It is given that $y={{\mathrm{e}}^{{{\sin }^{-1}}x}}$, for $-1 Show that $\left( 1-{{x}^{2}} \right)

Question

It is given that $y={{\mathrm{e}}^{{{\sin }^{-1}}x}}$, for $-1 Show that $\left( 1-{{x}^{2}} \right)\frac{{{\mathrm{d}}^{2}}y}{\mathrm{d}{{x}^{2}}}=x\frac{\mathrm{d}y}{\mathrm{d}x}+y$. [4] Find the first $4$ terms of the Maclaurin expansion of ${{\mathrm{e}}^{{{\sin }^{-1}}x}}$. [5]

Accepted Answer

$(1-x^2)\dfrac{d^2y}{dx^2} = x\dfrac{dy}{dx} + y$; $e^{\sin^{-1}x} = 1 + x + \dfrac{1}{2}x^2 + \dfrac{1}{3}x^3 + \cdots$

N2021 P1 Q7

N2021 P1 Q7

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