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Explain solution RD Sharma class 12 Chapter 10 Differential Equation Exercise 21.10 question 16

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

Hint: To solve this equation we use $e\int f dx$ formula.

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

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

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

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

$I f=e^{\int P d x} \\$

$=e^{\int \frac{1}{1+x^{2}} d x} \\$

$=e^{\tan ^{-1} x} \\$

$=I f=e^{\tan ^{-1} x} \\$

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

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

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

$=y e^{\tan ^{-1} x}=\int e^{t} t d t \\$

$=y e^{\tan ^{-1} x}=t \int e^{t} d t-\int \frac{d}{d t}(t) \int e^{t} d t d x+C \\$

$=t e^{t}-\int e^{t} d t+C$

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

$=y e^{\tan ^{-1} x}=e^{t}(t-1)+C \\$

$=y e^{\tan ^{-1} x}=e^{t}(t-1)+C \\$

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

$=y=\tan ^{-1} x-1+\frac{C}{\tan ^{-1} x} \\$

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

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