RD Sharma Class 12 Exercise 21.7 Differential Equations Solutions Maths

RD Sharma Class 12 Exercise 21.7 Differential Equation Solutions Maths - Download PDF Free Online

Updated on Jan 24, 2022 04:20 PM IST

The most preferred set of solution books by the students of CBSE board schools are the RD Sharma books. It helps the students prepare well for all their public exams in every subject and chapter. For the students who encounter lots of doubts and confusion while solving Differential Equations, the RD Sharma Class 12th Exercise 21.7 is the rescuer. A good guide is essential for the students who are preparing for their public exams.

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RD Sharma Class 12 Solutions Chapter21 Differential Equation - Other Exercise
Differential Equations Excercise:21.7
RD Sharma Chapter wise Solutions

RD Sharma Class 12 Solutions Chapter21 Differential Equation - Other Exercise

Differential Equations Excercise:21.7

Differential Equations exercise 21.7 question 1

Answer: $\log y + c = 2 x + 2 \log | x - 1 |$
Hint:Separate the terms of x and y and then integrate them.
Given: $(x - 1) \frac{d y}{d x} = 2 x y$
Solution:
$\begin{aligned} (x - 1) \frac{d y}{d x} = 2 x y \\ \frac{d y}{y} = (\frac{2 x}{x - 1}) d x \\ \frac{d y}{y} = 2 (1 + \frac{1}{x - 1}) d x \end{aligned}$
Integrating both sides

\log y + c = 2 x + 2 \log | x - 1 |

Differential Equations exercise 21.7 question 2

Answer: $C \sqrt{1 + x^{2}}$
Hint: Separate the terms of x and y and then integrate them.
Given: $(1 + x^{2}) d y = x y d x$
Solution: $(1 + x^{2}) d y = x y d x$
$\begin{aligned} \frac{d y}{y} = \frac{x d x}{1 + x^{2}} \\ Put 1 + x^{2} = t \\ 2 x d x = d t \end{aligned}$
$\begin{aligned} x d x = \frac{d t}{2} \\ \frac{d y}{y} = \frac{d t}{2 t} \end{aligned}$
Integrating both sides

\begin{aligned} \log | y | = \frac{1}{2} \log | t | + \log C \\ \log | y | = \frac{1}{2} \log | 1 + x^{2} | + \log C \end{aligned}

\begin{aligned} \log | y | = {(\log | 1 + x^{2} |)}^{\frac{1}{2}} + \log C \\ \log | y | = \log [C \sqrt{1 + x^{2}}] \\ y = C \sqrt{1 + x^{2}} \end{aligned}

Differential Equations exercise 21.7 question 3

Answer: $\log | y | = e^{x} + x + c$
Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x} = (e^{x} + 1) y$
Solution: $\frac{d y}{d x} = (e^{x} + 1) y$
$\frac{d y}{y} = (e^{x} + 1) d x$
Integrating on both sides
$\log | y | = e^{x} + x + c$

Differential Equations exercise 21.7 question 4 maths

Answer: $\log | y | = \frac{2}{3} x^{3} + x^{2} + 2 x + 2 \log | x - 1 | + c$
Hint: Separate the terms of x and y and then integrate them.
Given: $(x - 1) \frac{d y}{d x} = 2 x^{3} y$
Solution: $(x - 1) \frac{d y}{d x} = 2 x^{3} y$
$\begin{matrix} \frac{d y}{y} = \frac{2 x^{8}}{x - 1} d x \\ \frac{d y}{y} = \frac{2 ((x - 1) (x^{2} + x + 1) + 1)}{(x - 1)} d x \\ \frac{d y}{y} = 2 (x^{2} + x + 1 + \frac{1}{(x - 1)}) d x \end{matrix}$
Integrating both sides

\begin{aligned} \int \frac{d y}{y} = \int 2 (x^{2} + x + 1 + \frac{1}{(x - 1)}) d x \\ \log | y | = \frac{2 x^{3}}{3} + \frac{2 x^{2}}{2} + 2 x + 2 \log | x - 1 | + c \\ \log | y | = \frac{2 x^{3}}{3} + x^{2} + 2 x + 2 \log | x - 1 | + c \end{aligned}

Differential Equations exercise 21.7 question 5

Answer: $\frac{y^{3}}{3} + \frac{y^{2}}{2} = \frac{x^{2}}{2} + \log | x | + c$
Hint: Separate the terms of x and y and then integrate them.
Given: $x y (y + 1) d y = (x^{2} + 1) d x$
Solution: $x y (y + 1) d y = (x^{2} + 1) d x$
$\begin{aligned} y (y + 1) d y = \frac{(x^{2} + 1) d x}{x} \\ (y^{2} + y) d y = (\frac{x^{2}}{x} + \frac{1}{x}) d x \end{aligned}$
Integrating both sides

\begin{aligned} \int (y^{2} + y) d y = \int (x + \frac{1}{x}) d x \\ \frac{y^{3}}{3} + \frac{y^{2}}{2} = \frac{x^{2}}{2} + \log | x | + c \end{aligned}

Differential Equations exercise 21.7 question 6

Answer: $e^{y} = e^{x} + c$
Hint: Separate the terms of x and y and then integrate them.
Given: $e^{y - x} \frac{d y}{d x} = 1$
Solution: $e^{y - x} \frac{d y}{d x} = 1$
$\begin{aligned} e^{y} \cdot e^{- x} \frac{d y}{d x} = 1 \\ \frac{e^{y}}{e^{x}} d y = d x \\ e^{y} d y = e^{x} d x \end{aligned}$
Integrating both sides

\begin{aligned} \int e^{y} d y = \int e^{x} d x \\ e^{y} = e^{x} + c \end{aligned}

Differential Equations exercise 21.7 question 7

Answer: $\sin y = e^{x} \log x + c$
Hint:Separate the terms of x and y and then integrate them.
Given: $x \cos y d y = (x e^{x} \log x + e^{x}) d x$
Solution: $x \cos y d y = (x e^{x} \log x + e^{x}) d x$
$\cos y d y = [\frac{(x e^{x} \log x + e^{x})}{x}] d x$
Integrating both sides,

\begin{aligned} \int \cos y d y = \int \frac{(x e^{x} \log x + e^{x})}{x} d x \\ \sin y = \int (e^{x} \log x + \frac{e^{x}}{x}) d x \\ \sin y = \int e^{x} \log x d x + \int \frac{e^{x}}{x} d x \end{aligned}

[Integrating by parts]

\begin{aligned} \sin y = \log x \int e^{x} d x - \int \frac{1}{x} (\int e^{x} d x) d x + \int \frac{e^{x}}{x} d x \\ \sin y = \log x e^{x} - \int \frac{e^{x}}{x} d x + \int \frac{e^{x}}{x} d x + c \\ \sin y = e^{x} \log x + c \end{aligned}

Differential Equations exercise 21.7 question 8 maths

Answer: $- e^{- y} = e^{x} + \frac{x^{3}}{3} + 3 + c$
Hint:Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x} = e^{x + y} + x^{2} e^{y}$
Solution: $\frac{d y}{d x} = e^{x + y} + x^{2} e^{y}$
$\begin{aligned} \frac{d y}{d x} = e^{x} \cdot e^{y} + x^{2} e^{y} \\ d y = e^{y} [e^{x} + x^{2}] d x \\ \frac{d y}{e^{y}} = [e^{x} + x^{2}] d x \end{aligned}$
Integrating both sides

\begin{array}{r} \int e^{- y} d y = \int e^{x} d x + \int x^{2} d x \\ - e^{- y} = e^{x} + \frac{x^{3}}{3} + 3 + c \end{array}

Differential Equations exercise 21.7 question 9

Answer: $y - 1 = C x y$
Hint:Separate the terms of x and y and then integrate them.
Given: $x \frac{d y}{d x} + y = y^{2}$
Solution: $x \frac{d y}{d x} + y = y^{2}$

\begin{aligned} x \frac{d y}{d x} + y = y^{2} \\ x \frac{d y}{d x} = y^{2} - y \\ x d y = (y^{2} - y) d x \\ \frac{d y}{y (y - 1)} = \frac{d x}{x} \end{aligned}

Integrating on both sides

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

\begin{aligned} \int \frac{y - (y - 1)}{y (y - 1)} d y = \int \frac{d x}{x} \\ \int \frac{1}{(y - 1)} d y - \int \frac{1}{y} d y = \int \frac{d x}{x} \end{aligned}

We get,

\begin{aligned} - \log y + \log (y - 1) = \log x + \log c \\ \log \frac{(y - 1)}{y} = \log x C \\ \frac{y - 1}{y} = C x \\ y - 1 = C x y \end{aligned}

Differential Equations exercise 21.7 question 10

Answer: $(e^{y} + 1) \sin x = c$
Hint:Separate the terms of x and y and then integrate them.
Given: $(e^{y} + 1) \cos x d x + e^{y} \sin x d y = 0$
Solution: $(e^{y} + 1) \cos x d x + e^{y} \sin x d y = 0$
$e^{y} \sin x d y = - (e^{y} + 1) \cos x d x$
$\frac{- e^{y}}{(e^{y} + 1)} d y = \frac{\cos x}{\sin x} d x$
Integrating both sides

\begin{aligned} \int \frac{- e^{y}}{(e^{y} + 1)} d y = \int \frac{\cos x}{\sin x} d x \\ Put e^{y} + 1 = t \Rightarrow e^{y} d y = d t \\ \sin x = u \Rightarrow d x \cos x = d u \end{aligned}

\begin{aligned} - \int \frac{d t}{t} = \int \frac{d u}{u} \\ - \log | t | = \log | u | + \log c \\ - \log | e^{y} + 1 | = \log | \sin x | + \log c \end{aligned}

\begin{aligned} - \log | e^{y} + 1 | - \log | \sin x | = \log c \\ - [\log (e^{y} + 1) (\sin x)] = \log c \\ [\log (e^{y} + 1) (\sin x)] = - \log c \\ (e^{y} + 1) \sin x = c \end{aligned}

Differential Equations exercise 21.7 question 11

Answer: $x \tan x - y \tan y = \log | \sec x | - \log | \sec y | + c$
Hint: Separate the terms of x and y and then integrate them.
Given: $x \cos^{2} y d x = y \cos^{2} x d y$
Solution: $x \cos^{2} y d x = y \cos^{2} x d y$
$\begin{aligned} \frac{x}{\cos^{2} x} d x = \frac{y}{\cos^{2} y} d y \\ x \sec^{2} x d x = y \sec^{2} y d y \end{aligned}$
Integrating both sides

\int x \sec^{2} x d x = \int y \sec^{2} y d y

Using integration by parts

\begin{aligned} x \int \sec^{2} x d x = y \int \sec^{2} y d y \\ x \tan x - \int \tan x d x = y \tan y - \int \tan y d y \end{aligned}

Using identity,

\int \tan x d x = \log | \sec x |

\begin{aligned} x \tan (x) - \log | \sec x | = y \tan y - \log | \sec y | + c \\ x \tan x - y \tan y = \log | \sec x | - \log | \sec y | + c \end{aligned}

Differential Equations exercise 21.7 question 12 maths

Answer: $y - x = \log | x | - \log | y - 1 | + c$
Hint: Separate the terms of x and y and then integrate them.
Given: $x y d y = (y - 1) (x + 1) d x$
Solution: $x y d y = (y - 1) (x + 1) d x$
$\begin{aligned} \frac{y d y}{y - 1} = \frac{(x + 1)}{x} d x \\ (\frac{1 + y - 1}{y - 1}) d y = (1 + \frac{1}{x}) d x \\ (\frac{1}{y - 1} + \frac{(y - 1)}{(y - 1)}) d y = (1 + \frac{1}{x}) d x \end{aligned}$
Integrating both sides

\begin{aligned} \int \frac{1}{y - 1} d y + \int 1 d y = \int 1 d x + \int \frac{1}{x} d x \\ \log | y - 1 | + y = x + \log | x | + c \\ y - x = \log | x | - \log | y - 1 | + c \end{aligned}

Differential Equations exercise 21.7 question 13

Answer: $x c = \cos y$
Hint: Separate the terms of x and y and then integrate them.
Given: $x \frac{d y}{d x} + \cot y = 0$
Solution: $x \frac{d y}{d x} + \cot y = 0$
$\begin{aligned} x \frac{d y}{d x} = - \cot y \\ d y = - \cot y \frac{d x}{x} \\ \frac{d y}{- \cot y} = \frac{d x}{x} \end{aligned}$
Integrating both sides

\begin{aligned} - \tan y d y = \frac{d x}{x} \\ - \int \tan y d y = \int \frac{d x}{x} \end{aligned}

\begin{aligned} - [- \log | \cos y |] = \log | x | + \log C \\ \log | \cos y | = \log | x | + \log C \\ \cos y = x C \end{aligned}

Differential Equations exercise 21.7 question 14

Answer: $\sin y = e^{x} \log x + c$
Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x} = \frac{x e^{x} \log x + e^{x}}{x \cos y}$
Solution: $\frac{d y}{d x} = \frac{x e^{x} \log x + e^{x}}{x \cos y}$
$\cos y d y = \frac{x \cdot e^{x} \log x + e^{x}}{x} d x$
Integrating both sides

\begin{aligned} \int \cos y d y = \int e^{x} \log x d x + \int \frac{e^{x}}{x} d x \\ \sin y = \log x \int e^{x} d x - \int \frac{1}{x} (\int e^{x} d x) d x + \int \frac{e^{x}}{x} d x \end{aligned}

\begin{aligned} \sin y = e^{x} \log x - \int \frac{e^{x}}{x} d x + \int \frac{e^{x}}{x} d x + c \\ \sin y = e^{x} \log x + c \end{aligned}

Differential Equations exercise 21.7 question 15

Answer: $c = e^{x} + e^{- y} + \frac{x^{4}}{4}$
Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x} = e^{x + y} + e^{y} x^{3}$
Solution: $\frac{d y}{d x} = e^{x + y} + e^{y} x^{3}$
$\frac{d y}{d x} = e^{y} (e^{x} + x^{3})$
$\begin{aligned} \frac{d y}{e^{y}} = (e^{x} + x^{3}) d x \\ e^{- y} d y = (e^{x} + x^{3}) d x \end{aligned}$
Integrating both sides

\begin{aligned} \int e^{- y} d y = \int e^{x} d x + \int x^{3} d x \\ - e^{- y} + c = e^{x} + \frac{x^{4}}{4} \\ c = e^{x} + e^{- y} + \frac{x^{4}}{4} \end{aligned}

Differential Equations exercise 21.7 question 16 maths

Answer: $\sqrt{1 + y^{2}} + \sqrt{1 + x^{2}} + \frac{1}{2} \log | \frac{\sqrt{1 + x^{2}} - 1}{\sqrt{1 + x^{2}} + 1} | + \frac{1}{2} \log | \frac{\sqrt{1 + y^{2}} - 1}{\sqrt{1 + y^{2}} + 1} | = c$
Hint: Separate the terms of x and y and then integrate them.
Given: $y \sqrt{1 + x^{2}} + x \sqrt{1 + y^{2}} \frac{d y}{d x} = 0$
Solution: $y \sqrt{1 + x^{2}} + x \sqrt{1 + y^{2}} \frac{d y}{d x} = 0$
$\begin{aligned} x \sqrt{1 + y^{2}} d y = - y \sqrt{1 + x^{2}} d x \\ \frac{\sqrt{1 + y^{2}}}{- y} d y = \frac{\sqrt{1 + x^{2}}}{x} d x \end{aligned}$
$\begin{aligned} 1 + y^{2} = m^{2} \Rightarrow y = \sqrt{m^{2} - 1} \\ 2 y d y = 2 m d m \Rightarrow d y = \frac{m}{y} d m \\ And Put 1 + x^{2} = n^{2} \Rightarrow x = \sqrt{n^{2} - 1} \\ 2 x d x = 2 n d n \Rightarrow x d x = n d n \Rightarrow d x = \frac{n}{x} d n \end{aligned}$
Integrating both sides

\begin{aligned} => - \int \frac{\sqrt{1 + y^{2}}}{y} d y = \int \frac{\sqrt{1 + x^{2}}}{x} d x \\ => - \int \frac{m}{m^{2} - 1} d m . m = \int \frac{n}{n^{2} - 1} d n \cdot n \end{aligned}

=> - \int \frac{m^{2}}{m^{2} - 1} d m = \int \frac{n^{2}}{n^{2} - 1} d n

=> - \int \frac{m^{2} - 1 + 1}{m^{2} - 1} d m = \int \frac{n^{2} - 1 + 1}{n^{2} - 1} d n

\begin{aligned} = - \int 1 d m - \int \frac{1}{(m^{2} - 1)} d m = \int 1 d n + \int \frac{1}{n^{2} - 1} d n \\ => - m - \frac{1}{2} \log (\frac{m - 1}{m + 1}) = n + \frac{1}{2} \log (\frac{n - 1}{n + 1}) + c \end{aligned}

\begin{aligned} m = \sqrt{1 + y^{2}} & n = \sqrt{1 + x^{2}} \\ \Rightarrow - \sqrt{1 + y^{2}} - \frac{1}{2} \log (\frac{\sqrt{1 + y^{2}} - 1}{\sqrt{1 + y^{2}} + 1}) = \sqrt{1 + x^{2}} + \frac{1}{2} \log (\frac{\sqrt{1 + x^{2}} - 1}{\sqrt{1 + x^{2}} + 1}) + c \end{aligned}

c = \sqrt{1 + x^{2}} + \sqrt{1 + y^{2}} + \frac{1}{2} \log (\frac{\sqrt{1 + y^{2}} - 1}{\sqrt{1 + y^{2}} + 1}) + \frac{1}{2} \log (\frac{\sqrt{1 + x^{2}} - 1}{\sqrt{1 + x^{2}} + 1})

Differential Equations exercise 21.7 question 17

Answer: $\log (y + \sqrt{1 + y^{2}}) (x + \sqrt{1 + x^{2}}) = C$
Hint: Separate the terms of x and y and then integrate them.
Given: $\sqrt{1 + x^{2}} d y + \sqrt{1 + y^{2}} d x = 0$
Solution: $\sqrt{1 + x^{2}} d y + \sqrt{1 + y^{2}} d x = 0$
$\begin{aligned} - \sqrt{1 + x^{2}} d y = \sqrt{1 + y^{2}} d x \\ \frac{d y}{\sqrt{1 + y^{2}}} = - \frac{d x}{\sqrt{1 + x^{2}}} \end{aligned}$
Integrating both sides

\begin{aligned} \int \frac{d y}{\sqrt{1 + y^{2}}} = - \int \frac{d x}{\sqrt{1 + x^{2}}} \end{aligned}

Let I = \int \frac{d x}{\sqrt{1 + x^{2}}}

Put

x = \tan θ

d x = \sec^{2} θ d θ

\begin{aligned} I = \int \frac{1}{\sqrt{1 + \tan^{2} θ}} \sec^{2} θ d θ \\ = \int \frac{1}{\sqrt{\sec^{2} θ}} \sec^{2} θ d θ \\ (\sqrt{1 + \tan^{2} θ} = \sqrt{\sec^{2} θ}) \end{aligned}

\begin{aligned} I = \int \sec θ d θ \\ = \log | \sec θ + \tan θ | \\ \tan θ = x & \sec θ = \sqrt{1 + \tan^{2} θ} = \sqrt{1 + x^{2}} \\ I = \log | x + \sqrt{1 + x^{2}} | + c \end{aligned}

Similarly,

\int \frac{1}{\sqrt{1 + y^{2}}} = \log | y + \sqrt{1 + y^{2}} | + c

Hence,

\begin{aligned} \log | y + \sqrt{1 + y^{2}} | = - \log | x + \sqrt{1 + x^{2}} | + c \\ \log | y + \sqrt{1 + y^{2}} | + \log | x + \sqrt{1 + x^{2}} | = c \\ \log (y + \sqrt{1 + y^{2}}) (x + \sqrt{1 + x^{2}}) = c \end{aligned}

Differential Equations exercise 21.7 question 18

Answer: $\sqrt{1 + y^{2}} + \sqrt{1 + x^{2}} + \frac{1}{2} l og | \frac{\sqrt{1 + x^{2}} - 1}{\sqrt{1 + x^{2}} + 1} | + c = 0$
Hint: Separate the terms of x and y and then integrate them.
Given: $\sqrt{1 + x^{2} + y^{2} + x^{2} y^{2}} + x y \frac{d y}{d x} = 0$
Solution: $\sqrt{1 + x^{2} + y^{2} + x^{2} y^{2}} + x y \frac{d y}{d x} = 0$
$\begin{aligned} \Rightarrow x y \frac{d y}{d x} = - \sqrt{1 + x^{2} + y^{2} + x^{2} y^{2}} \\ \Rightarrow x y \frac{d y}{d x} = - \sqrt{(1 + x^{2}) (1 + y^{2})} \\ \Rightarrow \frac{y d y}{\sqrt{1 + y^{2}}} = \frac{- \sqrt{1 + x^{2}}}{x} \end{aligned}$
Integrating both sides

\Rightarrow \int \frac{y d y}{\sqrt{1 + y^{2}}} = \int \frac{- \sqrt{1 + x^{2}}}{x} d x

Let 1 + y^{2} = t \Rightarrow 2 y d y = d t \Rightarrow y d y = \frac{d t}{2}

1 + x^{2} = m^{2} \Rightarrow x = \sqrt{m^{2} - 1} \Rightarrow 2 x d x = 2 m d m \Rightarrow d x = \frac{m}{x} d m

\Rightarrow \frac{1}{2} \int \frac{1}{\sqrt{t}} d t = - \int \frac{m}{m^{2} - 1} d m \cdot m

\Rightarrow \frac{1}{2} \frac{t^{\frac{1}{2}}}{\frac{1}{2}} + \int \frac{m^{2}}{m^{2} - 1} d m = 0

\begin{aligned} \Rightarrow \sqrt{t} + \int \frac{m^{2} - 1 + 1}{m^{2} - 1} d m = 0 \\ \Rightarrow \sqrt{t} + \int \frac{m^{2} - 1}{m^{2} - 1} d m + \int \frac{1}{m^{2} - 1} d m = 0 \\ \Rightarrow \sqrt{t} + m + \frac{1}{2} \log \frac{(m - 1)}{m + 1} = 0 \end{aligned}

\dots \cdot \int \frac{1}{m^{2} - 1} d m = \frac{1}{2} \log \frac{(m - 1)}{m + 1}

\begin{aligned} Where m = \sqrt{1 + x^{2}}, 1 + y^{2} = t \\ \sqrt{1 + y^{2}} + \sqrt{1 + x^{2}} + \frac{1}{2} l og | \frac{\sqrt{1 + x^{2}} - 1}{\sqrt{1 + x^{2}} + 1} | + c = 0 \end{aligned}

Differential Equations exercise 21.7 question 19

Answer: $y^{2} \log y = e^{x} \sin^{2} x + c$
Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x} = \frac{e^{x} (\sin^{2} x + \sin 2 x)}{y (2 \log y + 1)}$
Solution: $\frac{d y}{d x} = \frac{e^{x} (\sin^{2} x + \sin 2 x)}{y (2 \log y + 1)}$
$\begin{aligned} \Rightarrow y (2 \log y + 1) d y = e^{x} (\sin^{2} x + \sin 2 x) d x \\ \Rightarrow (2 y \log y + y) d y = (e^{x} \sin^{2} x + e^{x} \sin 2 x) d x \\ \Rightarrow 2 y \log y d y + y d y = e^{x} \sin^{2} x d x + e^{x} \sin 2 x d x \end{aligned}$
Integrating both sides and using integrating by parts

\Rightarrow 2 [\log y \int y d y - \int {\frac{d}{d y} (\log y) \int y d y}] d y + \int y d y

= \sin^{2} x \int e^{x} d x - \int [\frac{d}{d x} \sin^{2} x \int (e^{x} d x)] d x + \int e^{x} \sin 2 x d x

\Rightarrow 2 [\log y (\frac{y^{2}}{2}) - \int (\frac{1}{y}) (\frac{y^{2}}{2}) d y] + y d y

= \sin^{2} x e^{x} - \int [2 \sin x \cos x e^{x}] d x + \int e^{x} \sin 2 x + c

\begin{aligned} \Rightarrow y^{2} \log y - \int y d y + \int y d y = - \int e^{x} \sin 2 x d x + \int e^{x} \sin 2 x d x + e^{x} \sin^{2} x \\ y^{2} \log y = e^{x} \sin^{2} x + c \end{aligned}

Differential Equations exercise 21.7 question 20 maths

Answer: $y \sin y = x^{2} \log x + c$
Hint: You must know about the rules of solving differential equation and integration
Given: $\frac{d y}{d x} = \frac{x (2 \log x + 1)}{\sin y + y \cos y}$
Solution: $\frac{d y}{d x} = \frac{x (2 \log x + 1)}{\sin y + y \cos y}$
$(\sin y + y \cos y) d y = x (2 \log x + 1) d x$
Integrating both sides

\int \sin y d y + \int y \cos y d y = 2 \int x \log x d x + \int x d x

\begin{aligned} - \cos y + y \int \cos y d y - \int [\frac{d}{d y} y \cdot \int \cos y d y] d y = 2 {\log x \int x - \int [\frac{d}{d x} \log x \cdot \int x d x] d x} \\ - \cos y + [y \sin y + \cos y] = 2 [\frac{x^{2}}{2} \log x - \frac{1}{2} \int x d x] + \frac{x^{2}}{2} \\ y \sin y = x^{2} \log x + c \end{aligned}

Differential Equations exercise 21.7 question 21

Answer: $- \log | y | + \log | y - 1 | = - \frac{1}{2} \log | 1 - x^{2} | + \log c$
Hint: Separate the terms of x and y and then integrate them.
Given: $(1 - x^{2}) d y + x y d x = x y^{2} d x$
Solution: $(1 - x^{2}) d y + x y d x = x y^{2} d x$
$\begin{aligned} (1 - x^{2}) d y = (x y^{2} - x y) d x \\ (1 - x^{2}) d y = x y (y - 1) d x \\ \frac{d y}{y (y - 1)} = \frac{x d x}{(1 - x^{2})} \end{aligned}$
Integrating both sides

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

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

Put

\begin{aligned} 1 - x^{2} = t \\ - 2 x d x = d t \\ - \log | y | + \log | y - 1 | = - \frac{1}{2} \int \frac{d t}{t} \end{aligned}

\begin{aligned} - \log | y | + \log | y - 1 | = - \frac{1}{2} \log | t | + \log c \\ - \log | y | + \log | y - 1 | = - \frac{1}{2} \log | 1 - x^{2} | + \log c \end{aligned}

Differential Equations exercise 21.7 question 22

Answer: $\tan x \times \sin x = c$
Hint: Separate the terms of x and y and then integrate them.
Given: $\tan y d x + \sec^{2} y \tan x d y = 0$
Solution: $\tan y d x + \sec^{2} y \tan x d y = 0$
$\begin{aligned} \Rightarrow \sec^{2} y \tan x d y = - \tan y d x \\ \Rightarrow \frac{\sec^{2} y}{\tan y} d y = - \frac{1}{\tan x} d x \\ \Rightarrow \frac{1}{\cos^{2} y} \times \frac{\cos y}{\sin y} d y = - \cot x d x \end{aligned}$
$\begin{aligned} \Rightarrow \frac{1}{\sin y \cos y} d y = - \cot x d x \\ \Rightarrow \frac{2}{\sin 2 y} = - \cot x d x \end{aligned}$
Integrating both sides

\begin{aligned} \Rightarrow 2 \int cosec 2 y d y = - \int \cot x d x \\ \Rightarrow \log \tan x = - \log \sin x + \log c \\ \Rightarrow \log \tan x + \log \sin x = \log c \\ \Rightarrow \log (\tan x \times \sin x) = \log c \\ \Rightarrow \tan x \times \sin x = c \end{aligned}

Differential Equations exercise 21.7 question 23

Answer: $\tan^{- 1} x + \tan^{- 1} y + \frac{1}{2} \log | 1 + x^{2} | + \frac{1}{2} \log | 1 + y^{2} | = c$
Hint: Separate the terms of x and y and then integrate them.
Given: $(1 + x) (1 + y^{2}) d x + (1 + y) (1 + x^{2}) d y = 0$
Solution:
$\begin{aligned} (1 + x) (1 + y^{2}) d x + (1 + y) (1 + x^{2}) d y = 0 \\ (1 + x) (1 + y^{2}) d x = - (1 + y) (1 + x^{2}) d y \\ \frac{(1 + x)}{1 + x^{2}} d x = - \frac{(1 + y)}{1 + y^{2}} d y \end{aligned}$
Integrating both sides

\begin{aligned} \int \frac{(1 + x)}{1 + x^{2}} d x = - \int \frac{(1 + y)}{1 + y^{2}} d y \\ \Rightarrow \int \frac{1}{1 + x^{2}} d x + \int \frac{x}{1 + x^{2}} d x = - \int \frac{1}{1 + y^{2}} d y - \int \frac{y}{1 + y^{2}} d y \end{aligned}

\begin{aligned} 1 + x^{2} = t \Rightarrow 2 x d x = d t \\ 1 + y^{2} = u \Rightarrow 2 y d y = d u \\ \int \frac{1}{1 + x^{2}} d x + \frac{1}{2} \int \frac{1}{t} d t = - \int \frac{1}{1 + y^{2}} d y - \frac{1}{2} \int \frac{1}{u} d u \end{aligned}

\begin{aligned} \tan^{- 1} x + \frac{1}{2} \log | t | = - \tan^{- 1} y - \frac{1}{2} \log | u | + c \\ \tan^{- 1} x + \frac{1}{2} \log | 1 + x^{2} | = - \tan^{- 1} y - \frac{1}{2} \log | 1 + y^{2} | + c \\ \tan^{- 1} x + \tan^{- 1} y + \frac{1}{2} \log | 1 + x^{2} | + \frac{1}{2} \log | 1 + y^{2} | = c \end{aligned}

Differential Equations exercise 21.7 question 24 maths

Answer: $\sec y = - 2 \cos x + c$
Hint: Separate the terms of x and y and then integrate them.
Given: $\tan y \frac{d y}{d x} = \sin (x + y) + \sin (x - y)$
Solution: $\tan y \frac{d y}{d x} = \sin (x + y) + \sin (x - y)$
$\frac{d y}{d x} = \frac{\sin (x + y) + \sin (x - y)}{\tan y}$
$= \frac{2 \sin x \cos y}{\tan y}$
$\begin{aligned} \Rightarrow \frac{\tan y}{\cos y} d y = 2 \sin x d x \\ \Rightarrow \frac{\sin y}{\cos^{2} y} d y = 2 \sin x d x \end{aligned}$
Integrating both sides

\Rightarrow \int \frac{\sin y}{\cos^{2} y} d y = \int 2 \sin x d x

\begin{aligned} \Rightarrow \sec y = - 2 \cos x + c \\ \Rightarrow \sec y = - 2 \cos x + c \end{aligned}

Differential Equations exercise 21.7 question 25

Answer: $\sin y = C \cos x$
Hint: Separate the terms of x and y and then integrate them.
Given: $\cos x \cos y \frac{d y}{d x} = - \sin x \sin y$
Solution: $\cos x \cos y \frac{d y}{d x} = - \sin x \sin y$
$\begin{aligned} \frac{\cos y}{\sin y} d y = \frac{- \sin x}{\cos x} d x \\ \cot y d y = - \tan x d x \end{aligned}$
Integrating both sides

\begin{aligned} \int \cot y d y = - \int \tan x d x \\ \log | \sin y | = - [- \log | \cos x |] + \log C \\ \log \sin y = \log \cos x + \log C \end{aligned}

\begin{aligned} \log \sin y = \log C (\cos x) \\ \sin y = C \cos x \end{aligned}

Differential Equations exercise 21.7 question 26

Answer: $\log | \sin y | = - \sin x + C$
Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x} + \frac{\cos x \sin y}{\cos y} = 0$
Solution: $\frac{d y}{d x} + \frac{\cos x \sin y}{\cos y} = 0$
$\begin{aligned} \frac{d y}{d x} = - \frac{\cos x \sin y}{\cos y} \\ \frac{\cos y}{\sin y} d y = - \cos x d x \\ \cot y d y = - \cos x d x \end{aligned}$
Integrating both sides

\begin{aligned} \int \cot y d y = - \int \cos x d x \\ \log | \sin y | = - \sin x + c \end{aligned}

Differential Equations exercise 21.7 question 27

Answer: $- \sqrt{1 - y^{2}} = \sqrt{1 - x^{2}} + c or \sqrt{1 - x^{2}} + \sqrt{1 - y^{2}} = c$
Hint: You must know about the rules of solving differential equation and integration
Given: $x \sqrt{1 - y^{2}} d x + y \sqrt{1 - x^{2}} d y = 0$
Solution:
$y \sqrt{1 - x^{2}} d y = - x \sqrt{1 - y^{2}} d x$
$\int \frac{y}{\sqrt{1 - y^{2}}} d y = \int \frac{- x}{\sqrt{1 - x^{2}}} d x$
We know

\frac{d}{d x} (\sqrt{1 - x^{2}}) = \frac{- x}{\sqrt{1 - x^{2}}}

Integration of

\frac{- x}{\sqrt{1 - x^{2}}} d x = \sqrt{1 - x^{2}}

Similarly Integration of

\frac{y}{\sqrt{1 - y^{2}}} d y = - \sqrt{1 - y^{2}}

\begin{aligned} \Rightarrow - \sqrt{1 - y^{2}} = \sqrt{1 - x^{2}} + c \\ \Rightarrow - \sqrt{1 - y^{2}} - \sqrt{1 - x^{2}} = c \\ \Rightarrow - [\sqrt{1 - x^{2}} + \sqrt{1 - y^{2}}] = c \end{aligned}

\begin{aligned} \Rightarrow \sqrt{1 - x^{2}} + \sqrt{1 - y^{2}} = - c \\ \Rightarrow \sqrt{1 - x^{2}} + \sqrt{1 - y^{2}} = c \end{aligned}

Differential Equations exercise 21.7 question 28 maths

Answer: $y = \log (c (1 + y) (e^{x} + 1))$
Hint: Separate the terms of x and y and then integrate them.
Given: $y (1 + e^{x}) d y = (y + 1) e^{x} d x$
Solution: $y (1 + e^{x}) d y = (y + 1) e^{x} d x$
$\frac{y d y}{(y + 1)} = \frac{e^{x}}{(1 + e^{x})} d x$
Integrating both sides

\begin{aligned} \Rightarrow y - \log | y + 1 | = \log | e^{x} + 1 | + \log c \\ \Rightarrow y = \log | y + 1 | + \log | e^{x} + 1 | + \log c \\ y = \log (c (1 + y) (e^{x} + 1)) \end{aligned}

Differential Equations exercise 21.7 question 29

Answer: $\log (x y) + x - \frac{y^{2}}{2} = C$
Hint: Separate the terms of x and y and then integrate them.
Given: $(y + x y) d x + (x - x y^{2}) d y = 0$
Solution:
$\begin{aligned} (y + x y) d x + (x - x y^{2}) d y = 0 \\ y (1 + x) d x + x (1 - y^{2}) d y = 0 \\ \frac{1 + x}{x} d x + \frac{(1 - y^{2})}{y} d y = 0 \end{aligned}$
Integrating both sides and also separate the terms

\begin{aligned} \int \frac{1}{x} d x + \int 1 d x + \int \frac{1}{y} d y - \int y d y = 0 \\ \log | x | + x + \log | y | - \frac{y^{2}}{2} = C \\ \log (x y) + x - \frac{y^{2}}{2} = C \end{aligned}

Differential Equations exercise 21.7 question 30

Answer: $\log (1 + y) = x - \frac{x^{2}}{2} + c$
Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x} = 1 - x + y - x y$
Solution: $\frac{d y}{d x} = 1 - x + y - x y$
$\begin{aligned} \frac{d y}{d x} = (1 - x) (1 + y) \\ \frac{d y}{(1 + y)} = (1 - x) d x \\ \int \frac{d y}{(1 + y)} = \int 1 d x - \int x d x \end{aligned}$
$\log (1 + y) = x - \frac{x^{2}}{2} + c$

Differential Equations exercise 21.7 question 31

Answer: $\tan^{- 1} y = \tan^{- 1} x + c$
Hint: Separate the terms of x and y and then integrate them.
Given: $(y^{2} + 1) d x - (x^{2} + 1) d y = 0$
Solution: $(y^{2} + 1) d x - (x^{2} + 1) d y = 0$
$(y^{2} + 1) d x = (x^{2} + 1) d y$
$\begin{aligned} \frac{d y}{(y^{2} + 1)} = \frac{d x}{(x^{2} + 1)} \\ \tan^{- 1} y = \tan^{- 1} x + c \end{aligned}$

Differential Equations exercise 21.7 question 32 maths

Answer: $\log (y + 1) + x + \frac{x^{2}}{2} = c$
Hint: Separate the terms of x and y and then integrate them.
Given: $d y + (x + 1) (y + 1) d x = 0$
Solution: $d y + (x + 1) (y + 1) d x = 0$
$\begin{aligned} d y = - (x + 1) (y + 1) d x \\ \frac{d y}{(y + 1)} = - (x + 1) d x \end{aligned}$
Integrating both sides

\begin{aligned} \int \frac{1}{y + 1} d y = \int - x d x - \int 1 d x \\ \Rightarrow \log (y + 1) = - \frac{x^{2}}{2} - x + c \\ \Rightarrow \log (y + 1) + x + \frac{x^{2}}{2} = c \end{aligned}

Differential Equations exercise 21.7 question 33

Answer: $\tan^{- 1} y = x + \frac{x^{3}}{3} + c$
Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x} = (1 + x^{2}) (1 + y^{2})$
Solution: $\frac{d y}{d x} = (1 + x^{2}) (1 + y^{2})$
$\frac{d y}{(1 + y^{2})} = (1 + x^{2}) d x$
Integrating both sides

\begin{aligned} \int \frac{1}{1 + y^{2}} d y = \int 1 d x + \int x^{2} d x \\ \tan^{- 1} y = x + \frac{x^{3}}{3} + c \end{aligned}

Differential Equations exercise 21.7 question 34

Answer: $\log | y | = \frac{2 x^{3}}{3} + x^{2} + 2 x + 2 \log (x - 1) + c$
Hint: Separate the terms of x and y and then integrate them.
Given: $(x - 1) \frac{d y}{d x} = 2 x^{3} y$
Solution: $(x - 1) \frac{d y}{d x} = 2 x^{3} y$
$\begin{aligned} \Rightarrow \frac{d y}{y} = \frac{2 x^{3}}{(x - 1)} d x \\ \Rightarrow \frac{d y}{y} = \frac{2 ((x - 1) (x^{2} + x + 1) + 1)}{(x - 1)} d x \\ \Rightarrow \frac{d y}{y} = 2 (x^{2} + x + 1 + \frac{1}{x - 1}) d x \end{aligned}$
Integrating both sides

\int \frac{d y}{y} = 2 [\int x^{2} d x + \int x d x + \int 1 d x + \int \frac{1}{x - 1}] d x

\begin{aligned} \log | y | = 2 [\frac{x^{3}}{3} + \frac{x^{2}}{2} + x + \log | x - 1 |] + c \\ \log | y | = \frac{2 x^{3}}{3} + x^{2} + 2 x + 2 \log (x - 1) + c \end{aligned}

Differential Equations exercise 21.7 question 35

Answer: $e^{- x} - e^{- y} - e^{x} = c$
Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x} = e^{x + y} + e^{- x + y}$
Solution: $\frac{d y}{d x} = e^{x + y} + e^{- x + y}$
$\begin{aligned} \frac{d y}{d x} = e^{x} e^{y} + e^{- x} e^{y} \\ \frac{d y}{d x} = e^{y} [e^{x} + e^{- x}] \\ \frac{d y}{e^{y}} = [e^{x} + e^{- x}] d x \end{aligned}$
Integrating both sides

\begin{aligned} \int e^{- y} d y = \int e^{x} d x + \int e^{- x} d x \\ - e^{- y} = e^{x} + [- e^{- x}] + c \\ - e^{- y} = e^{x} - e^{- x} + c \\ e^{- x} - e^{- y} - e^{x} = c \end{aligned}

Differential Equations exercise 21.7 question 36 maths

Answer: $\tan y = \frac{\sin 2 x}{2} + c$
Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x} = (\cos^{2} x - \sin^{2} x) \cos^{2} y$
Solution: $\frac{d y}{d x} = (\cos^{2} x - \sin^{2} x) \cos^{2} y$
$\begin{aligned} \frac{d y}{\cos^{2} y} = (\cos^{2} x - \sin^{2} x) d x \\ \frac{d y}{\cos^{2} y} = \cos 2 x d x \end{aligned}$
Integrating both sides

\begin{aligned} \int \sec^{2} y d y = \int \cos 2 x d x \\ \tan y = \frac{\sin 2 x}{2} + c \end{aligned}

Differential Equations exercise 21.7 question 37 (i)

Answer: $y^{2} + 2 = \frac{c^{1}}{x^{2} + 2}$
Hint: Separate the terms of x and y and then integrate them.
Given: $(x y^{2} + 2 x) d x + (x y^{2} + 2 y) d y = 0$
Solution: $(x y^{2} + 2 x) d x + (x y^{2} + 2 y) d y = 0$
$\begin{aligned} x (y^{2} + 2) d x + y (x^{2} + 2) d y = 0 \\ y (x^{2} + 2) d y = - x (y^{2} + 2) d x \\ \frac{y}{(y^{2} + 2)} d y = \frac{- x}{(x^{2} + 2)} d x \end{aligned}$
Integrating both sides

\begin{aligned} \Rightarrow \int \frac{y}{(y^{2} + 2)} d y = - \int \frac{x}{x^{2} + 2} d x \\ \Rightarrow \frac{1}{2} \int \frac{2 y}{y^{2} + 2} d y = - \frac{1}{2} \int \frac{2 x}{x^{2} + 2} d x \end{aligned}

\begin{aligned} \Rightarrow \frac{1}{2} \log | y^{2} + 2 | = - \frac{1}{2} \log | x^{2} + 2 | + \log c \\ \Rightarrow \frac{1}{2} [\log | y^{2} + 2 | + \log | x^{2} + 2 |] = \log c \end{aligned}

\begin{aligned} \Rightarrow \log (y^{2} + 2) (x^{2} + 2) = 2 \log c \\ \Rightarrow \log (y^{2} + 2) (x^{2} + 2) = \log c^{2} \\ \Rightarrow (y^{2} + 2) (x^{2} + 2) = c^{1} \\ \Rightarrow (y^{2} + 2) = \frac{c^{1}}{(x^{2} + 2)} \end{aligned}

Differential Equations exercise 21.7 question 37 (ii)

Answer: $- \frac{\log y}{y} - \frac{1}{y} = - [- x^{2} \cos x + 2 x \sin x + 2 \cos x] + c$
Hint: Separate the terms of x and y and then integrate them.
Given: $cosec x \log y \frac{d y}{d x} + x^{2} y^{2} = 0$
Solution: $cosec x \log y \frac{d y}{d x} + x^{2} y^{2} = 0$
$cosec x \log y \frac{d y}{d x} = - x^{2} y^{2}$
$\frac{\log y}{y^{2}} d y = - x^{2} \frac{1}{cosec x} d x$
Integrating both sides

\int \frac{\log y}{y^{2}} d y = \int - x^{2} \sin x d x

\int \frac{\log y}{y^{2}} d y = - [- x^{2} \cos x + 2 \int x \cos x d x]

{ using integration by parts}

\begin{aligned} - \frac{\log y}{y} - \frac{1}{y} = - [- x^{2} \cos x + 2 x \sin x - 2 \int \sin x d x] \\ - \frac{\log y}{y} - \frac{1}{y} = - [- x^{2} \cos x + 2 x \sin x + 2 \cos x] + c \end{aligned}

Differential Equations exercise 21.7 question 38 (i)

Answer: $(y - x) - (\log (x (1 + y))) = c$
Hint: Separate the terms of x and y and then integrate them.
Given: $x y \frac{d y}{d x} = 1 + x + y + x y$
Solution: $x y \frac{d y}{d x} = 1 + x + y + x y$
$\begin{aligned} x y \frac{d y}{d x} = (1 + x) + y (1 + x) \\ x y \frac{d y}{d x} = (1 + x) (1 + y) \\ \frac{y}{1 + y} d y = \frac{1 + x}{x} d x \end{aligned}$
Integrating both sides

\begin{aligned} \int 1 d y - \int \frac{1}{1 + y} d y = \int \frac{1}{x} d x + \int 1 d x \\ y - \log | 1 + y | = \log | x | + x + c \end{aligned}

\begin{aligned} y - x - \log | 1 + y | - \log | x | = c \\ (y - x) - \log [x (1 + y)] = c \end{aligned}

Differential Equations exercise 21.7 question 38 (ii) maths

Answer: $(1 + y^{2}) = \frac{c^{1}}{(1 - x^{2})}$
Hint: Separate the terms of x and y and then integrate them.
Given: $y (1 - x^{2}) \frac{d y}{d x} = x (1 + y^{2})$
Solution: $y (1 - x^{2}) \frac{d y}{d x} = x (1 + y^{2})$
$\frac{y d y}{(1 + y^{2})} = \frac{x}{(1 - x^{2})} d x$
Integrating both sides

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

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

\frac{1}{2} \log | 1 + y^{2} | = - \frac{1}{2} \log | 1 - x^{2} | + \log c

\begin{aligned} \frac{1}{2} [\log | 1 + y^{2} | + \log | 1 - x^{2} |] = \log c \\ \log (1 + y^{2}) (1 - x^{2}) = \log c^{2} \\ (1 + y^{2}) (1 - x^{2}) = c^{2} \\ (1 + y^{2}) = \frac{c^{1}}{1 - x^{2}} \end{aligned}

Differential Equations exercise 21.7 question 38 (iii)

Answer: $y = e^{\frac{x}{y}} + c$
Hint: Separate the terms of x and y and then integrate them.
Given: $y e^{\frac{x}{y}} d x = (x e^{\frac{x}{y}} + y^{2}) d y$
Solution: $y e^{\frac{x}{y}} d x = (x e^{\frac{x}{y}} + y^{2}) d y$
$\frac{d x}{d y} = \frac{x e^{\frac{x}{y}} + y^{2}}{y_{e}^{\frac{x}{y}}}$
$x = v y => \frac{d x}{d y} = v + y \cdot \frac{d v}{d y}$
$v + y \cdot \frac{d v}{d y} = \frac{v y \cdot e^{v} + y^{2}}{y e^{v}}$
$\begin{aligned} y \cdot \frac{d v}{d y} = \frac{v y \cdot e^{v} + y^{2}}{y e^{v}} - v \\ y \cdot \frac{d v}{d y} = \frac{v y \cdot e^{v} + y^{2} - v y e^{v}}{y e^{v}} \end{aligned}$
$\begin{aligned} e^{v} d v = \frac{y d y}{y} \\ e^{v} d v = d y \end{aligned}$
Integrating both sides

\begin{aligned} \int e^{y} d v = \int 1 d y \\ y = e^{v} + c \end{aligned}

Put

\begin{aligned} v = \frac{x}{y} \\ y = e^{\frac{x}{y}} + c \end{aligned}

Differential Equations exercise 21.7 question 38 (iv)

Answer: $\frac{1}{2} {(\tan^{- 1} x)}^{2} + \log | 1 + y^{2} | = c$
Hint: Separate the terms of x and y and then integrate them.
Given: $(1 + y^{2}) \tan^{- 1} x d x + 2 y (1 + x^{2}) d y = 0$
Solution: $(1 + y^{2}) \tan^{- 1} x d x + 2 y (1 + x^{2}) d y = 0$
$\begin{aligned} (1 + y^{2}) \tan^{- 1} x d x = - 2 y (1 + x^{2}) d y \\ \frac{\tan^{- 1} x d x}{(1 + x^{2})} = - \frac{2 y}{(1 + y^{2})} d y \end{aligned}$
Integrating both sides

\begin{aligned} \int \frac{\tan^{- 1} x d x}{(1 + x^{2})} = - \int \frac{2 y}{(1 + y^{2})} d y \\ \frac{1}{2} {(\tan^{- 1} x)}^{2} = - \log (1 + y^{2}) + c \\ \frac{1}{2} {(\tan^{- 1} x)}^{2} + \log | 1 + y^{2} | = c \end{aligned}

Differential Equations exercise 21.7 question 39

Answer: $y^{2} = 4 | \sec 2 x |$
Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x} = y \tan 2 x, y (0) = 2$
Solution: $\frac{d y}{d x} = y \tan 2 x, y (0) = 2$
$\frac{d y}{y} = \tan 2 x d x$
Integrating both sides

\begin{aligned} \int (\frac{d y}{y}) = \int \tan 2 x d x \\ \log y = \frac{1}{2} \log | \sec 2 x | + c \\ y (0) = 2 where, y = 2 & x = 0 \end{aligned}

\begin{aligned} \log (2) = \frac{1}{2} \log | \sec 2 (0) | + c \\ c = \log 2 \\ \log y = \frac{1}{2} \log | \sec 2 x | + \log 2 \end{aligned}

\begin{aligned} \log y = \log \sqrt{\sec 2 x} + \log 2 \\ y = 2 \sqrt{\sec 2 x} \\ y^{2} = 4 | \sec 2 x | \end{aligned}

Differential Equations exercise 21.7 question 40 maths

Answer: $3 \sqrt{\frac{y}{2}} = \sqrt{x}$
Hint: Separate the terms of x and y and then integrate them.
Given: $2 x \frac{d y}{d x} = 3 y, y (1) = 2$
Solution: $2 x \frac{d y}{d x} = 3 y$
$\frac{d x}{3 y} = \frac{d x}{2 x}$
Integrating both sides

\begin{aligned} \int \frac{d x}{3 y} = \int \frac{d x}{2 x} \\ \frac{1}{3} \log | y | = \frac{1}{2} \log | x | + c \end{aligned}

\begin{aligned} Put y (1) = 2 where, y = 2 & x = 1 \\ \frac{1}{3} \log | 2 | = \frac{1}{2} \log | 1 | + c \\ \frac{1}{3} \log | 2 | = c \end{aligned}

\begin{aligned} [\log 1 = 0] \\ \frac{1}{3} \log | y | = \frac{1}{2} \log | x | + \frac{1}{3} \log | 2 | \\ \frac{1}{3} \log | y | - \frac{1}{3} \log | 2 | = \frac{1}{2} \log | x | \end{aligned}

\begin{aligned} \frac{1}{3} \log \frac{y}{2} = \frac{1}{2} \log | x | \\ \log \sqrt[3]{\frac{y}{2}} = \log \sqrt{x} \\ \sqrt[3]{\frac{y}{2}} = \sqrt{x} \end{aligned}

Differential Equations exercise 21.7 question 41

Answer: $y = \log [\frac{(y + 2)^{2} x}{8}]$
Hint: Separate the terms of x and y and then integrate them.
Given: $x y \frac{d y}{d x} = y + 2, y (2) = 0$
Solution:
$\begin{aligned} x y \frac{d y}{d x} = y + 2 \\ \Rightarrow \frac{y d y}{y + 2} = \frac{d x}{x} \end{aligned}$
Integrating both sides

\begin{aligned} \int \frac{y}{y + 2} d y = \int \frac{d x}{x} \\ \Rightarrow \int \frac{(y + 2) - 2}{(y + 2)} d y = \int \frac{d x}{x} \end{aligned}

\begin{aligned} \Rightarrow \int (1 - \frac{2}{y + 2}) d y = \int \frac{d x}{x} \\ \Rightarrow y - 2 \log (y + 2) = \log | x | + c \end{aligned}

....................(1)

y (2) = 0 at x = 2, y = 0

Put in (1)

\begin{aligned} \Rightarrow 0 - 2 \log | 0 + 2 | = \log | 2 | + c \\ \Rightarrow - 2 \log 2 = \log 2 + c \\ \Rightarrow c = - 3 \log 2 \end{aligned}

Put in (1), we get

\begin{aligned} y - 2 \log (y + 2) = \log | x | - 3 \log 2 \\ \Rightarrow y = \log | y + 2 |^{2} + \log | x | - \log 2^{3} \\ \Rightarrow y = \log | y + 2 |^{2} + \log | x | - \log 8 \\ \Rightarrow y = \log [\frac{(y + 2)^{2} x}{8}] \end{aligned}

Differential Equations exercise 21.7 question 42

Answer: $4 \cdot y^{2} (2 - e^{x}) = 1$
Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x} = 2 e^{x} y^{3}, y (0) = \frac{1}{2}$
Solution:
$\begin{aligned} \frac{d y}{d x} = 2 e^{x} y^{3} \\ \frac{d y}{y^{3}} = 2 e^{x} d x \end{aligned}$
Integrating both sides

\begin{aligned} \int \frac{d y}{y^{3}} = 2 \int e^{x} d x \\ \Rightarrow \int y^{- 3} d y = 2 \int e^{x} d x \Rightarrow \frac{y^{- 3 + 1}}{- 3 + 1} = 2 e^{x} + c \\ [\int x^{n} d x = \frac{x^{n + 1}}{n + 1}] \end{aligned}

\begin{aligned} \Rightarrow \frac{y^{- 2}}{- 2} = 2 e^{x} + c \\ \Rightarrow \frac{- 1}{2 y^{2}} = 2 e^{x} + c \\ \Rightarrow - 1 = 2 y^{2} (2 e^{x} + c) \end{aligned}

..............(1)
Now given that

y (0) = \frac{1}{2} i.e. at x = 0 & y = \frac{1}{2}

Put in (1)

\begin{aligned} - 1 = 2 {(\frac{1}{2})}^{2} (2 e^{0} + c) \\ \Rightarrow - 1 = (\frac{1}{2}) (2 (1) + c) \Rightarrow - 2 = 2 + c \Rightarrow c = - 4 \end{aligned}

Put in (1) we get

\begin{aligned} \Rightarrow - 1 = 2 y^{2} (2 e^{x} - 4) \Rightarrow - 1 = - 2 (2 y^{2}) (2 - e^{x}) \\ \Rightarrow 4 y^{2} (2 - e^{x}) = 1 \end{aligned}

Differential Equations exercise 21.7 question 43

Answer: $r = r_{0} e^{\frac{- t^{2}}{2}}$
Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d r}{d t} = - r t, r (0) = r_{0}$
Solution:
$\begin{aligned} \frac{d r}{d t} = - r t \\ \frac{d r}{r} = - t d t \end{aligned}$
Integrating both sides

\begin{aligned} \Rightarrow \int \frac{d r}{r} = - \int t d t \\ \Rightarrow \log | r | = - \frac{t^{2}}{2} + c \end{aligned}

...............(1)
Given that

r (0) = r_{0} i.e. at t = 0, r = r_{0}

Put in (1)

\log r_{0} = 0 + c \Rightarrow c = \log r_{0}

Put in (1) we get

\begin{aligned} \log | r | = - \frac{t^{2}}{2} + \log r_{0} \Rightarrow \log | r | - \log r_{0} = - \frac{t^{2}}{2} \\ \Rightarrow \log \frac{r}{r_{0}} = - \frac{t^{2}}{2} \end{aligned}

\begin{aligned} [\begin{matrix} \log_{e} a = x \\ \Rightarrow a = e^{x} \end{matrix}] \\ \Rightarrow \frac{r}{r_{0}} = e^{\frac{- t^{2}}{2}} \\ \Rightarrow r = r_{0} e^{\frac{- t^{2}}{2}} \end{aligned}

Differential Equations exercise 21.7 question 44 maths

Answer: $y = e^{\sin^{2} x}$
Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x} = y \sin 2 x; y (0) = 1$
Solution:
$\begin{aligned} \frac{d y}{d x} = y \sin 2 x \\ \Rightarrow \frac{d y}{y} = \sin 2 x d x \end{aligned}$
Integrating both sides

\begin{aligned} \int \frac{d y}{y} = \int \sin 2 x d x \\ \Rightarrow \log | y | = \frac{- \cos 2 x}{2} + c \end{aligned}

...............(1)
Now Given that

at x = 0; y = 1 [y (0) = 1]

\begin{aligned} \log | 1 | = \frac{- \cos 2 (0)}{2} + c \\ \Rightarrow 0 = \frac{- 1}{2} + c \Rightarrow c = \frac{1}{2} \end{aligned}

Put in (1) we get

\begin{aligned} \log | y | = \frac{- \cos 2 x}{2} + \frac{1}{2} \\ \Rightarrow 2 \log | y | + \cos 2 x = 1 \Rightarrow 2 \log y = 1 - \cos 2 x \\ \Rightarrow \log y = \frac{2 \sin^{2} x}{2} \end{aligned}

\begin{aligned} [1 - \cos 2 x = 2 \sin^{2} x] \\ \Rightarrow \log y = \sin^{2} x \\ [\log_{ε} a = x \Rightarrow x = e^{x}] \\ \Rightarrow y = e^{\sin^{2} x} \end{aligned}

Differential Equations exercise 21.7 question 45 (i)

Answer: $z = \sec x$
Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x} = y \tan x, y (0) = 1$
Solution: $\frac{d y}{d x} = y \tan x, y (0) = 1$
$\Rightarrow \frac{d y}{y} = \tan x d x$
Integrating both sides

\begin{aligned} \Rightarrow \int \frac{1}{y} d y = \int \tan x d x \\ \Rightarrow \log | y | = \log | \sec x | + \log c \\ \Rightarrow \log | y | = \log | c \sec x | \\ \Rightarrow y = c \sec x \end{aligned}

...............(1)
Now Given that

y = 1 when x = 0

\begin{aligned} 1 = c \sec (0) \Rightarrow c = 1 [\sec 0 = 1] \\ y = (1) \sec x \Rightarrow y = \sec x \end{aligned}

Differential Equations exercise 21.7 question 45 (ii)

Answer: $\log | y | = \frac{5}{2} \log | x |$
Hint: Separate the terms of x and y and then integrate them.
Given: $2 x \frac{d y}{d x} = 5 y, y (1) = 1$
Solution:
$\begin{aligned} 2 x \frac{d y}{d x} = 5 y \\ \Rightarrow \frac{d y}{5 y} = \frac{d x}{2 x} \end{aligned}$
Integrating both sides

\Rightarrow \int \frac{d y}{5 y} = \int \frac{d x}{2 x} \Rightarrow \frac{1}{5} \log | y | = \frac{1}{2} \log | x | + c

...................(1)
Now given that

y (1) = 1 ∴ y = 1 at x = 1

∴ \frac{1}{5} \log | y | = \frac{1}{2} \log | x | + c \Rightarrow c = 0 [∴ \log 1 = 0]

Put in (1)

\frac{1}{5} \log | y | = \frac{1}{2} \log | x | + 0 \Rightarrow \log | y | = \frac{5}{2} \log | x |

Differential Equations exercise 21.7 question 45 (iii)

Answer: $y e^{2 x} + 1 = 0$
Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x} = 2 e^{2 x} y^{2}, y (0) = - 1$
Solution:
$\begin{aligned} \frac{d y}{d x} = 2 e^{2 x} y^{2} \\ \Rightarrow \frac{d y}{y^{2}} = 2 e^{2 x} d x \end{aligned}$
Integrating both sides

\begin{aligned} \int \frac{1}{y^{2}} d y = \int 2 e^{2 x} d x \\ \Rightarrow \int y^{- 2} d y = \int 2 e^{2 x} d x \end{aligned}

\Rightarrow \frac{y^{- 2 + 1}}{- 2 + 1} = \frac{2 e^{2 x}}{2} + c \Rightarrow \frac{- 1}{y} = e^{2 x} + c

..............(1)
Given that

y (0) = - 1 i.e at x = 0; y = - 1

\begin{aligned} \Rightarrow \frac{- 1}{- 1} = e^{2 (0)} + c \Rightarrow 1 = e^{0} + c \Rightarrow 1 = 1 + c \Rightarrow c = 0 \\ [∴ e^{0} = 1] \end{aligned}

Put in (1)

\Rightarrow \frac{- 1}{y} = e^{2 x} + 0 \Rightarrow y e^{2 x} + 1 = 0

Differential Equations exercise 21.7 question 45 (iv) maths

Answer: $\sin y = e^{x}$
Hint: Separate the terms of x and y and then integrate them.
Given: $\cos y \frac{d y}{d x} = e^{x}, y (0) = \frac{π}{2}$
Solution:
$\begin{aligned} \cos y \frac{d y}{d x} = e^{x} \\ \cos y d y = e^{x} d x \end{aligned}$
Integrating both sides

\begin{aligned} \int \cos y d y = \int e^{x} d x \\ \sin y = e^{x} + c \end{aligned}

..............(1)
Given that

y (0) = \frac{π}{2} i.e. at x = 0, y = \frac{π}{2}

\begin{aligned} ∴ \sin \frac{π}{2} = e^{0} + c [∴ o f (1)] \\ \Rightarrow 1 = 1 + c \Rightarrow c = 0 [∴ \sin \frac{π}{2} = 1, e^{0} = 1] \end{aligned}

Put in (1) we get

\sin y = e^{x} + 0 \Rightarrow \sin y = e^{x}

Differential Equations exercise 21.7 question 45 (v)

Answer: $x^{2} = \log y$
Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x} = 2 x y, y (0) = 1$
Solution:
$\begin{aligned} \frac{d y}{d x} = 2 x y \\ \Rightarrow \frac{d y}{y} = 2 x d x \end{aligned}$
Integrating both sides

\begin{aligned} \int \frac{1}{y} d y = \int 2 x d x \\ \Rightarrow \log y = \frac{2 x^{2}}{2} + c \Rightarrow \log y = x^{2} + c \end{aligned}

.............(1)
Given that at

x = 0, y = 1

∴ \log 1 = 0 + c \Rightarrow 0 = 0 + c \Rightarrow c = 0 [∴ \log 1 = 0]

Put in (1) we get

\log y = x^{2} + 0 \Rightarrow x^{2} = \log y

Differential Equations exercise 21.7 question 45 (vi)

Answer: $\tan^{- 1} y = \frac{x^{3}}{3} + x + \frac{π}{4}$
Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x} = 1 + x^{2} + y^{2} + x^{2} y^{2}, y (0) = 1$
Solution:
$\begin{aligned} \frac{d y}{d x} = 1 + x^{2} + y^{2} + x^{2} y^{2} \\ \Rightarrow \frac{d y}{d x} = (1 + x^{2}) + y^{2} (1 + x^{2}) \\ \Rightarrow \frac{d y}{d x} = (1 + x^{2}) (1 + y^{2}) \\ \Rightarrow \frac{d y}{1 + y^{2}} = (1 + x^{2}) d x \end{aligned}$
Integrating both sides

\begin{aligned} \int \frac{1}{1 + y^{2}} d y = \int (1 + x^{2}) d x \\ \Rightarrow \tan^{- 1} y = x + \frac{x^{3}}{3} + c \end{aligned}

...............(1)
Given that

y (0) = 1 i.e. at x = 0, y = 1

\begin{aligned} \tan^{- 1} (1) = 0 + \frac{0}{3} + c \Rightarrow c = \tan^{- 1} (1) = \tan^{- 1} (\tan \frac{π}{4}) \\ [∴ \tan \frac{π}{4} = 1] \\ \Rightarrow c = \frac{π}{4} \\ \tan^{- 1} y = x + \frac{x^{3}}{3} + \frac{π}{4} \end{aligned}

Differential Equations exercise 21.7 question 45 (vii)

Answer: $x + 2 + \log | \frac{(x y + 2 x)^{2}}{9} |$
Hint: Separate the terms of x and y and then integrate them.
Given: $x y \frac{d y}{d x} = (x + 2) (y + 2), y (1) = - 1$
Solution:
$\begin{aligned} x y \frac{d y}{d x} = (x + 2) (y + 2) \\ \Rightarrow \frac{y}{y + 2} d y = \frac{x + 2}{x} d x \end{aligned}$
Integrating both sides

\begin{aligned} \int \frac{y + 2 - 2}{y + 2} d y = \int 1 d x + \int \frac{2}{x} d x \\ \int (1 - \frac{2}{y + 2}) d y = \int 1 d x + \int \frac{2}{x} d x \end{aligned}

\begin{aligned} \Rightarrow \int 1 d y - 2 \int \frac{1}{y + 2} d y = \int 1 d x + 2 \int \frac{1}{x} d x \\ \Rightarrow y - 2 \log | y + 2 | = x + 2 \log | x | + c \\ \Rightarrow y - x = 2 \log | x | + 2 \log | y + 2 | + c \\ y = x + 2 \log | (x) (y + 2) | + c \end{aligned}

.................(1)
Given that

y (1) = - 1

i.e.

y = - 1

when

x = 1

\begin{aligned} ∴ - 1 = 1 + 2 \log | (1) (1 + 2) | + c \\ \Rightarrow 2 \log 3 + c = 2 \Rightarrow c = 2 - 2 \log 3 = 2 - \log 3^{2} \\ \Rightarrow c = 2 - \log 9 \end{aligned}

Put in (1)

\begin{aligned} y = x + 2 \log | (x) (y + 2) | + 2 - \log 9 \\ [\begin{matrix} ∴ \log m + \log n = \log m n \\ \log m - \log n = \log \frac{m}{n} \end{matrix}] \\ \Rightarrow y = x + 2 + \log | \frac{(x y + 2 x)^{2}}{9} | \end{aligned}

Differential Equations exercise 21.7 question 45 (viii) maths

Answer: $\tan^{- 1} y = \frac{x^{2}}{2} + x$
Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x} = 1 + x + y^{2} + x y^{2}, when y = 0, x = 0$
Solution:
$\begin{aligned} \frac{d y}{d x} = 1 + x + y^{2} + x y^{2} \\ \frac{d y}{d x} = (1 + x) + y^{2} (1 + x) = (1 + x) (1 + y^{2}) \\ \Rightarrow \frac{d y}{1 + y^{2}} = (1 + x) d x \end{aligned}$
Integrating both sides

\begin{aligned} \int \frac{1}{1 + y^{2}} d y = \int (1 + x) d x \\ \Rightarrow \tan^{- 1} y = x + \frac{x^{2}}{2} + c \end{aligned}

..............(1)
Given that when

y = 0, x = 0

∴ \tan^{- 1} (0) = 0 + \frac{0}{2} + c \Rightarrow c = 0

Put in (1) we get

\begin{aligned} \tan^{- 1} y = x + \frac{x^{2}}{2} + 0 \\ \Rightarrow \tan^{- 1} y = x + \frac{x^{2}}{2} \end{aligned}

Differential Equations exercise 21.7 question 45 (ix)

Answer: $y = \log | (y + 3)^{3} x^{2} | - 2$
Hint: Separate the terms of x and y and then integrate them.
Given: $2 (y + 3) - x y \frac{d y}{d x} = 0, y (1) = - 2$
Solution:
$\begin{aligned} 2 (y + 3) - x y \frac{d y}{d x} = 0 \\ \Rightarrow x y \frac{d y}{d x} = 2 (y + 3) \Rightarrow \frac{y d y}{y + 3} = \frac{2}{x} d x \end{aligned}$
Integrating both sides

\begin{aligned} \int \frac{y d y}{y + 3} = \int \frac{2}{x} d x \\ \Rightarrow \int \frac{y + 3 - 3}{y + 3} d y = 2 \int \frac{1}{x} d x \\ \Rightarrow \int (1 - \frac{3}{y + 3}) d y = 2 \int \frac{1}{x} d x \end{aligned}

\begin{aligned} \Rightarrow y - 3 \log | y + 3 | = 2 \log | x | + c \\ \Rightarrow y = 3 \log | y + 3 | + 2 \log | x | + c \\ \Rightarrow y = \log | (y + 3)^{3} | + \log x^{2} + c \\ \Rightarrow y = \log | (y + 3)^{3} x^{2} | + c \end{aligned}

..............(1)
Given that

y (1) = - 2 i.e. at x = 1, y = 2

\begin{aligned} ∴ - 2 = \log | (- 2 + 3)^{3} (1)^{2} | + c \\ - 2 = \log 1 + c \Rightarrow - 2 = 0 + c \Rightarrow c = - 2 \\ [∴ \log 1 = 0] \end{aligned}

Put in (1) we get

y = \log | (y + 3)^{3} x^{2} | - 2

Differential Equations exercise 21.7 question 45 (x)

Answer: $e^{x} - \tan y = 2$
Hint: Separate the terms of x and y and then integrate them.
Given: $: e^{x} \tan y d x + (2 - e^{x}) \sec^{2} y d y = 0; y (0) = \frac{π}{4}$
Solution:
$\begin{aligned} e^{x} \tan y d x + (2 - e^{x}) \sec^{2} y d y = 0 \\ \Rightarrow e^{x} \tan y d x = - (2 - e^{x}) \sec^{2} y d y \\ \Rightarrow e^{x} \tan y d x = (e^{x} - 2) \sec^{2} y d y \\ \Rightarrow \frac{e^{x}}{e^{x} - 2} d x = \frac{\sec^{2} y}{\tan y} d y \end{aligned}$
Integrating both sides

\begin{aligned} \int \frac{e^{x}}{e^{x} - 2} d x = \int \frac{\sec^{2} y}{\tan y} d y \\ \Rightarrow \log | e^{x} - 2 | = \log | \tan y | + \log | c | \\ [∴ \int \frac{1}{x} d x = \log | x | + c] \end{aligned}

\begin{aligned} \Rightarrow \log | e^{x} - 2 | - \log | \tan y | = \log | c | \\ \Rightarrow \log | \frac{e^{x} - 2}{\tan y} | = \log | c | \\ [∴ \log m - \log n = \log \frac{m}{n}] \end{aligned}

Given that

y (0) = \frac{π}{4} i.e when x = 0; y = \frac{π}{4}

\begin{aligned} \Rightarrow \log | \frac{e^{0} - 2}{\tan \frac{π}{4}} | = \log | c | \Rightarrow \log | \frac{1 - 2}{1} | = \log c \\ [∴ e^{0} = 1, \tan \frac{π}{4} = 1] \\ \Rightarrow \log 1 = \log c \Rightarrow \log c = 0 \end{aligned}

Put in (1)

\begin{aligned} \Rightarrow \log | \frac{e^{x} - 2}{\tan y} | = 0 \Rightarrow \frac{e^{x} - 2}{\tan y} = e^{0} \\ \Rightarrow e^{x} - 2 = \tan y (1) \\ \begin{matrix} [∴ e^{0} = 1] [\begin{matrix} ∴ \log_{e} a = x \\ a = e^{x} \end{matrix}] \\ \Rightarrow e^{x} - \tan y = 2 \end{matrix} \end{aligned}

Differential Equations exercise 21.7 question 46

Answer: $x = 2 \cos y$
Hint: Separate the terms of x and y and then integrate them.
Given: $x \frac{d y}{d x} + \cot y = 0, y = \frac{π}{4} when x = \sqrt{2}$
Solution:
$\begin{aligned} x \frac{d y}{d x} + \cot y = 0 \\ \Rightarrow x \frac{d y}{d x} = - \cot y \\ \Rightarrow \frac{d y}{\cot y} = - \frac{d x}{x} \end{aligned}$
Integrating both sides

\begin{aligned} \int \frac{1}{\cot y} d y = - \int \frac{1}{x} d x \\ - \int \frac{\sin y}{\cos y} d y = \int \frac{1}{x} d x \\ \Rightarrow \log | \cos y | = \log | x | + c \end{aligned}

\begin{aligned} [∴ \int \frac{1}{x} d x = \log | x | + c] \\ \Rightarrow \log | \cos y | - \log | x | = c \end{aligned}

...............(1)
Now, when

x = \sqrt{2}, y = \frac{π}{4}

\begin{aligned} ∴ \log | \cos \frac{π}{4} | - \log | \sqrt{2} | = c \\ \Rightarrow \log \frac{1}{\sqrt{2}} - \log \sqrt{2} = c \Rightarrow \log (2)^{- \frac{1}{2}} - \log (2)^{\frac{1}{2}} = c \end{aligned}

\begin{aligned} \Rightarrow - \frac{1}{2} \log 2 - \frac{1}{2} \log 2 = c \Rightarrow - 2 (\frac{1}{2} \log 2) = c \\ \Rightarrow - \log 2 = c \end{aligned}

Put in (1)

\begin{aligned} \Rightarrow \log | \cos y | - \log | x | = - \log 2 \\ \Rightarrow \log | \frac{\cos y}{x} | + \log 2 = 0 \end{aligned}

[\begin{array}{l} ∴ \log (m) + \log (n) = \log m n \\ \log (m) - \log (n) = \log \frac{m}{n} \end{array}]

\begin{aligned} \Rightarrow \log | \frac{2 \cos y}{x} | = 0 \\ [\begin{matrix} ∴ \log_{ε} a = x \\ x = a^{x} \end{matrix}] \\ \Rightarrow \frac{2 \cos y}{x} = e^{0} \end{aligned}

\begin{aligned} [e^{0} = 1] \\ \Rightarrow 2 \cos y = x (1) \\ \Rightarrow 2 \cos y = x \end{aligned}

Differential Equations exercise 21.7 question 47

Answer: $x + y + x y = 1$
Hint: Separate the terms of x and y and then integrate them.
Given: $(1 + x^{2}) \frac{d y}{d x} + (1 + y^{2}) = 0, y = 1 when x = 0$
Solution:
$\begin{aligned} (1 + x^{2}) \frac{d y}{d x} + (1 + y^{2}) = 0 \\ (1 + x^{2}) \frac{d y}{d x} = - (1 + y^{2}) \\ \Rightarrow \frac{d y}{1 + y^{2}} = - \frac{1}{1 + x^{2}} d x \end{aligned}$
Integrating both sides

\begin{aligned} \Rightarrow \int \frac{1}{(1 + y^{2})} d y = - \int \frac{1}{(1 + x^{2})} dx \\ \tan^{- 1} y = - \tan^{- 1} x + c \end{aligned}

................(1)
Given that

y = 1 when x = 0

∴ \tan^{- 1} (1) = - \tan^{- 1} (0) + c \Rightarrow \frac{π}{4} = 0 + c \Rightarrow c = \frac{π}{4}

Put in (1) we get

\begin{aligned} \Rightarrow \tan^{- 1} y = - \tan^{- 1} x + \frac{π}{4} \\ [∴ \tan^{- 1} x + \tan^{- 1} y = \tan^{- 1} (\frac{x + y}{1 - x y})] \\ \Rightarrow \tan^{- 1} x + \tan^{- 1} y = \frac{π}{4} \end{aligned}

\begin{aligned} \Rightarrow \tan^{- 1} (\frac{x + y}{1 - x y}) = \frac{π}{4} \\ \Rightarrow (\frac{x + y}{1 - x y}) = \tan \frac{π}{4} \end{aligned}

\begin{aligned} \Rightarrow (\frac{x + y}{1 - x y}) = 1 \\ \Rightarrow x + y = 1 - x y \\ \Rightarrow x + y + x y = 1 \end{aligned}

Differential Equations exercise 21.7 question 48 maths

Answer: $2 y \sin y = 2 x^{2} \log x + x^{2} - 1$
Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x} = \frac{2 x (\log x + 1)}{\sin y + y \cos y}, y = 0 when x = 1$
Solution:
$\begin{aligned} \frac{d y}{d x} = \frac{2 x (\log x + 1)}{\sin y + y \cos y} \\ \Rightarrow (\sin y + y \cos y) d y = 2 x (\log x + 1) d x \end{aligned}$
Integrating both sides

\begin{aligned} \Rightarrow \int (\sin y + y \cos y) d y = \int 2 x (\log x + 1) d x \\ \Rightarrow \int \sin y d y + \int y \cos y d y = \int 2 x \log x d x + \int 2 x d x \end{aligned}

[?Integration by parts]

\begin{aligned} \Rightarrow - \cos y + [y \sin y - \int \sin y d y] = 2 [\log x \cdot \frac{x^{2}}{2} - \int \frac{1}{x} \frac{x^{2}}{2} d x] \\ \Rightarrow - \cos y + y \sin y + \cos y = x^{2} \log x - \frac{x^{2}}{2} + c \\ \Rightarrow y \sin y = x^{2} \log x + \frac{x^{2}}{2} + c \end{aligned}

...........(1)
Given that

y = 0 when x = 1

\begin{aligned} \Rightarrow 0 \sin 0 = 1 \cdot \log 1 + \frac{1}{2} + c \\ \Rightarrow 0 = 0 + \frac{1}{2} + c \\ \Rightarrow c = - \frac{1}{2} \end{aligned}

Put in (1)

\begin{aligned} \Rightarrow y \sin y = x^{2} \log x + \frac{x^{2}}{2} - \frac{1}{2} \\ \Rightarrow 2 y \sin y = 2 x^{2} \log x + x^{2} - 1 \end{aligned}

Differential Equations exercise 21.7 question 49

Answer: $x \log (x + 1) - x + \log (x + 1) + 3$
Hint: Separate the terms of x and y and then integrate them.
Given: $e^{\frac{d y}{d x}} = x + 1 given that y = 3 when x = 0$
Solution: $e^{\frac{d y}{d x}} = x + 1$
Integrating both sides

\begin{aligned} \log e^{\frac{d y}{d x}} = \log (x + 1) \\ \Rightarrow \frac{d y}{d x} = \log (x + 1) d x [∵ \log_{e} e = 1] \\ \Rightarrow d y = \log (x + 1) d x \end{aligned}

Integrating both sides

\begin{aligned} \int d y = \int \log (x + 1) \cdot 1 d x \\ \Rightarrow y = \log (x + 1) x - \int \frac{1}{x + 1} x d x \\ \Rightarrow y = \log (x + 1) x - \int \frac{x + 1 - 1}{x + 1} d x \\ \Rightarrow y = x \log (x + 1) - \int (1 - \frac{1}{x + 1}) d x \end{aligned}

[Integration by parts]

\begin{aligned} \Rightarrow y = x \log (x + 1) - [x - \log | x + 1 |] + c \\ \Rightarrow y = x \log (x + 1) - x + \log | x + 1 | + c \end{aligned}

..............(1)
Now

y = 3

when

x = 0

\begin{aligned} 3 = 0 \cdot \log (0 + 1) - 0 + \log (0 + 1) + c \\ \Rightarrow 3 = 0 - 0 + 0 + c \Rightarrow c = 3 [∵ \log 1 = 0] \end{aligned}

Put in (1)

\Rightarrow y = x \log (x + 1) - x + \log | x + 1 | + 3

Differential Equations exercise 21.7 question 50

Answer: $\sin x + \log | \sin y | = 1$
Hint: Separate the terms of x and y and then integrate them.
Given: $\cos y d y + \cos x \sin y d x = 0, y = \frac{π}{2} when x = \frac{π}{2}$
Solution:
$\begin{aligned} \cos y d y + \cos x \sin y d x = 0 \\ \cos y d y = - \cos x \sin y d y \\ \Rightarrow \frac{- \cos y}{\sin y} d y = \cos x d x \end{aligned}$
Integrating both sides

\begin{aligned} \int \frac{- \cos y}{\sin y} d y = \int \cos x d x \\ \Rightarrow - \log | \sin y | = \sin x + c \end{aligned}

..............(1)

When $x = \frac{π}{2}, y = \frac{π}{2}$

$\begin{aligned} \Rightarrow - \log | \sin \frac{π}{2} | = \sin \frac{π}{2} + c \\ - \log | 1 | = 1 + c \Rightarrow 0 - 1 = c \Rightarrow c = - 1 \\ [∴ \log 1 = 0, \sin \frac{π}{2} = 1] \end{aligned}$
Put in (1) we get
$\begin{matrix} \Rightarrow - \log | \sin y | = \sin x - 1 \\ \Rightarrow \sin x + \log | \sin y | = 1 \end{matrix}$

Differential Equations exercise 21.7 question 51

Answer: $2 x^{2} y + y = 1$
Hint: Separate the terms of x and y and then integrate them.
Given: $\frac{d y}{d x} = - 4 x y^{2}, y = 1 when x = 0$
Solution:
$\begin{aligned} \frac{d y}{d x} = - 4 x y^{2} \\ \Rightarrow \frac{d y}{y^{2}} = - 4 x d x \end{aligned}$
Integrating both sides

\begin{aligned} \int \frac{1}{y^{2}} = - 4 \int x d x \\ \Rightarrow \frac{y^{- 2 + 1}}{- 2 + 1} = \frac{- 4 x^{2}}{2} + c \\ \Rightarrow \frac{- 1}{y} = - 2 x^{2} + c \\ \Rightarrow - \frac{1}{y} = - 2 x^{2} + c \end{aligned}

...............(1)
Now

y = 1

when

x = 0

\Rightarrow - \frac{1}{1} = - 2.0 + c => c = - 1

Put in (1)

\begin{aligned} \Rightarrow \frac{- 1}{y} & = - 2 x^{2} - 1 \end{aligned}

\Rightarrow 2 x^{2} y + y = 1

Differential Equations exercise 21.7 question 52 maths

Answer: $2 y = e^{x} (\sin x - \cos x) + 1$
Hint: Separate the terms of x and y and then integrate them.
Given: Curve passing through (0,0) with differential equation

$\frac{d y}{d x} = e^{x} \sin x$
Solution:
$\begin{aligned} \frac{d y}{d x} = e^{x} \sin x \\ \Rightarrow d y = e^{x} \sin x d x \end{aligned}$
Integrating both sides
$\int d y = \int e^{x} \sin x d x \Rightarrow y = \int e^{x} \sin x d x$ .................(*)
$\begin{aligned} y = \sin x e^{x} - \int \cos x e^{x} d x \\ \Rightarrow y = \sin x e^{x} - [\cos x e^{x} + \int \sin x e^{x} d x] + c \\ \Rightarrow y = \sin x e^{x} - \cos x e^{x} - y + c \\ \Rightarrow 2 y = e^{x} (\sin x - \cos x) + c \end{aligned}$ ..............(1)
The curve passes through (0,0) [∴ of given]
$\begin{aligned} 0 = e^{0} (\sin 0 - \cos 0) + c \\ \Rightarrow 0 = 1 (0 - 1) + c \Rightarrow c = 1 \\ Put in (1) \\ 2 y = e^{x} (\sin x - \cos x) + 1 \end{aligned}$

Differential Equations exercise 21.7 question 53

Answer: $y = x + \log {(\frac{x y + 2 x}{3})}^{2} + 2$
Hint: Separate the terms of x and y and then integrate them.
Given: $x y \frac{d y}{d x} = (x + 2) (y + 2); p t (1, - 1)$
Solution:
$\begin{aligned} x y \frac{d y}{d x} = (x + 2) (y + 2) \\ \Rightarrow \frac{y}{y + 2} d y = \frac{x + 2}{x} d x \end{aligned}$
Integrating both sides

\begin{aligned} \int \frac{y}{y + 2} d y = \int \frac{x + 2}{x} d x \\ \int \frac{y + 2 - 2}{y + 2} d y = \int (1 + \frac{2}{x}) d x \\ \Rightarrow \int (1 - \frac{2}{y + 2}) d y = \int (1 + \frac{2}{x}) d x \end{aligned}

\begin{aligned} \Rightarrow y - 2 \log | y + 2 | = x + 2 \log | x | + c \\ \Rightarrow y = x + 2 \log | x | + 2 \log | y + 2 | + c \\ \Rightarrow y = x + 2 \log | x (y + 2) | + c \\ \Rightarrow y = x + \log (x y + 2 x)^{2} + c \end{aligned}

..............(1)
It passes through pt(-1,1)

\begin{aligned} 1 = - 1 + \log | (- 1) (1) + 2 (- 1) |^{2} + c \\ \Rightarrow 2 = \log (- 1 - 2)^{2} + c \\ \Rightarrow 2 - \log 9 = c \end{aligned}

Put in (1) we get

\begin{aligned} \Rightarrow y = x + \log (x y + 2 x)^{2} + 2 - \log 9 \\ \Rightarrow y = x + \log {(\frac{x y + 2 x}{3})}^{2} + 2 \\ [∴ \log m - \log n = \log \frac{m}{n}] \end{aligned}

Differential Equations exercise 21.7 question 54

Answer: $(63 t + 27)^{\frac{1}{3}}$
Hint: Separate the terms of x and y and then integrate them.
Given: The volume of a spherical balloon being inflated changes at a constant rate. If initially it radius is 3 units and after 3 seconds it is 6 units. Find the radius of the balloon after t seconds
Solution: Let V be the volume of spherical balloon
Now it is being inflated changes at a constant rate

∵ \frac{d v}{d t} = k

where k is any constant ……………….(*)
New volume of spherical balloon

= \frac{4}{3} π r^{3}

, r is radius

\begin{aligned} \frac{d v}{d t} = \frac{4}{3} π 3 r^{2} \frac{d r}{d t} = 4 π r^{2} \frac{d r}{d t} \\ \Rightarrow k = 4 π r^{2} \frac{d r}{d t} \\ \Rightarrow d t = \frac{4 π}{k} r^{2} d r \end{aligned}

Integrating both sides

\begin{aligned} \int d t = \frac{4 π}{k} \int r^{2} d r \\ \Rightarrow t = \frac{4 π}{k} [\frac{r^{3}}{3}] + c \\ \Rightarrow \frac{r^{8}}{3} = \frac{k}{4 π} t + c \end{aligned}

...........(1)

Now given conditions are:
When

t = 0; r = 3

and when

t = 3, r = 6

We have to find r at

t = t

Put in (1) we get

\begin{aligned} At t = 0, r = 3 \Rightarrow \frac{3^{8}}{3} = \frac{k}{4 π} 0 + c \Rightarrow 9 = c \end{aligned}

?By(1)

\frac{r^{8}}{3} = \frac{k}{4 π} t + 9

..................(2)
Now at

t = 3, r = 6 \Rightarrow \frac{6^{3}}{3} = \frac{k}{4 π} 3 + 9

\Rightarrow \frac{6 \times 6 \times 6}{3} - 9 = \frac{3 k}{4 π}

\begin{aligned} \Rightarrow 72 - 9 = \frac{3 k}{4 π} \Rightarrow \frac{63 \times 4 π}{3} = k \\ \Rightarrow k = 84 π \end{aligned}

Put in (2)

\begin{aligned} \frac{r^{3}}{3} = \frac{84 π}{4 π} t + 9 \Rightarrow \frac{r^{3}}{3} = 21 t + 9 \\ \Rightarrow r^{3} = 63 t + 27 \\ \Rightarrow r = (63 t + 27)^{\frac{1}{3}} \end{aligned}

which is our required radius.

Differential Equations exercise 21.7 question 55

Answer: $6.931$
Hint: Separate the terms of x and y and then integrate them.
Given: In a bank principal increases at the rate of r% per year. We have to find the value of r if Rs.100 double itself in 10 years (log 2=0.6931)
Solution: Let Principal = P and rate = r
As principal increases at the rate of r% w.r.t time i.e. per year.

\begin{aligned} ∴ \frac{d P}{d t} = \frac{r}{100} \times P \\ \Rightarrow \frac{d P}{P} = \frac{r}{100} d t \end{aligned}

Integrating both sides

\Rightarrow \log P = \frac{r}{100} t + c

..................(1)
Suppose initially

t = 0, P = P_{0}

\begin{aligned} \Rightarrow \log P_{0} = \frac{r}{100} 0 + c \\ \Rightarrow c = \log P_{0} \end{aligned}

Put in (1)

\begin{aligned} \Rightarrow \log P = \frac{r}{100} t + \log P_{0} \\ \log P - \log P_{0} = \frac{r}{100} t \end{aligned}

..............(2)
According to given
When

t = 10, P = 2 * 100, P_{0} = 100

By (2)

\begin{aligned} \log 200 - \log 100 = \frac{r}{100} 10 \\ \Rightarrow \log (\frac{200}{100}) = \frac{r}{10} \\ \Rightarrow \log 2 = \frac{r}{10} \Rightarrow 10 \log_{e} 2 = r \\ \Rightarrow r = 10 \times 0.6931 = 6.931 \\ \Rightarrow r = 6.931 \end{aligned}

Differential Equations exercise 21.7 question 56 maths

Answer: $1648$
Hint: Separate the terms of x and y and then integrate them.
Given: In a bank principal increases increases 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.(e0.5 =1.648)
Solution: Let P be the Principal
As Principal increases at the rate of 5% w.r.t t

\begin{aligned} ∴ \frac{d P}{d t} = \frac{5}{100} \times P \\ \Rightarrow \frac{d P}{P} = \frac{1}{20} d t \end{aligned}

Integrating both sides

\int \frac{d P}{P} = \frac{1}{20} \int 1 d t

................(1)
Now initially P₀=1000; after 10 years P= P₁₀ also initially t =0 and after 10 years t = 10
By (1)

\begin{aligned} \int_{P_{0}}^{P_{0}} \frac{d P}{P} = \frac{1}{20} \int_{0}^{+} 1 d t \\ \Rightarrow [\log | P |]_{1000}^{P_{0}} = \frac{1}{20} [t]_{0}^{10} \\ \Rightarrow \log P_{10} - \log 1000 = \frac{1}{20} (10 - 0) \end{aligned}

\begin{aligned} \Rightarrow [\log | \frac{P_{10}}{1000} |] = \frac{1}{2} \\ [\log m - \log n = \log \frac{m}{n}] \\ \Rightarrow | \frac{P_{10}}{1000} | = e^{\frac{1}{2}} \end{aligned}

\begin{aligned} [\begin{matrix} \log_{a} e = x \\ a = e^{x} \end{matrix}] \\ \Rightarrow P_{10} = 1000 (1.648) \\ [e^{\frac{1}{2}} = 1.648] \\ \Rightarrow P_{10} = 1648 \end{aligned}

Principal after 10 years = 1648

Differential Equations exercise 21.7 question 57

Answer: $\frac{2 \log 2}{\log \frac{11}{10}}$
Hint: Separate the terms of x and y and then integrate them.
Given: In a culture the bacteria count is 100000. The number is increased by 10% in 2 hours. We have to find, in how many hours will the count reach 200000, if the rate of growth of bacteria is proportional to the number present.
Solution: Let the count of bacteria be N
As the rate of growth of bacteria is proportional to the no. present
$\begin{aligned} ∴ \frac{d N}{d t} α N \\ \Rightarrow \frac{d N}{d t} = K N \\ \Rightarrow \frac{d N}{N} = K d t \end{aligned}$
Integrating both sides

\begin{aligned} \int \frac{d N}{N} = K \int d t \\ \Rightarrow \log | N | = K t + c \end{aligned}

.................(1)
Initially

N = 100000; t = 0

By (1)

\begin{aligned} \log 100000 = K (0) + c \\ \Rightarrow c = \log 100000 \end{aligned}

Put in (1) we get

\log N = K t + \log 100000

...................(2)
According to given
When

\begin{aligned} t = 2, N & = 10 % of 100000 + 100000 \\ = 10000 + 100000 = 110000 \end{aligned}

\begin{aligned} By (2) \log 110000 = K (2) + \log 100000 \\ \log 110000 = K (2) + \log 100000 \\ \Rightarrow \log \frac{110000}{100000} = K (2) \\ [\log m - \log n = \log \frac{m}{n}] \end{aligned}

\begin{aligned} [\log m - \log n = \log \frac{m}{n}] \\ \Rightarrow K = \frac{1}{2} \log \frac{11}{10} \\ ∴ \log N = \frac{1}{2} \log \frac{11}{10} t + \log 100000 \end{aligned}

....................(3)
Now we have to find in how many hours i.e t¹ ; N=200000
By (3)

\begin{aligned} \log 200000 = \frac{1}{2} (\log \frac{11}{10}) t^{1} + \log 100000 \\ \Rightarrow \log | \frac{200000}{100000} | = \frac{1}{2} \log (\frac{11}{10}) t^{1} \\ \Rightarrow \log 2 = \frac{1}{2} \log (\frac{11}{10}) t^{1} \end{aligned}

\begin{aligned} \Rightarrow 2 \log 2 = \log (\frac{11}{10}) t^{1} \\ \Rightarrow \frac{2 \log 2}{\log (\frac{11}{10})} = t^{1} \\ \Rightarrow t^{1} = \frac{2 \log 2}{\log (\frac{11}{10})} \end{aligned}

Differential Equations exercise 21.7 question 58

Answer: $\frac{1}{3}$
Hint: Separate the terms of x and y and then integrate them.
Given: IF

y (x)

is a solution of the differential equation

(\frac{2 + \sin x}{1 + y}) \frac{d y}{d x} = - \cos x and y (0) = 1

then find

y (\frac{π}{2})

Solution:
$\begin{aligned} (\frac{2 + \sin x}{1 + y}) \frac{d y}{d x} = - \cos x \\ \Rightarrow \frac{d y}{1 + y} = \frac{- \cos x}{2 + \sin x} d x \end{aligned}$
Integrating both sides

\begin{aligned} \Rightarrow \int \frac{d y}{1 + y} = - \int \frac{\cos x}{2 + \sin x} d x \\ \Rightarrow \log | 1 + y | = - \log | 2 + \sin x | + \log c \\ [\begin{matrix} Put \\ \sin x = t \\ \cos x d x = d t \\ \int \frac{1}{t} d t = \log | t | + c \end{matrix}] \end{aligned}

\begin{aligned} \Rightarrow \log | y + 1 | + \log | 2 + \sin x | = \log c \\ \Rightarrow \log | (y + 1) (2 + \sin x) | = \log c \\ \Rightarrow (y + 1) (2 + \sin x) = c \end{aligned}

..........(1)
According to given:

y (0) = 1 a t x = 0, y = 1

(2) (2 + \sin 0) = c \Rightarrow c = 4

Put in (1)

\Rightarrow (y + 1) (2 + \sin x) = 4

.............(2)
Now we have to find

y (\frac{π}{2})

i.e. value of y at

x = \frac{π}{2}

Put in (2)

\begin{aligned} (y + 1) (2 + \sin \frac{π}{2}) = 4 \\ \Rightarrow (y + 1) (2 + 1) = 4 \\ [∴ \sin \frac{π}{2} = 1] \end{aligned}

\begin{aligned} \Rightarrow 3 (y + 1) = 4 \\ \Rightarrow 3 y + 3 = 4 \Rightarrow 3 y = 1 \Rightarrow y = \frac{1}{3} \\ y (\frac{π}{2}) = \frac{1}{3} \end{aligned}

Differential Equations exercise 21.7 question 59

Answer: $(1 + \log x)^{2} = \log {(1 - y^{2})}^{2} + 1$
Hint: Separate the terms of x and y and then integrate them.
Given: $(1 - y^{2}) (1 + \log x) d x + 2 x y d y = 0 given that y = 0 when x = 1$
Solution:
$\begin{aligned} (1 - y^{2}) (1 + \log x) d x = - 2 x y d y \\ \Rightarrow \frac{1 + \log x}{x} d x = \frac{- 2 y}{1 - y^{2}} d y \end{aligned}$
Integrating both sides

\begin{aligned} \int \frac{1 + \log x}{x} d x = \int \frac{- 2 y}{1 - y^{2}} d y \dots \dots \dots \dots . . (*) \\ \Rightarrow I_{1} = I_{2} where I_{1} = \int \frac{1}{x} (1 + \log x) d x and \\ I_{2} = \int \frac{- 2 y}{1 - y^{2}} d y \\ ∴ I_{1} = \int \frac{1}{x} (1 + \log x) d x \end{aligned}

Put

\begin{aligned} 1 + \log x = t \\ \frac{1}{x} d x = d t \\ = \int t d t \\ = \frac{t^{2}}{2} + c = \frac{(1 + \log x)^{2}}{2} + c \end{aligned}

Now,

I_{2} = \int \frac{- 2 y}{1 - y^{2}} d y

Put

\begin{aligned} 1 - y^{2} = t \\ - 2 y d y = d t \\ = \int \frac{d t}{t} \\ = \log | t | + c \\ = \log | 1 - y^{2} | + c \end{aligned}

Put the values in (*) we get

\frac{(1 + \log x)^{2}}{2} = \log | 1 - y^{2} | + c

................(1)
Now according to given

y = 0

when

x = 1

\begin{aligned} ∴ \frac{(1 + \log 1)^{2}}{2} = \log | 1 - 0 | + c \\ \Rightarrow \frac{(1 + 0)^{2}}{2} = \log (1) + c \\ [\log 1 = 0] \end{aligned}

\begin{aligned} \Rightarrow 1 = 2 [(0) + c] \\ \Rightarrow 1 = 2 c \Rightarrow c = \frac{1}{2} \end{aligned}

Put in (1) we get

\begin{aligned} \Rightarrow \frac{(1 + \log x)^{2}}{2} = \log | 1 - y^{2} | + \frac{1}{2} \\ \Rightarrow (1 + \log x)^{2} = 2 \log (1 - y^{2}) + 1 \\ \Rightarrow (1 + \log x)^{2} = \log {(1 - y^{2})}^{2} + 1 \end{aligned}

The Differential Equation chapter 21 in mathematics of class 12 has about eleven exercises. The seventh exercise, ex 21.7, has 72 questions, with some of them having subparts in the textbook. RD Sharma solutions The concept of these questions revolves around solving the differential equations, initial value problems, equations in variable separable form and word problems using differentiation. This exercise has questions only in the Level 1 category. Yet, the importance of RD Sharma Class 12 Chapter 21 Exercise 21.7 solution book has not been reduced.

As the RD Sharma books follow the NCERT pattern, the CBSE school students benefit from it. The previous exercises before ex 21.7 had sums wherein differentiation can be done directly. While in this exercise, the students must find the equation from the word problem and then solve it. Therefore, the RD Sharma Class 12th Exercise 21.7 reference material lends a helping hand to the students. All solutions are given in the exact order as present in the textbook for the convenience of the students.

The Class 12 RD Sharma Chapter 21 Exercise 21.7 Solution book has an abundance of additional practice questions that help them work out the concept more and understand it deeply. Each answer provided is the work of experts who have committed their time to create the best solution book for the class 12 students.

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Students who prepare for their public exam with the RD Sharma Class 12 Solutions Chapter 21 Ex 21.7 book are gradually getting ready to face their exams confidently. These best solutions books make the students cross their benchmark score and achieve more in their tests and exams.

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Frequently Asked Questions (FAQs)

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RD Sharma Class 12 Exercise 21.7 Differential Equation Solutions Maths - Download PDF Free Online

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