(b) $1+x+\frac{1}{2} x^2+\frac{1}{3} x^3+\cdots$

(a) $y=e^{\sin^{-1} x}$ —–(1)
$\frac{d y}{d x}=\frac{1}{\sqrt{1-x^2}} e^{\sin^{-1} x}=\frac{1}{\sqrt{1-x^2}} y$
$\sqrt{1-x^2} \frac{\mathrm{d} y}{\mathrm{~d} x}=y$. —– (2)
$\frac{\mathrm{d}}{\mathrm{d} x}\left(\left(1-x^2\right)^{\frac{1}{2}} \frac{\mathrm{d} y}{\mathrm{dx}}\right)=\frac{\mathrm{d}}{\mathrm{dx}}(y)$
$\left(1-x^2\right)^{\frac{1}{2}} \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}+\frac{\mathrm{d} y}{\mathrm{~d} x}\left(\frac{1}{2}\left(1-x^2\right)^{-\frac{1}{2}}(-2 x)\right)=\frac{\mathrm{d} y}{\mathrm{dx}}$
$\sqrt{1-x^2} \frac{d^2 y}{d x^2}-\frac{x}{\sqrt{1-x^2}} \frac{d y}{d x}=\frac{d y}{d x}$

Multiplying throughout by $\sqrt{1-x^2}$ :
$\left(1-x^2\right) \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}-x \frac{\mathrm{d} y}{\mathrm{~d} x}=\sqrt{1-x^2} \frac{\mathrm{d} y}{\mathrm{~d} x}$
$\left(1-x^2\right) \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}-x \frac{\mathrm{d} y}{\mathrm{~d} x}=y$ (from (2))
$\left(1-x^2\right) \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}=x \frac{\mathrm{d} y}{\mathrm{~d} x}+y \quad-\text { (3) }
$

(b) Differentiating (3) throughout with respect to $x$ :
$\left(1-x^2\right) \frac{\mathrm{d}^3 y}{\mathrm{~d} x^3}+\frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}(-2 x)=x \frac{\mathrm{d}^2 y}{\mathrm{~d} x^2}+\frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{\mathrm{d} y}{\mathrm{~d} x}$
When $x=0$, (1): $y=\mathrm{e}^{\sin ^{-1} 0}=\mathrm{e}^0=1$
(2): $\sqrt{1} \frac{\mathrm{d} y}{\mathrm{~d} x}=1 \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=1$
(3): $\frac{d^2 y}{d x^2}=1$
(4): $\frac{\mathrm{d}^3 y}{\mathrm{~d} x^3}=1+1=2$

By Maclaurin series,
$y =\mathrm{f}(0)+\mathrm{f}^{\prime}(0) x+\frac{\mathrm{f}^{\prime \prime}(0)}{2 !} x^2+\frac{\mathrm{f}^{\prime \prime \prime}(0)}{3 !} x^3+\cdots $
$=1+(1) x+\frac{1}{2} x^2+\frac{2}{6} x^3+\cdots $
$=1+x+\frac{1}{2} x^2+\frac{1}{3} x^3+\cdots$