Solve $\frac{dx}{x(c'_1y-2y^2)}=\frac{dy}{y(c'_1y-y^2-2x^3)}$

Question

A second family of characteristic curves comes from $$\frac{dx}{x(c'_1y-2y^2)}=\frac{dy}{y(c'_1y-y^2-2x^3)}$$ The solution of this ODE is :

$y=\frac{c'_1}{2}\pm \sqrt{x^3+\frac{(c'_1)^2}{4}+c_2}$

$\left(y-\frac{c'_1}{2}\right)^2-x^3-\frac{(c'_1)^2}{4}=c_2$

How do I solve this ODE? $$\frac{dy}{dx}=\frac{y(c'_1y-y^2-2x^3)}{x(c'_1y-2y^2)}$$

This equation is not exact.

This is not a homogeneous equation. So, I can't use the transformation $u=y/x.$

I used the Integrating formulas for $Mdx+Ndy=0$ equation $\frac{N_x-M_y}{M}$ and $\frac{M_y-N_x}{N}.$ It was not handy. Is there any analytical method to solve this?

Answer 1

I used $k$ for the constant. $$ky-y^2-2x^3=x(k-2y)y'$$ Complete the square on LHS: $$-(y-\dfrac k2)^2-2x^3+\dfrac {k^2}{4}=- 2x (y-\dfrac k2)(y-\frac k 2)'$$ Substitute $v=y-\dfrac k 2$ $$-v^2-2x^3+\dfrac {k^2}{4}=- 2x vv'$$ $$v^2+2x^3-\dfrac {k^2}{4}= x(v^2)'$$ Substitute $u=v^2$ and solve. $$xu'-u=2x^3-\dfrac {k^2}{4}$$ $$\left ( \dfrac ux \right)'=2x-\dfrac {k^2}{4x^2}$$ Integrate. $$u=x^3+\dfrac {k^2}{4}+cx$$ $$v^2=x^3+\dfrac {k^2}{4}+cx$$ Finally: $$\boxed {y(x)=\frac k 2\pm \sqrt {x^3+\dfrac {k^2}{4}+cx}}$$