Question

Solve $(1+x^2)\frac{dy}{dx}-x=ta{n}^{-1}x$

Answer 1

$(1+x^2)\frac{dy}{dx}-x={\tan }^{-1}x$

$\Rightarrow \frac{dy}{dx}-\frac{x}{1+x^2}=\frac{{\tan }^{-1}x}{1+x^2}$ on dividing both sides by $1+x^2$

$\Rightarrow dy=\frac{x}{1+x^2}dx+\frac{{\tan }^{-1}x}{1+x^2}dx$ on multiplying both sides by $dx$

$\Rightarrow \int dy=\int \frac{x}{1+x^2}dx+\int \frac{{\tan }^{-1}x}{1+x^2}dx$ on integrating both sides w.r.t $x$

Consider $\int \frac{x}{1+x^2}dx$

Take $t=1+x^2\Rightarrow dt=2xdx$

$\Rightarrow \frac{dt}{2}=xdx$

Now, $\int \frac{x}{1+x^2}dx=\int \frac{1}{2}dt=\frac{1}{2}\ln \left|t\right|=\frac{1}{2}\ln |1+x^2|$ where $t=1+x^2$

Consider $\int \frac{{\tan }^{-1}x}{1+x^2}dx$

Let $t={\tan }^{-1}x\Rightarrow dt=\frac{1}{1+x^2}dx$

Now, $\int \frac{{\tan }^{-1}x}{1+x^2}dx=\int tdt=\frac{t^2}{2}=\frac{{\left({\tan }^{-1}x\right)}^2}{2}$ where $t={\tan }^{-1}x$

Thus, $y=\frac{1}{2}\ln |1+x^2|+\frac{{\left({\tan }^{-1}x\right)}^2}{2}+c$ where $c$ is the constant of integration.