Question:
Find the integral curves of the equation:
$$\frac{dx}{y^2x-2x^4}=\frac{dy}{2y^4-x^3y}=\frac{dz}{2z(x^3-y^3)}$$
I could not find any similar example to understand this type of questions of solutions. Please show me how to solve this question. Thank you for helping.
Best Answer
I think your equation should be $$\frac{dx}{y^3x-2x^4}=\frac{dy}{2y^4-x^3y}=\frac{dz}{2z(x^3-y^3)}$$ ($y^3x$ is replaced by $y^2x$ in the first denumerator). In this case we have $$\frac{ydx+xdy}{3xy(y^3-x^3)}=\frac{dz}{2z(x^3-y^3)}$$ or $$\frac{ydx+xdy}{3xy}+\frac{dz}{2z}=0.$$ Substituting $t=xy$ and solving this equation we obtain $(xy)^{1/3}z^{1/2}=c$ or for simplicity $x^2y^2z^3=c_1=u(x, y, z)$. To find $v(x, y, z)= c_2$ we can solve the homogeneous ODE $$\frac{dy}{dx}=\frac{2y^4-x^3y}{y^3x-2x^4}$$ by setting $$s=\frac{y}{x}$$.