If `y=int_(0)^(x) f(t)sin{(k(x-t)}dt`, then `(d^(2)y)/(dx^(2))+k^(2)y=`

Options

A. f(x)
B. k f(x)
C. `k^(2)f(x)`
D. none of these

Answer

B

Explanation

We have,
`y=overset(x)underset(0)int f(t) sin {k(x-t)}dt`
Differentiating both sides w.r.to x, we obtain
`(dy)/(dx)=overset(x)underset(0)int (del)/(delx){f(t) sin k(x-t)} dt+(d(x))/(dx)xx{f(x) sin k (x-x)}-(d)/(dx)(0)xx{f(0) sin k(x-0)}`
`rArr (dy)/(dx)=k overset(x)underset(0)int f(t)cos k(x-t)dt+f(x)xx0-0`
`rArr (dy)/(dx)=koverset(x)underset(0)int f(t)cos k(x-t)dt`
Diffrerentiating both sides w.r. to x, we get
`(dy)/(dx)=k[overset(x)underset(0)int (del)/(delx){f(t) cos k(x-t)} dt+(d)/(dx)(x){f(x) cos k (x-x)}-(d)/(dx)(0){f(0)cos k (x-0)}]`
` rArr (d^(2)y)/(dx^(2))=k[-koverset(x)underset(0)int f(t)sin k (x-1) dt+f(x)-0]`
`rArr (d^(2)y)/(dx^(2))=-k^(2)y+kf(x) rArr(d^(2)y)/(dx^(2))+k^(2)y=kf(x)`