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In the following example verify that the given expression is a solution of the corresponding differential equation: y = xcxdydxxdydx(sin-1x)2+c;(1-x2)d2ydx2-xdydx=2 - Mathematics and Statistics

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प्रश्न

In the following example verify that the given expression is a solution of the corresponding differential equation:

y = `(sin^-1 "x")^2 + "c"; (1 - "x"^2) ("d"^2"y")/"dx"^2 - "x" "dy"/"dx" = 2`

बेरीज

उत्तर

y = `(sin^-1 "x")^2 + "c"`        .....(1)

Differentiating w.r.t. x, we get

`"dy"/"dx" = "d"/"dx" (sin^-1 "x")^2 + 0`

∴ `"dy"/"dx" = 2(sin^-1 "x") * "d"/"dx" (sin^-1 "x")`

`= 2 sin^-1 "x" xx 1/sqrt(1 - "x"^2)`

∴ `sqrt(1 - "x"^2)  "dy"/"dx" = 2 sin^-1 "x"`

∴ `(1 - "x"^2) ("dy"/"dx")^2 = 4(sin^-1 "x")^2`

∴ `(1 - "x"^2) ("dy"/"dx")^2 = 4("y - c")`    ....[By (1)]

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

`(1 - "x"^2) * "d"/"dx" ("dy"/"dx")^2 + ("dy"/"dx")^2 * "d"/"dx" (1 - "x"^2) = 4 "d"/"dx" ("y - c")`

∴ `(1 - "x"^2) * 2 "dy"/"dx" * ("d"^2"y")/"dx"^2 - 2"x" ("dy"/"dx")^2 = 4 ("dy"/"dx" - 0)`

Cancelling `2 "dy"/"dx"` throughout, we get

`(1 - "x"^2) ("d"^2"y")/"dx"^2 - "x" "dy"/"dx" = 2`

Hence, y = (sin-1 x)2 + c is a solution of the D.E.

`(1 - "x"^2) ("d"^2"y")/"dx"^2 - "x" "dy"/"dx" = 2`

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Formation of Differential Equations
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पाठ 6: Differential Equations - Exercise 6.3 [पृष्ठ २००]

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बालभारती Mathematics and Statistics 2 (Arts and Science) [English] 12 Standard HSC Maharashtra State Board
पाठ 6 Differential Equations
Exercise 6.3 | Q 1.2 | पृष्ठ २००

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