#### Explain solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 22 maths textbook solution.

Answer :  $2 x e^{\tan ^{-1} x}=e^{2 \tan ^{-1} y}+C$

Hint : To solve this equation we use $e^{\int f(x) d x}$ formula.

Give: $\left(1+y^{2}\right)+\left(x-e^{2 \tan ^{-1} y}\right) \frac{d y}{d x}=0$

Solution :

\begin{aligned} &x-e^{\tan ^{-1} y} \frac{d y}{d x}=-\left(1+y^{2}\right) \\ &=\left(e^{\tan ^{-1} y}-x\right) \frac{d y}{d x}=1+y^{2} \\ &=\left(e^{\tan ^{-1} y}-x\right) d y=\left(1+y^{2}\right) d x \end{aligned}

Put $\tan ^{-1}y=t$

\begin{aligned} &=\frac{1}{1+y^{2}} d y=d t \\ \end{aligned}

\begin{aligned} &=\left(e^{t}-x\right) d t=d x \\ \end{aligned}

\begin{aligned} &=\frac{d x}{d t}=e^{t}-x \\ \end{aligned}

\begin{aligned} &=\frac{d x}{d t}+x=e^{t} \\ \end{aligned}

\begin{aligned} &=\frac{d y}{d x}+P(x) y=Q x \end{aligned}

\begin{aligned} &=\frac{d x}{d t}+P(t) t=Q(x) \\ \end{aligned}

\begin{aligned} &P(t)=1, Q t=e^{t} \\ \end{aligned}

\begin{aligned} &I f=e^{\int P(t) d x} \\ \end{aligned}

\begin{aligned} &=e^{\int 1 d t} \\ \end{aligned}

\begin{aligned} &=e^{t} \end{aligned}

\begin{aligned} &=x e^{t}=\int e^{t} e^{t} d t+C \\ \end{aligned}

\begin{aligned} &=x e^{\tan ^{-1} y}=\frac{e^{2 \tan ^{-1} y}}{2}+C \\ \end{aligned}

\begin{aligned} &=2 x e^{\tan ^{-1} y}=e^{2 \tan ^{-1} y}+C \end{aligned}