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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.

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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 $$ u(x)=16x+x^4. $$

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  • $\begingroup$ 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

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