Just set x = 0 and y = 2
2 = (0 + k)^(1/3)
2 = k^(1/3)
cube both sides
8 = k
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 ?