If `x(t)` is a solution of `((1+t)dy)/(dx)-t y=1` and `y(0)=-1` then `y(1)` is

Options

A. 1
B. -1
C. `-1/2`
D. 0

Answer

c

Explanation

Given that , `(1+t) (dy)/(dt)- ty = 1 `
` rArr (dy)/(dt) = t/(1+t) y = 1/(1+t)`
which is a linear differential equation .
` :. IF = e^(-int1/(1+t)dt) = e^(-int(1-1/(1+t))dt) = e^({t - log (1+t)}`
` = e^(log (1+1)-t)= (1+t )*e^(-t)`
` rArr y* (1+t)^(e^(-t)) = int 1/((1+t) )* e^(-t) * dt + C `
` rArry ( 1+t) e^(-t) = int e^(-t) dt +C`
` rArr y (1+t) e^(-t) = -e^(-t) +C`
Now, ` y (0) = -1/(1+0) + (C*e^(0))/((1+0))= -1 [ :' y (0 = -1 ] `
` rArr -1 + C = -1 rArr C = 0`
From Eq. (i) , ` y (1+t) e^(-t) = -e^(-t)`
` rArr y = (-1)/(1+t)`
` :. y (1) = (-1)/2`