# Solve a differential equation for the initial conditions,

1. Sep 29, 2013

### mesa

1. The problem statement, all variables and given/known data

Solve y' = (y^3)(t^2) for the initial condition y(0)=0 and state in which interval in 't' this solution exists.

3. The attempt at a solution

First I divided both sides by y^3 and then subtracted t^2 from each as well,

-t^2 + y'(y^-3) = 0

then solved,

(-t^3)/3 + (y^-2)/-2 = C

and then uh oh for y = 0 so chugging ahead and leaving 'C' alone for now,

y^-2 = (C + (t^3)/3)/-2

so,

y = [-2/(C+(t^3)/3)]^(1/2)

Therefore (C + (t^3)/3) ≤ 0

I am sure (as usual) I am missing something simple :P

2. Sep 29, 2013

### Saitama

I cannot follow this. According to the question, y is zero when t is zero. What do you get for C?

3. Sep 29, 2013

### pasmith

For $y(0) \neq 0$ it is the case that the solution only exists when $3C + t^3 < 0$. But for $y(0) = 0$ that method doesn't work at all. And in fact $y(t) = 0$ is a solution of $y' = y^3t^2$ which satisfies $y(0) = 0$.

4. Sep 29, 2013

### mesa

Sorry, let me explain where I am coming from. We have,

(-t^3)/3 + (y^-2)/-2 = C

With the (1/-2y^2) term I get infinity for 'C'. That's my sticking point, so obviously I',ve goofed this up and there is another way to solve for 'C'?

5. Sep 29, 2013

### mesa

I'm sorry but I am not quite following what you are saying. Should I not solve for the general form?

6. Sep 29, 2013

### pasmith

You've done that, and shown that there is no choice of C for which y(0) = 0. At that point you need to find another approach.

In any event, if you have the initial condition $y(0) = D$, and it is obvious on inspection that $y' = 0$ when $y = D$, then the solution is $y(t) = D$ and you can save yourself some work.

7. Sep 29, 2013

### mesa

It's the not working in the general form that is bothering me but I think I have this one figured out.
Here is what I sent to my Professor,

I feel satisfied with this although I also like the explanation I just saw from the Professor.

Last edited: Sep 30, 2013
8. Sep 29, 2013

### smize

Edit:

What (s)he was trying to state is that y = 0 is an equilibrium solution of the differential equation. What would this imply about the equation?

Last edited: Sep 30, 2013
9. Sep 30, 2013

### mesa

I understand what they were getting at now, ignoring that we didn't have a solution for the general form in order to pursue the differential equation is where I was getting stuck. Both are solved at this point so I am satisfied unless someone else has something to add?