- #1

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## Homework Statement

Solve y' = (y^3)(t^2) for the initial condition y(0)=0 and state in which interval in 't' this solution exists.

## The Attempt at a Solution

First I divided both sides by y^3 and then subtracted t^2 from each as well,

-t^2 + y'(y^-3) = 0

then solved,

(-t^3)/3 + (y^-2)/-2 = C

and then uh oh for y = 0 so chugging ahead and leaving 'C' alone for now,

y^-2 = (C + (t^3)/3)/-2

so,

y = [-2/(C+(t^3)/3)]^(1/2)

Therefore (C + (t^3)/3) ≤ 0

I am sure (as usual) I am missing something simple :P