Solve the given Differential Equation by Undetermined Coefficients. y'' − y' + 1/4 y = 4 + ex/2
We will be solving this by finding the homogeneous and complementary solutions of the equation.
Answer: The general solution of differential equation y'' − y' + 1/4 y = 4 + ex/2 is c1ex/2 + c2xex/2 + 1/2 x2ex/2 + 16
Let's solve this step by step.
Given, differential equation: y'' − y' + 1/4 y = 4 + ex/2
First solve the corresponding homogeneous differential equation: y'' − y' + 1/4 y = 0
The characteristic equation is r2 - r + 1/4 = 0.
Let's find its roots by factorization.
(r - 1/2) (r - 1/2) = 0
r - 1/2 = 0
r = 1/2
Therefore, r = 1/2 is a repeated root thus one of the complimentary solutions yc is yc = c1ex/2 + c2xex/2
Now, find the remaining particular solution yp.
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
Substitute yp, y′p, and y′′p in given differential equation 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) = 4 + ex/2
(A/4 x2 + 2 Ax + 2A − A/2 x2 - 2 Ax + A/4 x2) ex/2 + B/4 = 4 + ex/2
ex/2 (2A) + B/4 = 4 + ex/2
On comparing, we get A = 1/2 and B = 16.
Therefore, yp = 1/2 x2ex/2 + 16
Now, combine complementary solution and particular solution together to arrive at the general solution.
y = yc + yp
y = c1ex/2 + c2xex/2 + 1/2 x2ex/2 + 16