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Find the particular solution of the following differential equation: x + 2ydydxy(x + 2y2)dydx=y, when x = 2, y = 1

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Question

Find the particular solution of the following differential equation:

`("x + 2y"^2) "dy"/"dx" = "y",` when x = 2, y = 1

Sum
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Solution

`("x + 2y"^2) "dy"/"dx" = "y"`

∴ `"dy"/"dx" = ("x + 2y"^2)/"y" = "x"/"y" + "2y"`

∴ `"dx"/"dy" - 1/"y" * "x" = "2y"`      ....(1)

This is the linear differential equation of the form

`"dx"/"dy" + "Px" = "Q"` where P = `- 1/"y"` and Q = 2y.

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

`= "e"^(- log "y") = "e"^(log (1/"y")) = 1/"y"`

∴ the solution of (1) is given by

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

∴ `"x" xx 1/"y" = int "2y" xx 1/"y" "dy" + "c"`

∴ `"x"/"y" = 2 int 1 "dy" + "c"`

∴ `"x"/"y" = 2"y" + "c"`

∴ x = 2y2 + cy

This is the general solution.

When x = 2, y = 1, we have

2 = 2(1)2 + c(1)

∴ c = 0

∴the particular solution is x = 2y2.

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Chapter 6: Differential Equations - Miscellaneous exercise 2 [Page 218]

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Balbharati Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
Chapter 6 Differential Equations
Miscellaneous exercise 2 | Q 6.2 | Page 218

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