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.