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Could someone please help me with the following?

I have a linear operator

$Ly=-x^{-2}(x^2y')'+y$, where $" ' "$ denotes $d\over dx$

I need to find the solution $y(x)$ to the forced equation $Ly=F(x)$

subject to boundary conditions $y(x)$ is bounded as $x\to 0, \infty$

where $$F(x) = \begin{cases}1 & x\in[0,x_0]\\0 &x>x_0\end{cases}$$

The solution, I was told, should have the form:

$$y(x) =\begin{cases} {\alpha\sinh x\over x}+1 &x\in [0, x_0]\\ {\beta e^{-x}\over x} &x>x_0 \end{cases}$$

Thanks in advance for any help!

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The [forcing] tag is for the set theory related technique known as Forcing. –  Asaf Karagila Nov 14 '11 at 9:38
    
Your equation is of 2nd order (linear ODE), but the answer involve only one arbitrary constants...? –  Tapu Nov 14 '11 at 9:44
    
@AsafKaragila, may be you should write a tag-wiki or at least a tag-description about forcing? That way innocent newcomers have a chance! It used to be that most questions carrying the division-algebra tag were about the problems high schoolers faced when dealing with dividing polynomials in an algebra class :-) I don't know whether my tag description played any role, but the situation is much improved now! –  Jyrki Lahtonen Nov 14 '11 at 9:46
    
@AsafKaragila: Thanks, I didn't know. :) –  Alfonso Catoia Nov 14 '11 at 9:47
    
@Swapan: You are right, I have stupidly forgotten to include the boundary conditions! :S They are added now. –  Alfonso Catoia Nov 14 '11 at 9:48

1 Answer 1

up vote 1 down vote accepted

You can define $z=xy$. By introducing this new variable in the equation I've find $-z''+z = x F(x)$.

Now you have a linear ODE of order 2 with constant coefficients, which is quite easy to solve.

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