अवकलन `(dx)/(dy) (x + y +1)/(x+y - 1)` समीकरण का हल क्या है ?

Options

A. `y - x + 4 In (x+y) = C`
B. `y + x + 2 In (x+y) = C`
C. `y - x +In (x + y) = C`
D. `y + x + 2 In (x + y) = C`

Explanation

`d/(dy) = (x+y+1)/(x+y-1)`
माना `x +y = u`
`(dx)/(dy) + 1 = (du)/(dy)`
`rArr (dx)/(dy) = (du)/(dy) - 1`
`rArr (du)/(dy) - 1 = (u+1)/(u-1)`
`rArr (du)/(dy) = (u+1)/(u-1) + 1`
`rArr (du)/(dy) = (u+1+u-1)/(u-1)`
`(dy)/(dy) = (2u)/(u - 1)`
`(u-1)/(u) du = 2dy`
समाकलन करने पर,
`int(1-1/u) du= int2dy`
`rArr u - log u = 2y + C [ :' u = x+y]`
`rArr x + y - log (x+ y) = 2 y + C`
`rArr x - log (x + y) = y+ C`
`rarr y -x + log (x + y) = C`