Answer to Question #296557 in Differential Equations for Ruhi

Question

(1 + 𝑥^2) (xdy+ydx) =-2yx^2

Answer 1

Solution:

The given DE is not solvable, we put dx on right side.

$(1 + 𝑥^2) (xdy+ydx) =-2yx^2 dx \\ \Rightarrow (1+x^2)(d(xy))=-2yx^2dx$

$\Rightarrow d(xy)=-\dfrac{2yx^2}{1+x^2} dx \\ \Rightarrow \dfrac{d(xy)}{xy}=-\dfrac{2x}{1+x^2}dx$

On integrating both sides,

$\ln (xy)=-\ln(1+x^2)+\ln C \\ \Rightarrow \ln (xy(1+x^2))=\ln C \\ \Rightarrow xy(1+x^2)= C$