Differential equation, perturbation method Consider the one dimensional ODE $\frac{dy}{dx}=\epsilon y^2 +x$, where $y=y(x,\epsilon)$.
Discuss the effect of changes in the values of parameter $\epsilon$ in the solution
(hint: assume initially $\epsilon_{0} =0$ and initial condition $y(0,0)=0$, then see how different $y(x,0)$ is from the $y(x,\epsilon)$.  
I first tried to solve the ODE, but I couldn't. How can I apply the perturbation method here?
 A: We start by expanding $y$ in powers of $\epsilon$, i.e.
$$ y(x;\epsilon) = y_{0}(x) + \epsilon y_{1}(x) + \epsilon^{2}y_{2}(x) + \ldots $$
Substitute this expansion into our differential equation:
$$ y_{0}' + \epsilon y_{1}' + \ldots = \epsilon\left(y_{0} + \epsilon y_{1} + \ldots \right)^{2} + x .$$
Now collect terms in increasing powers of epsilon and solve the resulting equations. Start with the $\epsilon^{0}$ equation.
$$\epsilon^{0}: y_{0}' = x$$
$$\implies y_{0} = \frac{1}{2}x^{2} + C_{0}$$
Apply the initial condition:
$$y(0) = 0 \implies C_{0} = 0.$$
Now solve the $\epsilon^{1}$ equation.
$$\epsilon^{1}:  y_{1}' = (y_{0})^{2}$$
$$y_{1}' = \frac{1}{4}x^{4}$$
$$\implies y_{1} = \frac{1}{20}x^{5} + C_{1}$$
Again, we apply the initial condition to see that $C_{1} = 0$.
Thus
$$y(x;\epsilon) = \frac{1}{2}x^{2} + \frac{\epsilon}{20}x^{5} + \text{higher order terms}.$$
(If you want to find the higher order terms you will have to substitute in and solve the $\epsilon^{2}$ equation, then the $\epsilon^{3}$ equation and so on, in a similar fashion as above.)
