The solution of `y'-y = 1 , y(0)=-1` is given by `y(x) =`

Options

A. `-exp(x)`
B. `-exp(-x)`
C. `-1`
D. `exp (x)-2`

Answer

C

Explanation

The given equation can be rewritten as
`(dy)/(1+y)=dx`
`rArr int(dy)/(1+y)=int 1dx`
`rArr log (1+y)=x+c`
`rArr 1+y=e^x.e^c`
At`x=0,y=-1`
`therefore 1-1=e^(0).e^(c)rArr e^(c)=0`
On putting this value in Eq.(i) , we get
`y=-1`