ADDITIONAL PRACTICE FOR CONSTANT COEFFICIENT LINEAR ODES
1. Summary of Algorithms
1.1. Identifying Homogeneous Solutions.
Finding the general solution yc (x) of Ly = P (D)y = 0:
Step 1: Form the auxiliary (characteristic) polynomial P (r) associated with the differential
equation.
Step 2: Find the roots of P (r), including their multiplicities.
Step 3: For each root:
• If a is a real root of multiplicity m, the corresponding solutions are:
eax , xeax , . . . , xm−1 eax .
• If a ± bi are complex conjugate roots with multiplicity m, the corresponding solutions
are:
eax cos(bx), xeax cos(bx), . . . , xm−1 eax cos(bx),
eax sin(bx), xeax sin(bx), . . . , xm−1 eax sin(bx).
Step 4: Write the general solution to the homogeneous equation, yc (x), as a linear combination
of all the linearly independent solutions identified in Step 3.
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ADDITIONAL PRACTICE FOR CONSTANT COEFFICIENT LINEAR ODES
2. Practice Problems
2.1. Identifying Homogeneous Solutions. For each of the following differential equations, write
the general solutions yc (x) to the homogeneous problem:
(1) y 00 + y 0 − 6y = 0
(2) y 00 + 4y 0 + 4y = 0
(3) P (D)y = (D2 + 4D + 13)y = 0
(4) P (D)y = (D − 1)3 (D + 2)y = 0
(5) P (D)y = (D − 1)(D2 + 4D + 13)2 y = 0
3. Solutions
3.1. Solutions Part I.
(1) The auxiliary polynomial is
P (r) = r2 + r − 6 = (r − 2)(r + 3).
This has roots r = 2, −3. Thus the general solution is:
yc (x) = c1 e2x + c2 e−3x .
(2) The auxiliary polynomial is
P (r) = r2 + 4r + 4 = (r + 2)2 .
This has one repeated root r = −2, with multiplicity m = 2. Thus the general solution is:
yc (x) = c1 e−2x + c2 xe−2x .
(3) The auxiliary polynomial is
P (r) = r2 + 4r + 13 = (r + 2 + 3i)(r − 2 − 3i).
This has complex conjugate roots r = −2 ± 3i. Thus the general solution is:
yc (x) = c1 e−2x cos(3x) + c2 e−2x sin(3x).
(4) The auxiliary polynomial is
P (r) = (r − 1)3 (r + 2)
This has real roots r = 1 with multiplicity m = 3 and r = −2 with multiplicity m = 1.
Thus the general solution is:
yc (x) = c1 ex + c2 xex + c3 x2 ex + c4 e−2x .
(5) The auxiliary polynomial is
P (r) = (r − 1)(r2 + 4r + 13)2 = (r − 1)(r + 2 + 3i)2 (r − 2 − 3i)2 .
(Notice the quadratic r2 + 4r + 13 is the same as above!). This has real roots r = 1 with
multiplicity m = 1 and complex conjugate roots r = −2 ± 3i with multiplicity m = 2. Thus
the general solution is:
yc (x) = c1 ex + c2 e−2x cos(3x) + c3 e−2x sin(3x) + c4 xe−2x cos(3x) + c5 xe−2x sin(3x).