Juan Lowe

2022-11-23

How to solve such fraction differential equation?
Here's my first-order differential equation:
$\left({x}^{3}-2x{y}^{2}\right)dx+3y{x}^{2}dy=xdy-ydx$
I've tried to make it fraction, but it isn't separable differential equation, also it isn't differential equation in total differentials, so after it I lose any clue for answer.

levraijournalk1o

Expert

$\left({x}^{3}-2x{y}^{2}\right)dx+3y{x}^{2}dy=xdy-ydx$
$\left({x}^{3}-2x{y}^{2}\right)dx+2y{x}^{2}dy+y{x}^{2}dy=xdy-ydx$
Rearrange terms:
$\left({x}^{3}dx+{x}^{2}ydy\right)-2xy\left(ydx-xdy\right)=xdy-ydx$
${x}^{2}\left(xdx+ydy\right)=\left(xdy-ydx\right)\left(1-2xy\right)$
$xdx+ydy=\left(1-2xy\right)d\left(\frac{y}{x}\right)$
$d\left({x}^{2}+{y}^{2}\right)=2\left(1-2xy\right)d\left(\frac{y}{x}\right)$
Divide by ${x}^{2}+{y}^{2}$ and substitute $u={x}^{2}+{y}^{2}$ and $v=\frac{y}{x}$
$\frac{1}{2}\frac{du}{u}=\left(\frac{1}{u}-\frac{2v}{{v}^{2}+1}\right)dv$
$\frac{du}{dv}=2-\frac{4uv}{{v}^{2}+1}$
${u}^{\prime }+\frac{4v}{{v}^{2}+1}u=2$
This is a first order linear DE. Try to integrate by integrating factor method:
$\mu \left(v\right)=\mathrm{exp}\int \frac{4v}{{v}^{2}+1}dv$
$\mu \left(v\right)=\mathrm{exp}\int \frac{2}{{v}^{2}+1}d{v}^{2}=\left({v}^{2}+1{\right)}^{2}$
The DE becomes:
$\left(u\left({v}^{2}+1{\right)}^{2}{\right)}^{\prime }=2\left({v}^{2}+1{\right)}^{2}$
$u\left({v}^{2}+1{\right)}^{2}=2\left(\frac{{v}^{5}}{5}+2\frac{{v}^{3}}{3}+v\right)+C$
Unsubstitute u and v:

Do you have a similar question?