Question

Find the particular solution of the differential equation $(1+x^2)\frac{dy}{dx}+2xy=\frac{1}{1+x^2}$ given that at $x=1,y=0$

Answer 1

We have,

$\left(1+x^2\right)\frac{dy}{dx}+2xy=\frac{1}{1+x^2}\frac{dy}{dx}+\frac{2x}{1+x^2}y=\frac{1}{{\left(1+x^2\right)}^2}$

Therefore,

$y.e^{\int \frac{2x}{1+x^2}dx}=\int e^{\int \frac{2x}{1+x^2}dx}\times \frac{1}{\left(1+x^2\right)}dxy\left(1+x^2\right)={\tan }^{-1}\left(x\right)+C0=\frac{\pi }{4}+CC=\frac{-\pi }{4}$

Therefore,

$y\left(1+x^2\right)={\tan }^{-1}x-\frac{\pi }{4}$

Hence, this is the answer.