Consider the differential equation y′′+αy′+βy=t+e6t. Suppose the form of the particular solution to this differential equation as prescribed

Question

Consider the differential equation y′′+αy′+βy=t+e6t. Suppose the form of the particular solution to this differential equation as prescribed by the method of undetermined coefficients is yp(t)=A1t2+A0t+B0te6t.

Determine the constants α and β.

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Maya 2 weeks 2021-11-18T10:57:37+00:00 1 Answer 0 views 0

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    2021-11-18T10:59:29+00:00

    The question is:

    Consider the differential equation

    y′′ + αy′ + βy = t + e^(6t).

    Suppose the form of the particular solution to this differential equation as prescribed by the method of undetermined coefficients is

    yp(t) = (A_1)t²+ (A_0)t + (B_0)te^(6t).

    Determine the constants α and β.

    Answer:

    α = 0

    β = -36

    Step-by-step explanation:

    Given the differential equation

    y” + αy’ + βy = t + e^(6t).

    We want to determine the constants α and β bt the method of undetermined coefficients.

    First, we differentiate

    yp(t) = (A_1)t²+ (A_0)t + (B_0)te^(6t)

    twice in succession, to obtain y’p(t) and y”p(t).

    y’p(t) = 2(A_1)t + (A_0) + 6(B_0)te^(6t) + (B_0)e^(6t).

    y”p(t) = 2(A_1) + 6(B_0)e^(6t) + 36(B_0)te^(6t) + 6(B_0)e^(6t)

    = 2(A_1) + 12(B_0)e^(6t) + 36(B_0)te^(6t)

    Substitute the values of y_p, y’_p, and y”_p into

    y”_p + αy’_p + βy_p = t + e^(6t).

    [2(A_1) + 12(B_0)e^(6t) + 36(B_0)te^(6t)] + α[2(A_1)t + (A_0) + 6(B_0)te^(6t) + (B_0)e^(6t)] + β[(A_1)t² + (A_0)t + (B_0)te^(6t)] = t + e^(6t)

    Collect like terms

    [2(A_1) + α(A_0)] + [12(B_0) + (B_0)]e^(6t) + [2α(A_1) + (A_0)]t + [36(B_0) + 6α(B_0) + β(B_0)] te^(6t) + β(A_1)t² = t + e^(6t)

    Compare coefficients, equating the coefficients of constants to constants, t to t, t² to t², and e^(6t) to e^(6t).

    We have the following equations.

    β(A_1) = 0…………………………………….(1)

    2α(A_1) + (A_0) = 1………………………(2)

    [36+ 6α + β](B_0) = 0………………….(3)

    13(B_0) = 1……………………………………(4)

    2(A_1) + α(A_0) = 0 …………………….(5)

    From (1): A_1 = 0

    Using this in (2):

    2α(0) + (A_0) = 1

    A_0 = 1

    Using these in (5):

    2(0) + α(1) = 0

    α = 0

    From (4), 13(B_0) = 1

    => B_0 = 1/13

    Using this in (3)

    36+ 6α + β = 0

    6α + β = -36

    But α = 0

    So, β = -36

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