How to solve this nonlinear differential equation? The nonlinear differential equation is:
$$y''+ a(\sin(y))y'+ b(\cos(y))=0$$
where $a,b \in \mathbb{R}$ and $b>0$.
How could i solve? Thank you in advance for any idea or proposed solution.
 A: As you can see that this does not contain $x$, you can put $v=y'$ so that:
$$
y''=\frac{dv}{dx}=\frac{dv}{dy}\frac{dy}{dx}=\frac{dv}{dy}v
$$
Hence the equation becomes,
$$
v\frac{dv}{dy}+a\sin(y)v+ b\cos(y)=0
$$
From this first order differential equation u can get $v=f(y)$ and hence you can get $y$ from it.
Hope this helps you......
Refer 2nd point of this link
A: You could assume you knew a power series for $y =\sum_{n=0}^{\infty} a_nx^n$ and take two formal derivatives and plug everything into the DE.   With a certain amount of tedium, you can get expressions for $\cos y$ and $\sin y$ and then $(\cos y)y^{\prime}.$  Keep grinding away and you can write out the first few terms of the power series for the left side.  The power series for the right side has zeros for all coefficients, so if you set the left-side coefficients equal to zero, you can successively solve for each $a_n$ in terms of $a_0$ and $a_1$.  You end up with $y= a_0y_1+a_1y_2$ where $y_1$ and $y_2$ are power series, with as many terms as you care to grind out.  
