For `x in R,x ne 0`, if y(x) is differentiable function such that `xint_(1)^(x) y(t)dt=(x+1)int_(1)^(x) t y (t) dt`, then y(x)equals (where C is a constant).

Options

A. `Cx^(3)e^(1//x)`
B. `(C )/(x^(2))e^(-1//x)`
C. `(C )/(x^(2))e^(-1//x)`
D. `(C )/(x^(3))e^(-1//x)`

Answer

D

Explanation

We have,
`xunderset(1)overset(x)int y(t)dt=(x+1)underset(1)overset(x)int t y (t) dt" "`...(i)
Differenting with respect to x,we get
`underset(1)overset(x)int y(t)dt+x y(x)=underset(1)overset(x)int t y (t) dt+(x+1) xy(x)`
`rArr underset(1)overset(x)int y(t)dt=underset(1)overset(x)int ty(t)dt+x^(2)y(x)`
Differentiating with respect to x, we get
`y(x)=xy(x)+x^(2)(dy)/(dx)(x)+xy(x)`
`rArr (1-3x)y(x)=x^(2)(d)/(dx)(y(x))`
`rArr ((1)/(x^(2))-(3)/(x))dx=(d(y(x)))/(y(x))`
Integrating, we obtain.
`-(1)/(x)-3logx=log y(x)+logk`
`rArr-(1)/(x)=log(kx^(3)y(x))`
`rArr kx^(3)y(x)=e^(-1//x)rArr y(x)=(1)/(kx^(3))e^(-1//x)rArry(x)=(C )/(x^(3))e^(-1//x)`