MINDANAO STATE UNIVERSITY – GENERAL SANTOS CITY
COLLEGE OF ENGINEERING
DEPARTMENT OF CIVIL ENGINEERING
Engineering Mathematics
Homogeneous Linear ODEs with Constant Coefficients
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Lesson Objectives:
1. Formulate an auxiliary equation from a given homogeneous second order linear ordinary differential equation
with constant coefficients.
2. Identify with case to use in solving these types of ODEs based on its auxiliary equation.
3. Solve homogeneous linear ordinary differential equations of second order with constant coefficients.
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In the previous module, homogeneous linear ODEs of second order were discussed. The method of reduction of
order was also introduced in solving these equations and in obtaining a basis of solutions given one solution. Now, we
shall consider second-order homogeneous linear ODEs with constant coefficients.
I. Homogeneous Linear ODEs with Constant Coefficients
Second-order homogeneous linear ODEs whose coefficients 𝑎 and 𝑏 are constant are written in the form,
𝑦′′ + 𝑎𝑦′ + 𝑏𝑦 = 0
(1)
To solve this equation, let’s try a solution of the form,
𝑦 = 𝑒 𝜆𝑥 .
(2)
Substituting (2) and its derivatives
𝑦′ = 𝜆𝑒 𝜆𝑥
and
𝑦′′ = 𝜆2 𝑒 𝜆𝑥
into equation the second-order homogeneous linear ODE (1)
𝜆2 𝑒 𝜆𝑥 + 𝑎𝜆𝑒 𝜆𝑥 + 𝑏𝑒 𝜆𝑥 = 0
(𝜆2 + 𝑎𝜆 + 𝑏)𝑒 𝜆𝑥 = 0
Since 𝑒 𝜆𝑥 is never zero for real values of 𝑥, it is apparent that the only way that this exponential function can
satisfy the differential equation (2) is to choose 𝜆 as a root of the quadratic equation
𝜆2 + 𝑎𝜆 + 𝑏 = 0
This equation is what we call the characteristic equation (or auxiliary equation). Now if 𝜆 is a solution of this
auxiliary equation, then the exponential function (2) is a solution of the ODE (1). Since the characteristic equation is a
quadratic equation, it will have two roots given by the quadratic formula
1
𝜆1 = (−𝑎 + √𝑎2 − 4𝑏)
2
1
𝜆2 = (−𝑎 − √𝑎2 − 4𝑏).
2
and
Therefore, the solutions of (1) will be
𝑦1 = 𝑒 𝜆1 𝑥
𝑦2 = 𝑒 𝜆2 𝑥
and
(3)
Note: Verify this by substituting (3) into (1).
From algebra, we further know that the auxiliary equation, being a quadratic equation, may have three kinds of
roots, depending on the sign of the discriminant, 𝑎2 − 4𝑏, namely,
CASE 1. Two real roots if 𝑎2 − 4𝑏 > 0.
CASE 2. A real double root if 𝑎2 − 4𝑏 = 0.
CASE 3. Complex conjugate roots if 𝑎2 − 4𝑏 < 0.
II. CASE 1. Two Distinct Real-Roots 𝜆1 and 𝜆2
In this case, a basis of solutions of (1) on any interval is,
𝑦1 = 𝑒 𝜆1 𝑥
𝑦2 = 𝑒 𝜆2 𝑥
and
because 𝑦1 and 𝑦2 are defined (and real) for all 𝑥 and their quotient is not constant, therefore they are linearly
independent. The corresponding general solution is
𝒚 = 𝒄𝟏 𝒆𝝀𝟏𝒙 + 𝒄𝟐 𝒆𝝀𝟐𝒙
Example:
1. Solve the initial value problem 𝑦 ′′ + 𝑦 ′ − 6𝑦 = 0,
𝑦(0) = 4, 𝑦 ′ (0) = −5.
Solution:
General Solution. The corresponding auxiliary equation for the given ODE is
𝜆2 + 𝜆 − 6 = 0.
Its roots are
1
𝜆1 = (−1 + √12 − 4(−6)) = 2
2
and
1
𝜆2 = (−1 − √12 − 4(−6)) = −3
2
So, its general solution is
𝒚 = 𝒄𝟏 𝒆𝟐𝒙 + 𝒄𝟐 𝒆−𝟑𝒙
Particular Solution. From the given initial conditions, and
𝑦′ = 2𝑐1 𝑒 2𝑥 − 3𝑐2 𝑒 −3𝑥 , we obtain the value of the two arbitrary constants.
𝑦 = 𝑐1 𝑒 2𝑥 + 𝑐2 𝑒 −3𝑥
4 = 𝑐1 𝑒 2(0) + 𝑐2 𝑒 −3(0)
𝟒 = 𝒄𝟏 + 𝒄𝟐
the
first
derivative
of
𝑦,
𝑦′ = 2𝑐1 𝑒 2𝑥 − 3𝑐2 𝑒 −3𝑥
−5 = 2𝑐1 𝑒 2(0) − 3𝑐2 𝑒 −3(0)
−𝟓 = 𝟐𝒄𝟏 − 𝟑𝒄𝟐
7
Hence, 𝑐1 = 5 and 𝑐2 =
13
.
5
Therefore, we have the particular solution
7
13
𝑦 = 𝑒 2𝑥 + 𝑒 −3𝑥
5
5
𝟓𝒚 = 𝟕𝒆𝟐𝒙 + 𝟏𝟑𝒆−𝟑𝒙
III. CASE 2. Real Double Roots 𝜆 = −𝑎/2
If the discriminant of the auxiliary equation is zero, the quadratic formula will give a single root 𝜆 = 𝜆1 = 𝜆2 =
−𝑎/2 . Hence we will only have one solution,
𝑦1 = 𝑒 −(𝑎/2)𝑥
To obtain a second independent solution 𝑦2 , we use the method of reduction of order discussed in the last module,
setting 𝑦2 = 𝑢𝑦1 . This will give us a result of 𝑦2 = 𝑥𝑦1 . Since these roots are not proportional, they are linearly
independent and form a basis. Therefore, in this case of double root a basis of solutions of (1) on any interval is
𝑦1 = 𝑒 −𝑎𝑥/2
and
𝑦2 = 𝑥𝑒 −𝑎𝑥/2
The corresponding general solution is
𝒚 = (𝒄𝟏 + 𝒄𝟐 𝒙)𝒆−𝒂𝒙/𝟐
Example:
1. Solve the initial value problem 𝑦 ′′ + 𝑦 ′ + 0.25𝑦 = 0,
𝑦(0) = 3, 𝑦 ′ (0) = −3.5.
Solution:
General Solution. The corresponding auxiliary equation for the given ODE is
𝜆2 + 𝜆 + 0.25 = 0.
Since its determinant [12 − 4(0.25)] is equal to zero, it has a double root equal and is equal to 𝜆 = −0.5.
So, its general solution is
𝒚 = (𝒄𝟏 + 𝒄𝟐 𝒙)𝒆−𝟎.𝟓𝒙
Particular Solution. From the given initial conditions, and the first derivative
𝑦 ′ = −0.5(𝑐1 + 𝑐2 𝑥)𝑒 −0.5𝑥 + 𝑐2 𝑒 −0.5𝑥 , we obtain the value of the two arbitrary constants.
𝑦 = (𝑐1 + 𝑐2 𝑥)𝑒 −0.5𝑥
3 = [𝑐1 + 𝑐2 (0)]𝑒 −0.5(0)
𝟑 = 𝒄𝟏
𝑦 ′ = −0.5(𝑐1 + 𝑐2 𝑥)𝑒 −0.5𝑥 + 𝑐2 𝑒 −0.5𝑥
of
𝑦,
−3.5 = −0.5[𝑐1 + 𝑐2 (0)]𝑒 −0.5(0) + 𝑐2 𝑒 −0.5(0)
−3.5 = −0.5(𝑐1 ) + 𝑐2
−𝟐 = 𝒄𝟐
Therefore, we have the particular solution
𝒚 = (𝟑 − 𝟐𝒙)𝒆−𝟎.𝟓𝒙
IV. CASE 3. Complex Roots
This case occurs when the discriminant of the auxiliary equation is negative. In this case the roots are the complex
𝜆=
1
− 𝑎
2
± 𝜔i. The basis of solutions are then given by,
𝑦1 = 𝑒 −(𝑎𝑥/2) cos 𝜔𝑥
and
𝑦2 = 𝑒 −(𝑎𝑥/2) sin 𝜔𝑥
1
2
where 𝜔 = √4𝑏 − 𝑎2 .
You can verify that these are solutions in the present case by substitution. Hence, we have the general solution for ODEs
falling under Case 3,
𝒚=𝒆
−(
𝒂𝒙
)
𝟐 (𝒄𝟏 𝐜𝐨𝐬 𝝎𝒙 + 𝒄𝟐 𝐬𝐢𝐧 𝝎𝒙)
Example:
1. Solve the initial value problem 4𝑦 ′′ + 4𝑦 ′ + 17𝑦 = 0,
𝑦(0) = −1, 𝑦 ′ (0) = 2.
Solution:
General Solution. The corresponding auxiliary equation for the given ODE is
𝜆2 + 𝜆 +
17
4
= 0.
𝑎
Using the quadratic formula, we obtain the roots −0.5 ± 2i. Hence, 𝜔 = 2 and − 2 = −0.5, and a general solution
is equal to
𝑦=𝑒
−(
𝑎𝑥
)
2 (𝑐1 cos 𝜔𝑥
+ 𝑐2 sin 𝜔𝑥)
𝒚 = 𝒆−𝟎.𝟓𝒙 (𝒄𝟏 𝐜𝐨𝐬 𝟐𝒙 + 𝒄𝟐 𝐬𝐢𝐧 𝟐𝒙)
Particular Solution. From the given initial condition 𝑦(0) = −1,
𝑦 = 𝑒 −0.5𝑥 (𝑐1 cos 2𝑥 + 𝑐2 sin 2𝑥)
−1 = 𝑒 −0.5(0) [𝑐1 cos 2(0) + 𝑐2 sin 2(0)]
−𝟏 = 𝒄𝟏
Now differentiating 𝑦 and then using 𝑦 ′ (0) = 2 and 𝑐1 = −1,
𝑦 ′ = 𝑒 −0.5𝑥 [−2𝑐1 sin(2𝑥) + 2𝑐2 cos(2𝑥)] − 0.5𝑒 −0.5𝑥 [𝑐1 cos(2𝑥) + 𝑐2 sin(2𝑥)]
2 = 𝑒 −0.5(0) [−2(−1) sin(2 ∙ 0) + 2𝑐2 cos(2 ∙ 0)] − 0.5𝑒 −0.5(0) [(−1) cos(2 ∙ 0) + 𝑐2 sin(2 ∙ 0)]
𝟑
= 𝒄𝟐
𝟒
Therefore, the particular solution of the IVP is
𝟑
𝒚 = 𝒆−𝟎.𝟓𝒙 (− 𝐜𝐨𝐬 𝟐𝒙 + 𝐬𝐢𝐧 𝟐𝒙)
𝟒