Solve the given Differential Equation by Undetermined Coefficients: 1/4 y'' + y' + y = x2 − 3x
We will be solving this by finding the homogeneous and complementary solutions of the equation.
Answer: The general solution of differential equation 1/4 y'' + y' + y = x2 − 3x is y = c1e-2x + c2xe-2x + x2 - 5x + 9/2
Let's solve this step by step.
Given, differential equation: 1/4 y'' + y' + y = x2 − 3x
First solve the corresponding homogeneous differential equation: 1/4 y'' + y' + y = 0
The characteristic equation is 1/4 r2 + r + 1 = 0
Let's find its roots by factorization.
(1/2 r + 1) (1/2 r + 1) = 0
1/2 r + 1 = 0
r = -2
Therefore, r = -2 is a repeated root thus one of the complimentary solutions yc is yc = c1e-2x + c2xe-2x
Now, find the remaining particular solution yp.
yp = Ax2 + Bx + C
y′p = 2Ax + B
y′′p = 2A
Substitute yp, y′p, and y′′p in given differential equation and solve for A, B and C.
1/4(2A) + (2Ax + B) + (Ax2 + Bx + C) = x2 − 3x
1/2A + 2Ax + B + Ax2 + Bx +C = x2 − 3x
Ax2 + x(2A + B) + (1/2A + B + C) = x2 − 3x
On comparing, we get A = 1, 2A + B = -3, 1/2A + B + C = 0
So, B = -5
1/2A + B + C = 0
1/2 - 5 + C = 0
C = 5 -1/2
C = 9/2
Therefore, yp = x2 - 5x + 9/2
Now, combine complementary solution and particular solution together to arrive at the general solution.
y = yc + yp
y = c1e-2x + c2xe-2x + x2 - 5x + 9/2