<s>[INST] Finding Taylor's series of the function: $\frac{e^{a \sin^{-1}x}}{\sqrt{1-x^2}}$ Show that $$\frac{e^{a \sin^{-1}x}}{\sqrt{1-x^2}}=1+\frac{ax}{1!}+\frac{(a^2+1^2)x^2}{2!}+\frac{a(a^2+2^2)x^3}{3!}+\frac{(a^2+1^2)(a^2+3^2)x^4}{4!}+\cdots$$
My attempt:
I integrated the function and got $\frac{e^{a \sin^{-1}x}}{a}$ then I wrote the series of $e^{a \sin^{-1}x}$ but it contained terms like $(\sin^{-1}x)^2$, $(\sin^{-1}x)^3$ and so on so I could not find the series. My idea was to find the series of the anti derivative of the function and then to derivate the obtained series. Any other way to do it?
 [/INST] Let
$$
f(x) = e^{a\arcsin x}  = \sum\limits_{n = 0}^\infty  {f_n x^n } .
$$
As you already observed, $f(x)$ satisfies the non-linear ODE $(1 - x^2 )(f'(x))^2  = a^2f^2 (x)$. Differentiating this equation and dividing through by $2f'(x)$ yields
$$
(1 - x^2 )f''(x) - xf'(x) - a^2 f(x) = 0.
$$
Substituting the power series into this equation gives $f_0  = 1$, $f_1  = a$ (you can see from the definition that $f(0) = 1$, $f'(0) = a$) and
$$
f_{n + 2}  = \frac{{a^2+n^2}}{{(n + 1)(n + 2)}}f_n 
$$
for $n\geq 0$. The power series expansion you are asking for then follows since
$$
\frac{{e^{a\arcsin x} }}{{\sqrt {1 - x^2 } }} = \frac{1}{a}f'(x) = \sum\limits_{n = 0}^\infty  {\frac{{(n + 1)f_{n + 1} }}{a}x^n } .
$$
</s>