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.

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