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Homework Help: System of Differential equation to solve.

  1. Apr 14, 2013 #1
    1. The problem statement, all variables and given/known data
    Solve the following differential equation.

    y'= {{2,-1},{3,-2}}y + {{1},{-1}}(e)[itex]^{x}[/itex]

    If it's not clear, I made an image for it.


    2. Relevant equations

    [itex]y{g}[/itex] = [itex]y{h}[/itex] + [itex]y{p}[/itex]

    3. The attempt at a solution

    So basically, I am looking for a homogeneous solution and a particular solution.

    I started by looking at the eigen values and eigen vectors.

    I found [itex]\lambda[/itex]1 = 1 and [itex]\lambda[/itex]2 = -1

    And the vectors that go with them : [itex]\stackrel{\rightarrow}{V1}[/itex] = (1, 1) and [itex]\stackrel{\rightarrow}{V2}[/itex] = (1,3). with give me the homogeneous solution

    [itex]y{h}[/itex] = C1(e)[itex]^{x}[/itex](1, 1) + C2(e)[itex]^{-x}[/itex](1, 3)

    But I have a hard time looking for a particular solution..

    I supposed that Yp was something like [itex]\stackrel{\rightarrow}{a}[/itex] * x * (e)[itex]^{x}[/itex] + [itex]\stackrel{\rightarrow}{b}[/itex] * (e)[itex]^{x}[/itex] and... I'm having a lot of trouble from there.. I found Y'p and. it start to get pretty complicated.. is there an easier way? I'm I doing this right?


  2. jcsd
  3. Apr 14, 2013 #2

    Simon Bridge

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    The built-in equation editor is so clumsy isn't it?
    Here's how to write that out in latex: ##\renewcommand{\dyx}[1]{\frac{d #1}{dx}}## $$\dyx{\vec{y}} = \begin{pmatrix} 2 & -1 \\ 3 & -2 \end{pmatrix} \vec{y} + \begin{pmatrix} 1 \\ -1 \end{pmatrix} e^x = \text{A}\vec{y}+\vec{w}e^x$$ (Ill do this all the way through just to make sure I have properly understood you.)
    You mean $$y_g=y_h+y_p$$ ??
    ##y_h## would be commonly referred to as the "complimentary solution".
    You got eigenvalues ##\lambda_1=1## and ##\lambda_2=-1## with corresponding eigenvectors $$\vec{v}_1= \begin{pmatrix} 1 \\ 1 \end{pmatrix} \; ;\; \vec{v}_2= \begin{pmatrix} 1 \\ 3 \end{pmatrix} \; \Rightarrow \vec{y}_h=C_1 \vec{v}_1 e^x + C_2 \vec{v}_2 e^{-x}$$

    method of undertermined coefficients?
    maybe: ##\vec{y}_p = \vec{a}e^x + \vec{b}xe^{x}## ?
    then $$\vec{y}^\prime_p = (\vec{a}+\vec{b})e^x + \vec{b}xe^{x} = \text{A}(\vec{a}e^x + \vec{b}xe^{x}) + \vec{w}e^x$$ ... so that $$\vec{a}+\vec{b}=\text{A}\vec{a}+\vec{w}\\ \vec{b} = \text{A}\vec{b}$$... something like that?
    Where do you get stuck?

    Last edited: Apr 14, 2013
  4. Apr 15, 2013 #3
    Haha sorry for my bad English, I'm still learning all the terms since I'm a french student.

    The method i'm doing is pretty much the same thing, But I'm just maybe confused on how it actually works..

    I don't get how a + b = Aa + w and b = Ab.. But i'm going to do the problem again and maybe i'll wake up haha
  5. Apr 15, 2013 #4


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    That would be complementary.
  6. Apr 15, 2013 #5
    @DaxInvader, you get these equations by matching up the vector terms containing ##e^x## and ##xe^x## on both sides.
  7. Apr 15, 2013 #6


    Staff: Mentor

    Yeah - they don't give them away for free. :tongue:
  8. Apr 15, 2013 #7

    Simon Bridge

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    Why thank you, that's nice of you to say so...
    (There is no i in complementary huh... it's always about the other person...)

    Your English is better than a lot of native speakers, I don't want you to think I was correcting you when I was checking my understanding ;)
    From grouping like terms as SithsNGiggles points out. It's just the same as you are used to only the coefficients are vectors. I was wondering if that was the trouble.
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