# Solve the given differential equation by undetermined coefficients. y'' - y' + 1/4 y = 8 + e^{x/2}

**Solution:**

y'' - y' + 1/4 y = 8 + e^{x/2} [Given]

Solve the corresponding homogeneous differential equation

y'' - y' + 1/4 y = 0

r^{2} - 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 y_{c}

y_{c} = c_{1}e^{x/2} + c_{2}xe^{x/2}

Then determine the remaining particular solution

y_{p} = Ax^{2}e^{x/2} + B

y′_{p }= A/2 x^{2}e^{x/2 }+ 2 Axe^{x/2}

y′′_{p} = A/4 x^{2}e^{x/2 }+ 2 Axe^{x/2} + 2 Ae^{x/2}

In the given differential equation, substitute y_{p}, y’_{p} and y’’_{p} and solve for A and B

A/4 x^{2}e^{x/2 }+ Axe^{x/2} + 2 Ae^{x/2} − A/2 x^{2}e^{x/2 } - 2 Axe^{x/2} + 1/4 (Ax^{2}e^{x/2} + B) = 8 + e^{x/2}

(A/4 x^{2 }+ 2 Ax + 2A − A/2 x^{2 } - 2 Ax + A/4 x^{2}) e^{x/2} + B/4 = 8 + e^{x/2}

e^{x/2} (2A) + B/4 = 8 + e^{x/2}

By comparison

A = 1/2 and B = 32

Let us combine complementary and particular solution to find the general solution

y = y_{c} + y_{p}

y = c_{1}e^{x/2} + c_{2 }xe^{x/2} + 1/2 x^{2}e^{x/2} + 32

Therefore, by solving the differential equation by undetermined coefficients we get y = c_{1}e^{x/2} + c_{2 }xe^{x/2} + 1/2 x^{2}e^{x/2} + 32

## Solve the given differential equation by undetermined coefficients. y'' - y' + 1/4 y = 8 + e^{x/2}

**Summary:**

By solving the given differential equation by undetermined coefficients we get y = c_{1}e^{x/2} + c_{2 }xe^{x/2} + 1/2 x^{2}e^{x/2} + 32