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