Question

Solve `dy/dx = (x+y+1)/(x+y-1)  when  x = 2/3 and y = 1/3`

Answer 1

`dy/dx = (x+y+1)/(x+y-1)`  ...(i)

Put x + y = t  ...(ii)

∴ y = t - x

Differentiating w.r.t.x, we get

∴ `dy/dx = dt/dx -1` ...(iii)

Substituting (ii) and (iii) in (i), we get

`dt/dx -1 = (t+1)/(t-1)`

∴ `dt/dx =(t+1)/(t-1) + 1 = (t+1+t-1)/(t-1)`

∴ `dt/dx = (2t)/(t-1)`

∴ `((t-1)/t)dt = 2dx`

∴ `(1-1/t)dt = 2dx`

Integrating on both sides, we get

`int (1-1/t) dt = 2 int dx`

∴ t - log | t | = 2x + c

∴ x + y -log |x + y| = 2x + c

x - 2x + y - log (x + y) = c

y - x - log (x + y) = c

y - x - c = log (x + y)   ...(i)

Putting `x = 2/ 3 and y = 1/ 3` , we get 

 ∴ from (i)

`1/3 - 2/3 - c = log (2/3 + 1/3)`

`(-1)/3 - c = log`   (i)

`(-1)/3 - c = 0`

∴ c = `(-1)/3`

Put c = `1/ 3` in (i)

∴ c = `1/3`

`y - x - (-1/3)` = log (x + y)

`y - x +1/3` = log (x + y)