Calculus Of One Real Variable
By Pheng Kim Ving
Chapter 16: Differential Equations
Group 16.2: Second-Order Linear Homogeneous Equations
Section 16.2.1: Equations With Constant Coefficients – Characteristic Equation

16.2.1
Equations With Constant Coefficients –
Characteristic Equation

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1. Second-Order Linear Homogeneous Differential Equations With
Constant Coefficients

In this section, we'll use the abbreviations:

“CE”	for	“characteristic equation”,
“DE”	for	“differential equation”,
“GS”	for	“general solution”, and
“PS”	for	“particular solution”.

In the differential equation:

y'' + b(x)y' + c(x)y = q(x),

For example, the equations:

y'' + 3y' – 2y = 0 and y'' – 5x²y' + 3e^5xy = 0

are linear, while the equations:

y'' + 3( y')³ – 2y² = 0 and y''y – 5x²y' + 3e^5xy = 0

are non-linear.

The differential equation y'' = f (x), as discussed in Section 16.1.1 Part 3, is a second-order linear one with constant
coefficients. For this equation, b(x) = c(x) = 0 and q(x) = f (x).

Definition 1.1 – Second-Order Linear Homogeneous Differential Equations With Constant
Coefficients

A differential equation of the form:

where b and c are constants, is called a second-order linear homogeneous differential equation with constant coefficients.

When The Coefficient Of y'' Isn't 1

The DE:

ay'' + b₁y' + c₁y = 0,

where a isn't 1 or 0, can be transformed into one of the form of Eq. [1.1] by dividing both sides by a:

So y₁ is also a solution of ay'' + b₁ y' + c₁ y = 0. Thus to solve ay'' + b₁ y' + c₁ y = 0, where a isn't 1 or 0, we first divide both
sides of it by a to get the form of Eq. [1.1].

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2. General Solution

A Particular Solution

Of course the second-order linear homogeneous DE with constant coefficients:

always has y = 0 as a PS (if y = 0 then y' = 0 and y'' = 0 and thus y'' + by' + cy = 0). Of course we're interested in finding
non-0 solutions of this DE.

By Section 16.1.3 A Particular Case, the GS of the first-order linear DE y' + by = 0 is y = Ce^–bx, where C is an arbitrary
constant. So a PS is y = e^–bx. Thus it's natural to wonder if Eq. [2.1] has a PS of the form:

Characteristic Equations

The quadratic equation:

Definition 2.1 – The Characteristic Equation

Given the differential equation:

is called the characteristic equation of the differential equation [2.2].

Solutions Obtained From Two Solutions

So the sum c₁y₁ + c₂y₂ is also a solution. We've proved the following theorem.

Theorem 2.1 – Solutions Obtained From Two Solutions

Suppose y₁ and y₂ are solutions of:

Then for any constants c₁ and c₂, the function:

y₃ = c₁ y₁ + c₂ y₂

is also a solution of Eq. [2.4].

General Solution

Some PSs of the DE y' = 2x are y = x², y = x² + 10, or y = x² – 5/4. Its GS is y = x² + C, where C is an arbitrary constant.
Recall that the GS of a DE is the solution that represents all the solutions, ie, every PS can be obtained from the GS by giving
appropriate value(s) to the constant(s) present in the GS. This definition of course also applies to the GS of a second-order
linear homogeneous DE with constant coefficients.

Linear Independence And General Solution

Let y₁ and y₂ be 2 different solutions of Eq. [2.4]. Then, by Theorem 2.1, for any constants c₁ and c₂ the function c₁y₁ + c₂y₂ is
also a solution. That is, if a function is of the form c₁y₁ + c₂y₂ then it's a solution. We wonder if the converse is true: if a
function is a solution, is it of the form c₁y₁ + c₂y₂? That is, is every solution of the form c₁y₁ + c₂y₂? That means, is c₁y₁ + c₂y₂
the GS? That is, can every solution be obtained from c₁ y₁ + c₂ y₂ by giving appropriate particular values to c₁ and c₂?

Suppose y₁, y₂, and y₃ are solutions where y₂ is a constant multiple of y₁ but y₃ isn't a constant multiple of y₁. So y₂ = ky₁,
where k is a constant other than 0 and 1. We have:

c₁y₁ + c₂y₂ = c₁y₁ + c₂(ky₁) = (c₁ + c₂k)y₁ = Ky₁,

where K = c₁ + c₂k, and any solution obtained from y₁ and y₂ can actually be obtained from y₁ alone. Now c₁y₁ + c₃y₃ can't be
reduced to the form Ly₁, where L is constant, and any solution obtained from y₁ and y₃ must indeed be obtained from both y₁
and y₃; it can't be obtained just from y₁ alone.

Thus a PS that's obtained from y₁ and y₃ where y₃ isn't a constant multiple of y₁ can't be obtained from y₁ alone, ie it can't be
obtained from y₁ and y₂ where y₂ is a constant multiple of y₁. Consequently c₁y₁ + c₂y₂ isn't the GS if y₂ is a constant multiple
of y₁.

Also recall that if two vectors v₁ and v₂ in a plane are linearly independent, then any vector v₃ in the same plane can be
expressed as a linear combination of v₁ and v₂: v₃ = k₁v₁ + k₂v₂ for some constants k₁ and k₂. That is, the linear combination
c₁v₁ + c₂v₂, where c₁ and c₂ are arbitrary constants, represents all the vectors in that plane.

Analogously, two functions f (x) and g(x) are said to be linealy independent if none is a constant multiple of the other, ie if
there doesn't exist a constant k such that f (x) = kg(x) for all x where both f (x) and g(x) are defined, ie if the ratio f (x)/g(x)
isn't a constant, or if they have the following property: if c₁ f (x) + c₂ g(x) = 0, where c₁ and c₂ are constants, then c₁ = c₂ = 0.

The answer to our wondering is as follows. If y₁ and y₂ are particular linearly independent solutions of Eq. [2.4], then the GS
of Eq. [2.4] is y = c₁y₁ + c₂y₂, where c₁ and c₂ are arbitrary constants. The GS is a linear combination of the 2 particular
linearly independent solutions. The proof of the following theorem is given in a course on differential equations and is omitted
here.

Theorem 2.2 – General Solution

If y₁ and y₂ are linearly independent particular solutions of the HE:

then the general solution of this equation is:

y = c₁ y₁ + c₂ y₂,

where c₁ and c₂ are arbitrary constants.

Remark 2.1

How about y = c₁ y₁ + c₂ y₂ + C, where C is also an arbitrary constant? Shouldn't it instead be the general solution? Let's find
out:

So y = c₁ y₁ + c₂ y₂ + C isn't the general solution of y'' + by' + cy = 0. The reason is that the general solution of a
second-order equation involves only two arbitrary constants, not three, and y = c₁ y₁ + c₂ y₂ already involves two, which are c₁
and c₂ .

Remark 2.2

Combining Theorems 2.1 and 2.2 we see that if y₁ and y₂ are solutions of y'' + by' + cy = 0, then for any constants c₁ and c₂,
c₁ y₁ + c₂ y₂ is also a solution, and additionally if y₁ and y₂ are linearly independent, then for arbitrary constants c₁ and c₂ , c₁ y₁
+ c₂ y₂ is the GS.

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3. The Method Of Characteristic Equation

Recall that to solve a DE is to find its GS. Let's discuss the solving of our equation:

General Solution If b² – 4c > 0

If b² – 4c > 0, then the CE of Eq. [3.1] has 2 distinct real roots:

where c₁ and c₂ are arbitrary constants.

General Solution If b² – 4c = 0

If b² – 4c = 0, then the CE of Eq. [3.1] has 2 equal real roots:

divide both sides by u'y₁:

y₂ = uy₁ = (Cx + K)e^–(b/2)x.

Let's check that y₂ is indeed a solution of Eq. [3.1]:

y = c₁e^–(b/2)x + c₂xe^–(b/2)x,

where c₁ and c₂ are arbitrary constants.

General Solution If b² – 4c < 0

If b² – 4c < 0, then the CE of Eq. [3.1] has 2 imaginary roots:

where c₁ and c₂ are arbitrary constants.

In Section 15.3 Problem & Solution 8, we established Euler formula:

e^ix = cos x + i sin x.

Thus the GS can be written as:

where c₁ and c₂ are arbitray constants.

General Solution Of y'' + by' + cy = 0

All the above discussion shows that we've proved the following theorem. The above procedure relies on the characteristic
equation, and thus is called the method of characteristic equation. Recall that the GS of a DE is such that every PS of the
DE can be obtained from it by assigning appropriate values to the constants present in it.

Theorem 3.1 – General Solution Of y'' + by' + cy = 0

Consider the second-order linear homogeneous differential equation with constant coefficients:

and let c₁ and c₂ be arbitrary constants. By the method of characteristic equation:

and the general solution of the differential equation is:

Remarks 3.1

a. Once we've found 2 linearly independent solutions of Eq. [3.2], we've essentially found all the solutions of that equation.

b. The GS of Eq. [3.2] always involves the exponential function e^kx, where k is a real number.
c. Eq. [3.2] has infinitely many solutions.

Example 3.1

Find the GS of the DE y'' – 3y' + 2y = 0.

Solution
(–3)² – 4(2) = 1,

the GS is:

y = c₁e^2x + c₂e^x,

where c₁ and c₂ are arbitrary constants.
EOS

Example 3.2

Note

Recall that to solve a DE is to find its GS.

Solution

where c₁ and c₂ are arbitrary constants.
EOS

Remark 3.2

When the letter x is used for the function, of course we can't use it also for the variable, we must use another letter. Usually
when the function is x, it's customary that the variable is t.

Example 3.3

Solve the equation w'' + w' + w = 0.

Solution
1² – 4(1) = –3,

where c₁ and c₂ are arbitrary constants.
EOS

Remark 3.3

When a letter other than x, whatever it is, is used for the function, we can always use the letter x for the variable.

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4. For Initial-Value Problems

As seen in Section 16.1.1 Remarks 4.1, an initial-value problem involving a first-order DE specifies 1 initial condition. As seen in
the same section Remarks 4.2, an initial-value problem involving a second-order DE specifies 2 initial conditions. The DE y'' +
by' + cy is of second order, and so an intial-value problem involving it specifies 2 initial conditions. This fact is also clear from
the form of the equation's GS: y = c₁y₁ + c₂y₂, where there are 2 constants c₁ and c₂ to be determined to get a PS. To solve an
initial-value problem is to find the PS of the DE that satisfies the conditions.

Example 4.1

Solve the initial-value problem:

Solution
3² – 4(–10) = 49,

EOS

Remark 4.1

In Example 3.1, we differentiate the GS, use the condition y'(0) = 3 to get an equation in c₁ and c₂, use the condition y(0) = 1
to get another equation in c₁ and c₂, obtain a system of 2 simultaneous equations in 2 unknowns c₁ and c₂, solve it, obtain a
unique set of solutions, and thus obtain a unique PS of the given initial-value problem.

In general, the second-order linear DE y'' + by' + cy = 0 has infinitely many PSs, but when 2 initial conditions are imposed,
there's only 1 PS that satisfies the conditions. Note that for a second-order equation, there are 2 initial conditions, which are on
y' and y (= y⁽⁰⁾).

Theorem 4.1 – Existence And Uniqueness Of A Solution

The solution of the initial-value problem:

where x₀, k₀, and k₁ are real numbers, exists and is unique.

This theorem is a special case where n = 2 of the general theorem that asserts that the solution of an initial-value problem
involving an nth-order linear DE:

y⁽ⁿ⁾ + p_n–1(x)y^(n–1) + p_n–2(x)y^(n–2) + ... + p₂(x)y'' + p₁(x)y' + p₀(x)y = q(x)

and n initial conditions:

y(x₀) = k₀, y'(x₀) = k₁, y''(x₀) = k₂, ..., y^(n–2)(x₀) = k_n–2, y^(n–1)(x₀) = k_n–1,

exists and is unique provided the functions p_i(x)'s and q(x) are continuous on a neighborhood of x₀. This general theorem is
presented and proved in a course on differential equations. Remark that for an nth-order equation, there are n conditions,
which are on y (= y⁽⁰⁾) thru to y^(n–1).

Example 4.2

Solve the following initial-value problem:

Solution
0² – 4(1) = – 4,

c₁ = –1,

the desired PS is y = (–1) cos x + (0) sin x, or y = – cos x.
EOS

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Problems & Solutions

1. Find the GS of the DE 4y'' + 12y' + 5y = 0.

Solution

y = c₁e^–(1/2)x + c₂e^–(5/2)x.

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2. Solve the DE x'' – 8x' + 16x = 0.

Solution

(–8)² – 4(16) = 0,

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3. Solve the equation y'' + 3y = 0.

Solution

0² – 4(3) = – 12,

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4. Find the solution of the initial-value problem:

Solution

(–5)² – 4(6) = 1,

y = c₁e^3x + c₂e^2x,
1 = y(0) = c₁e⁰ + c₂e⁰ = c₁ + c₂, c₁ + c₂ = 1, [1]
y' = 3c₁e^3x + 2c₂e^2x,
–1 = y'(0) = 3c₁e⁰ + 2c₂e⁰ = 3c₁ + 2c₂, 3c₁ + 2c₂ = –1, [2]
[1] x (–3): – 3c₁ – 3c₂ = –3, [3]
[2] + [3]: – c₂ = – 4,
c₂ = 4,
[1]: c₁ = 1 – c₂ = 1 – 4 = –3,
y = – 3e^3x + 4e^2x, or y = 4e^2x – 3e^3x.

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5. Solve the following initial-value problem:

Solution

4v'' + 20v' + 25v = 0,

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6. Solve this initial-value problem:

Solution

0² – 4(4) = –16,

c₂ = –4/2 = –2,

y = 3 cos 2x – 2 sin 2x.

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7. Consider the DE:

y' + y² + cy + d = 0,

where c and d are constants. It's a first-order non-linear equation. Observe that the exponents of y decrease from 2 to 1 to
0 and that the coefficient of y² is 1.

a. Transform the given y-equation into a second-order linear homogeneous z-equation with constant coefficients by letting z
be a function such that z'/z = y.

b. Prove that if z is the GS of the z-equation found in part a and y = z'/z, then y is the GS of the y-equation.

c. Prove that the GS of the y-equation, which is of first-order, indeed involves only 1 arbitrary constant.

Solution

a. Let z be a function such that z'/z = y. Then:

Thus, by part a, w is a solution of the z-equation, and, by part b, w is a PS of the z-equation, for if it's the GS of the
z-equation then v would be the GS of the y-equation, a contradiction. Since z is the GS of the z-equation which is
second-order linear homogeneous with constant coefficients, we have:

z = c₁z₁ + c₂z₂,

where z₁ and z₂ are linearly independent PSs of the z-equation and c₁ and c₂ are arbitrary constants. As a consequence:

w = k₁z₁ + k₂z₂,

where k₁ and k₂ are particular constants. Then:

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