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

##### 1 Answer

# 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

# 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^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) #