Q:

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.

A:

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

