Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \left(1+y^2\right)+\left(x-e^{{\tan }^{-1}y}\right)\frac{dy}{dx}=0 \\ \Rightarrow \left(x-e^{{\tan }^{-1}y}\right)\frac{dy}{dx}=-\left(1+y^2\right) \\ \Rightarrow \frac{dy}{dx}=-\frac{\left(1+y^2\right)}{\left(x-e^{{\tan }^{-1}y}\right)} \\ \Rightarrow \frac{dx}{dy}=-\frac{x-e^{{\tan }^{-1}y}}{1+y^2} \\ \Rightarrow \frac{dx}{dy}+\frac{x}{1+y^2}=\frac{e^{{\tan }^{-1}y}}{1+y^2}.....\left(1\right) \\ \mathrm{Clearly},\mathrm{it}\mathrm{is}\mathrm{a}\mathrm{linear}\mathrm{differential}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{form} \\ \frac{dx}{dy}+Px=Q \\ \mathrm{where} \\ P=\frac{1}{1+y^2} \\ Q=\frac{e^{{\tan }^{-1}y}}{1+y^2} \\ \therefore \mathrm{I}.\mathrm{F}.=e^{\int Pdy} \\ =e^{\int \frac{1}{1+y^2}dy} \\ =e^{{\tan }^{-1}y} \\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{of}\left(1\right)\mathrm{by}{e}^{{\tan }^{-1}y},\mathrm{we}\mathrm{get} \\ {e}^{{\tan }^{-1}y}\left(\frac{dx}{dy}+\frac{x}{1+y^2}\right)=e^{{\tan }^{-1}y}\frac{e^{{\tan }^{-1}y}}{1+y^2} \\ \Rightarrow e^{{\tan }^{-1}y}\frac{dx}{dy}+\frac{x{e}^{{\tan }^{-1}x}}{1+y^2}=\frac{e^{2{\tan }^{-1}y}}{1+y^2} \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides}\mathrm{with}\mathrm{respect}\mathrm{to}y,\mathrm{we}\mathrm{get} \\ x{e}^{{\tan }^{-1}y}=\int \frac{e^{2{\tan }^{-1}y}}{1+y^2}dy+C \\ \Rightarrow x{e}^{{\tan }^{-1}y}=I+C.....\left(2\right) \\ \mathrm{Here}, \\ I=\int \frac{e^{2{\tan }^{-1}y}}{1+y^2}dy \\ \mathrm{Putting}{\tan }^{-1}y=t,\mathrm{we}\mathrm{get} \\ \frac{1}{1+y^2}dy=dt \\ \therefore I=\int e^{2t}dt \\ =\frac{e^{2t}}{2} \\ =\frac{e^{2{\tan }^{-1}y}}{2} \\ \mathrm{Putting}\mathrm{the}\mathrm{value}\mathrm{of}I\mathrm{in}\left(2\right),\mathrm{we}\mathrm{get} \\ x{e}^{{\tan }^{-1}y}=\frac{e^{2{\tan }^{-1}y}}{2}+C \\ \Rightarrow 2x{e}^{{\tan }^{-1}y}=e^{2{\tan }^{-1}y}+2C \\ \Rightarrow 2x{e}^{{\tan }^{-1}y}=e^{2{\tan }^{-1}y}+k\left(\mathrm{where}k=2C\right) \\ \mathrm{Hence},2x{e}^{{\tan }^{-1}y}=e^{2{\tan }^{-1}y}+k\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$