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Differential equations with eigenvalues.

  1. Jul 30, 2017 #1
    1. The problem statement, all variables and given/known data
    Find all solutions of the given differential equations: ## \frac{dx}{dt} =
    \begin{bmatrix}
    6 & -3 \\
    2 & 1
    \end{bmatrix} x ##

    2. Relevant equations


    3. The attempt at a solution
    So, we just take the determinate of A-I##\lambda## and set it equal to 0 to get the eigenvalues of 3, and 4. Then, we plug our eigenvalues one at a time into A-I##\lambda## and solve it to equal the 0 vector.

    For 3, I get ##
    \begin{bmatrix}
    3 & -3 \\
    2 & -2
    \end{bmatrix} ## Which implies that ##x_1 = x_2## so I can just pick
    ##\begin{bmatrix}
    1 \\
    1
    \end{bmatrix} ## to be my vector? This is the part im confused about, I'm not sure that's what I actually do! Do I just say for ##\lambda## = 3, my solution is ## e^{3t} \begin{bmatrix}
    1 \\
    1
    \end{bmatrix} ##

    Similarly for ##\lambda = 4 ## i find my matrix to be
    ##
    \begin{bmatrix}
    2 & -3 \\
    2 & -3
    \end{bmatrix} ## which i see that ##2x_1=3x_2## this, i can just pick
    ##\begin{bmatrix}
    2 \\
    3
    \end{bmatrix} ## Thus, my solution for ##\lambda = 4## is just ##e^{4t}
    \begin{bmatrix}
    2 \\
    3
    \end{bmatrix} ##

    Thanks for all the help, I'm just unsure this is the method I should be going about to solve these types of problems! (Also, if anyone knows of an algebraic calculator in which I could check my solutions for these types of problems, I would also appreciate that)
     
  2. jcsd
  3. Jul 30, 2017 #2

    vela

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    Yup, that's right.

    You got the 2 and 3 backwards.

    So now that you have to two solutions, what's the general solution?
     
  4. Jul 30, 2017 #3

    Ray Vickson

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    Homework Helper

    You are correct.

    However, when in doubt, check to see if your alleged "solutions" actually satisfy the DE---a step you should always do, anyway, as part of good working practice.
     
  5. Jul 30, 2017 #4
    What you want to do is merge those two eigenvectors into a matrix, and left-multiply it by the solution to the diagonal system of differential equations to get the general solution for your original matrix.
     
  6. Jul 30, 2017 #5
    Should just be the two of the added together? So, ## x(t) = C_1 e^{3t} \begin{bmatrix}
    1 \\
    1
    \end{bmatrix} + C_2 e^{4t}
    \begin{bmatrix}
    3 \\
    2
    \end{bmatrix} ## where C is a constant.

    Do I check them vs the original equations, or just the new "eigenedited" matrices? I.E for ## \lambda = 4## I'd use the matrix ##
    \begin{bmatrix}
    2 & -3 \\
    2 & -3
    \end{bmatrix} ## and multiple it by
    ##\begin{bmatrix}
    3 \\
    2
    \end{bmatrix} ## to get
    ##\begin{bmatrix}
    6+-6 \\
    6+-6
    \end{bmatrix} ## =
    ##\begin{bmatrix}
    0 \\
    0
    \end{bmatrix} ## Thus it works.
     
  7. Jul 30, 2017 #6

    vela

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    Staff Emeritus
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    Yes, that's correct.


    You want to see if your solution ##x(t)## satisfies the differential equation, so plug it into the differential equation.
     
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