1. The problem statement, all variables and given/known data 2y * y'(t) = 3t^2 such that y(0) = 9 2. Relevant equations g(y)y'(t) = h(t) 3. The attempt at a solution So I have done many of these seperable ones in homework that did not require a parameter so now I got lost. This is what I did; Integral of 2y dy = integral of 3t^2 dt y^2 + C = t^3 Then since y(0) = 9 I assumed I plug in 0 for y and that gives me 0 + C= t^3 so I figured that means that C= 9 then solving for y I get y= sqrt (t^3 - 9) but that is not the answer. It is sqrt(t^3 +81) now that sort of is similar because -9^2 is 81 but I dont see any correlation to any mistakes. Any help?