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