Help Solving Diff. Eq. Integration: y^3 = 1/2y^2+t+C

[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 separated, 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?

 

[Mystic998]

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

 

[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..

 

[sutupidmath]

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..

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

 

[Tom1]

I'm an idiot, thank you.

 

[sutupidmath]

I'm an idiot, thank you.

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