Question

Solve the following differential equation.

`(x + y) dy/dx = 1`

Answer 1

`(x + y) dy/dx = 1`

∴ `dy/dx = 1/(x + y)`

∴ `dx/dy = x + y`

∴ `dx/dy − x = y`

∴ `dx/dy + (− 1)x = y`                 ...(I)

The given equation is of the form `dx/dy + Px = Q`

where, P = − 1 and Q = y

∴ `"I.F." = e int ^"pdy" = e int ^("(−1)dy") = e^-"y"`

∴ Solution of the given equation is

`"x (I.F.)" = int "Q (I.F.) dy" + C`

∴ `"xe"^(−"y") = ubrace(int y.e^(−y))_(("I"))   dy + C`        ...(II)

Let I = `int y. e^(−y)  dy`

Using Integration by parts,

I = `y int e^(−y) dy - int [ d/dy y int e^(−y)  dy] dy`

I =  `y (e^(−y))/(-1) − int 1. (e^(−y))/(-1)  dy`

I = `− y. e^(−y) + int e^(−y)  dy`

I = `− y. e^(−y) + e^(−y)/(- 1)  dy`

I = `− y. e^(−y) − e^(−y)`

Putting value of I in (2),

∴ `"xe"^(−"y") = int y.e^(−y)   dy + C`

∴ `"xe"^(−"y") = − y. e^(−y) − e^(−y)  + C`

Dividing by e^−y,

∴ x = − y − 1 + Ce^y

∴  x + y + 1 = Ce^y