I have this problem:

With the aid of differentiation under the integral sign, show that the function

y=integral from 0 to x of h(t)sin(x-t)dt

satisfies the differential equation

y''+y=h(t).

Ok. I have the following equation to help me:

the derivative of the integral from a(x) to b(x) of f(x,t)dt is equal to

the integral from a(x) to b(x) of the partial derivative of x in f(x,t)dt + f(x,b(x))*b'(x)-f(x,a(x))*a'(x).

So I note that

a(x)=0

b(x)=x

a'(x)=0

b'(x)=1

f(x,t)=h(t)sin(x-t)

the partial derivative of x of f(x,t)=h(t)cos(x-t)

f(x,b(x))=0

f(x,a(x))=h(0)sin(x)

Thus, the derivative of the integral from a(x) to b(x) of f(x,t)dt is equal to

the integral from 0 to x of h(t)cos(x-t)dt.

This is where I get stuck. I'm pretty sure that if you plug in 0 and x into the above you get the second derivative, but I'm not positive.

Any help would be appreciated.