Finding a so that e^(-ax^2) solves y''+xy'+y=0

[swebonny]

Homework Statement

Is it possible to give a a value so that e^(-ax^2) becomes a solution to the differential equation y''(x)+x*y'(x)+y = 0

Homework Equations

Already given.

The Attempt at a Solution

Hello!

I'm new to this forum and doesn't really know how the fancy Latex stuff works, but I hope you'll understand me. Anyhow, I have been sitting with this problem for a while.
Should I take the derivative of e^(-ax^2) and put it into the equation and in that way (somehow) find out x and a?

After the I have taken the derivative and put it into the equation I get this, which I'm fairly sure is correct :/

4a^2*(e^(ax^2))+2a(e^(ax^2))+2a(e^(ax^2))*x+(e^(ax ^2))= 0

What am I supposed to do next?

I hope you understand me.

Thanks. (nice forum btw)

 

[rock.freak667]

e^-ax2 ≥ 0 for all x, so you can divide throughout by it.

 

[Char. Limit]

e^-ax2 ≥ 0 for all x, so you can divide throughout by it.

If such a condition were true, then you could not divide throughout by it. However, e^ax2 > 0 for all x, not ≥ 0 for all x.

So you can divide by e^ax2, but I wanted to correct that.