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.

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