MHB Jaganmaya's question at Yahoo Answers regarding a first order ODE

[MarkFL]

Here is the question:

Solve the differential equation (y^4 - 2x^3y)dx + (x^4 - 2xy^3)dy = 0 ?

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Re: Jaganmaya's question ay Yahoo! Answers regarding a first order ODE

Hello Jaganmaya,

We are given to solve:

$$\left(y^4-2x^3y \right)\,dx+\left(x^4-2xy^3 \right)\,dy=0$$

This equation is in the differential form:

$$M(x,y)\,dx+N(x,y)\,dy=0$$

Rather than test for exactness (which will in fact not lead to an integrating factor in one variable), let's put the equation in the form:

$$\frac{dy}{dx}=f(x,y)$$

and we get:

$$\frac{dy}{dx}=\frac{2x^3y-y^4}{x^4-2xy^3}=\frac{2\left(\dfrac{y}{x} \right)-\left(\dfrac{y}{x} \right)^4}{1-2\left(\dfrac{y}{x} \right)^3}$$

We now have a homogeneous equation, and may use the substitution:

$$v=\frac{y}{x}\implies y=vx\,\therefore\,\frac{dy}{dx}=v+x\frac{dv}{dx}$$

And so, our ODE becomes:

$$v+x\frac{dv}{dx}=\frac{2v-v^4}{1-2v^3}$$

$$x\frac{dv}{dx}=\frac{2v-v^4}{1-2v^3}-v$$

$$x\frac{dv}{dx}=\frac{2v-v^4-v\left(1-2v^3 \right)}{1-2v^3}=\frac{v^4+v}{1-2v^3}$$

Separating variables, we obtain:

$$\frac{1-2v^3}{v^4+v}\,dv=\frac{1}{x}\,dx$$

Note: we have just lost the trivial solution $$y\equiv0$$.

$$\frac{v^3+1-3v^3}{v\left(v^3+1 \right)}\,dv=\frac{1}{x}\,dx$$

$$\left(\frac{1}{v}-\frac{3v^2}{v^3+1} \right)\,dv=\frac{1}{x}\,dx$$

Integrating, we obtain:

$$\ln\left|\frac{v}{v^3+1} \right|=\ln|Cx|$$

And this implies:

$$\frac{v}{v^3+1}=Cx$$

Back-substitute for $v$:

$$\frac{\dfrac{y}{x}}{\dfrac{y^3}{x^3}+1}=Cx$$

$$\frac{x^2y}{x^3+y^3}=Cx$$

And so the solution is given implicitly by:

$$x^2y=Cx\left(x^3+y^3 \right)$$