Question

If x = cos t, y = e^mt, show that `(1 - x^2)(d^2y)/(dx^2) - x"dy"/"dx" - m^2y` = 0.

Answer 1

x = cos t, y = e^mt 
∴ t = cos^–1x and y = `e^(mcos^-1x)`           ...(1)

∴ `"dy"/"dx" = "d"/"dx"(e^(mcos^-1x))`

= `e^(mcos^-1x)."d"/"dx"(mcos^-1x)`

= `e^(mcos^-1x) xx m xx (-1)/sqrt(1 - x^2)`

∴ `sqrt(1 - x^2)."dy"/"dx"` = – my            ...[By (1)]

∴ `(1 - x^2)(dy/dx)^2` = m²y2
Differentiating again w.r.t. x, we get

`(1 - x^2)."d"/"dx"(dy/dx)^2 + (dy/dx)^2."d"/"dx"(1 - x^2) = m^2."d"/"dx"(y^2)`

∴ `(1 - x^2).2"dy"/"dx".(d^2y)/(dx^2) + (dy/dx)^2 (0 - 2x) = m^2 xx 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.