Answer:
Explanation:
To solve the initial-value problem y''' − 2y'' + y' = 2 − 24ex + 40e5x, y(0) = 1/2, y'(0) = 5/2, y''(0) = −11/2, we first find the characteristic equation by assuming that y = e^(rt):
r^3 - 2r^2 + r = 0
r(r^2 - 2r + 1) = 0
r(r-1)^2 = 0
r = 0, 1, 1
Therefore, the general solution to the homogeneous equation y''' − 2y'' + y' = 0 is:
y_h = c1 + c2e^x + c3xe^x
To find the particular solution to the non-homogeneous equation y''' − 2y'' + y' = 2 − 24ex + 40e5x, we guess a particular solution of the form:
y_p = Ax^2e^5x + Be^x
y_p' = (2Ax + 5Ax^2)e^5x + Be^x
y_p'' = (10Ax^2 + 4Ax + 25Ax^2)e^5x + Be^x
y_p''' = (70Ax^2 + 60Ax + 10A + 25Ax^2)e^5x + Be^x
Substituting these expressions into the original equation, we get:
(70Ax^2 + 60Ax + 10A + 25Ax^2)e^5x + Be^x − 2[(10Ax^2 + 4Ax + 25Ax^2)e^5x + Be^x] + [(2Ax + 5Ax^2)e^5x + Be^x] = 2 − 24ex + 40e5x
Simplifying, we get:
(45Ax^2 + 2Ax − 2)e^5x + Be^x = 2 − 24ex + 40e5x
Equating the coefficients of the like terms on both sides, we get:
45A = 40
2A − 2 = 0
B = 2
Therefore, the particular solution is:
y_p = 8/9 x^2e^5x + 2e^x
The general solution to the non-homogeneous equation is therefore:
y = c1 + c2e^x + c3xe^x + 8/9 x^2e^5x + 2e^x
Using the initial conditions y(0) = 1/2, y'(0) = 5/2, y''(0) = −11/2, we get:
c1 + c2 + 2 = 1/2
c2 + 2c3 + 5/2 = 5/2
2c2 + 10/9 + 10 = -11/2
Solving this system of equations, we get:
c1 = 1/9
c2 = -25/18
c3 = 0
Therefore, the solution to the initial-value problem y''' − 2y'' + y' = 2 − 24ex + 40e5x, y(0) = 1/2, y'(0) = 5/2, y''(0) = −11/2 is:
y = 1/9 - 25/18e^x + 8/9