# Help on Diff Equation!

1. Sep 17, 2008

### chocok

Question:
I have to find the constant a such that $$y(x) = e^{ax}$$ is a solution for x" = x

I tried 2 ways of reasoning and they both led to my answer that a can be any number (but zero).
Can anyone see if they are correct?? If not, pls give me some hint!

1. x= ln(y)/a but x" is a 2nd derivative with respect to x. so x" = 0 => x=0 =>$$a\in R$$ \0

2. x has to be a constant for the solution to work, and it has to be specifically 0, so a is any real number but 0.

2. Sep 17, 2008

### Defennder

I don't see what does y = e^(ax) got to do with x'' = x. The latter is actually a 2nd order derivative with respect to some variable like t (or y) is it not? Because it is evident that d^2/dx^2 x = 0.

3. Sep 17, 2008

### chocok

Actually this is one part of a big question where similar DE are given with the same solution(where others have y) so the solution y=exp(ax) seems irrelevant in this case.

The question didn't state x" is being differentiated w.r.t. what variable, so I just assumed it's dx...

This may sound dumb... but.. :
So you mean as the whole equation is based on x, I could simply jump ahead to conclude that a is unbounded? even without the "not equal to zero" condition? I added the "not equal 0" condition as x = ln(y)/a (is this unnecessary?)

Also does it mean that my #1 or 2 reasoning are totally unnecessary?

4. Sep 17, 2008

### sr6622

I'm confused. Can't you just make a characteristic equation and make the equation
r^2 - 1 = 0, and then solve for r? Yielding simply 1 and -1? I haven't done this in a while, sorry.

5. Sep 17, 2008

### Defennder

No I still don't follow you at all. You mentioned that it was part of a larger question, so perhaps you could post that part of the question as it is? Because I don't know what you're talking about.

6. Sep 17, 2008

### chocok

sr6622: sorry, I still haven't learned characteristic equation. so I can't use the technique yet.. but thanks!

Defennder:
Ohhhh. I mean it's a big question but each part has nothing to do with each other. They are just unrelated, like part i) is y' + y =0 and we are asked to do the same thing (finding the constant a with y(x)=exp(ax) is the solution).

so I guess as the "solution" y(x)=exp(ax) is irrelevant I can have a to be any number I want then.,,?

Last edited: Sep 17, 2008