Use WKB method to find the approximate solution of $\epsilon y''+2y'+2y=0$, for $0<x<1$ where y(0)=0 and y(1)=1. My attempt: Using WKB method, the asymptotic expansion of a solution is $y \sim e^{\theta(x)/\epsilon^\alpha}[y_0(x)+\epsilon^\alpha y_1(x)+ ... ...]$. Then substituting this to given equation we get $ \epsilon e^{\theta(x)/\epsilon^\alpha}[\epsilon^{-2\alpha}\theta^2_x y_0+ \epsilon^{-\alpha} (\theta_{xx} y_0+ 2 \theta y_0'+ \theta_x^2 y_1)+... ...]+ 2 e^{\theta(x)/\epsilon^\alpha} (\epsilon^{-\alpha} \theta' y_0 +y_0' + \theta' y_1+... ...)+ 2 e^{\theta(x)/\epsilon^\alpha}[y_0(x)+\epsilon^\alpha y_1(x)+ ... ...]=0$.
Then Balancing the terms we get $\alpha=1$. That is the equation becomes $ e^{\theta(x)/\epsilon^\alpha}[\epsilon^{-1}\theta^2_x y_0+ \theta_{xx} y_0+ 2 \theta y_0'+ \theta_x^2 y_1+... ...+2\epsilon^{-1} \theta' y_0 +2 y_0' +2 \theta' y_1+... ...+2y_0(x)+2\epsilon y_1(x)+ ... ... ]=0$. Then the $O(\epsilon^{-1}): \theta_x^2+ 2 \theta'=0$ ...(1).
$O(1): \theta_{xx} y_0+ 2 \theta_x y_0'+ \theta_x^2 y_1+2 y_0' +2 \theta' y_1+2y_0(x)=0 $
i.e. $O(1): \theta_{xx} y_0+ 2 \theta_x y_0'+2 y_0'+2y_0(x)=0 $. I am struck here. Please help me to find the solution.
You should consider the $O(\frac{1}{\epsilon})$ equation instead of the $O(1)$ equation as we must consider the largest terms first. Since $\alpha = 1$ then the largest terms are those proportional to $\epsilon^{-1}$.
$$O(\epsilon^{-1}): \left (\theta'^2 + 2\theta' \right )y_0 = 0$$ and since we want nontrivial $y_0$ then we consider $\theta'^2 + 2\theta' = 0$ and continue from this.