2y * y'(t) = 3t^2 such that y(0) = 9
g(y)y'(t) = h(t)
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;[/B]
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?