Solving an ordinary differential equation with initial conditions Can someone please help me with this ODE problem? Here is the question:
Consider the ODE 
$ {d^2 U\over dx^2} - [{s^2\over c^2}]U=e^{{-sx\over v}}.  U(0) = 0, U(x)$ is bounded as $x$ goes to infinity.  $s, v, c$ are positive constants and that $C \neq V.$ Solve for $U$.
I think that I am to guess that $U=Ae^{{-sx\over v}}$ and solve for $A$, for the homogeneous equation and then use that to satisfy the given initial conditions.
 A: Related problems: (I) First solve the homogeneous equation which gives the following fundamental set of solutions

$$ \left\{ e^{-\frac{s}{c}x}, e^{\frac{s}{c}x} \right\}. $$

Since $ c,s > 0 $ you should pick up the solution $U_h= \alpha \, e^{-\frac{s}{c}x}$, for some $\alpha$ to be determined later using the initial condition, to satisfy your condition of boundedness. Your choice for the particular solution $U_p=A e^{-\frac{s}{v}x}$ is correct as long as $c\neq v$. I believe you know how to determine $A$. The solution will have the form

$$ U(x) = \alpha \, e^{-\frac{s}{c}x} + U_p(x). $$

You need to determine $\alpha$ using the initial condition.    
A: Hint:  Do $U(x) = e^{\lambda x}$ for homogeneous equations.
A: @Mhenni (or others),
I believe solving for A requires method of undetermined coefficients.  What I got was:
Pick $Ae^{-{sx\over v}}$.  From the given ODE, we have:
$[Ae^{-{sx\over v}}]^{{\prime}{\prime}}$ - ${s^2\over c^2}[Ae^{-{sx\over v}}] = e^{-{sx\over v}}$.
Deriving twice, get:
${s^2\over v^2}[Ae^{-{sx\over v}}]$-${s^2\over c^2}[Ae^{-{sx\over v}}]$ = $e^{-{sx\over v}}$
Cancel terms:
${s^2\over v^2}A$-${s^2\over c^2}A$ = $1$
$A = {1\over{{s^2\over v^2}} - {s^2\over c^2}}$
Thus,
$U_p = {e^{-{ss\over v}}\over{1\over {{s^2\over v^2}} - {s^2\over c^2}}} $
Then, using the initial conditions, we solve for $\alpha$
From $U(0) = 0$, we have:
$\alpha e^0+{1\over{{s^2\over v^2}} - {s^2\over c^2}} e^0 = 0$.  And, 
$\alpha = - {1\over{{s^2\over v^2}} - {s^2\over c^2}} $
Finally, $U(x) = - {1\over{{s^2\over v^2}} - {s^2\over c^2}} e^{-{sc\over v}} + {1\over{1\over {{s^2\over v^2}} - {s^2\over c^2}}} e^{-{sc\over v}} $ .
Did I finish this problem correctly?
