Differential Equation Help Please!

Solve giving y as a function of t please :)

3y dy/dt = 2t where y = 1 when t = 0.

Thanks a lot :)

∫ 3y dy = ∫ 2t dt

(3/2)y^2 = t^2 + c

y^2 = (2/3) t^2 + c

y = sqrt([2/3] t^2) + c

y = sqrt(2/3) * t + c

At y = 1, t = 0

So:

1 = sqrt(2/3) * 0 + c

c = 1

So:

y = sq