# Probably simple

1. Feb 24, 2008

### Tom1

I am working on a diff. eq. assignment, and have encountered the simple, seperable differential equation:

dy/dt = y^3 with the initial condition y(0)=1

I am posting in this forum, because I feel my trouble is with my integration.

When seperated, I am integrating this equation as $$\int1/y^3$$=$$\int1dt$$.

After doing so, I end up with: -1/2y^2=t+C

After solving for y, I come up with: y=$$\sqrt{1/2t}$$+C

The only problem with this is I cannot solve for C, since the initial condition would require division by zero.

Anyone have an clue as to where I went wrong?

2. Feb 24, 2008

### Mystic998

Just solve for the constant here: -1/2y^2=t+C.

3. Feb 24, 2008

### Tom1

If I do it then, I end up with C=-1/2y^2-1 which when plugged back into the general solution, results in y disappearing.

Can't be right..

4. Feb 24, 2008

### sutupidmath

Look, you are not followint Mystic998's point and suggestion at all!
YOu have the initial condition y(0)=1, right?
so as Mystic998 suggested

-1/2y^2=t+C.=> -1/2 *1^2=c=> c=-1/2
and remember y(0)=1, means the value of y when t=0

5. Feb 24, 2008

### Tom1

I'm an idiot, thank you.

6. Feb 24, 2008

### sutupidmath

Oh no, i am sure you are not, it happens to all of us sometimes not to notice these minor things!!