Question

Solve : $(x^2-4xy-2y^2)dx+(y^2-4xy-2x^2)dy=0$

Answer 1

Let, $M=x^2-4xy-2y^{2}N=y^2-4xy-2x^2$

$\therefore \frac{\partial M}{\partial y}=-4x-4y$

and $\frac{\partial N}{\partial x}=-4y-4x$

Here, $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$

So, the equation is exact.

$\therefore \int Mdx=\int (x^2-4xy-2y^2)dx=\frac{x^3}{3}-\frac{4x^2}{2}y-2y^{2}x=\frac{x^3}{3}-2x^{2}y-2y^{2}x$

And, $N$ free of $x=y^2$

$\therefore \int y^{2}dy=\frac{y^3}{3}$

$\therefore$ the solution is

$\frac{x^3}{3}-2x^{2}y-2y^{2}x+\frac{y^3}{3}=c.$