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Maths Differential Equations

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

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Vishal Baghel

Contributor-Level 10

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,

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

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Maths Binomial Theorem

2. a⁵b⁷in (a – 2b)¹².

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6 months ago

Maths Differential Equations

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

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Vishal Baghel

Contributor-Level 10

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

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

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Maths Continuity and Differentiability

117. Examine the applicability of Mean Value Theorem for all three functions given in the above exercise 2.

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alok kumar singh

Contributor-Level 10

117. Solution :
Mean Value Theorem states that for a function f[a,b] →R, if

(a) f is continuous on [a, b]

(b) f is differentiable on (a, b)

then, there exists some c ∈ (a, b) such that $f' (c) = \frac{f (b) - f (a)}{b - a}$

Therefore, Mean Value Theorem is not applicable to those functions that do not satisfy any of the two conditions of the hypothesis.

$f (x) = [x]$ for $x \in [5, 9]$

It is evident that the given function f (x) is not continuous at every integral point.

In particular, f(x) is not continuous at x = 5 and x = 9

⇒ f (x) is not continuous in [5, 9].

The differentiability of f in (5, 9) is checked as follows.

Let n be an integer such that n ∈ (5, 9).

$\underset{h \to 0}{l i m} \frac{f (n + h) - f (n)}{h} = \underset{h \to 0}{l i m} \frac{[n + h] - [n]}{h} = \underset{h \to 0}{l i m} \frac{n - 1 - n}{h} = \underset{h \to 0}{l i m} \frac{- 1}{h} = \infty$

The righ

...more

117. Solution :
Mean Value Theorem states that for a function f[a,b] →R, if

(a) f is continuous on [a, b]

(b) f is differentiable on (a, b)

then, there exists some c ∈ (a, b) such that $f' (c) = \frac{f (b) - f (a)}{b - a}$

Therefore, Mean Value Theorem is not applicable to those functions that do not satisfy any of the two conditions of the hypothesis.

$f (x) = [x]$ for $x \in [5, 9]$

It is evident that the given function f (x) is not continuous at every integral point.

In particular, f(x) is not continuous at x = 5 and x = 9

⇒ f (x) is not continuous in [5, 9].

The differentiability of f in (5, 9) is checked as follows.

Let n be an integer such that n ∈ (5, 9).

The right hand limit of f at x=n is,

$\underset{h \to 0}{l i m} \frac{f (n + h) - f (n)}{h} = \underset{h \to 0}{l i m} \frac{[n + h] - [n]}{h} = \underset{h \to 0}{l i m} \frac{n - n}{h} = \underset{h \to 0}{l i m} 0 = 0$

Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n

∴f is not differentiable in (5, 9).

It is observed that f does not satisfy all the conditions of the hypothesis of Mean Value Theorem.

Hence, Mean Value Theorem is not applicable for $f (x) = [x]$ for $x \in [5, 9]$ .

(ii) $f (x) = [x]$ $i n x \in [- 2, 2]$

It is evident that the given function f (x) is not continuous at every integral point.

In particular, f(x) is not continuous at x = −2 and x = 2

⇒ f (x) is not continuous in [−2, 2].

The differentiability of f in (−2, 2) is checked as follows.

Let n be an integer such that n ∈ (−2, 2).

The right hand limit of f at x=n is,

$\underset{h \to 0}{l i m} \frac{f (n + h) - f (n)}{h} = \underset{h \to 0}{l i m} \frac{[n + h] - [n]}{h} = \underset{h \to 0}{l i m} \frac{n - n}{h} = \underset{h \to 0}{l i m} 0 = 0$

Since the left and right hand limits of f at x = n are not equal, f is not differentiable at x = n

∴f is not differentiable in (−2, 2).

It is observed that f does not satisfy all the conditions of the hypothesis of Mean Value Theorem.

Hence, Mean Value Theorem is not applicable for $f (x) = [x]$ for $x \in [- 2, 2]$

(iii) $f (x) = x^{2} - 1$ for $x \in [1, 2]$

It is evident that f, being a polynomial function, is continuous in [1, 2] and is differentiable in (1, 2).

It is observed that f satisfies all the conditions of the hypothesis of Mean Value Theorem.

Hence, Mean Value Theorem is applicable for $f (x) = x^{2} - 1$ for $x \in [1, 2]$

It can be proved as follows.

$\begin{array}{l} f (1) = 1^{2} - 1 = 0, f (2) = 2^{2} - 1 = 3 \\ ∴ \frac{f (b) - f (a)}{b - a} = \frac{f (2) - f (1)}{2 - 1} = \frac{3 - 0}{1} = 3 \\ f' (x) = 2 x \\ ∴ f' (c) = 3 \\ \Rightarrow 2 c = 3 \\ \Rightarrow c = \frac{3}{2} = 1.5, w h e r e, 1.5 \in (1, 2) \end{array}$

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6 months ago

Maths Differential Equations

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

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Vishal Baghel

Contributor-Level 10

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.

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Maths Differential Equations

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

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Vishal Baghel

Contributor-Level 10

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.

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Maths Continuity and Differentiability

116. Verify Mean Value Theorem, if $f (x) = x^{3} - 5 x^{2} - 3 x$ in the interval [a, b], where a = 1 and b = 3. Find all c ∈ (1, 3) for which f'(c) = 0

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alok kumar singh

Contributor-Level 10

116. Solution:
The given function f is $f (x) = x^{3} - 5 x^{2} - 3 x$ f, being a polynomial function, is continuous in [1, 3] and is differentiable in (1, 3) whose derivative is 3x2 − 10x − 3.

$\begin{array}{l} f (1) = 1^{3} - 5 * 1^{2} - 3 * 1, f (3) = 3^{3} - 5 * 3^{2} - 3 * 3 = - 27 \\ ∴ \frac{f (b) - f (a)}{b - a} = \frac{f (3) - f (1)}{3 - 1} = \frac{- 27 - (- 7)}{3 - 1} = - 10 \end{array}$

Mean Value Theorem states that there exists a point c ∈ (1, 3) such that f'(c) = - 10

$\begin{array}{l} f' (c) = - 10 \\ \Rightarrow 3 c^{2} - 10 c - 3 = 10 \\ \Rightarrow 3 c^{2} - 10 c + 7 = 0 \\ \Rightarrow 3 c^{2} - 3 c - 7 c + 7 = 0 \\ \Rightarrow 3 c (c - 1) - 7 (c - 1) = 0 \\ \Rightarrow (c - 1) (3 c - 7) = 0 \\ \Rightarrow c = 1, \frac{7}{3}, w h e r e, c = \frac{7}{3} \in (1, 3) \end{array}$

Hence, Mean Value Theorem is verified for the given function and c = 7/3 ∈ (1,3) is the only point for which f'(c) = 0

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6 months ago

Maths Differential Equations

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

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Vishal Baghel

Contributor-Level 10

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.

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Maths Differential Equations

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

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Vishal Baghel

Contributor-Level 10

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 $c_{1} = \frac{1}{c}$

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Maths Continuity and Differentiability

115. Verify Mean Value Theorem, if $f (x) = x^{2} - 4 x - 3$ in the interval [a,b], where a = 1 and b = 4.

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alok kumar singh

Contributor-Level 10

115.

Solution :
The given function is $f (x) = x^{2} - 4 x - 3$ f, being a polynomial function, is continuous in [1, 4] and is differentiable in (1, 4) whose derivative is 2x − 4.

$\begin{array}{l} f (1) = 1^{2} - 4 * 1 - 3, f (4) = 4^{2} - 4 * 4 - 3 \\ ∴ \frac{f (b) - f (a)}{b - a} = \frac{f (4) - f (1)}{4 - 1} = \frac{- 3 - (- 6)}{3} = \frac{3}{3} = 1 \end{array}$

Mean Value Theorem states that there is a point c ∈ (1, 4) such that f' (c) = 1

$\begin{array}{l} f' (c) = 1 \\ \Rightarrow 2 c - 4 = 1 \\ \Rightarrow c = \frac{5}{2}, w h e r e, c = \frac{5}{2} \in (1, 4) \end{array}$

Hence, Mean Value Theorem is verified for the given function.

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