3
$\begingroup$

I've been attempting to solve this ODE the past few days: $$(1): [e^{-x^2}u'(x)]' = u(x) + xu'(x).$$ What I did was differentiate the left hand side and move things around a bit to obtain $$(2):e^{-x^2}u''(x) - [2x*e^{-x^2} + x]u'(x) - u(x) = 0.$$ We can see that we are dealing with a homogenous second-order linear differential equation. The problem: I simply just don't know how to solve this. Observing the equation, Cauchy-Euler is not possible. Since we're not dealing with constant coefficients, Principle of Superposition doesn't work as is. And, since this equation is homogenous, we cannot use the method of Undetermined Coefficients or Variation on Parameters.

I've attempted to to use v-substitution on (1), setting v(x)=u'(x), but this just hit dead ends. I came to a solution that I realized was absolutely preposterous (I had forgotten v was a function of x and integrated incorrectly so I got a very wonky answer). Anyone able to give a hand?

$\endgroup$
5
  • $\begingroup$ Can you check the equation $(1)$? $\endgroup$ – Valerin May 31 '14 at 3:08
  • $\begingroup$ Is $u'+xu'$ or $u+xu'$? $\endgroup$ – Valerin May 31 '14 at 3:11
  • $\begingroup$ Thank you for the heads up. Just fixed the equation. $\endgroup$ – user30625 May 31 '14 at 3:11
  • $\begingroup$ it's supposed to be u + xu' $\endgroup$ – user30625 May 31 '14 at 3:11
  • $\begingroup$ I think your equation is easy to solve :) $\endgroup$ – Valerin May 31 '14 at 3:16
1
$\begingroup$

Hint: The right hand side is precisely $(ux)'$ then integrating both sides we have $$e^{-x^2}u'=ux+C$$ can you continue the problem from here?

$\endgroup$
1
  • $\begingroup$ Your welcome!!! $\endgroup$ – Valerin May 31 '14 at 3:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.