Question

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

Answer 1

We have,
$\left(1+y^2\right)+\left(x-e^{-{\tan }^{-1}y}\right)\frac{dy}{dx}=0$
$$\begin{gathered} \Rightarrow \frac{dx}{dy}=\frac{e^{-{\tan }^{-1}y}-x}{1+y^2} \\ \Rightarrow \frac{dx}{dy}+\frac{x}{1+y^2}=\frac{e^{-{\tan }^{-1}y}}{1+y^2} \end{gathered}$$

$$\begin{gathered} \mathrm{Comparing}\mathrm{with}\frac{dx}{dy}+Px=Q,\mathrm{we}\mathrm{get} \\ P=\frac{1}{1+y^2} \\ Q=\frac{e^{-{\tan }^{-1}y}}{1+y^2} \\ \mathrm{Now}, \\ \mathrm{I}.\mathrm{F}.=e^{\int \frac{1}{1+y^2}dy}=e^{{\tan }^{-1}y} \\ \mathrm{So},\mathrm{the}\mathrm{solution}\mathrm{is}\mathrm{given}\mathrm{by} \\ x\times {\mathrm{e}}^{{\tan }^{-1}\mathrm{y}}=\int \frac{e^{-{\tan }^{-1}y}}{1+y^2}\times {\mathrm{e}}^{{\tan }^{-1}\mathrm{y}}dy+C \\ \Rightarrow x\times {\mathrm{e}}^{{\tan }^{-1}\mathrm{y}}=\int \frac{1}{1+y^2}dy+C \\ \Rightarrow x{\mathrm{e}}^{{\tan }^{-1}\mathrm{y}}={\tan }^{-1}y+C \end{gathered}$$