# 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 r^{2} - 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 = c_{1}e^{x/2} + c_{2}xe^{x/2}

Determine the remaining particular solution y_{p}

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}

By substituting y_{p}, y′_{p}, and y′′_{p }in given differential equation 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) = 4 + 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 = 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 x^{2}e^{x/2} + 16

By combining complementary solution and particular solution to get 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} + 16

Therefore, the general solution is y = c_{1}e^{x/2} + c_{2}xe^{x/2} + 1/2 x^{2}e^{x/2} + 16.

## 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 = c_{1}e^{x/2} + c_{2}xe^{x/2} + 1/2 x^{2}e^{x/2} + 16.

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