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Math Help - linear differential equation

  1. #1
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    linear differential equation

    Hello

    im stuck on the following differential equation. I understand since it linear we find the general solution by finding the complementary function and adding it together with a particular integral. Im stuck because in this example the RHS has both a linear function and an exponential and im unsure of how to go about doing the trail function subsitution.

    Find a solution to the following differential equation

    dy/dx + 4y = 4x + e^x

    with y(0)=0


    iv gotten as far as the CF = Ae^-4x
    any help most appreciated
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  2. #2
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    Hello, jordanrs!

    Find a solution to the following differential equation:
    . . \frac{dy}{dx} + 4y \:=\: 4x + e^x\;\;\text{ with }y(0)=0
    You already found the homogeneous solution: . y \:=\:ke^{-4x}

    We can conjecture that the particular solution has the form:
    . . y \;=\;Ax + B + Ce^x

    . . \begin{array}{ccccc}\text{Then: }& \dfrac{dy}{dx} &=& A + Ce^x \\ \\[-4mm] \text{and:} & 4y &=& 4Ax + 4B + 4Ce^x \end{array}

    Add the equations: . \frac{dy}{dx} + 4y \;=\;4Ax + (A+4B) + 5Ce^x

    This is supposed to equal 4x + e^x

    . . So we have: . 4Ax + (A+4B) + 5Ce^x \;=\;4x+ 0+ 1\!\cdot\!e^x

    Equate the coefficients: . \begin{Bmatrix}4A &=& 4 \\ A+4B &=& 0 \\ 5C &=& 1 \end{Bmatrix}

    . . and we get: . A = 1,\;B = \text{-}\tfrac{1}{4},\;C = \tfrac{1}{5}

    Hence, the particular solution is: . y \;=\;x - \tfrac{1}{4} + \tfrac{1}{5}e^x


    So far, the solution is: . y \;=\;ke^{-4x} + x - \tfrac{1}{4} + \tfrac{1}{5}e^x

    Since y(0) = 0, we have: . ke^0 + 0 - \tfrac{1}{4} + \tfrac{1}{5}e^0 \;=\;0 \quad\Rightarrow\quad k - \tfrac{1}{4} + \tfrac{1}{5} \:=\:0 \quad\Rightarrow\quad k \:=\:\tfrac{1}{20}


    Therefore, the solution is: . \boxed{y \;=\;\frac{1}{20}e^{-4x} + x - \frac{1}{4} + \frac{1}{5}e^x}

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  3. #3
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    Thumbs up

    yep that makes alot of sense using the 2 combined trial solutions in one and then differentiating and solveing through

    really appreciate the help

    thanks alot
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