Differential Equation: Am I missing a trick? I am trying to solve the following calculus problem:

Show the function
$$\displaystyle y(x)=\int_{0}^x \sin(x-t)f(t)dt$$
solves the differential equation
$$y''+y=f(x)$$

I have put $$\sin(x-t)=\sin(x)\cos(t)-\sin(t)\cos(x)$$
I assume I'm supposed to use the fundamental theorem of calculus to find $y''$ and then $y$ will fall out, however I do not see how to derive $y''$ from $y$. I would appreciate any help.
 A: It is sort of hard to explain the Leibniz's rule without giving an example. 
Let's use a full answer of this as an example.
Leibniz's rule of differentiation under integral sign states that

If you have a function $\varphi(x)$ defined by an integral where the upper limit, lower limit and integrand depends on $x$, i.e.
  $$\varphi(x) = \int_{a(x)}^{b(x)} G(x,t) dt$$
  then the derivative of $\varphi(x)$ contains 3 pieces:
  $$\frac{d\varphi(x)}{dx} = 
\underbrace{G(x,b(x))b'(x)}_{I_1} - 
\underbrace{G(x,a(x))a'(x)}_{I_2} + 
\underbrace{\int_{a(x)}^{b(x)} \frac{\partial G(x,t)}{\partial x} dt}_{I_3}
$$
  These 3 pieces are contribution due to changes in upper limit, lower limit and the integrand respectively.
  
  
*
  
*$I_1$ alone is really a combination of first fundamental theorem of calculus together with the usual chain rule of computing the derivative of composition of two functions. 
  
*$I_3$ alone is the contribution from usual differentiation under integral sign.
  

Using our $y(x)$ as an example, we have
$$b(x) = x,\quad a(x) = 0,\quad\text{ and }\quad G(x,t) = \sin(x-t)f(t)$$
and hence
$$
\begin{align}
I_1 &= G(x,b(x))b'(x) = \sin(x-b(x))f(b(x))b'(x) = \sin(x-x)f(x) \times 1 = 0\\[10pt]
I_2 &= G(x,a(x))a'(x) = \sin(x-a(x))f(a(x))a'(x) = \sin(x-0)f(0) \times 0 = 0\\[10pt]
I_3 &= \int_{a(x)}^{b(x)} \frac{\partial G(x,t)}{\partial x} dt =
\int_0^x \frac{\partial}{\partial x}\big[\sin(x-t)f(t)\big] dt =
\int_0^x \cos(x-t)f(t) dt
\end{align}
$$
This give us
$$
y'(x) = I_1 - I_2 + I_3 = \int_0^x \cos(x-t) f(t)dt
$$
Apply Leibniz's rule one more time, we get
$$\begin{align}y''(x) 
&= \cos(x-x)f(x) + \int_0^x\frac{\partial}{\partial x}\big[\cos(x-t)f(t)\big] dt\\
&= f(x) - \int_0^x \sin(x-t)f(t)\\
&= f(x) - y(x)
\end{align}$$
This is essentially what we are asked to prove.
A: use the Laplace transform and check the transform are bot equal
for the integral $$ Y(s)=\frac{F(s)}{s^{2}+1} $$
for teh differnetial equation  $$ (s^{2}+1)Y(s)=F(s)$$
since they have both the same transform :D of laplace the functions are equal and the integral equation and the differential equation are equal too 
A: HINT: The best way is to think $y(x)$ as
$$y(x)=F(x,x)\text{,}$$
where
$$F(x,y)=\int_0^x\sin(y-t)f(t)\,dt\text{.}$$
Then apply usual rules for derivation in several variables, more concretely the chain rule.
