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प्रश्न
The general solution of the differential equation (ex + 1) ydy = (y + 1) exdx is ______.
विकल्प
(y + 1) = k(ex + 1)
y + 1 = ex + 1 + k
y = log {k(y + 1)(ex + 1)}
y = `log{("e"^x + 1)/(y + 1)} + "k"`
उत्तर
The general solution of the differential equation (ex + 1) ydy = (y + 1) exdx is y = log {k(y + 1)(ex + 1)}.
Explanation:
The given differential equation is (ex + 1) ydy = (y + 1) exdx
⇒ `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)|`
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