Question

Solve the different equation:- $({\tan }^{-1}y-x)dy=(1+y^2)dx$.

Answer 1

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

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

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

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

Compare it with $\frac{dx}{dy}+px=q$

$p=\frac{1}{1+y^2}q=\frac{{\tan }^{-1}y}{1+y^2}$

Integrating factor $=e^{\int pdy}$

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

$I.F=e^{{\tan }^{-1}y}$

standard differential equation of this is

$x\times I.F=\int q\times I.F+C$

$x\times e^{{\tan }^{-1}y}=\int \frac{{\tan }^{-1}y}{1+y^2}\times e^{{\tan }^{-1}y}dy+c$...........(1)

$\int \frac{{\tan }^{-1}y}{1+y^2}{e}^{{\tan }^{-1}ydy}=\int {\tan }^{-1}y\times e^{{\tan }^{-1}y}\times \left(\frac{1}{1+y^2}dy\right)$

put ${\tan }^{-1}y=t\Rightarrow \frac{1}{1+y^2}dy=dt$

on substituting $=\int t{e}^{t}dt$

$=t\int e^{t}dt-\int e^t.1dt=t{e}^t-e^t$

$=e^t(t-1)$

put $t={\tan }^{-1}y$

we get $=e^{{\tan }^{-1}y}[{\tan }^{-1}y-1]$

put in equation (1)

$x{\tan }^{-1}y=e^{{\tan }^{-1}y}[{\tan }^{-1}y-1]+c$