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Provide Solution for RD Sharma Class 12 Chapter 21 Differential Equation Exercise 21.10 Question 8

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Answer: $y\left(x^{2}+1\right)^{2}=-x+C$

Hint: To solve this equation we will use differentiate both terms.

Give: $\frac{d y}{d x}+\frac{4 x}{x^{2}+1} y+\frac{1}{\left(x^{2}+1\right)^{2}}=0$

Solution: $\frac{d y}{d x}+\frac{4 x}{x^{2}+1} y=\frac{-1}{\left(x^{2}+1\right)^{2}}$

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

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

$=e^{\int \frac{4 x}{x^{2}+1} d x}$ $\quad\left[\text { Let } x^{2}+1=u, 2 x d x=d u\right.$

$\begin{aligned} &=2 e^{\int \frac{d u}{u}} \\ & \end{aligned}$

$=e^{2} \ln \left|x^{2}+1\right| \\$

$=\left(x^{2}+1\right)^{2} \\$

$y\, I\! f=\int Q \, I\! f d x \\$

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

$y=\int-d x \\$

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

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