# A boundary value problem

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 $$u(x)=16x+x^4.$$
• 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 $A=0$. – LutzL Aug 14 at 6:40