Bounding a solution of an ODE with a small source I have an ODE of the form
$$ f''(x) + f(x) = \epsilon g(x)$$
with initial conditions
$$ f(0) = f'(0) = 0 $$
$g(x)$ is $O(1)$ as $\epsilon \to 0$, and $g(x)$ is as smooth as necessary.  Is there a way I can bound $f(x)$?  In particular, I would like to be able to claim that $f(x)$ must be $O(\epsilon)$ as $\epsilon \to 0$ for all $x>0$, but I'm not sure if this is true or how to approach this.  Any hints/advice would be greatly appreciated.
EDIT: I also know that $g(0) = g'(0) = 0$, if that helps with anything.
EDIT2: Upon seeing the answers, I think we can see that the values of $g$ and $g'$ at $0$ are irrelevant as long as $g(x) = O(1)$.
 A: You can rewrite the equation as a system of two linear differential equations. There is a closed form for the solution of such system. In particular, you have to:


*

*Find the solutions to the homogeneous problem and stack them in a matrix $X$

*Compute the particular solution as $X\int X^{-1} \underline{f}(x)$, where $\underline{f}$ is the right hand side in the linear system formulation.


Now, if you can compute the integral, then you have an explicit formula for the solution, and you can find a bound. Otherwise, you need to bound the integral, to find a bound for the solution.
Note: since both the initial conditions are zero, the particular solution is THE solution to your problem.
A: Put $f_1 = f$, $f_2 = f'$. Then $f_1' = f_2$ and the equation gives $$f_2' = -f_1 + \epsilon g.$$ Thus we have a system $$\frac{d}{dx} \left( \begin{matrix}f_1 \\ f_2 \end{matrix} \right) = \left( \begin{matrix}0 & 1\\ -1 & 0 \end{matrix} \right)\left( \begin{matrix}f_1 \\ f_2 \end{matrix} \right) + \epsilon \left( \begin{matrix} 0 \\ g \end{matrix} \right), \,\,\,\,\,\,\,\,\,\,\,    \left( \begin{matrix}f_1(0) \\ f_2(0) \end{matrix} \right) = \left( \begin{matrix}0 \\ 0 \end{matrix} \right).$$ You can solve this systems explicitly using an integrating factor (and matrix exponentials) then look for a bound on $f_1$ since $f_1 = f$.
