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Question
For the following differential equation find the particular solution satisfying the given condition:
`("x" + 1) "dy"/"dx" - 1 = 2"e"^-"y" , "y" = 0`, when x = 1
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Solution
`("x" + 1) "dy"/"dx" - 1 = 2"e"^-"y"`
∴ `("x" + 1) "dy"/"dx" = 2/"e"^"y" + 1 = (2 + "e"^"y")/"e"^"y"`
∴ `"e"^"y"/(2 + "e"^"y") "dy" = 1/("x" + 1)`dx
Integrating both sides, we get
`int "e"^"y"/(2 + "e"^"y") "dy" = int 1/("x" + 1)`dx
∴ log |2 + ey| = log |x + 1| + log c ......`[∵ "d"/"dy" (2 + "e"^"y") = "e"^"y" and int("f"'("y"))/("f"("y")) "dy" = log |"f"("y")| + "c"]`
∴ log |2 + ey| = log |c (x + 1)|
∴ 2 + ey = c(x + 1)
This is the general solution.
Now, y = 0, when x = 1
∴ 2 + e0 = c (1 + 1)
∴ 3 = 2c
∴ c = `3/2`
∴ the particular solution is `2 + "e"^"y" = 3/2("x" + 1)`
∴ 2(2 + ey) = 3(x + 1)
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