Juan Lowe

Answered

2022-11-23

How to solve such fraction differential equation?

Here's my first-order differential equation:

$({x}^{3}-2x{y}^{2})dx+3y{x}^{2}dy=xdy-ydx$

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.

Here's my first-order differential equation:

$({x}^{3}-2x{y}^{2})dx+3y{x}^{2}dy=xdy-ydx$

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.

Answer & Explanation

levraijournalk1o

Expert

2022-11-24Added 10 answers

$({x}^{3}-2x{y}^{2})dx+{3y{x}^{2}dy}=xdy-ydx$

$({x}^{3}-2x{y}^{2})dx+{2y{x}^{2}dy+y{x}^{2}dy}=xdy-ydx$

Rearrange terms:

$({x}^{3}dx+{x}^{2}ydy)-2xy(ydx-xdy)=xdy-ydx$

${x}^{2}(xdx+ydy)=(xdy-ydx)(1-2xy)$

$xdx+ydy=(1-2xy)d\left({\displaystyle \frac{y}{x}}\right)$

$d({x}^{2}+{y}^{2})=2(1-2xy)d\left({\displaystyle \frac{y}{x}}\right)$

Divide by ${x}^{2}+{y}^{2}$ and substitute $u={x}^{2}+{y}^{2}$ and $v={\displaystyle \frac{y}{x}}$

$\frac{1}{2}}{\displaystyle \frac{du}{u}}=({\displaystyle \frac{1}{u}}-{\displaystyle \frac{2v}{{v}^{2}+1}})dv$

$\frac{du}{dv}}=2-{\displaystyle \frac{4uv}{{v}^{2}+1}$

${u}^{\prime}+{\displaystyle \frac{4v}{{v}^{2}+1}}u=2$

This is a first order linear DE. Try to integrate by integrating factor method:

$\mu (v)=\mathrm{exp}\int {\displaystyle \frac{4v}{{v}^{2}+1}}dv$

$\mu (v)=\mathrm{exp}\int {\displaystyle \frac{2}{{v}^{2}+1}}d{v}^{2}=({v}^{2}+1{)}^{2}$

The DE becomes:

$(u({v}^{2}+1{)}^{2}{)}^{\prime}=2({v}^{2}+1{)}^{2}$

$u({v}^{2}+1{)}^{2}=2({\displaystyle \frac{{v}^{5}}{5}}+2{\displaystyle \frac{{v}^{3}}{3}}+v)+C$

Unsubstitute u and v:

$u={x}^{2}+{y}^{2}\text{and}v={\displaystyle \frac{y}{x}}$

$({x}^{3}-2x{y}^{2})dx+{2y{x}^{2}dy+y{x}^{2}dy}=xdy-ydx$

Rearrange terms:

$({x}^{3}dx+{x}^{2}ydy)-2xy(ydx-xdy)=xdy-ydx$

${x}^{2}(xdx+ydy)=(xdy-ydx)(1-2xy)$

$xdx+ydy=(1-2xy)d\left({\displaystyle \frac{y}{x}}\right)$

$d({x}^{2}+{y}^{2})=2(1-2xy)d\left({\displaystyle \frac{y}{x}}\right)$

Divide by ${x}^{2}+{y}^{2}$ and substitute $u={x}^{2}+{y}^{2}$ and $v={\displaystyle \frac{y}{x}}$

$\frac{1}{2}}{\displaystyle \frac{du}{u}}=({\displaystyle \frac{1}{u}}-{\displaystyle \frac{2v}{{v}^{2}+1}})dv$

$\frac{du}{dv}}=2-{\displaystyle \frac{4uv}{{v}^{2}+1}$

${u}^{\prime}+{\displaystyle \frac{4v}{{v}^{2}+1}}u=2$

This is a first order linear DE. Try to integrate by integrating factor method:

$\mu (v)=\mathrm{exp}\int {\displaystyle \frac{4v}{{v}^{2}+1}}dv$

$\mu (v)=\mathrm{exp}\int {\displaystyle \frac{2}{{v}^{2}+1}}d{v}^{2}=({v}^{2}+1{)}^{2}$

The DE becomes:

$(u({v}^{2}+1{)}^{2}{)}^{\prime}=2({v}^{2}+1{)}^{2}$

$u({v}^{2}+1{)}^{2}=2({\displaystyle \frac{{v}^{5}}{5}}+2{\displaystyle \frac{{v}^{3}}{3}}+v)+C$

Unsubstitute u and v:

$u={x}^{2}+{y}^{2}\text{and}v={\displaystyle \frac{y}{x}}$

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