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
