Question

What is a general solution to the differential equation `y'=5x^(2/3)y^4`?

Answer 1

`y = root(3)(1/(C -9x^(5/3)))`

Explanation:

`dy/dx = 5x^(2/3)y^4`

This is a First Order separable DE, so collecting like terms and "separating the variables" we get;

`int y^-4dy = int5x^(2/3) dx`

Integrating we get:

`y^-3/-3 = 5x^(5/3)/(5/3) + C_1`
`y^-3 = -9x^(5/3) - 3C_1`
`1/y^3 = -9x^(5/3) + C`
`y^3 = 1/(C -9x^(5/3))`
`y = root(3)(1/(C -9x^(5/3)))`