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Form the differential equation having y = (sin–1x)2 + Acos–1x + B, where A and B are arbitrary constants, as its general solution. - Mathematics

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Question

Form the differential equation having y = (sin–1x)2 + Acos–1x + B, where A and B are arbitrary constants, as its general solution.

Sum
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Solution

Given equation is y = (sin–1x)2 + Acos–1x + B

`"dy"/"dx" = 2 sin^-1x * 1/sqrt(1 - x^2) + "A" * ((-1)/sqrt(1 - x^2))`

Multiplying both sides by `sqrt(1 - x^2)`, we get

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

Again differentiating w.r.t x, we get

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

⇒ `sqrt(1 - x^2) ("d"^2y)/("d"x^2) - x/sqrt(1 - x^2) "dy"/"dx" * 2/sqrt(1 - x^2)`

Multiplying both sides by `sqrt(1 - x^2)`, we get

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

Which is the required differential equation.

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Chapter 9: Differential Equations - Exercise [Page 194]

APPEARS IN

NCERT Exemplar Mathematics [English] Class 12
Chapter 9 Differential Equations
Exercise | Q 13 | Page 194

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