## Consider the differential equation y′′+αy′+βy=t+e6t. Suppose the form of the particular solution to this differential equation as prescribed

Question

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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2021-11-18T10:57:37+00:00
2021-11-18T10:57:37+00:00 1 Answer
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## Answers ( )

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