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30 June, 03:18

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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  1. 30 June, 03:43
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    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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