# Solve the Initial Value problem

1. Feb 6, 2012

### lmanri

dy/dt=t^(2)y^(3) , y(0)=-1

I need help solving this

I put the integral (dy/y^3)= integral (t^2)dt
but idk what to do after that or if thats even right

Last edited: Feb 6, 2012
2. Feb 6, 2012

### ehild

Hi Imanri,

You dropped an integral sign...

$$\int{\frac{dy}{y^3}}=\int{t^2dt}$$

Do the integration on both sides.

ehild

3. Feb 6, 2012

### lmanri

Thank you, but after I find that what do I do with y(0)=1

4. Feb 6, 2012

### ehild

You have an undetermined constant after integration. Substitute t=0 and y=-1 in your integral and solve for c.

ehild

5. Feb 6, 2012

Thank you

6. Feb 6, 2012