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प्रश्न
If y = `e^(mtan^-1x)`, show that `(1 + x^2)(d^2y)/(dx^2) + (2x - m)"dy"/"dx"` = 0.
उत्तर
y = `e^(mtan^-1x)` ...(1)
∴ `"dy"/"dx" = "d"/"dx" (e^(mtan^-1x))`
= `e^(mtan^-1x)."d"/"dx"(mtan^-1x)`
= `e^(mtan^-1x) xx m xx (1)/(1 + x^2)`
∴ `(1 + x^2)"dy"/"dx"` = my ...[By (1)]
Differentiaitng again w.r.t. x, we get
`(1 + x^2)."d"/"dx"("dy"/"dx") + "dy"/"dx"."d"/"dx"(1 + x^2) = m"dy"/"dx"`
∴ `(1 + x^2)(d^2y)/(dx^2) + "dy"/"dx"(0 + 2x) = m"dy"/"dx"`
∴ `(1 + x^2)(d^2y)/(dx^2) + 2x."dy"/"dx" = m"dy"/"dx"`.
∴ `(1 + x^2)(d^2y)/(dx^2) + (2x - m)"dy"/"dx"` = 0.
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