# If $y=\int_0^x f(t)\sin[k(x-t)]dt$, then calculate $\frac{d^2y}{dx^2}+k^2y$

If $y=\displaystyle \int_0^x f(t)\sin[k(x-t)]dt$, then calculate $\dfrac{d^2y}{dx^2}+k^2y$

$y=\displaystyle \int_0^x f(t)\sin[k(x-t)]dt$ $\implies$ $\dfrac{dy}{dx}=\displaystyle \int_0^x kf(t)\cos[k(x-t)]dt$

$\dfrac{d^2y}{dx^2}=\displaystyle \int_0^x -k^2f(t)\sin[k(x-t)]dt$

So, $\dfrac{d^2y}{dx^2}+k^2y$ should be zero. But the given answer shows $kf(x)$

How can a new term appear during differentiation when there is only one function of x here ?

• The integral depends on $x$ and so your computation of $\dfrac{dy}{dx}$ is wrong. – voldemort Jan 21 '15 at 0:55
• Look into Leibniz. Taking the derivative of that integral is simply what's in the integrand but with a change of variables to $x$ instead of $t$. – bjd2385 Jan 21 '15 at 0:57
You should notice that the bound of the integral depends on x, so the derivation of $y$ by $x$ is not simply computed by derivating the integrand. In fact, $$\dfrac{dy}{dx}= \int_0^x kf(t)\cos[k(x-t)]dt + f(x)\cdot \sin[k(x-x)] = \int_0^x kf(t)\cos[k(x-t)]dt,$$ and \begin{align} \dfrac{d^2y}{dx^2} &=\int_0^x -k^2f(t)\sin[k(x-t)]dt + kf(x)\cdot \cos[k(x-x)] \\ &= \int_0^x -k^2f(t)\sin[k(x-t)]dt + kf(x) \end{align}