Solve: (dydx)+xy=x(sinx+logx) .

This is a Long Answer Type Questions as classified in NCERT Exemplar Sol: T h e g i v e n d i f f e r e n t i a l e q u a t i o n i s y + d d x ( x y ) = x ( s i n x + l o g x ) ⇒ y + x . d y d x + y = x ( s i n x + l o g x ) ⇒ x d y d x = x ( s i n x + l o g x ) − 2 y ⇒ d y d x = ( s i n x + l o g x ) − 2 y x ⇒ d y d x + 2 x y = ( s i n x + l o g x ) H e r e , P = 2 x a n d Q = ( s i n x + l o g x ) I n t e g r a t i n g f a c t o r I . F = e ∫ P d x = e ∫ 2 x . d x = e 2 l o g x = e l o g x 2 = x 2 ∴ S o l u t i o n o f t h e e q u a t i o n i s y × I . F . = ∫ Q × I . F . d x + c ⇒ y . x 2 = ∫ ( s i n x + l o g x ) x 2 d x + c … ( 1 ) L e t I = ∫ ( s i n x + l o g x ) x 2 d x = ∫ x 2 I s i n x I I d x + ∫ x 2 I I l o g x I d x = [ x 2 . ∫ s i n x d x − ∫ ( D ( x 2 ) . ∫ s i n x d x ) d x ] + [ l o g x . ∫ x 2 d x − ∫ ( D ( l o g x ) . ∫ x 2 d x ) d x ] = [ x 2 ( − c o s x ) − 2 ∫ − x c o s x d x ] + [ l o g x . x 3 3 − ∫ 1 x . x 3 3 d x ] = [ − x 2 c o s x + 2 ( x s i n x − ∫ 1 . s i n x d x ) ] + [ x 3 3 l o g x − 1 3 ∫ x 2 d x ] = − x 2 c o s x + 2 x s i n x + 2 c o s x + x 3 3 l o g x − 1 9 x 3 N o w f r o m e q ( 1 ) w e g e t , y . x 2 = − x 2 c o s x + 2 x s i n x + 2 c o s x + x 3 3 l o g x − 1 9 x 3 + c ∴ y = − c o s x + 2 s i n x x + 2 c o s x x 2 + x l o g x 3 − 1 9 x + c . x − 2 H e n c e , t h e r e q u i r e d s o l u t i o n i s y = − c o s x + 2 s i n x x + 2 c o s x x 2 + x l o g x 3 − 1 9 x + c . x − 2

Solve: dydx=cos(x+y)+sin(x+y) . [Hint: Substitute x+y=z ]

This is a Long Answer Type Questions as classified in NCERT Exemplar Sol: G i v e n t h a t d y d x = c o s ( x + y ) + s i n ( x + y ) P u t x + y = v , o n d i f f e r e n t i a t i n g w . r . t . x , w e g e t , 1 + d y d x = d v d x ∴ d y d x = d v d x − 1 ∴ d v d x − 1 = c o s v + s i n v ⇒ d v d x = c o s v + s i n v + 1 ⇒ d v c o s v + s i n v + 1 = d x I n t e g r a t i n g b o t h s i d e s , w e h a v e ∫ d v c o s v + s i n v + 1 = ∫ 1 . d x ⇒ ∫ d v ( 1 − t a n 2 v 2 1 + t a n 2 v 2 + 2 t a n v 2 1 + t a n 2 v 2 + 1 ) = ∫ 1 . d x ⇒ ∫ 1 + t a n 2 v 2 ( 1 − t a n 2 v 2 + 2 t a n v 2 + 1 ) d v = ∫ 1 . d x ⇒ ∫ s e c 2 v 2 2 + 2 t a n v 2 d v = ∫ 1 . d x P u t 2 + 2 t a n v 2 = t 2 . 1 2 s e c 2 v 2 d v = d t ⇒ s e c 2 v 2 d v = d t ⇒ ∫ d t 2 = ∫ 1 . d x ⇒ l o g | t | = x + c ⇒ l o g | 2 + 2 t a n v 2 | = x + c ⇒ l o g | 2 + 2 t a n ( x + y 2 ) | = x + c ⇒ l o g 2 [ 1 + t a n ( x + y 2 ) ] = x + c ⇒ l o g 2 + l o g [ 1 + t a n ( x + y 2 ) ] = x + c ⇒ l o g [ 1 + t a n ( x + y 2 ) ] = x + c − l o g 2 H e n c e , t h e r e q u i r e d s o l u t i o n i s l o g [ 1 + t a n ( x + y 2 ) ] = x + K [ c − l o g 2 = K ]

Let R be a relation from the set {1,2,3,.....,60} to itself such that R = {(a,b):b=pq} where p, q ≥ 3 are prime numbers. Then, the number of elements in R is:

R = { (a, b): b = pq, where p, q ≥3 are prime} ⇒60×11=660 p, q ∈ {3, 5, 7, 11, 13, 17, 19, 23, 29, 3, 37, 41, 43, 47, 53, 59} total 16 p, q ∈ {3, 5, 7, 11, 13, 17, 19} {3, 5, 7, 11, 13, 17, 19} 7 + 3 + 1 = 11

If z = 2 + 3 i, then z5 + (z¯)5 is equal to:

z5+ (z¯)5 = (2+3i)5+ (2−3i)5 =2 (32−720+810)=244

Maths NCERT Exemplar Solutions Class 12th Chapter Nine: Overview, Questions, Preparation

Q: Let A and B be two 3 × 3 non-zero real matrices such that AB is a zero matrix. Then

A3×3, B3×3&AB=0 ≠0≠0⇒|A|=0 &|B|=0 only .

Updated on Jul 11, 2025 15:09 IST

By Vishal Baghel, Executive Content Operations

Table of content

Differential Equations Questions and Answers
JEE Mains Solution 2022

Differential Equations Questions and Answers

Q1.Solve: $(\frac{d y}{d x}) + x y = x (s i n x + l o g x)$ .

Sol:

$\begin{array}{l} T h e g i v e n d i f f e r e n t i a l e q u a t i o n i s y + \frac{d}{d x} (x y) = x (s i n x + l o g x) \\ \Rightarrow y + x . \frac{d y}{d x} + y = x (s i n x + l o g x) \\ \Rightarrow x \frac{d y}{d x} = x (s i n x + l o g x) - 2 y \\ \Rightarrow \frac{d y}{d x} = (s i n x + l o g x) - \frac{2 y}{x} \\ \Rightarrow \frac{d y}{d x} + \frac{2}{x} y = (s i n x + l o g x) \\ H e r e, P = \frac{2}{x} a n d Q = (s i n x + l o g x) \\ I n t e g r a t i n g f a c t o r I . F = e^{\int P d x} = e^{\int \frac{2}{x} . d x} = e^{2 l o g x} = e^{l o g x^{2}} = x^{2} \\ ∴ S o l u t i o n o f t h e e q u a t i o n i s \\ y \times I . F . = \int Q \times I . F . d x + c \\ \Rightarrow y . x^{2} = \int (s i n x + l o g x) x^{2} d x + c \dots (1) \\ L e t I = \int (s i n x + l o g x) x^{2} d x \\ = \int \underset{I}{x^{2}} \underset{I I}{s i n x} d x + \int \underset{I I}{x^{2}} \underset{I}{l o g x} d x \\ = [x^{2} . \int s i n x d x - \int (D (x^{2}) . \int s i n x d x) d x] + [l o g x . \int x^{2} d x - \int (D (l o g x) . \int x^{2} d x) d x] \\ = [x^{2} (- c o s x) - 2 \int - x c o s x d x] + [l o g x . \frac{x^{3}}{3} - \int \frac{1}{x} . \frac{x^{3}}{3} d x] \\ = [- x^{2} c o s x + 2 (x s i n x - \int 1 . s i n x d x)] + [\frac{x^{3}}{3} l o g x - \frac{1}{3} \int x^{2} d x] \\ = - x^{2} c o s x + 2 x s i n x + 2 c o s x + \frac{x^{3}}{3} l o g x - \frac{1}{9} x^{3} \\ N o w f r o m e q (1) w e g e t, \\ y . x^{2} = - x^{2} c o s x + 2 x s i n x + 2 c o s x + \frac{x^{3}}{3} l o g x - \frac{1}{9} x^{3} + c \\ ∴ y = - c o s x + \frac{2 s i n x}{x} + \frac{2 c o s x}{x^{2}} + \frac{x l o g x}{3} - \frac{1}{9} x + c . x^{- 2} \\ H e n c e, t h e r e q u i r e d s o l u t i o n i s \\ y = - c o s x + \frac{2 s i n x}{x} + \frac{2 c o s x}{x^{2}} + \frac{x l o g x}{3} - \frac{1}{9} x + c . x^{- 2} \end{array}$

Q2.Find the general solution of $(1 + t a n y) (d x- d y) + 2 x d y = 0$ .

Sol:

$\begin{array}{l} G i v e n t h a t (1 + t a n y) (d x - d y) + 2 x d y = 0 \\ \Rightarrow (1 + t a n y) d x - (1 + t a n y) d y + 2 x d y = 0 \\ \Rightarrow (1 + t a n y) d x - (1 + t a n y - 2 x) d y = 0 \\ \Rightarrow (1 + t a n y) \frac{d x}{d y} = (1 + t a n y - 2 x) \Rightarrow \frac{d x}{d y} = \frac{1 + t a n y - 2 x}{1 + t a n y} \\ \Rightarrow \frac{d x}{d y} = 1 - \frac{2 x}{1 + t a n y} \Rightarrow \frac{d x}{d y} + \frac{2 x}{1 + t a n y} = 1 \\ H e r e, P = \frac{2}{1 + t a n y} a n d Q = 1 \\ I n t e g r a t i n g f a c t o r I . F = e^{\int P d y} = e^{\int \frac{2}{1 + t a n y} . d y} = e^{\int \frac{2 c o s y}{s i n y + c o s y} . d y} \\ = e^{\int \frac{s i n y + c o s y - s i n y + c o s y}{s i n y + c o s y} . d y} = e^{\int (1 + \frac{c o s y - s i n y}{s i n y + c o s y}) . d y} \\ = e^{\int 1 . d y} . e^{\int \frac{c o s y - s i n y}{s i n y + c o s y} . d y} = e^{y} . e^{l o g (s i n y + c o s y)} = e^{y} . (s i n y + c o s y) \\ ∴ S o l u t i o n o f t h e e q u a t i o n i s \\ x \times I . F . = \int Q \times I . F . d y + c \\ \Rightarrow x . e^{y} . (s i n y + c o s y) = \int 1 . e^{y} . (s i n y + c o s y) d y + c \\ \Rightarrow x . e^{y} . (s i n y + c o s y) = e^{y} . s i n y + c [∵ \int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + c] \\ \Rightarrow x (s i n y + c o s y) = s i n y + c . e^{- y} \\ H e n c e, t h e r e q u i r e d s o l u t i o n i s x (s i n y + c o s y) = s i n y + c . e^{- y} \end{array}$

Q3.Solve: $\frac{d y}{d x} = c o s (x + y) + s i n (x + y)$ . [Hint: Substitute $x + y = z$ ]

Sol:

$\begin{array}{l} G i v e n t h a t \frac{d y}{d x} = c o s (x + y) + s i n (x + y) \\ P u t x + y = v, o n d i f f e r e n t i a t i n g w . r . t . x, w e g e t, \\ 1 + \frac{d y}{d x} = \frac{d v}{d x} \\ ∴ \frac{d y}{d x} = \frac{d v}{d x} - 1 \\ ∴ \frac{d v}{d x} - 1 = c o s v + s i n v \\ \Rightarrow \frac{d v}{d x} = c o s v + s i n v + 1 \\ \Rightarrow \frac{d v}{c o s v + s i n v + 1} = d x \\ I n t e g r a t i n g b o t h s i d e s, w e h a v e \\ \int \frac{d v}{c o s v + s i n v + 1} = \int 1 . d x \\ \Rightarrow \int \frac{d v}{(\frac{1 - t a n^{2} \frac{v}{2}}{1 + t a n^{2} \frac{v}{2}} + \frac{2 t a n \frac{v}{2}}{1 + t a n^{2} \frac{v}{2}} + 1)} = \int 1 . d x \\ \Rightarrow \int \frac{1 + t a n^{2} \frac{v}{2}}{(1 - t a n^{2} \frac{v}{2} + 2 t a n \frac{v}{2} + 1)} d v = \int 1 . d x \\ \Rightarrow \int \frac{s e c^{2} \frac{v}{2}}{2 + 2 t a n \frac{v}{2}} d v = \int 1 . d x \\ P u t 2 + 2 t a n \frac{v}{2} = t \\ 2 . \frac{1}{2} s e c^{2} \frac{v}{2} d v = d t \Rightarrow s e c^{2} \frac{v}{2} d v = d t \\ \Rightarrow \int \frac{d t}{2} = \int 1 . d x \\ \Rightarrow l o g | t | = x + c \\ \Rightarrow l o g | 2 + 2 t a n \frac{v}{2} | = x + c \\ \Rightarrow l o g | 2 + 2 t a n (\frac{x + y}{2}) | = x + c \\ \Rightarrow l o g 2 [1 + t a n (\frac{x + y}{2})] = x + c \\ \Rightarrow l o g 2 + l o g [1 + t a n (\frac{x + y}{2})] = x + c \\ \Rightarrow l o g [1 + t a n (\frac{x + y}{2})] = x + c - l o g 2 \\ H e n c e, t h e r e q u i r e d s o l u t i o n i s l o g [1 + t a n (\frac{x + y}{2})] = x + K [c - l o g 2 = K] \end{array}$

Q4.Find the general solution of $\frac{d y}{d x} - 3 y = s i n 2 x$ .

Sol:

$\begin{array}{l} G i v e n e q u a t i o n i s \frac{d y}{d x} - 3 y = s i n 2 x \\ H e r e, P = - 3 a n d Q = s i n 2 x \\ I n t e g r a t i n g f a c t o r I . F = e^{\int P d x} = e^{\int - 3 . d x} = e^{- 3 x} \\ ∴ S o l u t i o n o f t h e e q u a t i o n i s \\ y \times I . F . = \int Q \times I . F . d x + c \\ \Rightarrow y . e^{- 3 x} = \int s i n 2 x . e^{- 3 x} d x + c \\ L e t I = \int \underset{I}{s i n 2 x} . \underset{I I}{e^{- 3 x}} d x \\ \Rightarrow I = s i n 2 x . \int e^{- 3 x} d x - \int (D (s i n 2 x) . \int e^{- 3 x} d x) d x \\ \Rightarrow I = s i n 2 x . \frac{e^{- 3 x}}{- 3} - \int 2 c o s 2 x . \frac{e^{- 3 x}}{- 3} d x \\ \Rightarrow I = \frac{e^{- 3 x}}{- 3} s i n 2 x + \frac{2}{3} \int \underset{I}{c o s 2 x} . \underset{I I}{e^{- 3 x}} d x \\ \Rightarrow I = \frac{e^{- 3 x}}{- 3} s i n 2 x + \frac{2}{3} [c o s 2 x . \int e^{- 3 x} d x - \int (D (c o s 2 x) . \int e^{- 3 x} d x) d x] \\ \Rightarrow I = \frac{e^{- 3 x}}{- 3} s i n 2 x + \frac{2}{3} [c o s 2 x . \frac{e^{- 3 x}}{- 3} - \int 2 s i n 2 x . \frac{e^{- 3 x}}{- 3} d x] \\ \Rightarrow I = \frac{e^{- 3 x}}{- 3} s i n 2 x - \frac{2}{9} c o s 2 x . e^{- 3 x} - \frac{4}{9} \int s i n 2 x . e^{- 3 x} d x \\ \Rightarrow I = \frac{e^{- 3 x}}{- 3} s i n 2 x - \frac{2}{9} c o s 2 x . e^{- 3 x} - \frac{4}{9} I \\ \Rightarrow I + \frac{4}{9} I = \frac{e^{- 3 x}}{- 3} s i n 2 x - \frac{2}{9} e^{- 3 x} c o s 2 x \\ \Rightarrow \frac{13}{9} I = - \frac{1}{9} [3 e^{- 3 x} s i n 2 x + 2 e^{- 3 x} c o s 2 x] \\ \Rightarrow I = - \frac{1}{13} e^{- 3 x} [3 s i n 2 x + 2 c o s 2 x] \\ ∴ T h e e q u a t i o n b e c o m e s \\ \Rightarrow y . e^{- 3 x} = - \frac{1}{13} e^{- 3 x} [3 s i n 2 x + 2 c o s 2 x] + c \\ ∴ y = - \frac{1}{13} [3 s i n 2 x + 2 c o s 2 x] + c . e^{3 x} \\ H e n c e, t h e r e q u i r e d s o l u t i o n i s y = - [\frac{3 s i n 2 x + 2 c o s 2 x}{13}] + c . e^{3 x} \end{array}$

Commonly asked questions

Solve: $(\frac{d y}{d x}) + x y = x (s i n x + l o g x)$ .

Find the general solution of $(1 + t a n y) (d x- d y) + 2 x d y = 0$ .

Solve: $\frac{d y}{d x} = c o s (x + y) + s i n (x + y)$ . [Hint: Substitute $x + y = z$ ]

JEE Mains Solution 2022

Commonly asked questions

Let R be a relation from the set ${1, 2, 3, . . . . ., 60}$ to itself such that R = ${(a, b) : b = p q}$ where p, q $\geq$ 3 are prime numbers. Then, the number of elements in R is:

If z = 2 + 3 i, then z5 + ${(\bar{z})}^{5}$ is equal to:

Let A and B be two 3 × 3 non-zero real matrices such that AB is a zero matrix. Then

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Maths NCERT Exemplar Solutions Class 12th Chapter Nine: Overview, Questions, Preparation

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Commonly asked questions

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