Question

If $y={\int }_0^{x}f\left(t\right)\sin \left[k(x-t)\right]dt$ then $\frac{d^{2}y}{d{x}^2}+k^{2}y$ equals

• A. $kf\left(x\right)$
• B. $k^{2}f\left(x\right)$
• C. $f\left(x\right)$
• D. a constant

Answer 1

The correct option is A $kf\left(x\right)$

$y={\int }_0^{x}f\left(t\right)\sin \left[k\right(x-t\left)\right]dty={\int }_0^{x}f\left(t\right)(\sin kx\cos kt-\cos kx\sin kt)dty={\int }_0^{x}f\left(t\right)\sin kx\cos ktdt-{\int }_0^{x}f\left(t\right)\cos kx\sin ktdt{y}^{\prime }=k\cos kx{\int }_0^{x}f\left(t\right)\cos ktdt+\sin kx\left[f\right(x\left)\cos kx\right]+k\sin kx{\int }_0^{x}f\left(t\right)\sin ktdt-\cos kx\left[f\right(x\left)\sin kx\right]y^{\prime }=k\cos kx{\int }_0^{x}f\left(t\right)\cos ktdt+k\sin kx{\int }_0^{x}f\left(t\right)\sin ktdt{y}^{\prime \prime }=-k^2\sin kx{\int }_0^{x}f\left(t\right)\cos ktdt+k\sin kx\left[f\right(x\left)\sin kx\right]+k^2\cos kx{\int }_0^{x}f\left(t\right)\sin ktdt+k\cos kx\left[f\right(x\left)\cos kx\right]y^{\prime \prime }=-k^{2}y+kf\left(x\right)$