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

  1. Jul 16, 2007 #1
    1. The problem statement, all variables and given/known data

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

    [tex]\left(1\,+\,\epsilon\,y\right)\,\frac{dy}{dx}\,+\,y\,=\,0[/tex]

    subject to

    [tex]x\,=\,1;\,y\,=\,1[/tex]

    Show that the asymptotic solution is of the form

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



    2. Relevant equations

    Perturbation methods.



    3. The attempt at a solution

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

    [tex]\frac{dy_0}{dx}\,+\,y_0\,=\,0[/tex]

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

    [tex]\frac{dy_0}{dx}\,=\,-y_0\,=\,-1[/tex]

    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

    [tex]y\,=\,y_0\,+\,\epsilon\,y_1\,+\,\epsilon^2\,y_2\,+\,\dots[/tex]

    Differentiating that approximation

    [tex]\frac{dy}{dx}\,=\,\frac{dy_0}{dx}\,+\,\epsilon\,\frac{dy_1}{dx}\,+\,\epsilon^2\,\frac{dy_2}{dx}\,+\,\dots[/tex]

    Expanding the original equation given

    [tex]\frac{dy}{dx}\,+\,\epsilon\,y\,\frac{dy}{dx}\,+\,y\,=\,0[/tex]

    and substituting the approximations into it

    [tex]\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[/tex]

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


    Now I start matching terms according to their order.

    O(1): [tex]\frac{dy_0}{dx}\,=\,-y_0[/tex]

    O([itex]\epsilon[/itex]): [tex]\frac{dy_1}{dx}\,=\,-y_0\,\frac{dy_0}{dx}\,-\,y_1[/tex]

    O([itex]\epsilon^2[/itex]): [tex]\frac{dy_2}{dx}\,=\,-2\,y_0\,\frac{dy_1}{dx}\,-\,y_2[/tex]

    Solving for order one

    [tex]\int\,\frac{dy_0}{dx}\,=\,\int\,-y_0[/tex]

    [tex]y_0\,=\,K_0[/tex]

    OR using the base case that I am unsure about

    [tex]y_0\,=\,1[/tex]

    But which one is it?

    Thanks in advance for your help!
     
  2. jcsd
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