Ask Question
19 May, 03:35

Consider the differential equation

y′′+αy′+βy=t+e^ (4t).

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

yp (t) = A1t^2+A0t+B0te^ (4t).

Determine the constants α and β.

+2
Answers (1)
  1. 19 May, 06:22
    0
    Yp (t) = A1 t^2 + A0 t + B0 t e (4t)

    => y ' = 2A1t + A0 + B0 [e^ (4t) + 4 te^ (4t) ]

    y ' = 2A1t + A0 + B0e^ (4t) + 4B0 te^ (4t)

    => y '' = 2A1 + 4B0e (4t) + 4B0 [ e^ (4t) + 4te^ (4t)

    y '' = 2A1 + 4B0e^ (4t) + 4B0e^ (4t) + 16B0te^ (4t)

    Now substitute the values of y ' and y '' in the differential equation:

    y′′+αy′+βy=t+e^ (4t)

    2A1 + 4B0e^ (4t) + 4B0e^ (4t) + 16B0te^ (4t) + α{2A1t + A0 + B0e^ (4t) + 4B0 te^ (4t) } + β{A1 t^2 + A0 t + B0 t e (4t) } = t + e^ (4t)

    Next, we equate coefficients

    1) Constant terms of the left side = constant terms of the right side:

    2A1 + 2αA0 = 0 ... eq (1)

    2) Coefficients of e^ (4t) on both sides

    8B0 + αB0 = 1 = > B0 (8 + α) = 1 ... eq (2)

    3) Coefficients on t

    2αA1 + βA0 = 1 ... eq (3)

    4) Coefficients on t^2

    βA1 = 0 ... eq (4)

    given that A1 ≠ 0 = > β = 0

    5) terms on te^ (4t)

    16B0 + 4αB0 + βB0 = 0 = > B0 (16 + 4α + β) = 0 ... eq (5)

    Given that B0 ≠ 0 = > 16 + 4α + β = 0

    Use the value of β = 0 found previously

    16 + 4α = 0 = > α = - 16 / 4 = - 4.

    Answer: α = - 4 and β = 0
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “Consider the differential equation y′′+αy′+βy=t+e^ (4t). Suppose the form of the particular solution to this differential equation as ...” in 📙 Physics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers