Question

If y = sin (m cos^–1x), then show that `(1 - x^2)(d^2y)/(dx^2) - x"dy"/"dx" + m^2y` = 0.

Answer 1

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.