#### Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 16 textbook solution.

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

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

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

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

$=\frac{d y}{d x}+\frac{y}{\left(1+x^{2}\right)}=\frac{\tan ^{-1} x}{1+x^{2}}$                        ....(i)

\begin{aligned} &P=\frac{1}{1+x^{2}}, \cos x=\frac{\tan ^{-1} x}{1+x^{2}} \\ \end{aligned}

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

\begin{aligned} &=e^{\int \frac{1}{1+x^{2}} d x} \\ \end{aligned}

\begin{aligned} &=e^{\tan ^{-1} x} \\ \end{aligned}

\begin{aligned} &=I f=e^{\tan ^{-1} x} \end{aligned}

$=4 e^{\tan ^{-1} x}=\int e^{\tan ^{-1} x} \frac{\tan ^{-1} x}{1+x^{2}} d x+C$

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

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

\begin{aligned} &=y e^{\tan ^{-1} x}=\int e^{t} t d t \\ \end{aligned}

\begin{aligned} &=y e^{\tan ^{-1} x}=t \int e^{t} d t-\int \frac{d}{d t}(t) \int e^{t} d t d x+C \\ \end{aligned}

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

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

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

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

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

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

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

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