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For the following differential equation find the particular solution satisfying the given condition: xdydxeyy(x+1)dydx-1=2e-y,y=0, when x = 1 - Mathematics and Statistics

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प्रश्न

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

योग
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उत्तर

`("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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अध्याय 6: Differential Equations - Exercise 6.3 [पृष्ठ २०१]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 6 Differential Equations
Exercise 6.3 | Q 3.5 | पृष्ठ २०१

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