If `dy/dx=(e^y-x)^(-1),` where `y(0)=0` , then `y` is expressed explicitly as

Options

A. `1/2log_(e)(1+x^(2))`
B. `log_(e)(1+x^(2))`
C. `log_(e)(x+sqrt(1+x^(2)))=c`
D. None of these

Answer

C

Explanation

We have `(dy)/(dx) = (e^(y)-x)^(-1) rArr (dx)/(dy)=e^(y)-x`
`rArr (dx)/(dy)+x=e^(y)`
`rArr x=e^(y)/2+Ce^(-y)`
As `y(0)=0, C=-1/2` ,
`therefore x=e^(y)/2-1/2e^(-y)`
`rArr e^(2y)-2xe^(y)-1=0`
`rArr 2e^(y)=2x+-sqrt(4x^(2)+4)`.
But `e^(y)=x-sqrt(x^(2)+1)` (Rejected)
Hence `y=log_(e)(x+sqrt(x^(2)+1))`