given D.E. we get,

$L . H . S . = y^{|} + s i n x = - s i n x + s i n x = 0 = R . H . S$

$∴$ The given $f x^{n}$ is a solution of the given D.E.

16. y = √1 + x² ; y/= $\frac{x y}{1 + x^{2}}$

Given, y= √1 + x²

17. y = Ax : xy′ = y (x ≠ 0)

Given, $y = A x :$

So, $y^{|} = \frac{A d x}{d x} = A$

Putting value of $y^{|}$ in L.H.S. of the given D.E.

L.H.S= $x y^{|} = x A = A x = y$ =R.H.S

$∴$ The given $f x^{n}$ is a solution of the given D.E.

19. xy = log y + C :

$y' = \frac{y^{2}}{1 - x y} (x y \neq 1)$

Given, $x y = l o g y + c$

Differentiate w.r.t. x we have

$\begin{array}{l} x \frac{d y}{d x} + y \frac{d x}{d x} = \frac{d}{d x} l o g y + \frac{d}{d x} C \\ \Rightarrow x \frac{d y}{d x} + y = \frac{1}{y} \frac{d y}{d x} + 0 \\ \Rightarrow x \frac{d y}{d x} - \frac{1}{y} \frac{d y}{d x} = - y \\ \Rightarrow \frac{d y}{d x} [x - \frac{1}{y}] = - y \\ \Rightarrow \frac{d y}{d x} [\frac{x y - 1}{y}] = - y \\ \Rightarrow \frac{d y}{d x} = \frac{- y^{2}}{x y - 1} = \frac{(- 1) \times - y^{2}}{(- 1) \times (x y - 1)} = \frac{y^{2}}{1 - x y} \\ ∴ y^{|} = \frac{y^{2}}{1 - x y} \end{array}$

Hence, y is a Solution of the given D.E

20. y – cos y = x : (y sin y + cos y + x) y′ = y

Given $y - c o s y = x$

Differentiate w.r.t ‘x’ we get

$\begin{array}{l} \frac{d y}{d x} - (- s i n y) \frac{d y}{d x} = \frac{d x}{d x} \\ \Rightarrow \frac{d y}{d x} [1 + s i n y] = 1 \\ \Rightarrow \frac{d y}{d x} = \frac{1}{1 + s i n y} = y^{|} \end{array}$

So, L.H.S of given D.E $= (y s i n y + c o s y + x) y^{|}$

$\begin{array}{l} = (y s i n y + c o s y + y - c o s x) [\frac{1}{1 + s i n y}] \\ = \frac{y (1 + s i n y)}{(1 + s i n y)} = y = R . H . S \end{array}$

$∴$ The given $f x^{n}$ is a solution of the given D.E.

23. The number of arbitrary constants in the general solution of a differential equation of fourth order are:

(A) 0

(B) 2

(C) 3

(D) 4

The number of arbitrary constant is general solution of D.E of 4^th order is four.

$∴$ Option (D) is correct.

24. The number of arbitrary constants in the particular solution of a differential equation of third order are:

(A) 3

(B) 2

(C) 1

(D) 0

In a particular solution, there are no arbitrary constant.

Hence, option (D) is correct.

26. $y^{2} = a (b^{2} - x^{2})$

Given: Equation of the family of curves $y^{2} = a (b^{2} - x^{2})$

Differentiating both sides with respect to x, we get:

$\begin{array}{l} 2 y \frac{d y}{d x} = a (- 2 x) \\ \Rightarrow 2 y y^{'} = - 2 a x \\ \Rightarrow y y^{'} = - a x . . . . . . . . . . (1) \end{array}$

Again, differentiating both sides with respect to x, we get:

$\begin{array}{l} y^{'} . y^{'} + y y^{"} = - a \\ \Rightarrow {(y^{'})}^{2} + y y^{"} = - a . . . . . . . . . . (2) \end{array}$

Dividing equation (2) by equation (1), we get:

$\begin{array}{l} \frac{{(y^{'})}^{2} + y y^{"}}{y y^{'}} = \frac{- a}{- a x} \\ \Rightarrow x y y^{"} + x {(y^{'})}^{2} - y y^{"} = 0 \end{array}$

This is the required differential equation of the given curve.

27. $y = a e^{3 x} + b e^{- 2 x}$

Given: Equation of the family of curves $y = a e^{3 x} + b e^{- 2 x} . . . . . . . . . . (i)$

Differentiating both sides with respect to x, we get:

$y^{'} = 3 a e^{3 x} - 2 b e^{- 2 x} . . . . . . . . . . (i i)$

Again, differentiating both sides with respect to x, we get:

$y^{"} = 9 a e^{3 x} - 4 b e^{- 2 x} . . . . . . . . . . (i i i)$

Multiplying equation (i) with (ii) and then adding it to equation (ii), we get:

$\begin{array}{l} (2 a e^{3 x} + 2 b e^{- 2 x}) + (3 a e^{3 x} - 2 b e^{- 2 x}) = 2 y + y^{'} \\ \Rightarrow 5 a e^{3 x} = 2 y + y^{'} \\ \Rightarrow a e^{3 x} = \frac{2 y + y^{'}}{5} \end{array}$

Now, multiplying equation (i) with (iii) and subtracting equation (ii) from it, we get:

$\begin{array}{l} (3 a e^{3 x} + 2 b e^{- 2 x}) - (3 a e^{3 x} - 2 b e^{- 2 x}) = 3 y - y^{'} \\ \Rightarrow 5 b e^{- 2 x} = 3 y - y^{'} \\ \Rightarrow b e^{- 2 x} = \frac{3 y - y^{'}}{5} \end{array}$

Substituting the values of $a e^{3 x}$ and $b e^{- 2 x}$ in equation (iii), we get:

$\begin{array}{l} y^{"} = 9 . \frac{(2 y - y^{'})}{5} + 4 \frac{(3 y - y^{'})}{5} \\ \Rightarrow y^{"} = \frac{18 y + 9 y^{'}}{5} + \frac{12 y - 4 y^{'}}{5} \\ \Rightarrow y^{"} = \frac{30 y + 5 y^{'}}{5} \\ \Rightarrow y^{"} = 6 y + y^{'} \\ \Rightarrow y^{"} - y^{'} - 6 y = 0 \end{array}$

This is the required differential equation of the given curve.

Kindly Consider the following

28. $y = e^{2 x} (a + b x)$

Given: Equation of the family of curves $y = e^{2 x} (a + b x) . . . . . . . . . . (1)$

Differentiating both sides with respect to x, we get:

$\begin{array}{l} y^{'} = 2 e^{2 x} (a + b x) + e^{2 x} . b \\ \Rightarrow y^{'} = e^{2 x} (2 a + 2 b x + b) . . . . . . . . . . (2) \end{array}$

Multiplying equation (1) with (2) and then subtracting it from equation (2), we get:

$\begin{array}{l} y^{'} - 2 y = e^{2 x} (2 a + 2 b x + b) - e^{2 x} (2 a + 2 b x) \\ \Rightarrow y^{'} - 2 y = b e^{2 x} . . . . . . . . . . (3) \end{array}$

Differentiating both sides with respect to x, we get:

$y^{"} - 2 y^{'} = 2 b e^{2 x} . . . . . . . . . . (4)$

Dividing equation (4) by equation (3), we get:

$\begin{array}{l} \frac{y^{"} - 2 y^{'}}{y^{'} - 2 y} = 2 \\ \Rightarrow y^{"} - 2 y^{'} = 2 y^{'} - 4 y \\ \Rightarrow y^{"} - 4 y^{'} + 4 y = 0 \end{array}$

This is the required differential equation of the given curve.

29. $y = e^{x} (a c o s x + b s i n x)$

Given: Equation of the family of curves $y = e^{x} (a c o s x + b s i n x) . . . . . . . . . . (i)$

Differentiating both sides with respect to x, we get:

$\begin{array}{l} y^{'} = e^{x} (a c o s x + b s i n x) + e^{x} (- a c o s x + b s i n x) \\ \Rightarrow y^{'} = e^{x} [(a + b) c o s x - (a - b) s i n x] . . . . . . . . . . (2) \end{array}$

Again, differentiating with respect to x, we get:

$\begin{array}{l} y^{"} = e^{x} [(a + b) c o s x - (a - b) s i n x] + e^{x} [- (a + b) s i n x - (a - b) c o s x] \\ y^{"} = e^{x} [2 b c o s x - 2 a s i n x] \\ y^{"} = 2 e^{x} (b c o s x - a s i n x) \\ \Rightarrow \frac{y^{"}}{2} = e^{x} (b c o s x - a s i n x) . . . . . . . . . . (3) \end{array}$

Adding equations (1) and (3), we get:

$\begin{array}{l} y + \frac{y^{"}}{2} = e^{x} [(a + b) c o s x - (a - b) s i n x] \\ \Rightarrow y + \frac{y^{"}}{2} = y^{'} \\ \Rightarrow 2 y + y^{"} = 2 y^{'} \\ \Rightarrow y^{"} - 2 y^{'} + 2 y = 0 \end{array}$

This is the required differential equation of the given curve.

30. Form the differential equation of the family of circles touching the y-axis at the origin.

The centre of the circle touching the y-axis at origin lies on the x-axis.

Let (a, 0) be the centre of the circle.

Since it touches the y-axis at origin, its radius is a.

Now, the equation of the circle with centre (a, 0) and radius (a) is

$\begin{array}{l} {(x - a)}^{2} + y^{2} = a^{2} . \\ \Rightarrow x^{2} + y^{2} = 2 a x . . . . . . . . . . (1) \end{array}$

Differentiating equation (1) with respect to x, we get:

$\begin{array}{l} 2 x + 2 y y^{'} = 2 a \\ \Rightarrow x + y y^{'} = a \end{array}$

Now, on substituting the value of a in equation (1), we get:

$\begin{array}{l} x^{2} + y^{2} = 2 (x + y y^{'}) x \\ \Rightarrow x^{2} + y^{2} = 2 x^{2} + 2 x y y^{'} \\ \Rightarrow 2 x y y^{'} + x^{2} = y^{2} \end{array}$

This is the required differential equation.

31. Find the differential equation of the family of parabolas having vertex at origin and axis along positive y-axis.

The equation of the parabola having the vertex at origin and the axis along the positive y-axis is:

$x^{2} = 4 a y$

Differentiating equation (1) with respect to x, we get:

$2 x = 4 a y'$

Dividing equation (2) by equation (1), we get:

$\begin{array}{l} \frac{2 x}{x^{2}} = \frac{4 a y^{'}}{4 a y} \\ \Rightarrow \frac{2}{x} = \frac{y^{'}}{y} \\ \Rightarrow x y^{'} = 2 y \\ \Rightarrow x y^{'} - 2 y = 0 \end{array}$

This is the required differential equation.

32. Form the differential equation of family of ellipse having foci on y-axis and centre at the origin.

The equation of the family of ellipses having foci on the y-axis and the centre at origin is as follows:

$\frac{x^{2}}{b^{2}} + \frac{y^{2}}{a^{2}} = 1 . . . . . . . . . . (1)$

Differentiating equation (1) with respect to x, we get:

$\begin{array}{l} \frac{2 x}{b^{2}} + \frac{2 y y^{'}}{b^{2}} = 0 \\ \Rightarrow \frac{x}{b^{2}} + \frac{y y^{'}}{a^{2}} . . . . . . . . . . (2) \end{array}$

Again, differentiating with respect to x, we get:

$\begin{array}{l} \Rightarrow \frac{1}{b^{2}} + \frac{y^{'} . y^{'} + y . y^{"}}{a^{2}} = 0 \\ \Rightarrow \frac{1}{b^{2}} + \frac{1}{a^{2}} ({y^{'}}^{2} + y y^{"}) = 0 \\ \Rightarrow \frac{1}{b^{2}} = - \frac{1}{a^{2}} ({y^{'}}^{2} + y y^{"}) \end{array}$

Substituting this value in equation (2), we get:

$\begin{array}{l} x [- \frac{1}{a^{2}} ({(y^{'})}^{2} + y y^{"})] + \frac{y y^{"}}{a^{2}} = 0 \\ \Rightarrow - x {(y^{'})}^{2} - x y y^{"} + y y^{'} = 0 \\ \Rightarrow x y y^{"} + x {(y^{'})}^{2} = 0 \end{array}$

This is the required differential equation

33. Form the differential equation of the family of hyperbolas having foci on x-axis and centre at the origin.

The equation of the family of hyperbolas with the centre at origin and foci along the x-axis is:

$\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 . . . . . . . . . . (1)$

Differentiating both sides of equation (1) with respect to x, we get:

$\begin{array}{l} \frac{2 x}{a^{2}} - \frac{2 y y^{'}}{b^{2}} = 0 \\ \Rightarrow \frac{x}{a^{2}} - \frac{y y^{'}}{b^{2}} = 0 . . . . . . . . . . (2) \end{array}$

Again, differentiating both sides with respect to x, we get:

$\begin{array}{l} \frac{1}{a^{2}} - \frac{y^{'} . y^{'} + y . y^{"}}{b^{2}} = 0 \\ \Rightarrow \frac{1}{a^{2}} + \frac{1}{b^{2}} ({(y^{'})}^{2} + y y^{"}) = 0 \end{array}$

Substituting the value of $\frac{1}{a^{2}}$ in equation (2), we get:

$\begin{array}{l} \frac{x}{b^{2}} ({(y^{'})}^{2} + y y^{"}) - \frac{y y^{'}}{b^{2}} = 0 \\ \Rightarrow x {(y^{'})}^{2} + x y y^{"} - y y^{'} = 0 \\ \Rightarrow x y y^{"} + x {(y^{'})}^{2} - y y^{'} = 0 \end{array}$

This is the required differential equation.

34. Form the differential equation of the family of circles having centres on y-axis and radius 3 units.

Let the centre of the circle on y-axis be (0, b).

The differential equation of the family of circles with centre at (0, b) and radius 3 is as follows:

$\begin{array}{l} x^{2} + {(y - b)}^{2} = 3^{2} \\ \Rightarrow x^{2} + {(y - b)}^{2} = 9 . . . . . . . . . . (1) \end{array}$

Differentiating equation (1) with respect to x, we get:

$\begin{array}{l} 2 x + 2 (y - b) . y^{'} = 0 \\ \Rightarrow (y - b) . y^{'} = - x \\ \Rightarrow y - b = \frac{- x}{y^{'}} \end{array}$

Substituting the value of $(y - b)$ in equation (1), we get:

$\begin{array}{l} x^{2} + {(\frac{- x}{y^{'}})}^{2} = 9 \\ \Rightarrow x^{2} [1 + \frac{1}{{(y^{'})}^{2}}] = 9 \\ \Rightarrow x^{2} ({(y^{'})}^{2} + 1) = 9 {(y^{'})}^{2} \\ \Rightarrow (x^{2} - 9) {(y^{'})}^{2} + x^{2} = 0 \end{array}$

This is the required differential equation.

35. Which of the following differential equation has $y = c_{1} e^{x} + c_{2} e^{- x}$ as the general solution:

$\begin{array}{l} (A) \frac{d^{2} y}{d x^{2}} + y = 0 \\ (B) \frac{d^{2} y}{d x^{2}} - y = 0 \\ (C) \frac{d^{2} y}{d x^{2}} + 1 = 0 \\ (D) \frac{d^{2} y}{d x^{2}} - 1 = 0 \end{array}$

Given: $y = c_{1} e^{x} + c_{2} e^{- x} . . . . . . . . . (1)$

Differentiating with respect to x, we get:

$\frac{d y}{d x} = c_{1} e^{x} - c_{2} e^{- x}$

Again, differentiating with respect to x, we get:

$\begin{array}{l} \frac{d^{2} y}{d x^{2}} = c_{1} e^{x} + c_{2} e^{- x} \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = y \\ \Rightarrow \frac{d^{2} y}{d x^{2}} - y = 0 \end{array}$

This is the required differential equation of the given equation of curve.

Hence, the correct answer is B.

36. Which of the following differential equations has y = x as one of its particular solutions:

$\begin{array}{l} (A) \frac{d^{2} y}{d x^{2}} - x^{2} \frac{d y}{d x} + x y = x \\ (B) \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + x y = x \\ (C) \frac{d^{2} y}{d x^{2}} - x^{2} \frac{d y}{d x} + x y = 0 \\ (D) \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + x y = 0 \end{array}$

The given equation of curve is $y = x$ .

Differentiating with respect to x, we get:

$\frac{d y}{d x} = 1 . . . . . . . . . . (1)$

Again, differentiating with respect to x, we get:

$\frac{d^{2} y}{d x^{2}} = 0 . . . . . . . . . . (2)$

Now, on substituting the values of y, $\frac{d^{2} y}{d x^{2}}$ and $\frac{d y}{d x}$ from equation (1) and (2) in each of the given alternatives, we find that only the differential equation given in alternative C is correct

$\begin{array}{l} \frac{d^{2} y}{d x^{2}} - x^{2} \frac{d y}{d x} + x y = 0 - x^{2} . 1 + x . x \\ = - x^{2} + x^{2} \\ = 0 \end{array}$

Therefore, option (C) is correct.

48. $x (x^{2} - 1) \frac{d y}{d x} = 1; y = 0 w h e n x = 2$

The given D.E. is

$\begin{array}{l} x (x^{2} - 1) \frac{d y}{d x} = 1 \\ \Rightarrow d y = \frac{d x}{x (x^{2} - 1)} \end{array}$

Integrating both sides,

$\begin{array}{l} \int d y = \int \frac{d x}{x (x^{2} - 1)} \\ \Rightarrow y = \int \frac{d x}{x (x^{2} - 1) (x + 1)} d x + c . \end{array}$

Let, $\frac{1}{x (x - 1) (x + 1)} = \frac{A}{x} + \frac{B}{x - 1} + \frac{c}{x + 1}$

$\begin{array}{l} 1 = A (x - 1) (x + 1) + B (x) (x + 1) + C (x) (x - 1) \\ = A (x^{2} - 1) + B x^{2} + B x + C x^{2} - C x \\ = A x^{2} - A + B x^{2} + B x + C x^{2} - C x \\ = (A + B + C) x^{2} + (B - C) x - A \end{array}$

Comparing the coefficient,

$\begin{array}{l} - A = 1 \\ \Rightarrow A = - 1 - - - - - - - - - - - - (1) \\ \Rightarrow A + B + C = 0 - - - - - - - - - (2) \\ \Rightarrow B - C = 0 \\ \Rightarrow B = C - - - - - - - - - - - - - (3) \end{array}$

Putting equation (1) & (2) in (1) we get,

$\begin{array}{l} - 1 + B + B = 0 \\ \Rightarrow - 1 + 2 B = 0 \\ \Rightarrow B = \frac{1}{2} = C \end{array}$

So, $\frac{1}{x (x - 1) (x + 1)} = - \frac{1}{x} + \frac{\frac{1}{2}}{x - 1} + \frac{\frac{1}{2}}{x + 1}$

$= - \frac{1}{x} + \frac{1}{2 (x - 1)} + \frac{1}{2 (x + 1)}$

Integrating becomes,

$\begin{array}{l} y = \int - \frac{1}{x} d x + \int \frac{1}{2 (x - 1)} d x + \int \frac{1}{2 (x + 1)} d x + c \\ = - l o g (x) + \frac{1}{2} l o g (x - 1) + \frac{1}{2} l o g (x + 1) + c \\ = \frac{1}{2} [- 2 l o g (x) + l o g (x - 1) + l o g (x + 1)] + c \\ = \frac{1}{2} [- l o g x^{2} + l o g (x + 1) (x - 1)] + c \\ = \frac{1}{2} l o g \frac{x^{2} - 1}{x^{2}} + c \end{array}$

Given, $y = 0 w h e n x = 2 .$

Then, $0 = \frac{1}{2} l o g \frac{2^{2} - 1}{2^{2}} + c$

$\begin{array}{l} \Rightarrow 0 = \frac{1}{2} l o g \frac{3}{4} + c \\ \Rightarrow c = - \frac{1}{2} l o g \frac{3}{4} \end{array}$

$∴$ The required particular solution is

$y = \frac{1}{2} l o g \frac{x^{2} - 1}{x^{2}} - \frac{1}{2} l o g \frac{3}{4}$

49. $c o s (\frac{d x}{d y}) = a (a \in R); y = 1 w h e n x = 0$

Given, D.E. is

$\begin{array}{l} c o s \frac{d y}{d x} = a \\ \Rightarrow \frac{d y}{d x} = c o s^{- 1} (a) \\ \Rightarrow d y = c o s^{- 1} (a) d x \end{array}$

Integrating both sides,

$\begin{array}{l} \int d y = \int c o s^{- 1} (a) d x \\ \Rightarrow y = c o s^{- 1} (a) \times x + c \\ \Rightarrow y = x c o s^{- 1} (a) d x \end{array}$

Given, $y = 1, a t x = 0$

Then, $1 = 0 c o s^{- 1} (a) + c$

$\Rightarrow c = 1$

$∴$ The required particular solution is

50. $\frac{d y}{d x} = y t a n x; y = 1 w h e n x = 0$

Given, $\frac{d y}{d x} = y t a n x$

$\Rightarrow \frac{d y}{y} = t a n x d x$

Integrating both sides we get,

$\begin{array}{l} \int \frac{d y}{y} = \int t a n x d x \\ \Rightarrow l o g y = l o g | s e c x | + l o g c \\ \Rightarrow l o g y = l o g | c s e c x | \\ \Rightarrow y = c_{1} s e c x (w h e r e, c_{1} = \pm c) \end{array}$

As, $y = 1, a t, x = 0$ we have,

$1 = c_{1} s e c (0) = c \Rightarrow c = 1$

$∴$ The required particular solution is $y = s e c x$ .

51. Find the equation of the curve passing through the point (0, 0) and whose differential equation is y' = e^x sin x

The given D.E. is $y^{1} = e^{x} s i n x$

$d y = e^{x} s i n x d x$

Integrating both sides,

$\begin{array}{l} \int d y = \int e^{x} s i n x d x \\ \Rightarrow y = I + c \end{array}$

Where, $I = \int e^{x} s i n x d x$

$\begin{array}{l} = s i n x \int e^{x} d x - \int \frac{d}{d x} s i n x \int e^{x} d x . d x \\ = s i n x . e^{x} - \int c o s x e^{x} d x \\ = s i n x e^{x} - {c o s x \int e^{x} d x - \int \frac{d}{d x} (c o s x) . \int I = \int e^{x} x d x} \\ = s i n x . e^{x} - {c o s x e^{x} + \int s i n x e^{x} d x} \\ = s i n x . e^{x} - c o s x e^{x} - I \\ \Rightarrow I + I = e^{x} (s i n x - c o s x) \\ \Rightarrow I = \frac{e^{x}}{2} (s i n x - c o s x) + c \end{array}$

Hence, $y = \frac{e^{x}}{2} (s i n x - c o s x) + c$

When the curve passed point (0,0),

$\begin{array}{l} y = 0, a t, x = 0 \\ \Rightarrow 0 = \frac{e^{x}}{2} (s i n 0 - c o s 0) + c \\ \Rightarrow \frac{e^{0}}{2} (0 - 1) = c \\ \Rightarrow c = \frac{1}{2} \end{array}$

$∴$ The required equation of the curve is $y = \frac{e^{x}}{2} (s i n x - c o s x) + \frac{1}{2}$

$\begin{array}{l} \Rightarrow 2 y = e^{x} (s i n x - c o s x) + 1 \\ \Rightarrow 2 y - 1 = e^{x} (s i n x - c o s x) \end{array}$

52. For the differential equation $x y \frac{d y}{d x} = (x + 2) (y + 2)$ find the solution curve passing through the point (1,-1)

The Given D.E is

$\begin{array}{l} x y \frac{d y}{d x} = (x + 2) (y + 2) \\ \Rightarrow y \frac{d y}{y + 2} = \frac{(x + 2)}{2} d x \\ \Rightarrow \frac{y + 2 - 2}{y + 2} d y = (\frac{x}{x} + \frac{2}{x}) d x \\ \Rightarrow (1 - \frac{2}{y + 2}) d y = (1 + \frac{2}{x}) d y d x \end{array}$

Integrating both sides,

$\begin{array}{l} \int (1 - \frac{2}{y + 2}) d y = \int (1 + \frac{2}{x}) d y d x \\ \Rightarrow y - 2 l o g | y + 2 | = x + 2 l o g | x | + c \\ \Rightarrow y - l o g {(y + 2)}^{2} = x + l o g x^{2} + c \\ \Rightarrow y - x = l o g {(y + 2)}^{2} + l o g x^{2} + c \\ \Rightarrow y - x = l o g [{(y + 2)}^{2} . x^{2}] + c \end{array}$

A the curve passes through (-1,1) then $y = - 2, a t, x = 1$

So, $- 1 - 1 = l o g {(- 1 + 2)}^{2} . {(1)}^{2} + c$

$\begin{array}{l} \Rightarrow - 2 = l o g 1 + c \\ \Rightarrow c = - 2 \end{array}$

$∴$ The required equation of curve is,

$y - x = l o g [{(y + 2)}^{2} x^{2}] - 2$

71. x² dy + (xy + y² ) dx = 0; y = 1 when x = 1

The given D.E. is

. $\begin{array}{l} x^{2} d y + (x y + y^{2}) d x = 0 \\ = x^{2} d y = - (x y + y^{2}) d x \\ = \frac{d y}{d x} = - (\frac{x y + y^{2}}{x^{2}}) = - [\frac{y}{x} + (\frac{y^{2}}{x})] = f (\frac{y}{x}) \end{array}$

i.e, the D.E is homogenous.

Let, $y = v x = v = \frac{y}{x}$ so that $\frac{d y}{d x} = v + x \frac{d v}{d x}$ in the given D.E.

Then, $v + x \frac{d v}{d x} = - [v + v^{2}] = - v - v^{2}$

$\begin{array}{l} = x \frac{d v}{d x} = - 2 v - v^{2} = - v (2 + v) \\ = \frac{d v}{v (2 + v)} = - \frac{d x}{x} \end{array}$

Integrating both sides we get,

$\begin{array}{l} \int \frac{d v}{v (2 + v)} = - \int \frac{d x}{x} \\ = \frac{1}{2} \int \frac{2 d v}{2 (v + 2)} = - \int \frac{d x}{x} \\ = \frac{1}{2} \int \frac{v + 2 - v}{v (v + 2)} d v = \int - \frac{d x}{x} \\ = \frac{1}{2} {\int \frac{1}{v} d v - \frac{1}{v + 2} d v} = \int - \frac{d x}{x} \end{array}$

$\begin{array}{l} = \frac{1}{2} [l o g v - l o g | v + 2 |] = - l o g x + l o g c \\ = \frac{1}{2} l o g (\frac{v}{v + 2}) = l o g \frac{c}{x} \\ = l o g {(\frac{v}{v + 2})}^{\frac{1}{2}} = l o g \frac{c}{x} \\ = {(\frac{v}{v + 2})}^{\frac{1}{2}} = \frac{c}{x} \\ = \frac{v}{v + 2} = {(\frac{c}{x})}^{2} \end{array}$

Putting back $v = \frac{y}{x}$ we get,

$\begin{array}{l} = \frac{\frac{y}{x}}{\frac{y}{x} + 2} = {(\frac{c}{x})}^{2} \\ = \frac{y}{y + 2 x} = {(\frac{c}{x})}^{2} \\ = \frac{x^{2} y}{y + 2 x} = c^{2} \end{array}$

Given, y = 1 when x = 1

So, $= \frac{1}{1 + 2} = c^{2} = c^{2} = \frac{1}{3}$

Hence, the required particular solution is,

$\begin{array}{l} = \frac{x^{2} y}{y + 2 x} = \frac{1}{3} \\ = 3 x^{2} y = y + 2 x \end{array}$

72. $[x s i n^{2} (\frac{y}{x} - y)] d x + x d y = 0; y = \frac{π}{4} w h e n x = 1$

The given D.E.is

$\begin{array}{l} [x s i n^{2} (\frac{y}{x}) - y] d x + x d y = 0 \\ = [x s i n^{2} (\frac{y}{x}) - y] d x = - x d y \\ = \frac{d y}{d x} = \frac{[x s i n^{2} (\frac{y}{x}) - y]}{- x} \\ = - [s i n^{2} (\frac{y}{x}) - \frac{y}{x}] = f (\frac{y}{x}) \end{array}$

i.e, the given D.E is homogenous.

Let, $y = v x = \frac{y}{x} = v$ so that, $\frac{d y}{d x} = v x \frac{d v}{d x}$ in the D.E.

$\begin{array}{l} = v + \frac{d v}{d x} = - [s i n^{2} v - v] = v - s i n^{2} v \\ = \frac{d v}{d x} = - s i n^{2} v \\ = \frac{d v}{s i n^{2} v} = - d x \end{array}$

Integrating both sides we get,

$\begin{array}{l} = \int c o s e c^{2} v d v = \int - d x \\ = - c o t v = - l o g | x | + c \\ = c o t v = l o g | x | - c \end{array}$

Putting back $v = \frac{y}{x}$ we have,

$c o t \frac{y}{x} = l o g | x | - c$

Then, $y = \frac{π}{4}$ when, $x = 1$

$\begin{array}{l} c o t \frac{π}{4} = l o g | 1 | - c \\ = c = - 1 \end{array}$

$∴$ The required particular solution is,

$\begin{array}{l} c o t \frac{y}{x} = l o g | x | + 1 = l o g | x | + l o g c {? l o g e = 1} \\ = c o t \frac{y}{x} = l o g | e x | \end{array}$

73. $\frac{d y}{d x} - \frac{y}{x} + c o s e c (\frac{y}{x}) = 0; y = 0 w h e n x = 1$

The given D.E.is

$\begin{array}{l} \frac{d y}{d x} - \frac{y}{x} + c o s e c (\frac{y}{x}) = 0 \\ = \frac{d y}{d x} = \frac{y}{x} - c o s e c (\frac{y}{x}) = f (\frac{y}{x}) \end{array}$

i.e, the given D.E. is homogenous.

Let, $y = v x = \frac{y}{x} = v$ So that, $\frac{d y}{d x} = v + x \frac{d v}{d x}$ in the D.E

Then, $v + x \frac{d v}{d x} = v - c o s e c v$

$\begin{array}{l} = x \frac{d v}{d x} = - c o s e c v \\ = \frac{d v}{c o s e c v} = - \frac{d x}{x} \\ = s i n v d v = - \frac{d x}{x} \end{array}$

Integrating both sides we get,

$\begin{array}{l} \int s i n v d v = - \int \frac{d x}{x} \\ = - c o s v = - l o g | x | + c \\ = c o s v = l o g | x | - c \end{array}$

Putting back $v = \frac{y}{x}$ we get,

$= c o s \frac{y}{x} = l o g | x | - c$

Given, $y = 0, w h e n, x = 1$

$\begin{array}{l} = c o s 0 = l o g 1 - c \\ = c = - 1 \end{array}$

$∴$ The required particular solution is

$\begin{array}{l} c o s (\frac{y}{x}) = l o g | x | + 1 = l o g | x | + l o g | c | \\ = c o s (\frac{y}{x}) = l o g | c x | \end{array}$

74. $2 x y + y^{2} - 2 x^{2} \frac{d y}{d x} = 0; y = 2 w h e n x = 1$

The given D.E. is

$\begin{array}{l} 2 x y + y^{2} - 2 x^{2} \frac{d y}{d x} = 0 \\ = 2 x^{2} \frac{d y}{d x} = 2 x y + y^{2} \\ = \frac{d y}{d x} = \frac{2 x y + y^{2}}{2 x^{2}} = \frac{y}{x} + \frac{1}{2} {(\frac{y}{x})}^{2} = f (\frac{y}{x}) \end{array}$

i.e, the given is homogenous.

Let, $y = v x$ $= \frac{y}{x} = v$ so that $\frac{d y}{d x} = v + x \frac{d v}{d x}$ is the D.E.

Then, $v + x \frac{d v}{d x} = v + \frac{1}{2} v^{2}$

$\begin{array}{l} = x \frac{d v}{d x} = \frac{1}{2} v^{2} \\ = \frac{d v}{v^{2}} = \frac{d x}{2 x} \end{array}$

Now, $= \int \frac{d v}{v^{2}} = \int \frac{d x}{2 x}$

$\begin{array}{l} = \frac{v - 2 + 1}{- 2 + 1} = \frac{1}{2} l o g | x | + c \\ = \frac{- 1}{v} = \frac{1}{2} l o g | x | + c \end{array}$

Putting back $\frac{y}{x} = v$ we get,

$= - \frac{x}{y} = \frac{1}{2} l o g | x | + c$

$G i v e n \frac{y}{x} = v w h e n x = 1$ and y= 2

$\begin{array}{l} = - \frac{1}{2} = \frac{1}{2} l o g | 1 | + c \\ = c = - \frac{1}{2} \end{array}$

$∴$ The particular solution is,

$\begin{array}{l} = - \frac{x}{y} = \frac{1}{2} l o g | x | - \frac{1}{2} \\ = - \frac{2 x}{y} = l o g | x | - 1 \\ = y = \frac{- 2 x}{l o g | x | - 1} = \frac{2 x}{1 - l o g | x |} \end{array}$

75. A homogeneous differential equation of the form $\frac{d x}{d y} = h (\frac{x}{y})$ can be solved by making the substitution:

(A) y = vx

(B) v = yx

(C) x = vy

(D) x = v

For a homogenous D.E. of the formula $f (\frac{y}{x})$

We put, $\frac{x}{y} = 0 = x = v y$

$∴$ Option (c) is correct.

76. Which of the following is a homogeneous differential equation:

(A) (4x +6y +5) dy – (3y + 2x + 4) dx = 0

(B) (xy) dx – x³ + y³) dy = 0

(C) (x² + 2y²) dx + (2xy + dy) = 0

(D) y² dx + (x²– xy + y²) dy = 0

In (D), each of the terms has a degree 2.

Hence, (D) is homogenous

$∴$ Option (D) is correct.

77. Kindly Consider the following

$\frac{d y}{d x} + 2 y = 2 s i n x$

The given D.E. is

$\frac{d y}{d x} + 2 y = 2 s i n x$ which is of form $\frac{d y}{d x} + P x = Q$

We have, P = 2

$Q = s i n x$

So, I.F. $= e^{\int P d x} = e^{\int 2 d x} = e^{2 x}$

The solution is $y \times I . F = \int Q \times (I . F) d x + c$

$\begin{array}{l} y e^{2 x} = \int s i n x e^{2 x} d x + c \\ = y e^{2 x} = I + c - - - - - - - - - (1) \\ W h e r e, I = \int s i n x e^{2 x} \\ = s i n x \int e^{2 x} - \int (\frac{d}{d x} s i n x \int e^{2 x} d x) d x \\ = s i n x \frac{e^{2 x}}{2} - \frac{1}{2} [\int c o s x e^{2 x} d x] \\ = \frac{e^{2 x} s i n x}{2} - \frac{1}{2} [c o s \int e^{2 x} d x - \int (\frac{d}{d x} c o s x \int e^{2 x} d x) d x] \\ = \frac{e^{2 x} s i n x}{2} - \frac{1}{2} [c o s x \frac{e^{2 x}}{2} \frac{1}{2} \int (- c o s x) e^{2 x} d x] \\ = \frac{e^{2 x} s i n x}{2} - \frac{e^{2 x} c o s x}{4} - \frac{1}{4} \int s i n x e^{2 x} d x \end{array}$

$\begin{array}{l} = I = e^{2 x} \frac{s i n x}{2} - e^{2 x} \frac{c o s x}{4} - \frac{1}{4} 1 \\ = I + \frac{1}{4} I = 2 e^{2 x} \frac{s i n x - e^{2 x} c o s x}{4} \\ = \frac{5}{4} I = e^{2 x} \frac{(2 s i n x - c o s x)}{4} \\ = I = e^{2 x} \frac{(2 s i n x - c o s x)}{5} \end{array}$

Hence, equation, (1) becomes,

$\begin{array}{l} y e^{2 x} = \frac{e^{2 x}}{5} (2 s i n x - c o s x) + c \\ = y = \frac{(2 s i n x - c o s x)}{5} + c e^{2 x} \end{array}$

Is the required solution.

78. $\frac{d y}{d x} + 3 y = e^{- 2 x}$

Given, D.E. is

$\frac{d y}{d x} + 3 y = e^{- 2 x}$ which is of the form

$\frac{d y}{d x} + P y = Q$

Where $\begin{array}{l} P = 3 \\ & Q = e^{- 2 x} \end{array}$

So, I.F $= e^{\int P d x} = e^{\int 3 d x} = e^{3 x}$

So, the solution is $= y \times I . F = \int e^{- 2 x} (I . F) . d x + c$

$\begin{array}{l} = y \times e^{3 x} = \int e^{- 2 x} . e^{3 x} d x + c \\ = e^{3 x} y = \int e^{x} d x + c \\ = e^{3 x} y = e^{x} + c \\ = y = \frac{e^{x}}{e^{3 x}} + \frac{c}{e^{3 x}} \\ = y = e^{- 2 x} + c e^{- 3 x} \end{array}$

Is the required general solution.

79. Kindly Consider the following

$\frac{d y}{d x} + \frac{y}{x} = x^{2}$

The given D.E. is

$\frac{d y}{d x} + \frac{y}{x} = x^{2}$ Which is in the form $\frac{d y}{d x} + P y = Q$

So, $P = \frac{1}{x} = & Q = x^{2}$

$∴ I . F = e^{\int P d x} = e^{\int \frac{1}{2} d x} = e^{l o g x} = x {? e^{l o g x} = x}$

Thus, the general solution is

$\begin{array}{l} y \times I . F = \int Q \times I . F d x + c \\ y . x = \int x^{2} . x d x + c \\ = x y = \int x^{3} d x + c \\ = x y = \frac{x^{4}}{4} + c \end{array}$

80. $\frac{d y}{d x} + s e c x y = t a n x (0 \leq x \leq \frac{π}{2})$

The given D.E.is

$\frac{d y}{d x} + (s e c x) y = t a n x$ Which is in the form $\frac{d y}{d x} + P y = Q$

So, $P = s e c x & Q = t a n x$

$∴$ $∴ I . F = e^{\int P d x} = e^{\int s e c x d x} = e^{l o g | s e c x + t a n x |} = s e c x + t a n x$

Thus, the general solution is ,

$\begin{array}{l} y \times I . F = \int Q \times I . F d x + c \\ = y \times (s e c x + t a n x) = \int t a n x (s e c x + t a n x) d x + c \end{array}$

$\begin{array}{l} = \int (t a n x s e c x + t a n^{2} x) d x + c \\ = \int (t a n x s e c x + s e c^{2} - 1) d x + c \\ = s e c + t a n x - x + c \end{array}$

$= (s e c x + t a n x) y = s e c x + t a n x - x + c {? s e c^{2} x = t a n^{2} x + 1}$

81. $c o s^{2} x \frac{d y}{d x} + y = t a n x (0 \leq x \leq \frac{π}{2})$

The given D.E.is

$\begin{array}{l} c o s^{2} x \frac{d y}{d x} + y = t a n x \\ = \frac{d y}{d x} + \frac{1}{c o s^{2} x} y = \frac{t a n x}{c o s^{2} x} \end{array}$

$= \frac{d y}{d x} + s e c^{2} x y = s e c^{2} x t a n x$ Which is of form $\frac{d y}{d x} + P y = Q$

So, $P = s e c^{2} x & Q = s e c^{2} x t a n x$

$∴ I . F = e^{\int P d x} = e^{\int s e c^{2} d x} = e^{t a n x}$

Thus, the general solution is of the form.

$y . e^{t a n x} = \int s e c^{2} x t a n x . e^{t a n x} d x + c$

Let, $t a n x = t = s e c^{2} x d x = d t$

$\begin{array}{l} = y e^{t} = \int t . e^{t} d t + c \\ = t \int e^{t} d t - \int \frac{d}{d t} t \int e^{t} d t . d t + c \\ = t e^{t} - \int e^{t} d t + c \\ = t e^{t} - e^{t} + c \\ = e^{t} (t - 1) + c \end{array}$

$\begin{array}{l} \Rightarrow y e^{t a n x} = e^{t a n x} (t a n x - 1) + c \\ y = (t a n x - 1) + c e^{- t a n x} \end{array}$

82. Kindly Consider the following

$x \frac{d y}{d x} + 2 y = x^{2} l o g x$

The given D.E. is

$x \frac{d y}{d x} + 2 y = x^{2} l o g x$

$\Rightarrow \frac{d y}{d x} + \frac{2}{x} . y = x l o g x$ which is of form $\frac{d y}{d x} + P y = Q$

So, $P = \frac{2}{x} & Q = x l o g x$

$∴ I . F = e^{\int P d x} = e^{\int \frac{2}{x} d x} = e^{2 l o g x} = e^{l o g x^{2}} = x^{2}$

Thus, the general solution is of the form.

$y \times x^{2} = \int x l o g x . x^{2} d x + c$

$= \int l o g x x^{3} d x + c$

$\begin{array}{l} = l o g x \int x^{3} d x - \int \frac{d}{d x} l o g x \int x^{3} d x d x + c \\ = \frac{l o g x . x^{4}}{4} - \int \frac{1}{4} \times \frac{x^{4}}{4} d x + c \end{array}$

$\begin{array}{l} = y x^{2} = \frac{x^{4}}{4} l o g x - \frac{x^{4}}{16} + c \\ = y = \frac{x^{2}}{4} l o g x - \frac{x^{2}}{16} + \frac{c}{x^{2}} \end{array}$

83. $x l o g x \frac{d y}{d x} + y = \frac{2}{x} l o g x$

The given D.E. is

$x l o g x \frac{d y}{d x} + y = \frac{2}{x} l o g x$

$= \frac{d y}{d x} + \frac{y}{x l o g x} = \frac{2}{x^{2}}$ Which is of form $\frac{d y}{d x} + P y = Q$

So, $P = \frac{1}{x l o g x} & Q = \frac{2}{x^{2}}$

$∴ I . F = e^{\int P d x} = e^{\int \frac{1}{x} l o g x d x} = e^{\int \frac{\frac{1}{x}}{l o g x} d x} = e^{l o g | l o g x |} = l o g x$

Thus, the general solution is of the form

$\begin{array}{l} y \times l o g x = \int \frac{2}{x^{2}} \times l o g x d x + c \\ y . l o g x = 2 [l o g x \int \frac{1}{x^{2}} d x - \int \frac{d}{d x} l o g x \int \frac{1}{x^{2}} d x . d x] + c \\ = 2 [l o g x \times (\frac{x^{- 1}}{- 1}) - \int \frac{1}{x} (\frac{x^{- 1}}{- 1}) d x] + c \\ = 2 [- \frac{l o g x}{x} + \int x^{- 2} d x] + c \\ = 2 [- \frac{l o g x}{x} - (\frac{x^{- 1}}{- 1})] + c \end{array}$

$= y l o g x = \frac{- 2}{x} [l o g x + 1 + c]$

84. $(1 + x^{2}) d y + 2 x y d x = c o t x d x (x \neq 0)$

The given D.E. is

$\begin{array}{l} {(1 + x)}^{2} d y + 2 x y d x = c o t x d x \\ = {(1 + x)}^{2} \frac{d y}{d x} + 2 x y = c o t x \end{array}$

$= \frac{d y}{d x} + \frac{2 x}{1 + x^{2}} \times y = \frac{c o t x}{1 + x^{2}}$ Which is of form $\frac{d y}{d x} + P y = Q$

So , $P = \frac{2 x}{1 + x^{2}} & Q = \frac{c o t x}{1 + x^{2}}$

$∴ I . F = e^{\int P d x} = e^{\int \frac{2 x}{1 + x^{2}} d x} = e^{l o g | 1 + x^{2} |} = 1 + x^{2}$

Thus the solution is of the form,

$\begin{array}{l} \Rightarrow y (1 + x^{2}) = \int \frac{c o t x}{1 + x^{2}} \times (1 + x^{2}) d x + c \\ \Rightarrow y (1 + x^{2}) = \int c o t x d x + c \end{array}$

$= l o g | s i n x | + c$

$\begin{array}{l} \Rightarrow y = \frac{l o g | s i n x |}{1 + x^{2}} + \frac{c}{1 + x^{2}} \\ = {(1 + x^{2})}^{- 1} l o g | s i n x | + {(1 + x^{2})}^{- 1} c \end{array}$

85. $x \frac{d y}{d x} + y - x + x y c o t x = 0 (x \neq 0)$

The given D.E is

$= \frac{d y}{d x} + y \frac{(1 + x c o t x)}{x} = 1$ Which is of form $\frac{d y}{d x} + P y = Q$

So, $P = \frac{(1 + x c o t x)}{x} & Q = 1$

$\int P d x = \int (\frac{1}{x} + \frac{x c o t x}{x}) d x = l o g | x | + l o g | s i n x | = l o g | x s i n x |$

$∴ I . F = e^{\int P d x} - e^{\int l o g | x s i n x |} x s i n x$

Thus the solution is of the form.

$\begin{array}{l} y \times x s i n x = \int 1 . x s i n x d x + c \\ = x \int s i n x - \int \frac{d}{d x} \int s i n x d x d x + c \\ = - 2 c o s x + \int c o s x d x + c \\ = y \times x s i n x = - x c o s x + s i n x + c \\ = y = \frac{- x c o s x}{s i n x} + \frac{s i n x}{x s i n x} + \frac{c}{x s i n x} \\ = y = - c o t x + \frac{1}{x} + \frac{c}{x s i n x} \end{array}$

86. $(x + y) \frac{d y}{d x} = 1$

The given D.E is

$\begin{array}{l} (x + y) \frac{d y}{d x} = 1 \\ = x + y = \frac{d x}{d y} \end{array}$

$= \frac{d x}{d y} - x = y$ Which is of form $= \frac{d x}{d y} + P x = Q$

So, $P = - 1 & Q = y$

$∴ I . F = e^{\int P d y} = e^{\int - 1 d y} = e^{y}$

Thus the general solution is of the form, $x \times (I . F) = \int Q (I . F) d y + c$

87. $y d x + (x - y^{2}) d y = 0$

The given D.E. is

$\begin{array}{l} y d x + (x - y^{2}) d y = 0 \\ = y \frac{d x}{d y} + x - y^{2} = 0 \\ = \frac{d x}{d y} + \frac{x}{y} - y = 0 \end{array}$

$= \frac{d x}{d y} + \frac{1}{y} . x = y$ Which is of form.

$\frac{d x}{d y} + P x = Q$

So, $P = \frac{1}{y} & Q = y$

$∴ I . F = e^{\int P d y} = e^{\int \frac{1}{y} d y} = e^{l o g y} = y$

Thus the general solution is of form, $x \times (I . F) = \int Q \times (I . F) d y + c$

$\begin{array}{l} x . y = \int y . y d y + c \\ = x y = \int y^{2} d y + c \\ = x y = \frac{y^{3}}{3} + c \\ = x = \frac{y^{2}}{3} + \frac{c}{y} \end{array}$

88. $(x + 3 y^{2}) \frac{d y}{d x} = y (y > 0)$

The given D.E. is

$\begin{array}{l} (x + 3 y^{2}) \frac{d y}{d x} = y \\ \Rightarrow (x + 3 y^{2}) d y = y d x \\ \Rightarrow y \frac{d x}{d y} = x + 3 y^{2} \\ \Rightarrow \frac{d x}{d y} = \frac{x}{y} + 3 y \end{array}$

$\Rightarrow \frac{d x}{d y} - \frac{1}{y} . x = 3 y$ Which is form $\frac{d x}{d y} + P x = Q$

So, $P = - \frac{1}{y} & Q = 3 y$

$∴ I . F = e^{\int P d y} = e^{\int - \frac{1}{y} d y} = e^{- l o g | y |} = e^{l o g y^{- 1}} = y^{- 1} = \frac{1}{y}$

Thus the solution is of the form.

$\begin{array}{l} x \times \frac{1}{y} = \int 3 y . \frac{1}{y} d y + c \\ = \frac{x}{y} = \int 3 d y + c \\ = \frac{x}{y} = 3 y + c \\ = x = 3 y^{2} + c y \end{array}$

For each of the differential equations given in Questions 13 to 15, find a particular solution satisfying the given condition:

89. $\frac{d y}{d x} + 2 y t a n x = s i n x; y = 0 w h e n x = \frac{π}{3}$

The given D.E. is

$\frac{d y}{d x} + 2 y t a n x = s i n x$

$\Rightarrow \frac{d y}{d x} + (2 t a n x) y = s i n x$ Which is of form $\frac{d y}{d x} + P x = Q$

So, $P = 2 t a n x & Q = s i n x$

$∴ I . F = e^{\int P d y} = e^{\int 2 t a n x d x} = e^{2 l o g | s e c x |} = e^{l o g s e c^{2}} = s e c^{2}$

Thus the solution is of the form $y \times (I . F) = \int Q . (I . F) d x + c$

$\begin{array}{l} \Rightarrow y . s e c^{2} x = \int s i n x . s e c^{2} x d x + c \\ = \frac{s i n x}{c o s^{2}} d x + c \\ = \int t a n x . s e c x d x + c \\ = y s e c^{2} = s e c x + c \\ = y = \frac{1}{s e c x} + \frac{c}{s e c^{2} x} = c o s x + c c o s^{2} \\ = y = c o s x + c c o s^{2} x \end{array}$

Given, $y = 0, W h e n x = \frac{π}{3}$

$\begin{array}{l} = 0 = c o s \frac{π}{3} + c c o s^{2} \frac{π}{3} {\frac{c}{4} = - \frac{1}{2}, c = - \frac{4}{2}, c = - 2} \\ = 0 = \frac{1}{2} + c {(\frac{1}{2})}^{2} \\ = 0 = \frac{1}{2} + \frac{c}{4} \end{array}$

C = -2

$∴$ The particular solution is

$y = c o s x - 2 c o s^{2} x$

90. $(1 + x^{2}) \frac{d y}{d x} + 2 x y = \frac{1}{1 + x^{2}}; y = 0 w h e n x = 1$

The given D.E. is

$(1 + x^{2}) \frac{d y}{d x} + 2 x y = \frac{1}{1 + x^{2}}$

$\frac{d y}{d x} + P x = Q$

So, $P = \frac{2 x}{1 + x^{2}} & Q = \frac{1}{{(1 + x^{2})}^{2}}$

$\int P d x = \int \frac{2 x}{1 + x^{2}} d x = l o g | 1 + x^{2} |$

$∴ I . F = e^{\int P d x} = e^{l o g | 1 + x^{2} |} = 1 + x^{2}$

Thus the solution is if form,

$y \times (I . F) = \int Q . (I . F) d x + c$

$\begin{array}{l} y (1 + x^{2}) = \int \frac{1}{{(1 + x^{2})}^{2}} \times (1 + x^{2}) d x + c \\ = \int \frac{1}{(1 + x^{2})} d x + c \\ y \times (1 + x^{2}) = t a n^{- 1} x + c \end{array}$

Given, $y = 0, w h e n, x = 1$

$\begin{array}{l} 0 = t a n^{- 1} 1 + c \\ c = t a n^{- 1} 1 = - \frac{π}{4} \end{array}$

$∴$ The particular solution is

$y (1 + x^{2}) = t a n^{- 1} x - \frac{π}{4}$

91. $\frac{d y}{d x} - 3 y c o t x = s i n 2 x; y = 2 w h e n x = \frac{π}{2}$

The given D.E. is

$\frac{d y}{d x} - 3 y c o t x = s i n 2 x$

$\frac{d y}{d x} - 3 c o t x . y = s i n 2 x$ Which is of form $\frac{d y}{d x} + P y = Q$

So, $P = - 3 c o t x & Q = s i n 2 x$

$∴ I . F = e^{- \int P d x} = e^{\int - 3 c o t x d x} = e^{- 3 \int c o t x d x} = e^{- 3 l o g | c o t x d x |} = e^{l o g {(s i n)}^{- 3}} = \frac{1}{s i n^{3} x}$

Thus the solution is of the form.

$\begin{array}{l} y \times \frac{1}{s i n^{3} x} = \int s i n 2 x . \frac{1}{s i n^{3} x} d x + c \\ = \int \frac{2 s i n x c o s x}{s i n^{3} x} d x + c {? s i n 2 x = 2 s i n x c o s x} \\ = \int 2 c o s e c x c o t x d x + c \\ = - 2 c o s e c x + c \\ = \frac{y}{s i n^{3} x} = \frac{- 2}{s i n x} + c \\ = - 2 y = - 2 s i n^{2} x + c s i n^{3} x \end{array}$

Given, $y = 2, w h e n, x = \frac{π}{2}$

$\begin{array}{l} 2 = - 2 s i n^{2} \frac{1}{2} + c s i n^{3} \frac{π}{2} \\ = 2 = - 2 + c \\ = e = 2 + 2 = 4 \end{array}$

$∴$ The particular solution is, $y = - 2 s i n^{2} x + 4 s i n^{3} x$

92. Find the equation of the curve passing through the origin, given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of coordinates of that point.

We know the slope of tangent to curve is $\frac{d y}{d x}$ .

$∴$ $\frac{d y}{d x} = x + y$

$= \frac{d y}{d x} - y = x$ which has form $\frac{d y}{d x} + P y = Q$

So, $P = - 1 & Q = x$

$∴ I . F = e^{\int P d x} = e^{\int - d x} = e^{- x}$

Thus the solution is of the form .

$\begin{array}{l} y \times e^{- x} = \int x . e^{- x} d x + c \\ = x \int e^{- x} d x - \int \frac{d x}{d x} \int e^{- x} d x d x + c \\ = - x e^{- x} + \int e^{- x} d x + c \\ = y e^{- x} = - x e^{- x} - e^{- x} + c \\ = y = - x - 1 + \frac{c}{e^{- x}} \\ = y + x + 1 = c e^{x} \end{array}$

Given, the curve passes through origin (0,0) i.e, $y = 0, w h e n, x = 0$

$0 + 0 + 1 = c e^{0} = c = 1$

$∴$ Thus, equation of the curve is

$y + x + 1 = e^{x}$

93. Find the equation of the curve passing through the point (0, 2) given that the sum of the coordinates of any point on the curve exceeds the magnitude of the slope of the tangents to the curve at that point by 5.

We know that slope of tangent to the curve is $\frac{d y}{d x}$

$∴ x + y = \frac{d y}{d x} + 5$

$\frac{d y}{d x} - y = x - 5$ Which has form $\frac{d y}{d x} + P x = Q$

$w h e r e, P = 1 & Q = x - 5$

$∴ I . F = e^{\int P d x} = e^{\int - 1 d x} = e^{- x}$

Thus the solution has the form

$\begin{array}{l} y e^{- x} = \int (x - 5) e^{- x} d x + c \\ = \int x e^{- x} d x - \int 5 e^{- x} d x + c \\ = y e^{- x} = I + 5 e^{- x} + c \\ w h e r e, I = \int x e^{- x} d x \\ = x \int e^{- x} d x - \int \frac{d}{d x} x \int e^{- x} d x d x \\ = - x e^{- x} + \int e^{- x} d x \\ = - x e^{- x} - e^{- x} \end{array}$

$\begin{array}{l} ∴ y e^{- x} = - x e^{- x} - e^{- x} + 5 e^{- x} + c \\ = y e^{- x} = - x e^{- x} + 4 e^{- x} + c \\ = y = - x + 4 + \frac{c}{e^{- x}} \\ = y + x = 4 + c e^{x} \end{array}$

Given, the curve passes through (0,2) so y=2 when x=0

$\begin{array}{l} 2 + 0 = 4 c e^{0} \\ 2 - 4 = c \\ c = - 2 \end{array}$

$∴$ The particular solution is

$y + x = 4 - 2 e^{x}$

94. Choose the correct answer:

The integrating factor of the differential equation $\frac{d y}{d x} - y = 2 x^{2}$ is:

(A) e^–x

(B) e^–y

(C) $\frac{1}{x}$

(D) x

The given D.E is

$\begin{array}{l} x \frac{d y}{d x} - y = 2 x^{2} \\ \frac{d y}{d x} - \frac{1}{x} y = 2 x \end{array}$

Which is of form $\frac{d y}{d x} + P y = Q$

So, $P = - \frac{1}{x}$

$∴ I . E = e^{\int P d x} = e^{\int - \frac{1}{x} d x} = e^{l o g - x} = e^{l o g - x^{- 1}} = x^{- 1} = \frac{1}{x}$

$∴$ Option (c) is correct

95. Choose the correct answer:

The integrating factor of the differential equation $(1 - y^{2}) \frac{d x}{d y} + y x = a y (- 1 < < 1)$ .

The given D.E. is

$(1 - y^{2}) \frac{d x}{d y} + y x = \propto y (- 1 < y < 1)$

$\frac{d x}{d y} + \frac{y}{1 - y^{2}} \times x = \frac{\propto y}{1 - y^{2}}$ which is of form

$\begin{array}{l} \frac{d x}{d y} + P x = Q & P = \frac{y}{1 - y^{2}} & Q = \frac{\propto y}{1 - y^{2}} \\ \int p d x = \int \frac{y}{1 - y^{2}} d x = - \frac{1}{2} \int \frac{- 2 y}{1 - y^{2} d x} \end{array}$

$\begin{array}{l} = - \frac{1}{2} l o g | 1 - y^{2} | \\ = l o g {[1 - y^{2}]}^{- \frac{1}{2}} \end{array}$

$∴ I . F = e^{\int P d x} = e^{l o g {[1 - y^{2}]}^{- \frac{1}{2}}} = {[1 - y^{2}]}^{- \frac{1}{2}}$

$∴$ option (D ) is correct.

96. For each of the differential equations given below, indicate its order and degree (if defined):

$\begin{array}{l} (i) \frac{d^{2} y}{d x^{2}} + 5 x {(\frac{d y}{d x})}^{2} - 6 y = l o g x \\ (i i) {(\frac{d y}{d x})}^{3} - 4 {(\frac{d y}{d x})}^{2} + 7 y = s i n x \\ (i i i) \frac{d^{4} y}{d x^{4}} - s i n \frac{d^{3} y}{d x^{3}} = 0 \end{array}$

(i) Given: Differential equation $\frac{d^{2} y}{d x^{2}} + 5 x {(\frac{d y}{d x})}^{2} - 6 y = l o g x$

The highest order derivative present in this differential equation is $\frac{d^{2} y}{d x^{2}}$ and hence order of this differential equation if 2.

The given differential equation is a polynomial equation in derivatives and highest power of the highest order derivative $\frac{d^{2} y}{d x^{2}}$ is 1.

Therefore, Order = 2, Degree = 1

(ii) Given: Differential equation ${(\frac{d y}{d x})}^{3} - 4 {(\frac{d y}{d x})}^{2} + 7 y = s i n x$

The highest order derivative present in this differential equation is $\frac{d y}{d x}$ and hence order of this differential equation if 1.

The given differential equation is a polynomial equation in derivatives and highest power of the highest order derivative $\frac{d y}{d x}$ is 3.

Therefore, Order = 1, Degree = 3

(iii) Given: Differential equation $\frac{d^{4} y}{d x^{4}} - s i n \frac{d^{3} y}{d x^{3}} = 0$

The highest order derivative present in this differential equation is $\frac{d^{4} y}{d x^{4}}$ and hence order of this differential equation if 4.

The given differential equation is not a polynomial equation in derivatives therefore, degree of this differential equation is not defined.

Therefore, Order = 4, Degree not defined

97. For each of the exercises given below verify that the given function (implicit or explicit) is a solution of the corresponding differential equation:

$\begin{array}{l} (i) y a e^{2} + b e^{- x} + x^{2} : x \frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} - x y + x^{2} - 2 = 0 \\ (i i) y = e^{x} (a c o s x + b s i n x) : \frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} + 2 y = 0 \\ (i i i) y = x s i n 3 x : \frac{d^{2} y}{d x^{2}} + 9 y - 6 c o s 3 x = 0 \\ (i v) x^{2} = 2 y^{2} l o g y : (x^{2} + y^{2}) \frac{d y}{d x} - x y = 0 \end{array}$

(i) $y a e^{2} + b e^{- x} + x^{2}$

Differentiating both sides with respect to x, we get:

$\begin{array}{l} \frac{d y}{d x} = a \frac{d}{d x} (e^{x}) + b \frac{d}{d x} (e^{- x}) + \frac{d}{d x} (x^{2}) \\ \Rightarrow \frac{d y}{d x} = a e^{x} - b e^{- x} + 2 x \end{array}$

Again, differentiating both sides with respect to x, we get:

$\frac{d^{2} y}{d x^{2}} = a e^{x} - b e^{- x} + 2 x$

Now, on substituting the values of $\frac{d y}{d x}$ and $\frac{d^{2} y}{d x^{2}}$ in the differential equation, we get:

L.H.S

$\begin{array}{l} x \frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} - x y + x^{2} - 2 \\ = x (a e^{x} - b e^{- x} + 2) + 2 (a e^{x} - b e^{- x} + 2 x) - x (a e^{x} + b e^{- x} + x^{2}) + x^{2} - 2 \\ = (x a e^{x} - b x e^{- x} + 2 x) + (2 a e^{x} - 2 b e^{- x} + 4 x) - (a x e^{x} + b x e^{- x} + x^{3}) + x^{2} - 2 \\ = 2 a e^{x} - 2 b e^{- x} + x^{2} + 6 x - 2 \\ \neq 0 \end{array}$

Therefore, Function given by equation (i) is a solution of differential equation. (ii).

(ii) $y = e^{x} (a c o s x + b s i n x) = a e^{x} c o s x + b e^{x} s i n x$

Differentiating both sides with respect to x, we get:

$\begin{array}{l} \frac{d y}{d x} = a . \frac{d}{d x} (e^{x} c o s x) + b . \frac{d}{d x} (e^{x} s i n x) \\ \Rightarrow \frac{d y}{d x} = a (e^{x} c o s x - e^{x} s i n x) + b . (e^{x} s i n x + e^{x} c o s x) \\ \Rightarrow \frac{d y}{d x} = (a + b) e^{x} c o s x + (b - a) e^{x} s i n x \end{array}$

Again, differentiating both sides with respect to x, we get:

$\begin{array}{l} \frac{d^{2} y}{d x^{2}} = (a + b) . \frac{d}{d x} (e^{x} c o s x) (b - a) \frac{d}{d x} (e^{x} s i n x) \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = (a + b) . [e^{x} c o s x - e^{x} s i n x] + (b - a) [e^{x} s i n x + e^{x} c o s x] \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = e^{x} [(a + b) (c o s x - s i n x) + (b - a) (s i n x + c o s x)] \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = e^{x} [a c o s x - a s i n x + b c o s x - b s i n x + b s i n x + b c o s x - a s i n x - a c o s x] \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = [2 e^{x} (b c o s x - a s i n x)] \end{array}$

Now, on substituting the values of $\frac{d^{2} y}{d x^{2}}$ and $\frac{d y}{d x}$ in the L.H.S of the given differential equation, we get:

$\begin{array}{l} \frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} + 2 y \\ = 2 e^{x} (b c o s x - a s i n x) - 2 e^{x} [(a + b) c o s x + (b - a) s i n x] + 2 e^{x} (a c o s x + b s i n x) \\ = e^{x} [(2 b c o s x - 2 a s i n x) - (2 a c o s x + 2 b c o s x) - (2 b s i n x - 2 a s i n x) + (2 a c o s x + 2 b s i n x)] \\ = e^{x} [(2 b - 2 a - 2 b + 2 a) c o s x] + e^{x} [(- 2 a - 2 b + 2 a + 2 b) s i n x] \\ = 0 \end{array}$

Therefore, Function given by equation (i) is solution of differential equation (ii)

(iii) $y = x s i n 3 x$

Differentiating both sides with respect to x, we get:

$\begin{array}{l} \frac{d y}{d x} = \frac{d}{d x} (x s i n 3 x) = s i n 3 x + x . c o s 3 x . 3 \\ \Rightarrow \frac{d y}{d x} = s i n 3 x + 3 x c o s 3 x \end{array}$

Again, differentiating both sides with respect to x, we get:

$\begin{array}{l} \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (s i n 3 x) + 3 \frac{d}{d x} (x c o s 3 x) \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = 3 c o s 3 x + 3 [c o s 3 x + x (- s i n 3 x) . 3] \\ \Rightarrow \frac{d^{2} y}{d x^{2}} 6 c o s 3 x - 9 x s i n 3 x \end{array}$

Substituting the value of $\frac{d^{2} y}{d x^{2}}$ in the L.H.S. of the given differential equation, we get:

$\begin{array}{l} \frac{d^{2} y}{d x^{2}} + 9 y - 6 c o s 3 x \\ = (6 . c o s 3 x - 9 x s i n 3 x) + 9 x s i n 3 x - 6 c o s 3 x \\ = 0 \end{array}$

Therefore, Function given by equation (i) is a solution of differential equation (ii).

(iv) $x^{2} = 2 y^{2} l o g y$

Differentiating both sides with respect to x, we get:

$\begin{array}{l} 2 x = 2 . \frac{d}{d x} = [y^{2} l o g y] \\ \Rightarrow x = [2 y . l o g y . \frac{d y}{d x} + y^{2} . \frac{1}{y} . \frac{d y}{d x}] \\ \Rightarrow x = \frac{d y}{d x} (2 y l o g y + y) \\ \Rightarrow \frac{d y}{d x} = \frac{x}{y (1 + 2 l o g y)} \end{array}$

Substituting the value of $\frac{d y}{d x}$ in the L.H.S. of the given differential equation, we get:

$\begin{array}{l} (x^{2} + y^{2}) \frac{d y}{d x} - x y \\ = (2 y^{2} l o g y + y^{2}) . \frac{x}{y (1 + 2 l o g y)} - x y \\ = y^{2} (1 + 2 l o g y) . \frac{x}{y (1 + 2 l o g y)} - x y \\ = x y - x y \\ = 0 \end{array}$

Therefore, Function given by equation (i) is a solution of differential equation (ii).

98. Form the differential equation representing the family of curves ${(x - a)}^{2} + 2 y^{2} = a^{2}$ where a an arbitrary constant.

Equation of the given family of curves is ${(x - a)}^{2} + 2 y^{2} = a^{2}$

$\begin{array}{l} {(x - a)}^{2} + 2 y^{2} = a^{2} \\ \Rightarrow x^{2} + a^{2} - 2 a x + 2 y^{2} = a^{2} \\ \Rightarrow 2 y^{2} = 2 a x - x^{2} . . . . . . . . . . (1) \end{array}$

Differentiating with respect to x, we get:

$\begin{array}{l} 2 y \frac{d y}{d x} = \frac{2 a - 2 x}{2} \\ \Rightarrow \frac{d y}{d x} = \frac{a - x}{2 y} \\ \Rightarrow \frac{d y}{d x} = \frac{2 a - 2 x^{2}}{4 x y} . . . . . . . . . . (2) \end{array}$

From equation (*1), we get:

$2 a x = 2 y^{2} + x^{2}$

On substituting this value in equation (3), we get:

$\begin{array}{l} \frac{d y}{d x} = \frac{2 y^{2} + x^{2} - 2 x^{2}}{4 x y} \\ \Rightarrow \frac{d y}{d x} = \frac{2 y^{2} - x^{2}}{4 x y} \end{array}$

Hence, the differential equation of the family of curves is given as $\frac{d y}{d x} = \frac{2 y^{2} - x^{2}}{4 x y}$

99. Prove that $x^{2} - y^{2} = c {(x^{2} + y^{2})}^{2}$ is the general equation of the differential equation $(x^{3} - 3 x y^{2}) d x = (y^{3} - 3 x^{2} y) d y$ where c is a parameter.

$\Rightarrow \frac{d y}{d x} = \frac{x^{3} - 3 x y^{2}}{y^{3} - 3 x^{2} y} . . . . . . . . . . (1)$

This is a homogenous equation. To simplify it, we need to make the substitution as:

$\begin{array}{l} y = v x \\ \Rightarrow \frac{d}{d x} (y) = \frac{d}{d x} (v x) \\ \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x} \end{array}$

Substituting the values of y and $\frac{d v}{d x}$ in equation (1), we get:

$\begin{array}{l} v + x \frac{d v}{d x} = \frac{x^{3} - 3 x {(v x)}^{2}}{{(v x)}^{3} - 3 x^{2} (v x)} \\ \Rightarrow v + x \frac{d v}{d x} = \frac{1 - 3 v^{2}}{v^{3} - 3 v} \\ \Rightarrow x \frac{d v}{d x} = \frac{1 - 3 v^{2}}{v^{3} - 3 v} - v \\ \Rightarrow x \frac{d v}{d x} = \frac{1 - 3 v^{2} - v (v^{3} - 3 v)}{v^{3} - 3 v} \\ \Rightarrow x \frac{d v}{d x} = \frac{1 - v^{4}}{v^{3} - 3 v} \\ \Rightarrow (\frac{v^{3} - 3 v}{1 - v^{4}}) d v = \frac{d x}{x} \end{array}$

Integrating both sides, we get:

$\begin{array}{l} \int (\frac{v^{3} - 3 v}{1 - v^{4}}) d v = l o g x + l o g C^{'} . . . . . . . . . (2) \\ N o w, \int (\frac{v^{3} - 3 v}{1 - v^{4}}) d v = \int \frac{v^{3} d v}{1 - v^{4}} - 3 \int \frac{v d v}{1 - v^{4}} \\ \Rightarrow \int (\frac{v^{3} - 3 v}{1 - v^{4}}) d v = I_{1} - 3 I_{2}, W h e r e, I_{1} = \int \frac{v^{3} d v}{1 - v^{4}} a n d I_{2} = \int \frac{v d v}{1 - v^{4}} . . . . . . . . . . . (3) \end{array}$

$\begin{array}{l} L e t, 1 - v^{4} = t . \\ ∴ \frac{d}{d v} (1 - v^{4}) = \frac{d t}{d v} \\ \Rightarrow - 4 v^{3} = \frac{d t}{d v} \\ \Rightarrow v^{3} d v = - \frac{d t}{4} \\ N o w, I_{1} = \int - \frac{d t}{4} = - l o g t = - \frac{1}{4} l o g (1 - v^{4}) \end{array}$

$\begin{array}{l} A n d, I_{2} = \int \frac{v d v}{1 - v^{4}} = \int \frac{v d v}{1 - (v^{2})^{2}} \\ L e t, v^{2} = p . \\ ∴ \frac{d}{d v} (v^{2}) = \frac{d p}{d v} \\ \Rightarrow 2 v = \frac{d p}{d v} \\ \Rightarrow v d v = \frac{p}{2} \\ \Rightarrow I_{2} = \frac{1}{2} \int \frac{d p}{1 - p^{2}} = \frac{1}{2 \times 2} l o g | \frac{1 + p}{1 - p} | = \frac{1}{4} l o g | \frac{1 + v^{2}}{1 - v^{2}} | \end{array}$

Substituting the values of $I_{1}$ and $I_{2}$ in equation (3), we get:

$\int (\frac{v^{3} - 3 v}{1 - v^{4}}) d v = - \frac{1}{4} l o g (1 - v^{4}) - \frac{3}{4} l o g | \frac{1 + v^{2}}{1 - v^{2}} |$

Therefore, equation (2) becomes:

$\begin{array}{l} \frac{1}{4} l o g (1 - v^{4}) - \frac{3}{4} l o g | \frac{1 + v^{2}}{1 - v^{2}} | = l o g x + l o g C^{'} \\ \Rightarrow - \frac{1}{4} l o g [(1 - v^{4}) (\frac{1 + v^{2}}{1 - v^{2}})] = l o g C^{'} x \\ \Rightarrow \frac{{(1 + v^{2})}^{4}}{{(1 - v^{2})}^{2}} = {(C^{'} x)}^{- 4} \\ \Rightarrow \frac{{(1 + \frac{y^{2}}{x^{2}})}^{4}}{{(1 - \frac{y^{2}}{x^{2}})}^{2}} = \frac{1}{C^{' 4} x^{4}} \\ \Rightarrow \frac{{(x^{2} + y^{2})}^{4}}{x^{4} {(x^{2} - y^{2})}^{2}} = \frac{1}{C^{' 4} x^{4}} \\ \Rightarrow {(x^{2} - y^{2})}^{2} = C^{' 4} {(x^{2} + y^{2})}^{4} \\ \Rightarrow (x^{2} - y^{2}) = C^{' 2} {(x^{2} + y^{2})}^{2} \\ \Rightarrow x^{2} - y^{2} = C {(x^{2} + y^{2})}^{2}, w h e r e C = C^{' 2} \end{array}$

Hence, the given result is proved.

100. For the differential equation of the family of the circles in the first quadrant which touch the coordinate axes.

The equation of a circle in the first quadrant with centre (a, a) and radius (a) which touches the coordinate axes is:

${(x - a)}^{2} + {(y - a)}^{2} = a^{2} . . . . . . . . . . (1)$

Differentiating equation (1) with respect to x, we get:

$\begin{array}{l} 2 (x - a) + 2 (y - a) \frac{d y}{d x} = 0 \\ \Rightarrow (x - a) + (y - a) y^{'} = 0 \\ \Rightarrow x - a + y y^{'} - a y^{'} = 0 \\ \Rightarrow x + y y^{'} - a (1 + y^{'}) = 0 \\ \Rightarrow a = \frac{x + y y^{'}}{1 + y^{'}} \end{array}$

Substituting the value of a in equation (1), we get:

$\begin{array}{l} {[x - (\frac{x + y y^{'}}{1 + y^{'}})]}^{2} + {[y - (\frac{x + y y^{'}}{1 + y^{'}})]}^{2} = {(\frac{x + y y^{'}}{1 + y^{'}})}^{2} \\ \Rightarrow {[\frac{(x - a) y^{'}}{(1 + y^{'})}]}^{2} + {[\frac{y - x}{1 + y^{'}}]}^{2} = {[\frac{x + y y^{'}}{1 + y^{'}}]}^{2} \\ \Rightarrow {(x - y)}^{2} . y^{' 2} + {(x - y)}^{2} = (x + y y^{'})^{2} \\ \Rightarrow {(x - y)}^{2} [1 + (y^{'})^{2}] = (x + y y^{'})^{2} \end{array}$

Hence, the required differential equation of the family of circles is ${(x - y)}^{2} [1 + (y^{'})^{2}] = (x + y y^{'})^{2}$

101. Kindly Consider the following

Kindly go through the solution

102. Show that the general solution of the differential equation

$\frac{d y}{d x} + \frac{y^{2} + y + 1}{x^{2} + x + 1} = 0$ is given by $(x + y + 1) A (1 - x - y - 2 x y),$ where A is parameter.

Given: Differential equation $\frac{d y}{d x} + \frac{y^{2} + y + 1}{x^{2} + x + 1} = 0$

$\begin{array}{l} \frac{d y}{d x} + \frac{y^{2} + y + 1}{x^{2} + x + 1} = 0 \\ \Rightarrow \frac{d y}{d x} = - \frac{(y^{2} + y + 1)}{x^{2} + x + 1} \\ \Rightarrow \frac{d y}{y^{2} + y + 1} = \frac{- d x}{x^{2} + x + 1} \\ \Rightarrow \frac{d y}{y^{2} + y + 1} + \frac{d x}{x^{2} + x + 1} = 0 \end{array}$

Integrating both sides,

103. Find the equation of the curve passing through the point (0,π/4), whose differential equation is sin x cos y dx + cos x sin y dy = 0.

The differential equation of the given curve is:

$\begin{array}{l} s i n x c o s y d x + c o s x s i n y d y = 0 \\ \Rightarrow \frac{s i n x c o s y d x + c o s x s i n y d y}{c o s x c o s y} = 0 \\ \Rightarrow t a n x d x + t a n y d y = 0 \end{array}$

Integrating both sides, we get:

$\begin{array}{l} l o g (s e c x) + l o g (s e c y) = l o g C \\ l o g (s e c x . s e c y) = l o g C \\ \Rightarrow s e c x . s e c y = C . . . . . . . . . . (1) \end{array}$

The curve passes through point $(0, \frac{π}{4})$

$\begin{array}{l} ∴ 1 \times\sqrt2 = C \\ \Rightarrow C =\sqrt2 \end{array}$

On subtracting $C =\sqrt2$ in equation (10, we get:

$\begin{array}{l} s e c x . s e c y =\sqrt2 \\ \Rightarrow s e c x . \frac{1}{c o s y} =\sqrt2 \\ \Rightarrow c o s y = s e c x/√2 \end{array}$

104. Find the particular solution of the differential equation (1 + e2x ) dy + (1 + y2 ) ex dx = 0, given that y = 1 when x = 0.

$\begin{array}{l} (1 + e^{2 x}) d y + (1 + y^{2}) e^{x} d x = 0 \\ \Rightarrow \frac{d y}{1 + y^{2}} + \frac{e^{x} d x}{1 + e^{2 x}} = 0 \end{array}$

Integrating both sides, we get:

$\begin{array}{l} t a n^{- 1} y + \int \frac{e^{x} d x}{1 + e^{2 x}} = C . . . . . . . . . . (1) \\ L e t, e^{x} = t \Rightarrow e^{2 x} = t^{2} \\ \Rightarrow \frac{d}{d x} (e^{x}) = \frac{d t}{d x} \\ \Rightarrow e^{x} = \frac{d t}{d x} \\ \Rightarrow e^{x} d x = d t \end{array}$

Substituting these values in equation (1), we get:

$\begin{array}{l} t a n^{- 1} y + \int \frac{d t}{1 + t^{2}} = C \\ \Rightarrow t a n^{- 1} y + t a n^{- 1} t = C \\ \Rightarrow t a n^{- 1} y + t a n^{- 1} (e^{x}) = C . . . . . . . . . . (2) \\ N o w, y = 1, a t, x = 0 \end{array}$

Therefore, equation (2) becomes:

$\begin{array}{l} t a n^{- 1} 1 + t a n^{- 1} 1 = C \\ \Rightarrow \frac{π}{4} + \frac{π}{4} = C \\ \Rightarrow C = \frac{π}{2} \end{array}$

Substituting $C = \frac{π}{2}$ in equation (2), we get:

$t a n^{- 1} y + t a n^{- 1} (e^{x}) = \frac{π}{2}$

This is the required solution of the given differential equation.

105. Solve the differential equation: $y e^{\frac{x}{y}} d x = (x e^{\frac{x}{y}} + y^{2}) d y (y \neq 0)$

$\begin{array}{l} y e^{\frac{x}{y}} d x = (x e^{\frac{x}{y}} + y^{2}) d y \\ \Rightarrow y e^{\frac{x}{y}} \frac{d x}{d y} = x e^{\frac{x}{y}} + y^{2} \\ \Rightarrow e^{\frac{x}{y}} [y . \frac{d x}{d y} - x] = y^{2} \\ \Rightarrow e^{\frac{x}{y}} . \frac{[y . \frac{d x}{d y} - x]}{y^{2}} = 1 . . . . . . . . . . (1) \end{array}$

$L e t, e^{\frac{x}{y}} = z$

Differentiating it with respect to y, we get:

$\begin{array}{l} (e^{\frac{x}{y}}) = \frac{d z}{d y} \\ \Rightarrow e^{\frac{x}{y}} . \frac{d}{d y} (\frac{x}{y}) = \frac{d z}{d y} \\ \Rightarrow e^{\frac{x}{y}} . [\frac{y . \frac{d x}{d y} - x}{y^{2}}] = \frac{d z}{d y} . . . . . . . . . . (2) \end{array}$

From equation (1) and equation (2), we get:

$\begin{array}{l} \frac{d z}{d y} = 1 \\ \Rightarrow d z = d y \end{array}$

Integration both sides, we get:

$\begin{array}{l} z = y + C \\ \Rightarrow e^{\frac{x}{y}} y + C \end{array}$

106. Find a particular solution of the differential equation (x – y) (dx + dy) = dx – dy, given that y = –1, when x = 0. (Hint: put x – y = t)

$\begin{array}{l} (x - y) (d x + d y) = d x - d y \\ \Rightarrow (x - y + 1) d y = (1 - x + y) d x \\ \Rightarrow \frac{d y}{d x} = \frac{1 - x + y}{x - y + 1} \\ \Rightarrow \frac{d y}{d x} = \frac{1 - (x - y)}{1 + (x - y)} . . . . . . . . . . (1) \\ L e t, x - y = t \\ \Rightarrow \frac{d}{d x} (x - y) = \frac{d t}{d x} \\ \Rightarrow 1 - \frac{d y}{d x} = \frac{d t}{d x} \\ \Rightarrow 1 - \frac{d t}{d x} = \frac{d y}{d x} \end{array}$

Substituting the values of $x - y$ and $\frac{d y}{d x}$ in equation (1), we get:

$\begin{array}{l} 1 - \frac{d t}{d x} = \frac{1 - t}{1 + t} \\ \Rightarrow \frac{d t}{d x} = 1 - (\frac{1 - t}{1 + t}) \\ \Rightarrow \frac{d t}{d x} = \frac{(1 + t) - (1 - t)}{1 + t} \\ \Rightarrow \frac{d t}{d x} = \frac{2 t}{1 + t} \end{array}$

$\begin{array}{l} \Rightarrow (\frac{1 + t}{t} d t) = 2 d x \\ \Rightarrow (1 + \frac{1}{t}) d t = 2 d x . . . . . . . . . . (2) \end{array}$

Integrating both sides, we get:

$\begin{array}{l} t + l o g | t | = 2 x + C \\ \Rightarrow (x - y) + l o g | x - y | = 2 x + C \\ \Rightarrow l o g | x - y | = x + y + C . . . . . . . . . . (3) \end{array}$

$N o w, y = - 1, a t, x = 0$

Therefore, equation (3) becomes:

$l o g 1 = 0 - 1 + C$

$\Rightarrow C = 1$

Substituting $C = 1$ in equation (3), we get:

$o g | x - y | = x + y + 1$

This is the required particular solution of the given differential equation .

107. Kindly Consider the following

Kindly go through the solution

108. Find the particular solution of the differential equation $\frac{d y}{d x} + y c o t x = 4 x c o s e c x (x \neq 0)$

given that y = 0 when x = π/2

The given differential equation is:

$\frac{d y}{d x} + y c o t x = 4 x c o s e c x$

This equation is a linear equation of the form

$\begin{array}{l} \frac{d y}{d x} + P y = Q, w h e r e, p = c o t x & Q = 4 x c o s e c x \\ N o w, I . F = e^{\int P d x} = e^{\int c o t x d x} = e^{l o g | s i n x |} = s i n x \end{array}$

The general solution of the given differential equation is given by,

$y (I . F) = \int (Q \times I . F .) d x + C$

$\begin{array}{l} \Rightarrow y s i n x = \int (4 x c o s e c x . s i n x) d x + C \\ \Rightarrow y s i n x = 4 \int x d x + C \\ \Rightarrow y s i n x = 4 . \frac{x^{2}}{2} + C \\ \Rightarrow y s i n x = 2 x^{2} + C . . . . . . . . . . (1) \\ N o w, y = 0 a t, x = \frac{π}{2} \end{array}$

Therefore, equation (1) becomes:

$\begin{array}{l} 0 = 2 \times \frac{π}{2} + C \\ \Rightarrow C = - \frac{π^{2}}{2} \end{array}$

Substituting $C = - \frac{π^{2}}{2}$ in equation (1), we get:

$y s i n x = 2 x^{2} - \frac{π^{2}}{2}$

This is the required particular solution of the given differential equation.

109. Find the particular solution of the differential equation $(x + 1) \frac{d y}{d x} = 2 e^{- y} - 1$ given that y = 0 when x = 0

$\begin{array}{l} (x + 1) \frac{d y}{d x} = 2 e^{- y} - 1 \\ \Rightarrow \frac{d y}{2 e^{- y} - 1} = \frac{d x}{x + 1} \\ \Rightarrow \frac{e^{y} d y}{2 - e^{y}} = \frac{d x}{x + 1} \end{array}$

Integrating both sides, we get:

$\begin{array}{l} \int \frac{e^{y} d y}{2 - e^{y}} = l o g | x + 1 | + l o g C . . . . . . . . . . (1) \\ L e t 2 - e^{y} = t \\ ∴ \frac{d}{d y} (2 - e^{y}) = \frac{d t}{d y} \\ \Rightarrow - e^{y} = \frac{d t}{d y} \\ \Rightarrow e^{y} d y = - d t \end{array}$

Substituting this value in equation (1), we get:

$\begin{array}{l} \int \frac{- d t}{t} = l o g o g | x + 1 | + l o g C \\ \Rightarrow - l o g | t | = l o g | C (x + 1) | \\ \Rightarrow - l o g | 2 - e^{y} | = l o g | C (x + 1) | \\ \Rightarrow \frac{1}{2 - e^{y}} = C (x + 1) \\ \Rightarrow 2 - e^{y} = \frac{1}{C (x + 1)} . . . . . . . . . . (2) \end{array}$

Now, at x=0& y=0, equation (2) becomes:

$\begin{array}{l} \Rightarrow 2 - 1 = \frac{1}{C} \\ \Rightarrow C = 1 \end{array}$

Substituting $C = 1$ in equation (2), we get:

$\begin{array}{l} 2 - e^{y} = \frac{1}{x + 1} \\ \Rightarrow e^{y} = 2 - \frac{1}{x + 1} \\ \Rightarrow e^{y} = \frac{2 x + 2 - 1}{x + 1} \\ \Rightarrow e^{y} = \frac{2 x + 1}{x + 1} \\ \Rightarrow y = l o g | \frac{2 x + 1}{x + 1} |, (x \neq - 1) \end{array}$

This is the required particular solution of the given differential equation.

110. The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20,000 in 1999 and 25,000 in the year 2004, what will be the population of the village in 2009?

Let the population at any instant (t) be y.

It is given that the rate of increase of population is proportional to the number of inhabitants at any instant.

$∴ \frac{d y}{d t} α y$

$\Rightarrow \frac{d y}{d t} = k y$ (k is constant)

$\Rightarrow \frac{d y}{y} = k d t$

Integration both sides, we get:

$l o g y = k t + C . . . . . . . . . . (1)$

In the year $1999, t = 0 & y = 20000 .$

Therefore, we get:

$l o g 20000 = C . . . . . . . . . . (2)$

In the year $2004, t = 5 & y = 25000 .$

Therefore, we get:

$\begin{array}{l} \Rightarrow l o g 25000 = 5 k + l o g 20000 \\ \Rightarrow 5 k = l o g (\frac{25000}{20000}) = l o g (\frac{5}{4}) \\ \Rightarrow k = \frac{1}{5} l o g (\frac{5}{4}) . . . . . . . . . . (3) \end{array}$

In the year $2009, t = 10 y e a r s$

Now, on substituting the values of t, k, and C in equation (1), we get:

$\begin{array}{l} l o g y = 10 \times \frac{1}{5} l o g (\frac{5}{4}) + l o g (20000) \\ \Rightarrow l o g y = l o g [20000 \times {(\frac{5}{4})}^{2}] \\ \Rightarrow y = 20000 \times \frac{5}{4} \times \frac{5}{4} \\ \Rightarrow y = 31250 \end{array}$

Hence, the population of the village in 2009 will be 31250.

Choose the correct answer:

111. The general solution of the differential equation $\frac{y d x - x d y}{y} = 0$ is:

$\begin{array}{l} (A) x y = c \\ (B) x = c y^{2} \\ (C) y = c x \\ (D) y = c x^{2} \end{array}$

The given differential equation is:

$\begin{array}{l} \frac{y d x - x d y}{y} = 0 \\ \Rightarrow \frac{y d x - x d y}{x y} = 0 \\ \Rightarrow \frac{1}{x} d x - \frac{1}{y} d y = 0 \end{array}$

Integration both sides, we get:

$\begin{array}{l} l o g | x | - l o g | y | = l o g k \\ \Rightarrow l o g | \frac{x}{y} | = l o g k \\ \Rightarrow \frac{x}{y} = k \\ \Rightarrow y = \frac{1}{k} x \\ \Rightarrow y = C x, w h e r e, C = \frac{1}{k} \end{array}$

Therefore, option (C) is correct.

112. The general equation of a differential equation of the type $\frac{d x}{d y} + P_{1} x = Q, i s :$

$\begin{array}{l} (A) y e^{\int P_{1} d y} = \int (Q_{1} e^{\int P_{1} d y}) d y + C \\ (B) y . e^{\int P_{1} d x} = \int (Q_{1} e^{\int P_{1} d x}) d x + C \\ (C) x e^{\int P_{1} d y} = \int (Q_{1} e^{\int P_{1} d y}) d y + C \\ (D) x . e^{\int P_{1} d x} = \int (Q_{1} e^{\int P_{1} d x}) d x + C \end{array}$

The integrating factor of the given differential equation $\frac{d x}{d y} + P_{1} x = Q_{1}$

The general solution of the differential equation is given by,

$\begin{array}{l} x (I . F .) = \int (Q \times I . F .) d y + C \\ \Rightarrow x . e^{\int P_{1} d y} = \int (Q_{1} e^{\int P_{1} d y}) d y + C \end{array}$

Hence, the correct answer is C.

113. The general solution of the differential equation $e^{x} d y + (y e^{x} + 2 x) d x = 0$ is:

$\begin{array}{l} (A) x e^{y} + x^{2} = C \\ (B) x e^{y} + y^{2} = C \\ (C) y e^{x} + x^{2} = C \\ (D) y e^{y} + x^{2} = C \end{array}$

The given differential equation is:

$\begin{array}{l} e^{x} d y + (y e^{x} + 2 x) d x = 0 \\ \Rightarrow e^{x} \frac{d y}{d x} + y e^{x} + 2 x = 0 \\ \Rightarrow \frac{d y}{d x} + y = - 2 x e^{- x} \end{array}$

This is a linear differential equation of the form

$\begin{array}{l} \frac{d y}{d x} + P y = Q, w h e r e P = 1 & Q = - 2 x e^{- x} \\ N o w, I . F . = e^{\int P d x} = e^{\int d x} = e^{x} \end{array}$

The general solution of the given differential equation is given by,

$\begin{array}{l} y (I . F .) = \int (Q \times I . F .) d x + C \\ \Rightarrow y e^{x} = \int (- 2 x e^{- x} . e^{x}) d x + C \\ \Rightarrow y e^{x} = - \int 2 x d x + C \\ \Rightarrow y e^{x} = - x^{2} + C \\ \Rightarrow y e^{x} + x^{2} = C \end{array}$

Therefore, option (c) is correct.

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