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' + 3e5xy = 0

 

are linear, while the equations:

 

y'' + 3( y')3 – 2y2 = 0     and     y''y – 5x2y' + 3e5xy = 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:

 

eix = 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 ekx, 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 = c1e2x + c2ex,

 

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) + pn–1(x)y(n–1) + pn–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) = kn–2,     y(n–1)(x0) = kn–1,

 

exists and is unique provided the functions pi(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 = c1e3x + c2e2x,
1 = y(0) = c1e0 + c2e0 = c1 + c2,     c1 + c2 = 1,  [1]
y' = 3c1e3x + 2c2e2x,
–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 = – 3e3x + 4e2x, or y = 4e2x – 3e3x.

 

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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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