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
