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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`
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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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