0
$\begingroup$

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!

  • 0
    @Swapa$n$: You are right, I have stupidly forgotten to include the $b$oundary conditions! :$S$ They are added now.2011-11-14

1 Answers 1

1

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.