Question

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

Answer 1

On rearranging the equation becomes:

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

This is in the form $\frac{dx}{dy}+P\left(y\right).x=Q\left(y\right)$, so it is a non homogeneous linear differential equation of the first order.

the integrating factor is: $I.F.=e^{\int \frac{dy}{1+y^2}}=e^{{\tan }^{-1}y}$

Multiplying both sides of the equation with integrating factor, we get:

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

Integrating by parts, by taking ${\tan }^{-1}y$ as first function and $\frac{e^{{\tan }^{-1}y}}{1+y^2}$ as second function we get:

$\Rightarrow x{e}^{{\tan }^{-1}y}=\int e^{{\tan }^{-1}y}\frac{{\tan }^{-1}y}{1+y^2}dy={\tan }^{-1}y\int \frac{e^{{\tan }^{-1}y}}{1+y^2}dy-\int d\left({\tan }^{-1}y\right)(\int \frac{e^{{\tan }^{-1}y}}{1+y^2}dy)={\tan }^{-1}y.e^{{\tan }^{-1}y}-\int \frac{dy}{1+y^2}(\int \frac{e^{{\tan }^{-1}y}}{1+y^2}dy)\Rightarrow x{e}^{{\tan }^{-1}y}={\tan }^{-1}y.e^{{\tan }^{-1}y}-e^{{\tan }^{-1}y}+C\Rightarrow x={\tan }^{-1}y-1+C{e}^{-{\tan }^{-1}y}$