$y'' + (\sin x)y' + (\cos x)y = 0$; how to find solution using series analysis? I was reading about series solutions where I saw this question, and I am not sure how can I solve this question using series analysis.
 A: With a limited number of terms, it is not difficult. Use for example
$$\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}-\frac{x^7}{5040}+\frac{x^9}{362880}+O\left(x^{11}\right)$$
$$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+\frac{x^8}{40320}-\frac{x^{10}}{3628800}+O\left(x^{12}\right)$$
$$y=\sum_{i=0}^{10} a_i\,x^i$$
Replace and group terms of the same power. The beginning will be
$$0=(a_0+2 a_2)+(2 a_1+6 a_3) x+\left(-\frac{a_0}{2}+3 a_2+12 a_4\right)
   x^2+\left(-\frac{2 a_1}{3}+4 a_3+20 a_5\right) x^3+$$ $$\frac{1}{24} (a_0-20
   a_2+120 a_4+720 a_6) x^4+\left(\frac{a_1}{20}-a_3+6 a_5\right)
   x^5+$$ $$\frac{1}{720} (-a_0+42 a_2-840 a_4+5040 a_6) x^6+\cdots$$
Since you need two arbirary constants, epress all of them as functions of $a_0$ and $a_1$. Do one at the time and replace. This will give you
$$a_2=-\frac {a_0}2 \qquad a_3=-\frac {a_1}3\qquad a_4=\frac {a_0}6 \qquad a_5=\frac {a_1}{10} \qquad a_6=-\frac {31a_0}{720}$$
$$y=a_0 \left(1-\frac{x^2}{2}+\frac{x^4}{6}-\frac{31 x^6}{720}+\cdots \right)+a_1\left(x-\frac{x^3}{3}+\frac{x^5}{10}+\cdots\right)$$
If you look at the first term, you could notice that, multiplied by $e$, it is the start of the expansion of $e^{\cos(x)}$mentioned in @Physor's answer.
It would be harder to find the general solution for the whole set of coefficients but, with patience and a computer (!), it would be doable.
