# A differential equation

1. Dec 6, 2009

### manenbu

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

Solve:
y'''=(y'')2

2. Relevant equations

3. The attempt at a solution

Basically easy, use the substitution y'' = p and solve through the equations.
I end up having this:
y=-(x+c1)ln|x+c1| + c2x + c3, which is what I listed in the answers as well.
My question is about p≠0 which I get while solving the first integral.
p=y''=0 is a solution to the equation.
y''=0 means y'=c means y=cx+c
Is this included in the solution? I don't think so, and then it should be listed as a solution as well, but it's not. Am I missing something, or it's already in the first solution?

2. Dec 6, 2009

### LCKurtz

I don't see a choice of constants in your solution yielding y = cx + d. (You don't mean cx + c). But, yes, it is a perfectly good solution as you can test by plugging it into the equation. It should be listed as an answer. If you divided by p in the process of solving the equation, that would explain why the solution was missed. Good work noticing that.

3. Dec 6, 2009

### manenbu

yes, I meant y = cx + d of course.
Well, in that case that's another mistake in the answer sheet - for the DE:
2y y'' = (y')^2 + 1
they list y = c as an answer, when it is obviously not.

4. Dec 6, 2009

### LCKurtz

Yes. You're right about that too.