# Perturbation Methods for 1st order ODE - Find the asymptotic solution

1. Jul 16, 2007

### VinnyCee

1. The problem statement, all variables and given/known data

Apply the regular perturbation method to solve the following ordinary differential equation

$$\left(1\,+\,\epsilon\,y\right)\,\frac{dy}{dx}\,+\,y\,=\,0$$

subject to

$$x\,=\,1;\,y\,=\,1$$

Show that the asymptotic solution is of the form

$$y\,=\,e^{1\,-\,x}\,+\,\epsilon\,\left[e^{1\,-\,x}\,-\,e^{2\,(1\,-\,x)}\right]\,+\,\dots$$

2. Relevant equations

Perturbation methods.

3. The attempt at a solution

First, I get the base case by setting epsilon to zero.

$$\frac{dy_0}{dx}\,+\,y_0\,=\,0$$

Am I supposed to use the "subject to" conditions now?

$$\frac{dy_0}{dx}\,=\,-y_0\,=\,-1$$

I have no idea if that is what I am supposed to do, I hope so otherwise I am completely lost here.

Now, I assume that a solution is of the form

$$y\,=\,y_0\,+\,\epsilon\,y_1\,+\,\epsilon^2\,y_2\,+\,\dots$$

Differentiating that approximation

$$\frac{dy}{dx}\,=\,\frac{dy_0}{dx}\,+\,\epsilon\,\frac{dy_1}{dx}\,+\,\epsilon^2\,\frac{dy_2}{dx}\,+\,\dots$$

Expanding the original equation given

$$\frac{dy}{dx}\,+\,\epsilon\,y\,\frac{dy}{dx}\,+\,y\,=\,0$$

and substituting the approximations into it

$$\left[\frac{dy_0}{dx}\,+\,\epsilon\,\frac{dy_1}{dx}\,+\,\epsilon^2\,\frac{dy_2}{dx}\,+\,\dots\right]\,+\,\left[\epsilon\,y_0\,\frac{dy_0}{dx}\,+\,2\,\epsilon^2\,y_0\,\frac{dy_1}{dx}\,+\,\epsilon^3\,y_1\,\frac{dy_1}{dx}\,+\,\dots\right]\,+\,\left[y_0\,+\,\epsilon\,y_1\,+\,\epsilon^2\,y_2\,+\,\dots\right]\,=\,0$$

Please tell me if the above substitution and expansion are correct.

Now I start matching terms according to their order.

O(1): $$\frac{dy_0}{dx}\,=\,-y_0$$

O($\epsilon$): $$\frac{dy_1}{dx}\,=\,-y_0\,\frac{dy_0}{dx}\,-\,y_1$$

O($\epsilon^2$): $$\frac{dy_2}{dx}\,=\,-2\,y_0\,\frac{dy_1}{dx}\,-\,y_2$$

Solving for order one

$$\int\,\frac{dy_0}{dx}\,=\,\int\,-y_0$$

$$y_0\,=\,K_0$$

OR using the base case that I am unsure about

$$y_0\,=\,1$$

But which one is it?