Solve the given differential equation by undetermined coefficients. y'' - y' + 1/4 y = 4 + e^x/2

Solution:

y'' - y' + 1/4 y = 4 + e^x/2 [Given]

Let us solve the corresponding homogeneous differential equation

y'' − y' + 1/4 y = 0

Here the characteristic equation is r² - r + 1/4 = 0

By factorization let us find its roots

(r - 1/2) (r - 1/2) = 0

r - 1/2 = 0

r = 1/2

So r = ½ is the repeated root and one of the complimentary solutions y_c is

yc = c₁e^x/2 + c₂xe^x/2

Determine the remaining particular solution y_p

y_p = Ax²e^x/2 + B

y′_p = A/2 x²e^x/2 + 2 Axe^x/2

y′′_p = A/4 x²e^x/2 + 2 Axe^x/2 + 2 Ae^x/2

By substituting y_p, y′_p, and y′′_p in given differential equation and solve for A and B

A/4 x²e^x/2 + Axe^x/2 + 2 Ae^x/2 − A/2 x²e^x/2 - 2 Axe^x/2 + 1/4 (Ax²e^x/2 + B) = 4 + e^x/2

(A/4 x² + 2 Ax + 2A − A/2 x² - 2 Ax + A/4 x²) e^x/2 + B/4 = 4 + e^x/2

e^x/2 (2A) + B/4 = 4 + e^x/2

By comparing it, A = ½ and B = 16.

So y_p = 1/2 x²e^x/2 + 16

By combining complementary solution and particular solution to get the general solution

y = y_c + y_p

y = c₁e^x/2 + c₂xe^x/2 + 1/2 x²e^x/2 + 16

Therefore, the general solution is y = c₁e^x/2 + c₂xe^x/2 + 1/2 x²e^x/2 + 16.

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Solve the given differential equation by undetermined coefficients. y'' - y' + 1/4 y = 4 + e^x/2

Summary:

By solving the given differential equation by undetermined coefficients y'' - y' + 1/4 y = 4 + e^x/2 we get y = c₁e^x/2 + c₂xe^x/2 + 1/2 x²e^x/2 + 16.