Question

Solve $(1+y^2)+(x-e^{{\tan }^{-1}y})\frac{dy}{dx}=0$

Answer 1

$(1+y^2)+(x-e^{{\tan }^{-1}y})\frac{dy}{dx}=0$

$(1+y^2)\frac{dy}{dx}+x-e^{{\tan }^{-1}y}=0$

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

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

This is in form of $\frac{dy}{dx}+Px=Q$

$\therefore$Integrating Factor =I.F.$=e^{\int Pdy}$

$\therefore I.F=e^{\int \frac{dy}{1+y^2}}$

$I.F=e^{{\tan }^{-1}y}$

$\therefore x=\int Q.(I.F)dy$

$=\int \left(\frac{e^{{\tan }^{-1}y}}{1+y^2}\right).\left(e^{{\tan }^{-1}y}\right)dy$

$=\int \frac{e^{2{\tan }^{-1}y}}{1+y^2}dy;$ Let ${\tan }^{-1}y=t;\therefore \frac{dy}{1+y^2}=dt$

$=\int e^{2t}dt$

$=\frac{e^{2t}}{2}+C$

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