Mathematics - Differential Equation Question with Solution | TestHub

$\left(1+y^2\right)+\left(x-e^{{\tan }^{-1}y}\right)\frac{dy}{dx}=0$ $\left(1+y^2\right)\frac{dx}{dy}+x=e^{{\tan }^{-1}y}\Rightarrow \frac{dx}{dy}+\frac{x}{\left(1+y^2\right)}=\frac{e^{{\tan }^{-1}y}}{\left(1+y^2\right)}$ I.F. $=e^{\int \frac{1}{\left(1+y^2\right)}dy}=e{\tan }^{-1}y$ $x\left(e^{{\tan }^{-1}y}\right)=\int \frac{e^{{\tan }^{-1}y}}{1+y}{e}^{{\tan }^{-1}ydy}$ $x\left(e^{{\tan }^{-1}y}\right)=\frac{e^{2{\tan }^{-1}y}}{2}+C\therefore 2x{e}^{{\tan }^{-1}y}=e^{2{\tan }^{-1}y}+k$