<s>[INST] Help on differential equation $y''-2\sin y'+3y=\cos x$ $y''-2\sin y'+3y=\cos x$
I'm trying to solve it by power series, but I just can't find the way to get $\sin y'$. Is there any special way to find it?
 [/INST] Maple finds the two fundamental solutions: 
(with $y(0)=0$ and $y'(0)=b$)
$$\eqalign{ y \left( x \right) &=bx+ \left( \sin \left( b \right) +{\frac {1}{2}}
 \right) {x}^{2}+ \left( \frac23\,\cos \left( b \right) \sin \left( b
 \right) +\frac13\,\cos \left( b \right) -\frac12\,b \right) {x}^{3}\cr+& \left( \frac2
3\, \left( \cos \left( b \right)  \right) ^{2}\sin \left( b \right) -
\frac23\,\sin \left( b \right) -{\frac {1}{2}}-\frac14\,\cos \left( b \right) 
b+\frac12\, \left( \cos \left( b \right)  \right) ^{2} \right) {x}^{4}\cr+&
 \left( \frac45\, \left( \cos \left( b \right)  \right) ^{3}\sin \left( b
 \right) -{\frac {11}{10}}\,\cos \left( b \right) \sin \left( b
 \right) +{\frac {3}{20}}\,\sin \left( b \right) b-{\frac {13}{15}}\,
\cos \left( b \right)\right. \cr& \left.+\frac38\,b-\frac25\, \left( \cos \left( b \right) 
 \right) ^{2}b+\frac45\, \left( \cos \left( b \right)  \right) ^{3}
 \right) {x}^{5}\cr +&O \left( {x}^{6} \right) \cr}
$$
and (with $y(0)=a$ and $y'(0)=0$)
$$y \left( x \right) =a+ \left( -\frac32\,a+{\frac {1}{2}} \right) {x}^{2}+
 \left( -a+{\frac {1}{3}} \right) {x}^{3}-\frac18\,a{x}^{4}+ \left( \frac14\,a
-{\frac {1}{15}}+{\frac {9}{20}}\,{a}^{3}-{\frac {9}{20}}\,{a}^{2}
 \right) {x}^{5}+O \left( {x}^{6} \right) 
$$
Arbitrarily many terms of the series can be found (but they get complicated).
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