A First Course in Differential Equations

Lecture Notes Math 204

2015

Semester 1

Textbook

A First Course in Differential Equations Ninth Edition

By Dennis G. Zill

Prepared by

Dr Ebraheem Alzahrani Department of Mathematics,

Faculty of Science,

King Abdulaziz University,

Jeddah, Saudi Arabia

Chapter 1

Introduction to Differential

Equations

(1.1) Definitions and Terminology

DEFINITION 1.1.1 (Differential Equation)

An equation containing the derivatives of one or more dependent variables, with respect to one or more independent variables, is said to be a differential equation (DE).

To talk about them, we shall classify differential equations by type, order, and linearity.

CLASSIFICATION BY TYPE:

If an equation contains only ordinary derivatives of one or more dependent variables with respect to a single independent variable it is said to be an ordinary differential equation (ODE). For example,

{

𝑑𝑦

𝑑𝑥 + 5𝑦 = 𝑒

^𝑥

, 𝑑

²

𝑦

𝑑𝑥

²

− 𝑑𝑦

𝑑𝑥 + 6𝑦 = 0,

𝑑𝑥 𝑑𝑡 + 𝑑𝑦

𝑑𝑡

⏞

𝐀 𝐃𝐄 𝐜𝐚𝐧 𝐜𝐨𝐧𝐭𝐚𝐢𝐧 𝐦𝐨𝐫𝐞

𝐭𝐡𝐚𝐧 𝐨𝐧𝐞 𝐝𝐞𝐩𝐞𝐧𝐝𝐞𝐧𝐭

𝐯𝐚𝐫𝐢𝐚𝐛𝐥𝐞

= 2𝑥 + 𝑦

… … … (1)

are ordinary differential equations. On the other hand, an equation involving partial derivatives of one or more dependent variables of two or more independent variables is called a partial differential equation (PDE). For example,

{

𝜕

²

𝑢

𝜕𝑥

²

+ 𝜕

²

𝑢

𝜕𝑦

²

= 0,

𝜕

²

𝑢

𝜕𝑥

²

= 𝜕

²

𝑢

𝜕𝑡

²

− 2 𝜕𝑢

𝜕𝑡 ,

𝜕𝑢

𝜕𝑦 = − 𝜕𝑣

𝜕𝑥

… … … (2)

are partial differential equations.

Usually the ordinary derivatives are written by using

 Leibniz notation: ^𝑑𝑦

𝑑𝑥

,

^𝑑2𝑦

𝑑𝑥²

,

^𝑑3𝑦

𝑑𝑥³

,

^𝑑4𝑦

𝑑𝑥⁴

, …

 Prime notation:

𝑦

^′

, 𝑦

^′′

, 𝑦

^′′′

, 𝑦

⁽⁴⁾

, …

 Dot notation (Newton’s notation):

𝑦̇, 𝑦̈, 𝑦⃛, …

and for the partial derivatives, we use the subscript notation:

𝑦

_𝑥

, 𝑦

_𝑥𝑥

, 𝑦

_𝑥𝑥𝑥

, … .

CLASSIFICATION BY ORDER:

The order of a differential equation (either ODE or PDE) is the order of the highest derivative in the equation. For example,

second order ↘ ↙ first order

𝑑

²

𝑦

𝑑𝑥

²

+ 5 ( 𝑑𝑦 𝑑𝑥 )

3

− 4𝑦 = 𝑒

^𝑥

is a second-order ordinary differential equation.

CLASSIFICATION BY LINEARITY:

An 𝑛th-order ordinary differential equation is said to be linear if 𝐹 is linear in 𝑦, 𝑦^′, … , 𝑦^(𝑛). This means that an 𝑛th-order ODE is linear when

𝑎

_𝑛

(𝑥)𝑦

^(𝑛)

+ 𝑎

_𝑛−1

(𝑥)𝑦

^(𝑛−1)

+ ⋯ + 𝑎

₁

(𝑥)𝑦

^′

+ 𝑎

₀

(𝑥)𝑦 − 𝑔(𝑥) = 0

or

𝑎

_𝑛

(𝑥) 𝑑

^𝑛

𝑦

𝑑𝑥

^𝑛

+ 𝑎

_𝑛−1

(𝑥) 𝑑

^𝑛−1

𝑦

𝑑𝑥

^𝑛−1

+ ⋯ + 𝑎

₁

(𝑥) 𝑑𝑦

𝑑𝑥 + 𝑎

₀

(𝑥)𝑦 = 𝑔(𝑥). … … … . … (3)

Two important special cases of (3) are linear first-order (𝑛 = 1) and linear second order (𝑛 = 2) DEs:

𝑎

₁

(𝑥) 𝑑𝑦

𝑑𝑥 + 𝑎

₀

(𝑥)𝑦 = 𝑔(𝑥) and 𝑎

₂

(𝑥) 𝑑

²

𝑦

𝑑𝑥

²

+ 𝑎

₁

(𝑥) 𝑑𝑦

𝑑𝑥 + 𝑎

₀

(𝑥)𝑦 = 𝑔(𝑥). … … … (4)

In the additive combination on the left-hand side of equation (3) we see that the characteristic two properties of a linear ODE are as follows:

• The dependent variable 𝑦 and all its derivatives 𝑦, 𝑦^′, … , 𝑦^(𝑛) are of the first degree, that is, the power of each term involving 𝑦 is 1.

• The coefficients 𝑎₀, 𝑎₁, … , 𝑎_𝑛 of 𝑦, 𝑦^′, … , 𝑦^(𝑛) depend at most on the independent variable 𝑥. The equations

(𝑦 − 𝑥)𝑑𝑥 + 4𝑥𝑑𝑦 = 0, 𝑦

^′′

− 2𝑦

^′

+ 𝑦 = 0, and 𝑑

³

𝑦

𝑑𝑥

³

+ 𝑥 𝑑𝑦

𝑑𝑥 − 5𝑦 = 𝑒

^𝑥

are, in turn, linear first-, second-, and third-order ordinary differential equations. We have just demonstrated that the first equation is linear in the variable 𝑦 by writing it in the alternative form 4𝑥𝑦^′+ 𝑦 = 𝑥. A nonlinear ordinary differential equation is simply one that is not linear. Therefore, nonlinear term: nonlinear term: nonlinear term:

coefficient depends on 𝒚 nonlinear function of 𝒚 power not 𝟏 ↓ ↓ ↓

(1 − 𝑦)𝑦

^′

+ 2𝑦 = 𝑒

^𝑥

, 𝑑

²

𝑦

𝑑𝑥

²

+ sin 𝑦 = 0, and 𝑑

⁴

𝑦

𝑑𝑥

⁴

+ 𝑦

²

= 0

are examples of nonlinear first-, second-, and fourth-order ordinary differential equations, respectively.

SOLUTIONS:

DEFINITION 1.1.2 (Solution of an ODE)

Any function 𝜙, defined on an interval 𝐼 and possessing at least 𝑛 derivatives that are continuous on 𝐼, which when substituted into an 𝑛th-order ordinary differential equation reduces the equation to an identity, is said to be a solution of the equation on the interval.

INTERVAL OF DEFINITION:

You cannot think solution of an ordinary differential equation without simultaneously thinking interval.

The interval 𝐼 in Definition 1.1.2 is variously called the interval of definition, the interval of existence, the interval of validity, or the domain of the solution and can be an open interval (𝑎, 𝑏), a closed interval [𝑎, 𝑏], an infinite interval (𝑎, ∞), and so on.

(1.1) Definitions and Terminology

EXAMPLE 1 (Verification of a Solution)

Verify that the indicated function is a solution of the given differential equation on the interval (−∞, ∞).

(𝒂) 𝑑𝑦 𝑑𝑥⁄ = 𝑥𝑦^1/2 ; 𝑦 = 1

16𝑥⁴ (𝒃) 𝑦^′′− 2𝑦^′+ 𝑦 = 0 ; 𝑦 = 𝑥𝑒^𝑥 SOLUTION:

One way of verifying that the given function is a solution is to see, after substituting, whether each side of the equation is the same for every 𝑥 in the interval.

(a) From

𝑙𝑒𝑓𝑡 − ℎ𝑎𝑛𝑑 𝑠𝑖𝑑𝑒: 𝑑𝑦 𝑑𝑥 = 1

16(4. 𝑥³) =1 4𝑥³, 𝑟𝑖𝑔ℎ𝑡 − ℎ𝑎𝑛𝑑 𝑠𝑖𝑑𝑒: 𝑥𝑦¹² = 𝑥. (1

16𝑥⁴)

12

= 𝑥. (1

4𝑥²) = 1 4𝑥³,

we see that each side of the equation is the same for every real number 𝑥. Note that 𝑦^1/2 =¹

4𝑥² is, by definition, the nonnegative square root of ¹

16𝑥⁴.

(b) From the derivatives 𝑦^′ = 𝑥𝑒^𝑥 + 𝑒^𝑥 and 𝑦^′′ = 𝑥𝑒^𝑥 + 2𝑒^𝑥 we have, for every real number 𝑥,

𝑙𝑒𝑓𝑡 − ℎ𝑎𝑛𝑑 𝑠𝑖𝑑𝑒: 𝑦^′′− 2𝑦^′+ 𝑦 = (𝑥𝑒^𝑥 + 2𝑒^𝑥) − 2(𝑥𝑒^𝑥 + 𝑒^𝑥) + 𝑥𝑒^𝑥 = 0, 𝑟𝑖𝑔ℎ𝑡 − ℎ𝑎𝑛𝑑 𝑠𝑖𝑑𝑒: 0.

Note, too, that in Example 1 each differential equation possesses the constant solution 𝑦 = 0, −∞ < 𝑥 < ∞. A solution of a differential equation that is identically zero on an interval 𝐼 is said to be a trivial solution.

EXPLICIT AND IMPLICIT SOLUTIONS:

Explicit Solution: A solution in which the dependent variable is expressed only in terms of the independent variable and constants is said to be an explicit solution.

We can express this solution by either

𝑦 = 𝜙(𝑥) or 𝑦 = 𝜙(𝑥, 𝑐).

For instance, in the last example we found that 𝑦 =₁₆¹ 𝑥⁴ and 𝑦 = 𝑥𝑒^𝑥 are, in turn, explicit

solutions of 𝑑𝑦 𝑑𝑥⁄ = 𝑥𝑦^1/2 and 𝑦^′′− 2𝑦^′+ 𝑦 = 0. Moreover, the trivial solution 𝑦 = 0 is an explicit solution of the two equations.

Implicit Solution: Some methods of solution do not always lead directly to an explicit solution 𝑦 = 𝜙(𝑥) or 𝑦 = 𝜙(𝑥, 𝑐). This is particularly true when we attempt to solve nonlinear first-order

differential equations. Often we have to be content with a relation or expression 𝐺(𝑥, 𝑦) = 0 or 𝐺(𝑥, 𝑦, 𝑐) = 0 that defines a solution 𝜙 implicitly.

FAMILIES OF SOLUTIONS:

A solution containing an arbitrary constant represents a set 𝐺(𝑥, 𝑦, 𝑐) = 0 of solutions called a one- parameter family of solutions. When solving an 𝑛th-order differential equation

𝐹(𝑥, 𝑦, 𝑦

^′

, … , 𝑦

^(𝑛)

) = 0,

we seek an 𝒏-parameter family of solutions

𝐺(𝑥, 𝑦, 𝑐

₁

, 𝑐

₂

, … , 𝑐

_𝑛

) = 0.

GENERAL AND PARICULAR SOLUTIONS:

General Solution:

If a solution of a differential equation has at least one arbitrary constant appear in the solution, we call this solution is a general solution.

Particular Solution:

A solution of a differential equation that is free of arbitrary parameters is called a particular solution.

For example, the one-parameter family

𝑦 = 𝑐𝑥 − 𝑥 cos 𝑥

is an explicit solution of the linear first-order equation

𝑥𝑦

^′

− 𝑦 = 𝑥

²

sin 𝑥,

on the interval (−∞, ∞).

Figure 1: Some Solutions of 𝑥𝑦^′− 𝑦 = 𝑥²sin 𝑥

Figure 1, obtained by using graphing software, shows the graphs of some of the solutions in this family. The solution

𝑦 = −𝑥 cos 𝑥,

the blue curve in the figure, is a particular solution corresponding to 𝑐 = 0. Similarly, on the interval (−∞, ∞),

𝑦 = 𝑐

₁

𝑒

^𝑥

+ 𝑐

₂

𝑥𝑒

^𝑥

is a two-parameter family of solutions of the linear second-order equation 𝑦^′′− 2𝑦^′+ 𝑦 = 0 in Example 1. Some particular solutions of the equation are the trivial solution

𝑦 = 0, (𝑐₁ = 𝑐₂ = 0), 𝑦 = 𝑥𝑒^𝑥(𝑐₁ = 0, 𝑐₂ = 1), 𝑦 = 5𝑒^𝑥 − 2𝑥𝑒^𝑥 (𝑐₁ = 5, 𝑐₂ = −2), and so on.

Singular Solution:

Sometimes a differential equation possesses a solution that is not a member of a family of solutions of the equation —that is, a solution that cannot be obtained by specializing any of the parameters in the family of solutions. Such an extra solution is called a singular solution.

(1.1) Definitions and Terminology

EXAMPLE 2 (Particular and Singular Solutions)

Determine whether the solutions 𝑦 =₁₆¹ 𝑥⁴ and 𝑦 = 0 of the differential equation 𝑑𝑦 𝑑𝑥⁄ = 𝑥𝑦^1/2 on (−∞, ∞) represent particular or singular solutions?

SOLUTION:

In Section 2.2 we shall demonstrate, by actually solving the differential equation 𝑑𝑦 𝑑𝑥⁄ = 𝑥𝑦^1/2

which possesses one-parameter family of solutions 𝑦 = (1

4𝑥²+ 𝑐)

2

.

… … … . . … (5)

 For the solution 𝑦 =₁₆¹ 𝑥⁴. Substitute into (5), we obtain 1

16𝑥⁴ = (1

4𝑥²+ 𝑐)

2

Take the square root for both sides, we get 1

4𝑥² =1

4𝑥²+ 𝑐 ⇒ 𝑐 = 0 Hence, the solution 𝑦 = ₁₆¹ 𝑥⁴ is a particular solution.

 For the solution 𝑦 = 0. Substitute into (5), we obtain 0 = (1

4𝑥²+ 𝑐)

2

Take the square root for both sides, we get 0 =1

4𝑥²+ 𝑐 ⇒ 𝑐 = −1 4𝑥²

Hence, the solution 𝑦 = 0 is a singular solution because the arbitrary parameter 𝑐 must be constant.

EXAMPLE 3 (Particular and Singular Solutions)

Determine whether the solutions 𝑦 = 2 and 𝑦 = −2 of the differential equation 𝑑𝑦 𝑑𝑥⁄ = 𝑦²− 4 on (−∞, ∞) represent particular or singular solutions?

SOLUTION:

In Section 2.2 we shall demonstrate, by actually solving the differential equation 𝑑𝑦 𝑑𝑥⁄ = 𝑦²− 4

which possesses one-parameter family of solutions 𝑦 = 21 + 𝑐 𝑒^4𝑥

1 − 𝑐 𝑒^4𝑥.

… … … . . … (6)

 For the solution 𝑦 = 2. Substitute into (6), we obtain 2 = 21 + 𝑐 𝑒^4𝑥

1 − 𝑐 𝑒^4𝑥 ⇒ 1 =1 + 𝑐 𝑒^4𝑥

1 − 𝑐 𝑒^4𝑥 ⇒ 1 − 𝑐 𝑒^4𝑥 = 1 + 𝑐 𝑒^4𝑥 ⇒ 2𝑐 𝑒^4𝑥 = 0 ⇒ 𝑐 = 0

Hence, the solution 𝑦 = 2 is a particular solution.

 For the solution 𝑦 = −2. Substitute into (6), we obtain

−2 = 21 + 𝑐 𝑒^4𝑥

1 − 𝑐 𝑒^4𝑥 ⇒ −1 =1 + 𝑐 𝑒^4𝑥

1 − 𝑐 𝑒^4𝑥 ⇒ −1 + 𝑐 𝑒^4𝑥 = 1 + 𝑐 𝑒^4𝑥 ⇒ −1 = 1, which is wrong.

Hence, the solution 𝑦 = −2 is a singular solution because we couldn’t find an arbitrary parameter (𝑐).

EXAMPLE 4 (Using Different Symbols)

Verify that the combination of the functions 𝑥 = 𝑐₁cos 4𝑡 and 𝑥 = 𝑐₂sin 4𝑡, where 𝑐₁ and 𝑐₂ are arbitrary constants or parameters, are both solutions of the linear differential equation

𝑥^′′+ 16𝑥 = 0.

SOLUTION:

The linear combination of solutions, or the two-parameter family is 𝑥 = 𝑐₁cos 4𝑡 + 𝑐₂sin 4𝑡 So,

𝑥^′ = −4𝑐₁sin 4𝑡 + 4𝑐₂cos 4𝑡 𝑥^′′ = −16𝑐₁cos 4𝑡 − 16𝑐₂sin 4𝑡

𝐋. 𝐇. 𝐒. = 𝑥^′′+ 16𝑥 = −16𝑐₁cos 4𝑡 − 16𝑐₂sin 4𝑡 + 16(𝑐₁cos 4𝑡 + 𝑐₂sin 4𝑡) = −16𝑐₁cos 4𝑡 − 16𝑐₂sin 4𝑡 + 16𝑐₁cos 4𝑡 + 16𝑐₂sin 4𝑡 = 0 = 𝐑. 𝐇. 𝐒.

(1.1) Definitions and Terminology

Exercises 1.1: Page 10

(𝟐)

State the order of the given ordinary differential equation. Determine whether the equation is linear or nonlinear by matching it with equation (3) above:

𝑥 𝑑

³

𝑦

𝑑𝑥

³

− ( 𝑑𝑦 𝑑𝑥 )

4

+ 𝑦 = 0

SOLUTION:

The order of the given ordinary differential equation is of third order. The given ODE is nonlinear because of

(

^𝑑𝑦_𝑑𝑥

)

⁴

.

--- (𝟗)

Determine whether the given first-order differential equation is linear in the indicated dependent variable by matching it with the first differential equation given in (4) above:

(𝑦

²

− 1) 𝑑𝑥 + 𝑥 𝑑𝑦 = 0; in 𝑦; in 𝑥

SOLUTION:

We rewrite the given equation in terms of 𝑦, we obtain

𝑥 𝑑𝑦

𝑑𝑥 + 𝑦

²

= 1.

The resulting equation is nonlinear in 𝑦 because of 𝑦². However, if we rewrite the given differential equation in 𝑥, we have

(

𝑦² − 1

) 𝑑𝑥

𝑑𝑦 + 𝑥 = 0,

we see that the last ODE is linear in 𝑥.

Exercises 1.1: Page 10 (Homework)

In the following problems, state the order of the given ordinary differential equation. Determine whether the equation is linear or nonlinear by matching it with equation (5) above:

(𝟏) (1 − 𝑥)𝑦

^′′

− 4𝑥𝑦

^′

+ 5𝑦 = cos 𝑥 (𝟓) 𝑑

²

𝑦

𝑑𝑥

²

= √1 + ( 𝑑𝑦 𝑑𝑥 )

2

---

(𝟏𝟎)

Determine whether the given first-order differential equation is linear in the indicated dependent variable by matching it with the first differential equation given in (6) above:

𝑢 𝑑𝑣 + (𝑣 + 𝑢𝑣 − 𝑢𝑒

^𝑢

) 𝑑𝑢 = 0; in 𝑣; in 𝑢

---

In the following problems, verify that the indicated function is an explicit solution of the given differential equation. Assume an appropriate interval 𝐼 of definition for each solution.

(𝟏𝟐) 𝑑𝑦

𝑑𝑡 + 20𝑦 = 24; 𝑦 = 6 5 − 6

5 𝑒

^−20𝑡

(𝟏𝟒) 𝑦

^′′

+ 𝑦 = tan 𝑥 ; 𝑦 = −(cos 𝑥) ln(sec 𝑥 + tan 𝑥)