Answer:
Combining the method of undetermined coefficients with the method of variation of parameters, the solution to the differential equation
y'' - 2y' + y = 4x² - 6 + x^(-1)e^x
is
y = (C1 + C2x)e^x + 4x² + 16x + 18 -xe^x + xe^x lnx
Step-by-step explanation:
Given the differential equation:
y'' - 2y' + y = 4x² - 6 + x^(-1)e^x........(1)
Firstly, we solve the homogeneous part of (1)
y'' - 2y' + y = 0
Let the characteristic equation be
m² - 2m + 1 = 0
(m - 1)(m - 1) = 0
m = 1 twice.
The complementary function
y_c = (C1 + C2x)e^x...........................(2)
Now, consider the differential equation:
y'' - 2y' + y = 4x² - 6 ..........................(3)
Solve (3) using the method of UNDETERMINED COEFFICIENTS.
The nonhomogeneous part is 4x² - 6, so we assume a particular solution of the form
y_p = Ax² + Bx + C
y'_p = 2Ax + B
y''_p = 2A
Using these in (3), we have
y''_p - 2y'_p + y_p
= 2A - 2(2Ax + B) + Ax² + Bx + C = 4x² - 6
Ax² + Bx - 4Ax - 2B + 2A + C = 4x² - 6
Comparing the coefficients of various powers of x, we have that
A = 4
B - 4A = 0 => B = 16
2A - 2B + C = -6
=> C = -6 - 8 + 32 = 18
Therefore,
y_p = 4x² + 16x + 18 .....................(4)
Next, consider the differential equation:
y'' - 2y' + y = x^(-1)e^x ....................(3)
We solve using the method of VARIATION OF PARAMETERS.
With g(x) = x^(-1)e^x
Using the complementary solution, we have
y1 = e^x, and y2 = xe^x
Find the Wronskian of y1 and y2.
Let their Wronskian be W, then
W = |y1............y2|
........|y1'...........y2'|
= |e^x.....................xe^x|
...|e^x...........xe^x + e^x|
= xe^(2x) + e^(2x) - xe^(2x)
W = e^(2x)
y_q = Py1 + Qy2
Where P = integral of y2g(t)/W dx
= integral of xe^x.x^(-1)e^x/e^(2x) dx
= -x
Where Q = integral if y1g(t)/W dx
= integral of e^x.x^(-1)e^x/e^(2x) dx
= lnx
y_q = -xe^x + xe^x lnx
Finally, the general solution is
y = y_c + y_p + y_q
= (C1 + C2x)e^x + 4x² + 16x + 18 -xe^x + xe^x lnx
