Need help with differential equation of type y''+ay'+by=g(x)

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SUMMARY

The discussion centers on solving the differential equation y''(x) + x*y'(x) + y = 0 and determining if e^(-ax^2) can be a solution. The user attempts to derive the equation by substituting the function and its derivatives, resulting in the expression 4a^2*(e^(ax^2)) + 2a(e^(ax^2)) + 2a(e^(ax^2))*x + (e^(ax^2)) = 0. A critical point raised is the condition e^(-ax^2) ≥ 0, which allows for division by e^(ax^2), clarifying that e^(ax^2) is always positive for all x.

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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)
 
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e-ax2 ≥ 0 for all x, so you can divide throughout by it.
 
rock.freak667 said:
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, eax2 > 0 for all x, not ≥ 0 for all x.

So you can divide by eax2, but I wanted to correct that.
 

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