Question

Find the general solution of the following differential equation
$(1+y^2)+(x-e^{{\tan }^{-1}y})\frac{dy}{dx}=0$

Answer 1

Consider the given differential equation.

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

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

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

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

This is a differential equation of first order. Here,

$Q=\frac{e^{{\tan }^{-1}y}}{1+y^2}$

We know the solution of such differential equation is,

$x\cdot IF=\int Q\cdot IFdy$

Calculate $IF$.

$IF=e^{\int \frac{1}{1+y^2}dy}=e^{{\tan }^{-1}y}$

Therefore, the solution will be,

$x\cdot e^{{\tan }^{-1}y}=\int \frac{e^{{\tan }^{-1}y}}{1+y^2}{e}^{{\tan }^{-1}y}dy$

Put ${\tan }^{-1}y=t$.

$\frac{1}{1+y^2}dy=dt$

So,

$x\cdot e^t=\int e^t{e}^{t}dt$

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

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

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

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

Hence, this is the required solution.