How can I do a perturbation on this ODE?

Question

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?

Answer 1

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

Answer 2

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 $l < k$.