(2y^2+3x)dx+4xydy=0 Using Exact method of differential equations

Question

asked May 17, 2020 206k views

1 Answer

AlVelig · Answer 1 · 2020-05-23T05:34:57+0000

This ODE is exact because

$(\partial(2y^2+3x))/(\partial y)=4y$

$(\partial(4xy))/(\partial x)=4y$

So we're looking for a solution
$\Psi(x,y)=C$ such that

$(\partial\Psi)/(\partial x)=2y^2+3x$

$(\partial\Psi)/(\partial y)=4xy$

Integrate both sides of the first PDE with respect to
:

$\displaystyle\int(\partial\Psi)/(\partial x)=\int(2y^2+3x)\,\mathrm dx$

$\Psi=2xy^2+3x+g(y)$

Differentiate both sides with respect to
:

$(\partial\Psi)/(\partial y)=4xy+(\mathrm dg)/(\mathrm dy)=4xy$

$\implies(\mathrm dg)/(\mathrm dy)=0\imples g(y)=C$

So the solution to the ODE is

2xy^2+3x=C