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Obtain the differential equation by eliminating the arbitrary constants from the following equation: y = AexBexAe3x+1+Be-3x+1 - Mathematics and Statistics

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प्रश्न

Obtain the differential equation by eliminating the arbitrary constants from the following equation:

y = `"Ae"^(3"x" + 1) + "Be"^(- 3"x" + 1)`

बेरीज
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उत्तर

y = `"Ae"^(3"x" + 1) + "Be"^(- 3"x" + 1)`    ....(1)

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

`"dy"/"dx" = "Ae"^(3"x" + 1) * "d"/"dx" (3"x" + 1) + "Be"^(- 3"x" + 1) * "d"/"dx" (- 3"x" + 1)`

`= "Ae"^(3"x" + 1) xx (3 + 1 + 0) + "Be"^(- 3"x" + 1) xx (- 3 xx 1 + 0)`

`= 3"Ae"^(3"x" + 1) - 3 "Be"^(- 3"x" + 1)`

and `("d"^2 "y")/"dx"^2 = 3 "Ae"^(3"x" + 1) * "d"/"dx" ("3x" + 1) - 3 "Be"^(- 3"x" + 1) * "d"/"dx" (- 3"x" + 1)`

`= 3"Ae"^(3"x" + 1) xx (3 xx 1 + 0) - 3"Be"^(- 3"x" + 1) xx (- 3 xx 1 + 0)`

`= 9"Ae"^(3"x" + 1) - 9"Be"^(- 3"x" + 1)`

`= 9 ("Ae"^(3"x" + 1) + "Be"^(- 3"x" + 1))`

= 9y       ....[By (1)]

∴ `("d"^2"y")/"dx"^2 - 9"y" = 0`

This is the required D.E.

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पाठ 6: Differential Equations - Miscellaneous exercise 2 [पृष्ठ २१७]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
पाठ 6 Differential Equations
Miscellaneous exercise 2 | Q 3.5 | पृष्ठ २१७

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