Use Differentiation under the integral sign

show that the function,

y(x) = integral (0,x) h(t) sinh(x-t)dt

satisfies the differential equation

y'' - y = h(x)

What are the initial conditions at x=0?

Thanks! :)

Let y(x) = ∫(t = 0 to x) h(t) sinh(x - t) dt.

Differentiating under the integral sign yields
y' = h(x) sinh(x - x) + ∫(t = 0 to x) (∂/∂x) h(t) sinh(x - t) dt
....= ∫(t = 0 to x) h(t) cosh(x - t) dt

y'' = h(x) cosh(x - x) + ∫(t = 0 to x) (∂/∂x) h(t) cosh(x - t) dt
....= h(x) + ∫(t = 0 to x) h(t) sinh(x - t) dt.

Hence, y'' - y = h(t), as required.

Link:
http://en.wikipedia.org/wiki/Differentia…
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At x = 0, y(0) = ∫(t = 0 to 0) h(t) sinh(x - t) dt = 0.

I hope this helps!