$y''=2\arctan(y)+\frac{3}{2}y'-\frac{1}{2}\pi$ As part of a more difficult question, I need to solve this IVP:
$y''=2\arctan(y)+\frac{3}{2}y'-\frac{1}{2}\pi$
Where $y(0) = 1, y'(0) = 0$
I don't know how to solve this problem, I resorted to guessing and trial and error but not a clue.
 A: Let us assume that the class of functions in which we look for a soution is that of analytic functions in the vicinity of $x=0$. This assumption of analyticity is normally tacitly made in ODE problems.
Then consider the expansion of $y(x)$ in powers of $x$ about $x=0$.
From the initial conditions we have
$$y(0) = 1, y'(0) = 0$$
Now from the ODE 
$$y''(x) = 2 \text{arctan}(y(x)) +\frac{3}{2} y'(x) - \frac{\pi}{2}\tag{1}$$
we have at $x=0$
$$y''(0) = 2 \text{arctan}(y(0)) - \frac{\pi}{2} = \frac{\pi}{2} - \frac{\pi}{2} = 0$$
The third derivative is obtained by differentiating the ODE and is given by
$$y'''(x) = \frac{3 y''(x)}{2}+\frac{2 y'(x)}{y(x)^2+1}$$
and this vanishes at $x=0$ as well. 
This can be extended to all higher derivatives which all vanish (the proof is not difficult).
Hence the solution is $y(x) = 1$, as given previously by Fred.
Observation
The solution with the specific initial condition is very Special in that it is constant. Any other Initial condition gives rise to growing or oscillating solutions.
Consider the situation at large values of $y(x)$:
For $|y(x)| >> 1$ we have  $\text{arctan}(y(x)) \to \frac{\pi}{2}$ and the ODE simplifies approximateley to
$$y''(x) = \frac{3}{2} y(x)' + \frac{\pi}{2}\tag{2}$$
and the general solution is
$$y(x) = - x\frac{\pi}{3} + a \frac{2}{3} e^{x^{\frac{3}{2}}} + b\tag{3}$$
Where $a$ and $b$ are two constants.
Notice that for $x\gt 0$ the solution diverges over exponentially.
For negative $x$ we let $-x \to z>0$ and since $y''(z) = y''(x)$ and $y'(z) = - y'(x)$ (2) transforms to
$$y''(z) = -\frac{3}{2} y(z)' + \frac{\pi}{2}\tag{2a}$$
and the solution becomes
$$y(z) = z\frac{\pi}{3} + a \frac{2}{3} e^{-z^{\frac{3}{2}}} + b\tag{3a}$$
A: The IVP  has  a unique solution and the constant function $1$ is  a solution. .......
