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In the following example verify that the given function is a solution of the differential equation. xyaexbexxxdydxdydxxxyxy=aex+be-x+x2;xd2ydx2+2dydx+x2=xy+2 - Mathematics and Statistics

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Question

In the following example verify that the given function is a solution of the differential equation.

`"xy" = "ae"^"x" + "be"^-"x" + "x"^2; "x" ("d"^2"y")/"dx"^2 + 2 "dy"/"dx" + "x"^2 = "xy" + 2`

Sum
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Solution

`"xy" = "ae"^"x" + "be"^-"x" + "x"^2`

∴ `"xy" - "x"^2 = "ae"^"x" + "be"^-"x"`     ....(1)

Differentiating both sides w.r.t. x, we get

`"x" "dy"/"dx" + "y" * "d"/"dx" ("x") - "2x" = "ae"^"x" + "be"^-"x" xx (- 1)`

∴ `"x" "dy"/"dx" + "y" - 2"x" = "ae"^"x" - "be"^-"x"`

Differentiating again w.r.t. x, we get

`"x" * "d"/"dx" ("dy"/"dx") + "dy"/"dx" * "d"/"dx" ("x") + "dy"/"dx" - 2 xx 1 = "ae"^"x" - "be"^-"x" (- 1)`

∴ `"x" ("d"^2"y")/"dx"^2 + "dy"/"dx" xx 1 + "dy"/"dx" - 2 = "ae"^"x" + "be"^-"x"`

∴ `"x" ("d"^2"y")/"dx"^2 + 2 "dy"/"dx" - 2 = "xy" - "x"^2`        ....[By (1)]

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

Hence, xy = `"ae"^"x" - "be"^-"x" + "x"^2` is a solution of the D.E.

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

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Formation of Differential Equations
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Chapter 6: Differential Equations - Miscellaneous exercise 2 [Page 217]

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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 2.4 | Page 217

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