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Explain solution RD Sharma class 12 Chapter 21 Differential Equation exercise 21.10 question 22

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

Hint: To solve this equation we use $e\int f\left ( x \right )dx$ formula.

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

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

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

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

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

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

$=\left(e^{t}-x\right) d t=d x \\$

$=\frac{d x}{d t}=e^{t}-x \\$

$=\frac{d x}{d t}+x=e^{t} \\$

$=\frac{d y}{d x}+P(x) y=Q x$

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

$P(t)=1, Q t=e^{t} \\$

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

$=e^{\int 1 d t} \\$

$=e^{t} \\$

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

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

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

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