# Math Help - Initial Value Differential Equation

1. ## Initial Value Differential Equation

I have dy/dx = x^2/y^2

=y^2 dy = x^2 dx

y^2 dy = x^2 dx

y^3/3 = x^3/3+ C

y^3 = x^3 + 3C

3C= some constant k

y = (x^3+k)^(1/3)

I now have to solve the initial value problem of y(0)=2 ; kinda forgot how to do this part.

Do I just set it equal to 2 then plug in 0 so k=8 then do i multiply by 3 ?

2. ## Re: Initial Value Differential Equation

Just set x = 0 and y = 2

2 = (0 + k)^(1/3)
2 = k^(1/3)

cube both sides

8 = k