How can I do a perturbation on this ODE? I have an ode of the form $$\frac{dy}{dx}=y(x)+\sqrt{y(x)+\epsilon\cdot f(x)}$$
and I would like to do a perturbation up to first order $\epsilon$.
My advisor gave me an example for a simpler ODE, but I don't think it will work for a DE like this.  Any ideas on how I could do it?
 A: Basically, what you need to do is expand $y(x)$ as a formal power series in $\epsilon$:
$$y(x) = y_0(x) + y_1(x) \epsilon + y_2(x) \epsilon^2 + \cdots$$
Substitute this into the ODE and expand everything
$$\frac{d}{dx}\left[y_0 + y_1 \epsilon + y_2\epsilon^2 + \cdots\right] =
y_0 + y_1 \epsilon + y_2 \epsilon^2 + \cdots
+ \sqrt{ y_0 + ( y_1 + f) \epsilon + y_2 \epsilon^2 + \cdots}
$$
You then equate coefficients for each power of $\epsilon$ to obtain a bunch of equations.
$$\begin{align}
\frac{dy_0}{dx} &= y_0 + \sqrt{y_0}\\
\frac{d\color{red}{y_1}}{dx} &= \frac{(2y_0+\sqrt{y_0})\color{red}{y_1}+\sqrt{y_0} f}{2y_0}\\
\frac{d\color{blue}{y_2}}{dx} &= \frac{(8y_0^2+4y_0^{3/2})\color{blue}{y_2}-\sqrt{y_0}y_1^2-2\sqrt{y_0} f y_1 -\sqrt{y_0}f^2}{8y_0^2}\\
&\;\vdots
\end{align}
$$
Something you should notice are


*

*In the $1^{st}$ equation, only $y_0$ appears.

*In the $2^{nd}$ equation, only $y_0$ and $y_1$ appears. Furthermore, $y_1$ appears linearly.

*In the $3^{rd}$ equation, only $y_0, y_1, y_2$ appear and $y_2$ appears linearly.


What this means is aside from the $1^{st}$ equation, the rest are just a bunch of linear ODEs for each $y_k, k > 0$ whose coefficients has been determined by previous equations.
After you solve the $1^{st}$ non-linear ODE to get $y_0$, the solutions for the remaining $y_k$ can be formally written down as suitable integrals in terms of $y_l$ for $l < k$.
A: look please on this link,it is just approximation methods with several assumption,dont be confused because of root  
http://eaton.math.rpi.edu/faculty/Wahle/Courses/6620-S04/3-RegularODEs.pdf
you may need some knowledge of  solution of this kind of Ordinary differential equation
