Question

Solve the differential equation $({\tan }^{-1}x-y)dx=(1+x^2)dy$.

Answer 1

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

$\Rightarrow \frac{dy}{dx}=\frac{{\tan }^{-1}x-y}{1+x^2}$

$\Rightarrow \frac{dy}{dx}+\frac{y}{1+x^2}=\frac{{\tan }^{-1}x}{1+x^2}...\left(i\right)$

$\therefore$ $I.F.$ = $e^{\int \frac{1}{1+x^2}dx}=e^{{\tan }^{-1}x}$

Multiplying both sides of $\left(i\right)$ by $e^{{\tan }^{-1}x}$ , we have

$\frac{dy}{dx}.e^{{\tan }^{-1}x}+\frac{y}{1+x^2}.e^{{\tan }^{-1}x}=\frac{{\tan }^{-1}x}{1+x^2}.e^{{\tan }^{-1}x}$

Integrating both sides w.r.t. $x$, we have $y.e^{{\tan }^{-1}x}=\int \frac{{\tan }^{-1}x}{1+x^2}.e^{{\tan }^{-1}x}dx+c$

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

$\Rightarrow y.e^t=\int e^{t}tdt+C$

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