Question

If $x=g\left(y\right)e^{f\left(y\right)}$ is a solution of the differential equation $(1+y^2)+(x-e^{-{\tan }^{-1}y})\frac{dy}{dx}=0$, satisfying the condition $x=0$ at $y=0$, then the value of $f\left(\sqrt{2}\right)+g\left(\sqrt{2}\right)$ is

Answer 1

$(1+y{)}^2+(x-e^{-{\tan }^{-1}y})\frac{dy}{dx}=0$
$\Rightarrow \frac{dx}{dy}+\frac{x}{1+y^2}=\frac{e^{-{\tan }^{-1}y}}{1+y^2}$
$\Rightarrow P=\frac{1}{1+y^2},Q=\frac{e^{-{\tan }^{-1}y}}{1+y^2}$
$\therefore$ I.F. $=e^{\int \frac{1}{1+y^2}dy}=e^{{\tan }^{-1}y}$
$\Rightarrow x.e^{{\tan }^{-1}y}=\int \frac{\text{/}{e}^{{\tan }^{-1}y}.\text{/}{e}^{-{\tan }^{-1}y}}{1+y^2}dy$
$\Rightarrow x.e^{{\tan }^{-1}y}=\int \frac{1}{1+y^2}dy$
$\Rightarrow x.e^{{\tan }^{-1}y}={\tan }^{-1}y+c$
At $x=0,y=0\Rightarrow c=0$
$\Rightarrow x.e^{{\tan }^{-1}y}={\tan }^{-1}y$
$\Rightarrow x=e^{-{\tan }^{-1}y}.{\tan }^{-1}y$
$\because x=e^{f\left(y\right)}.g\left(y\right)$
$\therefore f\left(y\right)=-{\tan }^{-1}y$
$g\left(y\right)={\tan }^{-1}y$
$f\left(y\right)+g\left(y\right)=-{\tan }^{-1}y+{\tan }^{-1}y=0$
$\therefore f\left(\sqrt{2}\right)+g\left(\sqrt{2}\right)=0$