Solve the given differential equation by undetermined coefficients. y'' - y' + 1/4 y = 8 + ex/2
Solution:
y'' - y' + 1/4 y = 8 + ex/2 [Given]
Solve the corresponding homogeneous differential equation
y'' - y' + 1/4 y = 0
r2 - r + 1/4 = 0 is the characteristic equation
By factorization, let us find its roots
(r - 1/2) (r - 1/2) = 0
r - 1/2 = 0
r = 1/2 is a repeated root so one of the complementary solution yc
yc = c1ex/2 + c2xex/2
Then determine the remaining particular solution
yp = Ax2ex/2 + B
y′p = A/2 x2ex/2 + 2 Axex/2
y′′p = A/4 x2ex/2 + 2 Axex/2 + 2 Aex/2
In the given differential equation, substitute yp, y’p and y’’p and solve for A and B
A/4 x2ex/2 + Axex/2 + 2 Aex/2 − A/2 x2ex/2 - 2 Axex/2 + 1/4 (Ax2ex/2 + B) = 8 + ex/2
(A/4 x2 + 2 Ax + 2A − A/2 x2 - 2 Ax + A/4 x2) ex/2 + B/4 = 8 + ex/2
ex/2 (2A) + B/4 = 8 + ex/2
By comparison
A = 1/2 and B = 32
Let us combine complementary and particular solution to find the general solution
y = yc + yp
y = c1ex/2 + c2 xex/2 + 1/2 x2ex/2 + 32
Therefore, by solving the differential equation by undetermined coefficients we get y = c1ex/2 + c2 xex/2 + 1/2 x2ex/2 + 32
Solve the given differential equation by undetermined coefficients. y'' - y' + 1/4 y = 8 + ex/2
Summary:
By solving the given differential equation by undetermined coefficients we get y = c1ex/2 + c2 xex/2 + 1/2 x2ex/2 + 32
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