#### Please Solve R.D.Sharma class 12 Chapter 21 Differential Equations Exercise 21.8 Question 5 Maths textbook Solution.

Answer: $y-\tan ^{-1}\left ( x+y \right )=c$

Given:$\left ( x+y \right )^{2}\frac{dy}{dx}=1$

Hint : -  Use variable separable form by substitution.

Solution : -  We have,

$\left ( x+y \right )^{2}\frac{dy}{dx}=1$                            ......(i)

Let $x+y=v$

Differentiating with respect to x, we get,

$\frac{d}{dx}\left ( x+y \right )=\frac{dv}{dx}$

$\Rightarrow 1+\frac{dy}{dx}=\frac{dv}{dx}$

$\Rightarrow \frac{dy}{dx}=\frac{dv}{dx}-1$                                                .....(ii)

Now, substituting equation (ii) in equation (i), we get

$(\mathrm{x}+\mathrm{y})^{2}\left(\frac{d v}{d x}-1\right)=1 \\$

$\Rightarrow \mathrm{v}^{2}\left(\frac{d v}{d x}-1\right)=1 \\$

$\Rightarrow \frac{d v}{d x}-1=\frac{1}{v^{2}} \\$

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

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

Taking like variables on same side, we get,

$\Rightarrow \frac{v^{2}}{1+v^{2}}dv=dx$

Integrating on both the sides, we get,

$\int\left(\frac{v^{2}}{1+v^{2}}\right) d v=\int d x \\$

$\Rightarrow \int\left(\frac{v^{2}+1-1}{1+v^{2}}\right) d v=\int d x \\$

$\Rightarrow \int\left(\frac{v^{2}+1}{v^{2}+1}-\frac{1}{v^{2}+1}\right) d v=\int d x \\$

$\Rightarrow \int\left(1-\frac{1}{v^{2}+1}\right) d v=\int d x \\$

$\Rightarrow \int d v-\int \frac{1}{v^{2}+1} d v=\int d x \\$

$\Rightarrow v-\tan ^{-1} v=x+c \quad\left(\because \int\left(\frac{1}{x^{2}+1}\right) d x=\tan ^{-1} x\right)$

Putting $v=x+y,$ we have

$\Rightarrow \left ( x+y \right )-\tan ^{-1}\left ( x+y \right )=x+c$

$\Rightarrow y-\tan ^{-1}\left ( x+y \right )=c$

(This is the required solution).