$(2y^2–4x+5)dx=(y–2y^2–4xy)dy$

Solution:

$\left(2 y^{2}-4 x+5\right) d x=\left(y-2 y^{2}-4 x y\right) d y$

Compare with $\mathrm{M} \mathrm{d} \mathrm{x}+\mathrm{Ndy}=0$

$\therefore \mathrm{M}=\left(2 y^{2}-4 x+5\right)$

$\frac{\partial M}{\partial y}=4 y$

$\therefore \mathrm{N}=-\left(y-2 y^{2}-4 x y\right)$

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

$\therefore \quad \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$

The given diff. eqn is exact,

The solution of exact diff. eqn is given by,

$\int M d x+\int\left[N-\frac{\partial}{\partial y} M d x\right] d y=c$

$\int M d x=\int\left(\left(2 y^{2}-4 x+5\right)\right) d x=2 x y^{2}-2 x^{2}+5 x$

$\frac{\partial}{\partial y} \int M d x=4 x y$

$\int\left[N-\frac{\partial}{\partial y} M d x\right] d y=\int\left[4 x y-y+2 y^{2}-4 x y\right] d y=\frac{2}{3} y^{3}-\frac{y^{2}}{2}$

$ \mathbf{ \therefore 2 x y^{2}-2 x^{2}+5 x+\frac{2}{3} y^{3}-\frac{y^{2}}{2}=c }$