Question

Solve [2y^2 – 4x + 5]dx = [y – 2y^2 - 4xy]dy

Answer 1

Given,

$[2y^2 – 4x + 5]dx = [y – 2y^2 - 4xy]dy$

Compare with $Mdx+Ndy=0$

$M=2y^2-4x+5,N=-(y^2-2^2-4xy)$

$\dfrac{dM}{dy}=4y, \dfrac{dN}{dx}=4y$

$\therefore \dfrac{dM}{dy}=\dfrac{dN}{dx}=4y$

The given equation is exact diff. equation. So Integrating all terms of M keeping y as a constsant and integrating only those part of N which are free from x.

So

$\int (2y^2-4x+5)dx-\int (y-2y^2-4xy)dy=0 \\[9pt] 2xy^2-2x^2+5x-\dfrac{y^2}{2}-\dfrac{2}{3}^3-2xy^2+C=0$

So Solution is-

$2xy^2-2x^2+5x-\dfrac{y^2}{2}-\dfrac{2}{3}^3-2xy^2+C=0$