Question

The general solution of the differential equation (e^x + 1) ydy = (y + 1) e^xdx is ______.

Options

• (y + 1) = k(e^x + 1)

• y + 1 = e^x + 1 + k

• y = log {k(y + 1)(e^x + 1)}

• y = `log{("e"^x + 1)/(y + 1)} + "k"`

Answer 1

The general solution of the differential equation (e^x + 1) ydy = (y + 1) e^xdx is y = log {k(y + 1)(e^x + 1)}.

Explanation:

The given differential equation is (e^x + 1) ydy = (y + 1) e^xdx

⇒ `y/(y + 1) "d"y = "e"^x/("e"^x + 1) "d"x`

Integrating both sides, we get

`int y/(y + 1) "d"y = int "e"^x/("e"^x + 1)"d"x`

⇒ `int (y + 1 - 1)/(y + 1) "d"y = int "e"^x/("e"^x + 1) "d"x` 

⇒ `int 1. "d"y - int 1/(y + 1) "d"y = int "e"^x/("e"^x + 1) "d"x`

⇒ `y - log|y + 1| = log|"e"^x + 1| + log"k"`

⇒ y = `log|y + 1| + log|"e"^x + 1| + log "k"`

⇒ y = `log|"k"(y + 1)("e"^x + 1)|`