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Question
If y = sin (m cos–1x), then show that `(1 - x^2)(d^2y)/(dx^2) - x"dy"/"dx" + m^2y` = 0.
Solution
y = sin (m cos–1x)
∴ sin–1 y = m cos–1 x
Differentiating both sides w.r.t. x, we get
`(1)/sqrt(1 - y^2)."dy"/"dx" = m xx (-1)/sqrt(1 - x^2)`
∴ `sqrt(1 - x^2)."dy"/"dx" = -msqrt(1 - y^2)`
∴ `(1 - x^2)(dy/dx)^2 = m^2(1 - y^2)`
∴ `(1 - x^2)(dy/dx)^2` = m2 – m2y2
Differentiating both sides w.r.t. x, we get
`(1 - x^2)."d"/"dx"(dy/dx)^2 + (dy/dx)^2."d"/"dx"(1 - x^2) = 0 - m^2."d"/"dx"(y^2)`
∴ `(1 - x^2).2"dy"/"dx".(d^2y)/(dx^2) - 2x(dy/dx)^2 = -2m^2y"dy"/"dx"`
Cancelling `2"dy"/"dx"` throughtout, we get
`(1 - x^2)(d^2y)/(dx^2) - x"dy"/"dx"` = – m2y
∴ `( 1- x^2)(d^2y)/(dx^2) - x"dy"/"dx" + m^2y` = 0.
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