Question

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

Answer 1

`y = xroot(3)(C - 3ln x)`

Explanation:

We have:

`xy^2dy/dx=y^3-x^3`

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

`y = vx`

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

`dy/dx = v + x(dv)/dx`

Substituting into the initial ODE we get:

`x(vx)^2(v + x(dv)/dx) = (vx)^3 - x^3`

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

`v^2(v + x(dv)/dx) = v^3 -1`

`:. v + x(dv)/dx = v -1/v^2`

`:. x(dv)/dx = -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 \ v^2 \ dv = int \ -1/x \ dx`

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

`1/3 v^3 = - ln x + A`

`:. v^3 = 3A - 3ln x`

`:. v = root(3)(C - 3ln x) \ \ \`, say

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

`y/x = root(3)(C - 3ln x)`

`:. y = xroot(3)(C - 3ln x)`