`Ify=int_0^xf(t)sin{k(x-t)}dt ,t h e np rov et h a t(d^2y)/(dx^2)+k^2y=kf(x)dot`

Explanation

`y=int_(0)^(x)f(t)sin{k(x-t)}dt`
`=int_(0)^(x)f(t)[sinixcoskt-sinktcoskx]dt`
`=sinkx int_(0)^(x)f(t)cosktdt-coskx int_(0)^(x)f(t)sin kt dt`…………….1
`:.(dy)/(dt)=kcoskx int_(0)^(x)f(t)cosktdt+sinkx[f(x)cosx]`
`+ksin kx int_(0)^(x)f(t)sin ktdt-coskx[f(x)sinkx]`
`=kcoskx int_(0)^(x)f(t)cos ktdt+ksinkx int_(0)^(x)f(t)sin ktdt`..................2
Again differentiating equation (2) w.r.t `x` we get
`(d^(2)y)/(dx^(2))=-k^(2)sin kx int_(0)^(x)f(t)cos ktdt+cos kx[f(x)coskx]`
`+k^(2)coskx int_(0)^(x)f(t)sin tdt+k sin kx[f (x)sin kx]`
`=-k^(2)y+kf(x)`
`:.(d^(2)y)/(dx^(2))+k^(2)y=kf(x)`