- #1

Tom1

- 24

- 0

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 [tex]\int1/y^3[/tex]=[tex]\int1dt[/tex].

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

After solving for y, I come up with: y=[tex]\sqrt{1/2t}[/tex]+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?