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Consider the eigenvalue problem, $$x^2u_{xx}-2xu_x+2u=\lambda x^2u$$ for $0<x<1$, with boundary conditions $u_x(0)=0$ and $u(1)=u_x(1)$. Determine a function $M(x)$ so that, under the change of variable $u(x)=M(x)w(x)$ , it can be transformed to the form $$w_{xx}=-\lambda w$$ for $0<x<1$.

My attempt:

If $u(x)=M(x)w(x)$, then $u'=Mw'+wM'$ and $u''=Mw''+wM''+2w'M'$. Substituting this into the intial equation gives, \begin{align} x^2(Mw''+wM''+2w'M')-2x(Mw'+wM')+2wM&=\lambda x^2wM \\ x^2Mw''+x^2wM''+2x^2w'M'-2xMw'-2xwM'+2wm&=\lambda x^2wM. \end{align} Observing the coefficient of $w''$ indicates that $$M=\frac{1}{x^2}.$$ However, applying this value of $M$ will make the RHS positive, not negative. Substituting this value of $M$ in to the equation does not yield the required form.

I do not understand what to do. Any advice/hints are appreciated.

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  • $\begingroup$ Where is your $M''$? $\endgroup$ – xpaul Mar 13 at 1:27
  • $\begingroup$ Isn't it there? With coefficient $x^2w$. $\endgroup$ – Bell Mar 13 at 1:28
  • $\begingroup$ Sorry, I didn't see before. I see now. $\endgroup$ – xpaul Mar 13 at 1:32
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Let the coefficient of $w'(x)$ be zero, namely $$ 2x^2M'(x)-2xM(x)=0. $$ This gives $M(x)=x$ and hence $$ w''(x)=\lambda w(x). $$

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