Question

(1+y²)dx=(tan^-1 y-x)dy

Answer 1

$\text{Given, the differential equation,}\\ (1+y^2)dx=(tan^{-1} y-x)dy\\ \frac{dx}{dy}=\frac{tan^{-1}y}{1+y^2}-\frac{x}{1+y^2}\\ \frac{dx}{dy}+\frac{x}{1+y^2} =\frac{tan^{-1}y}{1+y^2}\\ \text{This is a linear differential equation in term of x.}\\ \text{ Integrating factor, }\\ IF=e^{ \int \frac{dy}{1+y^2}} \\ \text{Then, solution is given by} \\ xe^{tan^{-1}y}= \int \frac{tan^{-1}y}{1+y^2}e^{tan^{-1}y}dy\\ Put \\ tan^{-1}y=t\implies \frac{dy}{1+y^2}=dt\\ Therefore,\\ \int \frac{tan^{-1}y}{1+y^2}e^{tan^{-1}y}dy=-\int te^{t}dt\\ =(1-t)e^t\\ =(1-tan^{-1}y)e^{tan^{-1}y}+C\\ Then, the\space solution\\ xe^{tan^{-1}y}=(1-tan^{-1}y)e^{tan^{-1}y}+C\\$