Question

Solve the Differential equation 2(y-4x^2) dx + x dy = 0

Answer 1

Assume a solution of the form $\Psi(x,y)=C$. Taking the differential of both sides would give the ODE

$\Psi_x\,\mathrm dx+\Psi_y\,\mathrm dy=0$

where $\Psi_x=2y-8x^2$ and $\Psi_y=x$. The ODE is exact if $\Psi_{xy}=\Psi_{yx}$. We have

$\Psi_{xy}=(2y-8x^2)_y=2$

$\Psi_{yx}=(x)_x=1$

so the ODE is not exact. However, multiplying both sides by $x$ gives

$(2xy-8x^3)\,\mathrm dx+x^2\,\mathrm dy=0$

and now

$(x\Psi_x)_y=(2xy-8x^3)_y=2x$

$(x\Psi_y)_x=(x^2)_x=2x$

so the ODE is now exact.

Now,

$\Psi_x=2xy-8x^3\implies \Psi(x,y)=x^2y-2x^4+f(y)$

$\Psi_y=x^2=x^2+f_y\implies f_y=0\implies f(y)=C$

So the ODE has solution

$\Psi(x,y)=x^2y-2x^4+C=C$

or simply

$\boxed{x^2y-2x^4=C}$