Question

Solve the differential equation $(x^2-y^2)dx+2xydy=0$.

Answer 1

$\left(x^2-y^2\right)dx+2xydy=0$

Now,

$\begin{array}{c}\Rightarrow x^{2}dx=y^{2}dx-2xydy\\ \Rightarrow -dx=\frac{2xydy-y^{2}dx}{x^2}\\ \Rightarrow -dx=\frac{2xydy-y^{2}dx}{x^2}\\ \Rightarrow -dx=d\left(\frac{y^2}{x}\right)\\ \Rightarrow C-x=\frac{y^2}{x}\\ hence,weget\\ \frac{y^2}{x}+x=C\end{array}$