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    Working with Differential Equations

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    Hi,

    For this problem please state the method you used and show the work required to obtain the answer.

    Find the complete solution to this equation problem.

    y^3 - 2y^2 + y^1 = 2- 24e^(x) + 40e^(5x)

    © BrainMass Inc. brainmass.com December 24, 2021, 4:47 pm ad1c9bdddf
    https://brainmass.com/math/calculus-and-analysis/working-with-differential-equations-7027

    SOLUTION This solution is FREE courtesy of BrainMass!

    <br>y^3 - 2y^2 + y^1 = 2- 24e^(x) + 40e^(5x) ....(1)
    <br>
    <br>
    <br>First set the right side equal to 0 and find the solution as in the
    <br>case of a homogeneous equation (this is called complementary function
    <br>y1) and then by trial find a particular solution (particular integral
    <br>y2)
    <br>
    <br>Now the general solution is given by,
    <br>
    <br> Y = y1 + y2
    <br>
    <br>First evaluate y1
    <br>................
    <br>
    <br>y^3 - 2y^2 + y^1 = 0
    <br>
    <br>ie, y (y^2 - 2y +1) = 0 This gives y = 0, +1, +1
    <br>
    <br>Now our y1 is given by
    <br>
    <br>y1 = C1 e^(0 x) + (C2 x+C3) e^(1 x)
    <br>
    <br>= C1 + (C2 x + C3) e^x ........(2)
    <br>
    <br>
    <br>Now Find y2
    <br>...........
    <br>
    <br>We will use the method of undetermined coefficients to find y2,
    <br>
    <br>Let y2 = Ko + C1 e^x + C2 e^5x ......(3)
    <br>
    <br>(If, for example, a term 4x were present in the original equation, we
    <br>will have a K1 x term added with this and so on for higher powers of x)
    <br>
    <br>We will determine Ko, C1 and C2 in this by the above said method
    <br>For this find the derivatives y2^3, y2^2, y2^1 and substitute them in
    <br>our original equation.
    <br>
    <br>y2^1 = 0 + C1 e^x + 5 C2 e^5x
    <br>
    <br>y2^2 = C1 e^x + 25 C2 e^5x
    <br>
    <br>y2^3 = C1 e^x + 125 C2 e^5x
    <br>
    <br>Substitute back in (1)
    <br>
    <br>(C1 e^x + 125 C2 e^5x) -2(C1 e^x + 25 C2 e^5x) + C1 e^x + 5 C2 e^5x
    <br>
    <br> = 2- 24e^(x) + 40e^(5x)
    <br>
    <br>Simplify the left side and equate the coefficients of like terms and
    <br>find C1, K0 and C2
    <br>
    <br>We get, Ko = 0, C1 = 0, C2 = 40/80 = 1/2
    <br>
    <br>Give these values to (3) to get our original y2
    <br>
    <br>y2 = 0 + 0 + (1/2) e^(5x)
    <br>
    <br>Now the general solution of the given differential equation is,
    <br>
    <br>Y = y1 + y2
    <br>
    <br> = C1 + (C2 x + C3) e^x + (1/2) e^(5x) Answer
    <br>
    <br>
    <br>
    <br>......................................................................

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 4:47 pm ad1c9bdddf>
    https://brainmass.com/math/calculus-and-analysis/working-with-differential-equations-7027

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