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Identity (x-y)^2

If y=e^(msin-...

Question

If y=e^(msin-¹x) (where -1<=x<=1) then prove that (1-x²)d²y/dx²-xdy/dx=m²y

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y = e^{m {sin}^{- 1} x}, - 1 \leq x \leq 1

(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - a^{2} y = 0

y = e^{m {sin}^{- 1} x}

(1 - x^{2}) \frac{d^{2} y}{d x^{x}} - x \frac{d y}{d x} =

y = e^{m s i n^{- 1} x}

, then show that

(1 - x^{2}) \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} - m^{2} y = 0

y = x e^{- x}

then prove that

x \frac{d y}{d x} = (1 - x) y

y = \sqrt{x + 1} - \sqrt{x - 1}

(x^{2} - 1) \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} - \frac{1}{4} y = 0

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