# How can I solve this differential equation? : xy^2dy/dx=y^3-x^3

Feb 18, 2018

$y = x \sqrt[3]{C - 3 \ln x}$

#### Explanation:

We have:

$x {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = {y}^{3} - {x}^{3}$

Which is a First Order Nonlinear Ordinary Differential Equation. Let us attempt a substitution of the form:

$y = v x$

Differentiating wrt $x$ and applying the product rule, we get:

$\frac{\mathrm{dy}}{\mathrm{dx}} = v + x \frac{\mathrm{dv}}{\mathrm{dx}}$

Substituting into the initial ODE we get:

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

Then assuming that $v , x \ne 0$ this simplifies to:

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

$\therefore v + x \frac{\mathrm{dv}}{\mathrm{dx}} = v - \frac{1}{v} ^ 2$

$\therefore x \frac{\mathrm{dv}}{\mathrm{dx}} = - \frac{1}{v} ^ 2$

And we have reduced the initial ODE to a First Order Separable ODE, so we can collect terms and separate the variables to get:

$\int \setminus {v}^{2} \setminus \mathrm{dv} = \int \setminus - \frac{1}{x} \setminus \mathrm{dx}$

Both integrals are standard, so we can integrate to get:

$\frac{1}{3} {v}^{3} = - \ln x + A$

$\therefore {v}^{3} = 3 A - 3 \ln x$

$\therefore v = \sqrt[3]{C - 3 \ln x} \setminus \setminus \setminus$, say

Then, we restore the substitution, to get the General Solution:

$\frac{y}{x} = \sqrt[3]{C - 3 \ln x}$

$\therefore y = x \sqrt[3]{C - 3 \ln x}$