Question

Find the general solution to
(b) dy/dt = t^2y^3

Answer 1

this thing slow, or is it just me?

Answer 2

\[\frac{dy}{y^3}=t^2dt\]Integrate both sides to get:\[\frac{-1}{2y^2}=\frac{t^3}{3}+C\]Solve for y in terms of t.

Answer 3

separables:
\[\int \left(y^{-3}dy=t^2dt\right)\]

Answer 4

what does y =?

Answer 5

thats what solving it will actually get you.

If you cant remember how to integrate simple powers, then you might need to go back and review that.