If $y=\int\limits_0^x f(t) \sin \{k(x-t)\} d t$, then $\frac{d^2 y}{d x^2}+k^2 y=$

$k f(x)$

We have,

$y=\int\limits_0^x f(t) \sin \{k(x-t)\} d t$

Differentiating both sides w.r. to $x$, we obtain

$\frac{d y}{d x}=\int\limits_0^x \frac{\partial}{\partial x}\{f(t) \sin k(x-t)\} d t +\frac{d(x)}{d x} \times\{f(x) \sin k(x-x)\} -\frac{d}{d x}(0) \times\{f(0) \sin k(x-0)\}$

$\Rightarrow \frac{d y}{d x}=k \int\limits_0^x f(t) \cos k(x-t) d t+f(x) \times 0-0$

$\Rightarrow \frac{d y}{d x}=k \int\limits_0^x f(t) \cos k(x-t) d t$

Differentiating both sides w.r. to $x$, we get

$\frac{d^2 y}{d x^2}=k\left[\int\limits_0^x \frac{\partial}{\partial x} \{f(t) \cos k(x-t)\} d t + \frac{d}{d x}(x)\{f(x) \cos k(x-x)\}-\frac{d}{d x}(0)\{f(0) \cos k(x-0)\}\right]$

$\Rightarrow \frac{d^2 y}{d x^2}=k\left[-k \int\limits_0^x f(t) \sin k(x-t) d t+f(x)-0\right]$

$\Rightarrow \frac{d^2 y}{d x^2}=-k^2 y+k f(x) \Rightarrow \frac{d^2 y}{d x^2}+k^2 y=k f(x)$