Analytical approximation of the ODE: $\alpha y y'' = - y+1$ I would like to achieve an approximate solution to the following 2nd-order non-linear ODE of y(t):
$\alpha y y'' = - y +1$
Where $\alpha>0$ is a constant, and the initial conditions are as follows:
$y(0) = \beta>0$
$y'(0) = 0$
Solving it numerically with MATLAB yields a solution which reminds a cosinusoidal wave with an increasing amplitude. That is except for the case $\beta=1$, then the solution is simply $y(t) = 1$. From the numerical solution, it seems that the approximation will have the form:
$y = 1 + A cos(\omega t)$
Where the amplitude is $A = A(\beta,t)$ and $\omega = \omega(\beta)$.
Any help would be appreciated!
 A: multiply all the sides by $ y'(t)$ and you get
$ \alpha (y' (t))^{2}=-2y(t)+ 2 \log y(t) +C$
here C is a constant of integration
A: This is a follow up of my comment. 
First thing that one may notice is the fact that $\alpha$ plays an stretching role on the temporal coordinate. If one defines the rescaled time $\tau = t/\sqrt{\alpha}$, the problem reduces to
$$ y y'' = 1-y, \quad y(0) = \beta > 0, \quad y'(0) = 0 $$ 
As you mentioned, for $\beta = 1$, $y = 1$ is a solution to the problem. Now the primes denote differentiation with respect to $\tau$. If one is interested in values of $\beta$ close to $1$ (a similar analysis can be carried out for $\beta$ close to $0$, for instance), that is for $\beta = 1 + \epsilon$ with $\epsilon$ small, a solution of the form  
$$ y = 1 + \epsilon y_1 + \epsilon^2 y_2 + \ldots $$
may be sought. Note that $y_i'(0)=0$ and $y_1(0) = 1, \ y_{i>1}(0) = 0$ are the corresponding initial conditions for each new variable. By introducing this into the original equation and collecting powers of $\epsilon$ one arrives at 
\begin{align}
\epsilon^0: \quad & 1 = 1 \\
\epsilon^1: \quad & y_1'' + y_1 = 0 \\
\epsilon^2: \quad & y_2'' + y_2 = -y_1 y_1'' \\
\vdots
\end{align} The solution to these problems (with the corresponding initial conditions) is 
$$y_1(\tau) = \cos \tau , \quad y_2 = \frac{2}{3} ( 2 + \cos\tau ) \sin^2(\tau/2)$$
for which a first order correction of your solution is
$$ y = 1 + (\beta - 1) \cos \tau + \mathcal{O}[(\beta - 1)^2]$$
and a second order approximation is
$$ y = 1 + (\beta - 1) \cos \tau + \frac{2}{3}(\beta - 1)^2 ( 2 + \cos\tau ) \sin^2(\tau/2) +  \mathcal{O}[(\beta - 1)^3] $$
where I have put $\epsilon = \beta - 1$. Note that the first order correction to the solution works reasonably well even for values of $\epsilon$ up to $0.3$. Here's a plot (blue is the numerical solution and should be $\tau$ in the abscissa label)


I edited my comment because I missed a $y_1$ on the rhs of the equation for $y_2$, which completely changes the story...
Hope this helps!
