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

Given , ` (dy)/(dx) = (x+y+1)/(x+y-1) "Put " x + y = 1 `
` rArr 1+ (dy)/(dx) = (dt)/(dx) rArr (dy)/(dx) = (dt)/(dx) -1 `
` :. (dt)/(dx) -1 = (t+1) /(t-1)`
` rArr (dt)/(dx) = (t+1+t-1)/(t-1)`
` rArr (dt)/(dx) = (t+1+t-1)/(t-1)`
` rArr (dt)/(dx) = (2t)/(t-1) rArr ((t-1)/(2t))dt = dx`
` rArr (1/2 - 1/(2t))dt = dx `
On integrating , we get
`1/2 t -1/2 log t = x +C_(1)`
` rArr t - lg t = 2x + 2C_(1)`
` rArr x + y - log (x+y) = 2x +2C_(1)`
` rArr y = x + log (x+y) +C`