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Math Help - Dif EQ

  1. #1
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    Dif EQ

    So I'm trying to solve this partially decoupled system:

    \frac{dx}{dt} = 2x + 3y

    \frac{dy}{dt} = -4y

    1.) Solve it WITHOUT using matrices. (HINT: start with the second equation)

    2.) Solve it with using matrices.

    I can do #2... but #1 is so hard.

    For #2, I set up the system into a matrix so I get X' = AX, with A being:

    [[2,3],[0,-4]]

    Then, I found the eigenvalues:

    \lambda_1 = 2, \lambda_2 = -4

    And the corresponding eigenvectors (call them k_1, k_2, respectively:

    k_1 = [[1],[0]], k_2 = [[1/2],[1]

    So then the solution is:

    X = c_1e^{2t}k_1 + c_2e^{-4t}k_2

    Voila.

    Now for #1...

    So solving the 2nd equation first...:

    \frac{dy}{-4y} = dt \implies \frac{-ln(|y|)}{4} = t + C

    Solve for y (I ignored the absolute value so I don't get 2 diff soln's):

    y = e^{-4t-4c}

    So now plug this into 1st eq.:

    \frac{dx}{dt} = 2x + 3\cdot e^{-4t-4c}

    No idea how to solve this! Maybe I did something wrong.. no idea.
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  2. #2
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    Quote Originally Posted by caeder012 View Post
    :

    \frac{dx}{dt} = 2x + 3y

    \frac{dy}{dt} = -4y

    So solving the 2nd equation first...:

    \frac{dy}{-4y} = dt \implies \frac{-ln(|y|)}{4} = t + C

    Solve for y (I ignored the absolute value so I don't get 2 diff soln's):

    y = e^{-4t-4c}

    So now plug this into 1st eq.:

    \frac{dx}{dt} = 2x + 3\cdot e^{-4t-4c}

    No idea how to solve this! Maybe I did something wrong.. no idea.
    You can ignore the absolute value bars only so long as y is an always positive function. Since it is (barring an arbitrary constant making it negative) don't worry about it.

    \frac{dx}{dt} = 2x + 3Ae^{-4t}<-- Defining A = e^{-4c} for convenience

    \frac{dx}{dt} - 2x = 3Ae^{-4t}

    Solve the homogeneous equation first:
    \frac{dx_h}{dt} - 2x_h = 0

    The characteristic polynomial for this is m - 2 = 0, so
    x_h = Be^{2t}

    For the particular equation
    \frac{dx_p}{dt} - 2x_p = 3Ae^{-4t}
    we can try a solution
    x_p = De^{-4t}

    Using this gives:
    -4De^{-4t} - 2De^{-4t} = 3Ae^{-4t}

    -6D = 3A

    Thus
    x_p = -\frac{A}{2} e^{-4t}

    So the overall solution is
    x = x_h + x_p = Be^{2t} - \frac{A}{2} e^{-4t}

    which matches your other solution.

    -Dan
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by caeder012 View Post
    So now plug this into 1st eq.:

    \frac{dx}{dt} = 2x + 3\cdot e^{-4t-4c}

    No idea how to solve this! Maybe I did something wrong.. no idea.
    Lets assume your work up to this point is correct. Rewrite what you have as:

    \frac{dx}{dt} -2x = A\cdot e^{-4t}

    This is an inhomogeneous linear constant coefficient ODE, and its general
    solution is the sum of the general solution of the homogeneous equation:

    \frac{dx}{dt} -2x = 0

    and any particular integral of the original equation.

    I will assume you know how to solve the homogeneous equation, and that its
    solution is:

    x(t)=C e^{2t}

    Now to find a particular solution of the inhomegeneous equation you try a trial
    solution of the form x(t)=B~e^{-4t}.

    RonL
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  4. #4
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    Hello, caeder012!

    You're making the problem very messy . . .


    \frac{dx}{dt} = 2x + 3y .[1]
    \frac{dy}{dt} = -4y .[2]

    1) Solve it WITHOUT using matrices. (HINT: start with the second equation)
    From [2], we have: . \frac{dy}{y} \:=\:-4\,dt

    . . Integrate: . \ln y \:=\:-4t + c

    . . Then: . y \:=\:e^{-4t+c} \:=\:e^{-4t}\cdot e^c \quad\Rightarrow\quad\boxed{ y\:=\:C_1e^{-4t}}

    Substitute into [1]: . \frac{dx}{dt} \:=\:2x + 3\left(C_1e^{-4t}\right)

    . . Then we have: . \frac{dx}{dt} - 2x \:=\:3C_1e^{-4t}


    Integrating factor: . I \;=\;e^{\int(-2)\,dt} \;=\;e^{-2t}

    Multiply by I\!:\;\;e^{-2t}\!\cdot\!\frac{dx}{dt} - 2xe^{-2t} \;=\;3C_1e^{-6t}

    . . and we have: . \frac{d}{dt}\left(e^{-2t}\!\cdot\!x\right) \;=\;3C_1e^{-6t}

    Integrate: . e^{-2t}\!\cdot\!x \;=\;3C_1\int e^{-6t}dt \quad\Rightarrow\quad e^{-2t}x \;=\;-\frac{1}{2}C_1e^{-6t} + C_2

    Multiply by e^{2t}\!:\;\;\boxed{x \;=\;-\frac{1}{2}C_1e^{-4t} + C_2e^{2t}}

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