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The question is:

Show the differential equation $y''-k^2y=R(x)$ $k\neq0$ and real has the particular solution: $$ y=1/k \int^x_0R(t)\sinh(k(x-t))dt $$

I know I need to sub this into the equation above but I don't know how to differentiate it, especially as it has $x$ in the integral

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Use Leibniz rule: $$\frac{d}{dx} \int^x_0 f(x,t) dt = f(x,x) + \int^x_0 \frac{\partial f}{\partial x} (x,t) dt.$$

In your case, $$y'(x) = \frac 1 k R(x)\underbrace{\sinh(k(x-x))}_{\,\,\,\,\,\,\,\, = 0} + \frac 1 k \int^x_0 R(t) k \cosh(k(x-t)) dt = \int^x_0 R(t) \cosh(k(x-t))dt.$$ Then $$y''(x) = R(x) \underbrace{\cosh(k(x-x))}_{\,\,\,\,\,\,\,\, = 1} + \int^x_0 R(t)k \sinh(k(x-t))dt = R(x) + k^2 y(x)$$ so rearranging gives $$y'' - k^2 y = R(x).$$

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