# Separable differential equation

1. Oct 5, 2014

### TheKracken

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?

2. Oct 5, 2014

### RUber

It looks like you did everything right, until plugging in zero. zero is $t=0$ not $y=0$.

3. Oct 5, 2014

### Staff: Mentor

I would put the constant on the right side, but it should work this way, too.
No. The initial condition means that y = 9 when t = 0.
-9^2 means the negative of 92, or - 81. To square -9, write it as (-9)2, with parentheses.