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Math Help - System of first order differential equations

  1. #1
    Junior Member
    Joined
    Dec 2008
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    System of first order differential equations

    I'm having trouble solving this system of de's using eigenvalues and eigenvectors,

    <br />
\frac{dx}{dt} = x - 4y<br />
    <br />
\frac{dy}{dt} = x + y<br />
    where the initial conditions are; x(0)=3 and y(0)=-4

    I went through to calculate the eigenvalues and vectors to be;

    <br />
\lambda_1 = 1 + 2i<br />
    <br />
\lambda_2 = 1-2i<br />

    and v1 = [1 , -i/2]
    hence v2 = [1 , i/2]

    Using this the general solution should be ?

    x(t) = C1 . [1 , -i/2] e^{(1+2i)t} + C2 . [1 , i/2] e^{(1-2i)t}

    where C1 & c2 are complex constants.

    have i done this right so far?
    To get the actual solution with the initial conditions i'm a bit confused can anyone help me here?

    Many thanks.
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  2. #2
    Super Member Deadstar's Avatar
    Joined
    Oct 2007
    Posts
    722
    Quote Originally Posted by monster View Post
    I'm having trouble solving this system of de's using eigenvalues and eigenvectors,

    <br />
\frac{dx}{dt} = x - 4y<br />
    <br />
\frac{dy}{dt} = x + y<br />
    where the initial conditions are; x(0)=3 and y(0)=-4

    I went through to calculate the eigenvalues and vectors to be;

    <br />
\lambda_1 = 1 + 2i<br />
    <br />
\lambda_2 = 1-2i<br />

    and v1 = [1 , -i/2]
    hence v2 = [1 , i/2]

    Using this the general solution should be ?

    x(t) = C1 . [1 , -i/2] e^{(1+2i)t} + C2 . [1 , i/2] e^{(1-2i)t}

    where C1 & c2 are complex constants.

    have i done this right so far?
    To get the actual solution with the initial conditions i'm a bit confused can anyone help me here?

    Many thanks.
    A quick calc showed me your eigenvals are correct so I'll assume the vectors are as well.

    To finish this let's just take the x(0) = 3 condition first...

    This says that for this equation below.

    x(t) = C1 . [1 , -i/2] e^{(1+2i)t} + C2 . [1 , i/2] e^{(1-2i)t}

    When you set t=0, the 'x parts' of the LHS side should equal 3.

    By 'x parts' I mean the following...

    Consider that each eigenvector is of the form [ 'x', 'y'].

    So you only use the 'x' parts for equating x(0) = 3.

    For this example it would be...

    3 = x(0) = C_1 \cdot 1\cdot e^{(1+2i)\cdot 0} + C_2 \cdot 1 \cdot  e^{(1-2i)\cdot 0}

    I.e. 3 = C_1 + C_2

    Now do the same for y(0) = -4 and you should have 2 equations with 2 unknows ( C_1 and C_2) and you should be able to solve it from there!
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