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