COLLEGE of ENGINEERING – CIVIL ENGINEERING DEPARTMENT
HOMOGENEOUS LINEAR DIFFERENTIAL EQUATIONS
WITH CONSTANT COEFFICIENTS
The second-order differential equation of the form,
𝒂𝒚′′ + 𝒃𝒚′ + 𝒄𝒚 = 𝟎
is said to homogeneous with constant coefficients if 𝑎, 𝑏 and 𝑐 are constants. This special case
of differential equation can be solved by a solution method that only uses algebra with the aid
of the only nontrivial elementary function whose derivative is a constant multiple of itself –
exponential function, 𝒚 = 𝒆𝒎𝒙.
Finding an exponential solution,
𝑦 = 𝑒 𝑚𝑥
𝑦′ = 𝑚𝑒 𝑚𝑥
𝑦′′ = 𝑚2 𝑒 𝑚𝑥
Substituting to the second-order differential equation,
𝑎𝑦 ′′ + 𝑏𝑦 ′ + 𝑐𝑦 = 0
𝑎𝑚 𝑒 + 𝑏𝑚𝑒 𝑚𝑥 + 𝑐𝑒 𝑚𝑥 = 0
2 𝑚𝑥
𝑒 𝑚𝑥 (𝑎𝑚2 + 𝑏𝑚 + 𝑐) = 0
Since the domain of 𝑦 = 𝑒 𝑚𝑥 is (0, +∞), then 𝑒 𝑚𝑥 ≠ 0. It is apparent that the only way 𝑦 = 𝑒 𝑚𝑥 can
satisfy the differential equation is when 𝑚 is chosen as a root of the quadratic equation,
𝑎𝑚2 + 𝑏𝑚 + 𝑐 = 0
The quadratic equation above is called the auxiliary equation of the differential equation.
Three cases of general solution for Second-Order Equations:
CASE I. DISTINCT REAL ROOTS (𝑚1 ≠ 𝑚2 ), 𝑏 2 − 4𝑎𝑐 > 0 (real and unequal)
General Solution: 𝑦 = 𝐶1 𝑒 𝑚1𝑥 + 𝐶2 𝑒 𝑚2𝑥
CASE II. REPEATED REAL ROOTS (𝑚1 = 𝑚2 ), 𝑏 2 − 4𝑎𝑐 = 0 (real and equal)
General Solution: 𝑦 = 𝐶1 𝑒 𝑚1𝑥 + 𝐶2 𝑥𝑒 𝑚1𝑥
CASE III. CONJUGATE COMPLEX ROOTS (𝑚1 = 𝛼 + 𝛽𝑖, 𝑚2 = 𝛼 − 𝛽𝑖), 𝑏 2 − 4𝑎𝑐 < 0
(imaginary)
General Solution: 𝑦 = 𝑒 𝛼𝑥 (𝐶 cos 𝛽𝑥 + 𝐶 sin 𝛽𝑥)
1 2
Three cases of general solution for nth-Order Equations:
CASE I. DISTINCT REAL ROOTS (𝑚1 ≠ 𝑚2 ≠ 𝑚𝑛 )
General Solution: 𝑦 = 𝐶1 𝑒 𝑚1𝑥 + 𝐶2 𝑒 𝑚2𝑥 + ⋯ + 𝐶𝑛 𝑒 𝑚𝑛𝑥
CASE II. REPEATED REAL ROOTS (𝑚1 = 𝑚2 = 𝑚𝑛 )
General Solution: 𝑦 = 𝐶1 𝑒 𝑚1𝑥 + 𝐶2 𝑥𝑒 𝑚1𝑥 + ⋯ + 𝐶𝑛 𝑥 𝑛−1 𝑒 𝑚1𝑥
CASE III. CONJUGATE COMPLEX ROOTS (𝑚1 = 𝛼 + 𝛽𝑖, 𝑚2 = 𝛼 − 𝛽𝑖)
General Solution: 𝑦 = 𝑒 𝛼𝑥 (𝐶1 cos 𝛽𝑥 + 𝐶2 sin 𝛽𝑥)
COLLEGE of ENGINEERING – CIVIL ENGINEERING DEPARTMENT
EXAMPLES (Case I)
Find the general solution of the following differential equations.
Second-order DE
1. 𝟒𝐲 ′′ + 𝐲 ′ = 𝟎
No. of roots = 2
4D2 y + Dy = 0 Convert to differential operator.
(4D2 + D)y = 0 Factor out 𝑦.
Quantified expression is the
4D2 + D = 0 auxiliary equation of the given
differential equation. Replace 𝐷 to 𝑚.
Solve for the roots of the quadratic
4m2 + m = 0
equation.
1
m1 = 0, m2 = − Real and unequal roots (𝑚1 ≠ 𝑚2 )
4
y = C1 em1 x + C2 em2 x General solution
1
y = C1 e(0)x + C2 e(−4)x Substitute values of 𝑚1 and 𝑚2 .
1
y = C1 + C2 e−4x General solution of the DE.
Second-order DE
2. 𝐲 ′′ − 𝐲 ′ − 𝟔𝐲 = 𝟎
No. of roots = 2
D2 y − Dy − 6y = 0
(D2 − D − 6)y = 0
D2 − D − 6 = 0
m2 − m − 6 = 0
m1 = −2, m2 = 3 Real and unequal roots (𝑚1 ≠ 𝑚2 )
y = C1 em1 x + C2 em2 x
y = C1 e(−2)x + C2 e(3)x
y = C1 e−2x + C2 e3x General solution of the DE.
COLLEGE of ENGINEERING – CIVIL ENGINEERING DEPARTMENT
EXAMPLES (Case II)
Find the general solution of the following differential equations.
Second-order DE
1. 𝐲 ′′ + 𝟖𝐲 ′ + 𝟏𝟔𝐲 = 𝟎
No. of roots = 2
D2 y + 8Dy + 16y = 0 Convert to differential operator.
(D2 + 8D + 16)y = 0 Factor out 𝑦.
Quantified expression is the
D2 + 8D + 16 = 0 auxiliary equation of the given
differential equation. Replace 𝐷 to 𝑚.
Solve for the roots of the quadratic
m2 + 8m + 16 = 0
equation.
m1 = −4, m2 = −4 Real and equal roots (𝑚1 = 𝑚2 )
y = C1 em1 x + C2 xem1 x General solution
Substitute values of 𝑚1 or 𝑚2 since
y = C1 e(−4)x + C2 xe(−4)x
equal in value.
y = C1 e−4x + C2 xe−4x General solution of the DE.
Third-order DE
2. 𝐲 ′′′ − 𝟏𝟓𝐲′′ + 𝟕𝟓𝐲 ′ − 𝟏𝟐𝟓𝐲 = 𝟎
No. of roots = 3
D3 y − 15D2 y + 75Dy − 125y = 0
(D3 − 15D2 + 75D − 125)y = 0
D3 − 15D2 + 75D − 125 = 0
m3 − 15m2 + 75m − 125 = 0
m1 = 5, m2 = 5, m3 = 5 Real and equal roots (𝑚1 = 𝑚2 = 𝑚3 )
y = C1 em1 x + C2 xem1 x + ⋯ + Cn x n−1 em1 x General solution of nth-order
y = C1 em1 x + C2 xem1 x + C3 x 3−1 em1 x
y = C1 em1 x + C2 xem1 x + C3 x 2 em1 x
y = C1 e(5)x + C2 xe(5)x + C3 x 2 e(5)x
y = C1 e5x + C2 xe5x + C3 x 2 e5x General solution of the DE.
COLLEGE of ENGINEERING – CIVIL ENGINEERING DEPARTMENT
EXAMPLES (Case III)
Find the general solution of the following differential equations.
Second-order DE
1. 𝐲 ′′ − 𝟒𝐲 ′ + 𝟓𝐲 = 𝟎
No. of roots = 2
D2 y − 4Dy + 5y = 0 Convert to differential operator.
(D2 − 4D + 5)y = 0 Factor out 𝑦.
Quantified expression is the
D2 − 4D + 5 = 0 auxiliary equation of the given
differential equation. Replace 𝐷 to 𝑚.
Solve for the roots of the quadratic
m2 − 4m + 5 = 0
equation.
Imaginary roots (𝛼 ± 𝛽𝑖)
m1 = 2 + i, m2 = 2 − i
𝛼 = 2, 𝛽 = 1
y = eαx (C1 cos βx + C2 sin βx) General solution
y = e(2)x (C1 cos(1)x + C2 sin(1)x) Substitute values of 𝛼 or 𝛽.
y = e2x (C1 cos x + C2 sin x) General solution of the DE.
Second-order DE
2. 𝟑𝐲 ′′ + 𝟐𝐲 ′ + 𝐲 = 𝟎
No. of roots = 2
3D2 y + 2Dy + y = 0
(3D2 + 2D + 1)y = 0
3D2 + 2D + 1 = 0
3m2 + 2m + 1 = 0
1 √2 1 √2 Imaginary roots (𝛼 ± 𝛽𝑖)
m1 = − + i, m2 = − − i 1
𝛼 = − 3, 𝛽 =
√2
3 3 3 3 3
y = eαx (C1 cos βx + C2 sin βx)
1 √2 √2
(− )x
y=e 3 (C1 cos ( )x + C2 sin ( ) x)
3 3
1 √2 √2
y = e−3x (C1 cos ( ) x + C2 sin ( ) x) General solution of the DE.
3 3
COLLEGE of ENGINEERING – CIVIL ENGINEERING DEPARTMENT
NOTES:
In cases where a polynomial operator involves combination of three cases, i.e., a
polynomial having equal, unequal, and imaginary roots, obtain individual solutions for
each case and combine.
Example:
Suppose a differential equation has the following roots,
𝑚1 = 2, 𝑚2 = 3, 𝑚3 = 3, 𝑚4 = 2 + 𝑖, 𝑚5 = 2 − 𝑖
IMAGINARY ROOTS
REAL & EQUAL ROOTS
REAL & UNEQUAL ROOT
Get individual solutions,
Real and Unequal Root 𝐲 = 𝐂𝟏 𝐞𝟐𝐱
Real and Equal Roots y = C1 e3x + C2 xe3x
Imaginary Roots y = e2x (C1 cos x + C2 sin x)
Combining the solutions,
y = C1 e2x + C1 e3x + C2 xe3x + e2x (C1 cos x + C2 sin x)
Arbitrary parameters should not be repeated. Therefore, the general solution must be,
𝐲 = 𝐂𝟏 𝐞𝟐𝐱 + 𝐂𝟐 𝐞𝟑𝐱 + 𝐂𝟑 𝐱𝐞𝟑𝐱 + 𝐞𝟐𝐱 (𝐂𝟒 𝐜𝐨𝐬 𝐱 + 𝐂𝟓 𝐬𝐢𝐧 𝐱)
[End of Differential Equations]
Prepared by:
Engr. ARISTON C. TALOSIG
Lecturer, Differential Equations
