I am trying to solve the following boundary value problem:

$$ x^2 u'' + 2 x u' - 2u = 18x^4,\;\; 0 < x < 2, \\ u \text{ finite},\;\; x \rightarrow 0^+, \\ u' - u = 0, \;\; x = 2. $$

I'm not sure how to go about this. So far I've solved it for the homogeneous case and got $v = Ax^{-2} + Bx$ and not sure where to go from here. The answer in my texbook is $u = 16x + x^2$. Any help is appreciated.


Check homogeneous solution: $u=x^m$ gives the characteristic polynomial $0=m(m-1)+2m-2=(m+2)(m-1)$. Correct.

As the right side is not contained in the basis solutions, you can apply the method of undetermined coefficients and find a particular solution in the form $u_p=Cx^4$. Inserting leads to $$ Cx^4(12+8-2)=18x^4\implies C=1 $$

For a finite value at $x=0$ in $u(x)=Ax^{-2}+Bx+x^4$ we need $A=0$. Then $u'(x)-u(x)=B+4x^3-Bx-x^4$ and that is zero at $x=2$ if $B=32-16=16$. So the solution has indeed the cited coefficients, but different powers $$ u(x)=16x+x^4. $$

  • $\begingroup$ Thank you! Just a quick question - why does a finite value at x = 0 imply that A = 0? $\endgroup$ – scott Aug 14 at 6:37
  • $\begingroup$ Because otherwise there is a pole at $x=0$, that is, the solution does not exist there and can not be extended to there. One could also say that with $u(0)$ finite $x^2u(x)$ has the value $0$ there, directly implying $A=0$. $\endgroup$ – LutzL Aug 14 at 6:40

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.