General solution of the differential equation ` (dy)/(dx)=(x+y+1)/(x+y-1)` is given by

Options

A. `x+y=log|x+y|+c`
B. `x-y=log|x+y|+c`
C. `y=x+log|x+y|+c`
D. `y=x log|x+y|+c`

Answer

C

Explanation

`(dy)/(dx)=(x+y+1)/(x+y-1)`
Put ` x+y=t `
`implies 1+(dy)/(dx)=(dt)/(dx) `
`implies (dy)/(dx)=(dt)/(dx)-1`
`:. " "(dt)/(dt)=(t+1+t-1)/(t-1)`
`=(dt)/(dx)=(2t)/(t-1)`
`implies((t-1)/(2t))dt=dx `
`implies ((1)/(2)-(1)/(2t))dt=dx `
On integrating , we get
`(1)/(2)t-(1)/(2) log t=x +c_(1)`
`implies t-logt=2x+ 2c_(1)`
`implies x+y- log (x+y)=2x+2c_(1)`
`implies y=x+ log( x +y)+c `