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CHAPTER 1
Introduction to Differential Equations
1.1 Basic Terminology
Most of the phenomena studied in the sciences and engineering involve processes that change with time. For example, it is well known that the rate of decay of a radioactive material at time t is proportional to the amount of material present at time t. In mathematical terms this says that
dy dt = ky, k a negative constant (1)
where y = y(t) is the amount of material present at time t.
If an object, suspended by a spring, is oscillating up and down, then Newton’s Second Law of Motion (F = ma) combined with Hooke’s Law (the restoring force of a spring is proportional to the displacement of the object) results in the equation
d^2 y dt^2 +^ k
(^2) y = 0, k a positive constant (2)
where y = y(t) denotes the position of the object at time t.
The basic equation governing the diffusion of heat in a uniform rod of finite length L is given by
∂u ∂t = k^2 ∂^2 u ∂x^2
where u = u(x, t) is the temperature of the rod at time t at position x on the rod.
Each of these equations is an example of what is known as a differential equation.
DIFFERENTIAL EQUATION A differential equation is an equation that contains an unknown function together with one or more of its derivatives.
Here are some additional examples of differential equations.
Example 1.
(a) y′^ = x^2 y − y y + 1
(b) x^2 d^2 y dx^2 − 2 x dy dx
(c) ∂^2 u ∂x^2
∂^2 u ∂y^2 = 0 (Laplace’s equation)
(d) d^3 y dx^3
d^2 y dx^2
dy dx = 3e−x.
TYPE As suggested by these examples, a differential equation can be classified into one of two general categories determined by the type of unknown function appearing in the equation. If the
unknown function depends on a single independent variable, then the equation is an ordinary differ- ential equation; if the unknown function depends on more than one independent variable, then the equation is a partial differential equation. According to this classification, the differential equations (1) and (2) are ordinary differential equations, and (3) is a partial differential equation. In Exam- ple 1, equations (a), (b) and (d) are ordinary differential equations and equation (c) is a partial differential equation.
Differential equations, both ordinary and partial, are also classified according to the highest- ordered derivative of the unknown function.
ORDER The order of a differential equation is the order of the highest derivative of the unknown function appearing in the equation.
Equation (1) is a first order equation, and equations (2) and (3) are second order equations. In Example 1, equation (a) is a first order equation, (b) and (c) are second order equations, and equation (d) is a third order equation.
In general, the higher the order the more complicated the equation. In Chapter 2 we will consider some first order equations and in Chapter 3 we will study certain kinds of second order equations. Higher order equations and systems of equations will be considered in Chapter 6
The obvious question that we want to consider is that of “solving” a given differential equation.
SOLUTION A solution of a differential equation is a function defined on some interval I (in the case of an ordinary differential equation) or on some domain D in two or higher dimensional space (in the case of a partial differential equation) with the property that the equation reduces to an identity when the function is substituted into the equation.
Example 2. Given the second-order ordinary differential equation
x^2 y′′^ − 2 x y′^ + 2y = 4x^3 [Example 1 (b)]
show that:
(a) y(x) = x^2 + 2x^3 is a solution.
(b) z(x) = 2x^2 + 3x is not a solution.
SOLUTION (a) The first step is to calculate the first two derivatives of y.
y = x^2 + 2x^3 , y′^ = 2 x + 6x^2 , y′′^ = 2 + 12x.
Next, we substitute y and its derivatives into the differential equation.
x^2 (2 + 12x) − 2 x(2x + 6x^2 ) + 2(x^2 + 2x^3 ) = 4? x^3.
Simplifying the left-hand side, we get
2 x^2 + 12x^3 − 4 x^2 − 12 x^3 + 2x^2 + 4x^3 = 4? x^3 and 4 x^3 = 4x^3.
Exercises 1.
- Classify the following differential equations with respect to type (i.e., ordinary or partial) and order.
(a) (y′^ )^2 + xyy′^ = sin x.
(b) y′′^ + exy^ = tan x.
(c) ∂^2 u ∂x^2
∂^2 u ∂x∂y
∂^2 u ∂y^2
(d)
d^2 y dx^2
(e) y′′′^ − 5 xy′^ + y = ex^ − 1. (f) ∂u/∂x = k(∂u/∂y).
(g) d^2 y dx^2 −^2 y
dy dx +^ xy
(^2) = d^3 dx^3 [e
− 2 x].
For each differential equation determine whether or not the given functions are solutions.
y′′^ + 4y = 0; y(x) = sin 3x, z(x) = cos 2x + 2 sin 2x.
d^3 y dx^3
dy dx = ex; y(x) = 1 + sin x + 12 ex, z(x) = 2 cos x + 12 ex^.
xy′′^ + y′^ = 0; y 1 (x) = ln (1/x), y 2 (x) = x^2.
(x + 1)y′′^ + xy′^ − y = (x + 1)^2 ; y(x) = e−x^ + x^2 + 1, z(x) = x^2 + 1.
d^3 y dx^3
d^2 y dx^2
dy dx = 0; y(x) = c 1 e^2 x^ + c 2 e^3 x, c 1 , c 2 constants, z(x) = 2e^2 x^ + 3e^3 x^ + 4.
∂^2 u ∂x^2
∂^2 u ∂y^2 = 0; u 1 (x, y) = ln
x^2 + y^2 , u 2 (x, y) = x^3 − 3 xy^2.
- y′′^ − y = 2 − x; y(x) = e−x^ + x − 2 , z(x) = sinh x + x − 2.
∂u ∂t = k^2 ∂^2 u ∂x^2 ; u 1 (x, t) = e−k (^2) t cos x, u 2 (x, t) = e−k (^2) t sin 2πx.
Find the set of all solutions of each of the following differential equations.
y′^ = 2x + ln x
y′′^ = 32.
y′′^ = 6x + cos 2x.
dy dx = 3y.
- x dy dx +^ y^ = 0. Determine values of r, if possible, so that the given differential equation has a solution of the form y = erx.
- y′′^ − 4 y = 0.
- y′′^ + 2y′^ − 8 y = 0.
- y′′^ − 6 y′^ + 9y = 0.
- y′′′^ − 4 y′′^ + 5y′^ − 2 y = 0.
- y′′^ − 2 y′^ + 5y = 0.
Determine values of r, if possible, so that the given differential equation has a solution of the form y = xr^.
- x^2 d^2 y dx^2 − 2 x dy dx + 2y = 0.
- x^2 y′′^ + xy′^ − 9 y = 0.
- x^2 y′′^ − 3 xy′^ + 4y = 0.
and
y =
3 x^2 + 2e^2 x^ + C 1
dx = x^3 + e^2 x^ + C 1 x + C 2 , C 1 , C 2 arbitrary constants.
The set of functions y = x^3 + e^2 x^ + C 1 x + C 2
is a two-parameter family of solutions of the differential equation
y′′^ = 6x + 4e^2 x.
Again from calculus, we can conclude that this family actually represents the set of all solutions of the differential equation; there are no other solutions.
n-PARAMETER FAMILY OF SOLUTIONS The examples given above are very special cases. In general, to find a set of solutions of an n-th order differential equation we would expect, intuitively, to “integrate” n times, with each integration step producing an arbitrary constant of integration. As a result, we expect an n-th order differential equation to have an n-parameter family of solutions.
SOLVING A DIFFERENTIAL EQUATION To solve an n-th order differential equation means to find an n-parameter family of solutions. It is important to understand that the two n’s here are the same. For example, to solve a fourth-order differential equation we need to find a four-parameter family of solutions.
Example 3. Show that y = Cekx^ is a one-parameter family of solutions of
y′^ = ky, k a given constant. (Equation (1) in Section 1.1)
SOLUTION
y = Cekx
y′^ = kCekx
Substituting into the differential equation, we get
kCekx^ =? k
Cekx
kCekx^ = kCekx.
Thus y = Cekx^ is a one-parameter family of solutions. You were shown in calculus that y = Cekx represents the set of all solutions of the equation.
Example 4. Show that y = C 1 x^2 + C 2 x + 2x^3 is a two-parameter family of solutions of
x^2 y′′^ − 2 xy′^ + 2y = 4x^3.
SOLUTION We calculate the first two derivatives of y and then substitute into the differential equation: y = C 1 x^2 + c 2 x + 2x^3
y′^ = 2 C 1 x + C 2 + 6x^2 ,
y′′^ = 2 C 1 + 12x;
x^2 (2C 1 + 12x) − 2 x
2 C 1 x + C 2 + 6x^2
C 1 x^2 + C 2 x + 2x^3
= 4x^3.
Simplifying the left-hand side and re-arranging the terms, we get
C 1
2 x^2 − 4 x^2 + 2x^2
- C 2 (− 2 x + 2x) + 12x^3 − 12 x^3 + 4x^3 =? 4 x^3
C 1 (0) + C 2 (0) + 4x^3 =? 4 x^3
4 x^3 = 4 x^3
Thus, for any two constants C 1 , C 2 , the function y = C 1 x^2 + C 2 x + 2x^3 , is a solution of the differential equation. The set of functions y = C 1 x^2 + C 2 x + 2x^3 is a two-parameter family of solutions of the equation. In Chapter 3 we will see that this two-parameter family represents the set of all solutions of the equation.
GENERAL SOLUTION/SINGULAR SOLUTIONS For most of the equations that we will study in this course, an n-parameter family of solutions of a given n-th order equation will represent the set of all solutions of the equation. In such cases, the term general solution is often used in place of n-parameter family of solutions. Because it less cumbersome, we will use the term “general solution” rather than “n-parameter family of solutions” recognizing that there is possible imprecision in the use of the term; in some cases an n-parameter family of solutions may not be the set of all solutions.
Solutions of an n-th order differential equation which are not included in an n-parameter family of solutions are called singular solutions.
Example 5. Consider the differential equation
y′^ = 4x(y − 1)^1 /^2.
y = (x^2 + C)^2 + 1 is a one-parameter family of solutions (verify this). (In Section 2.2 you will learn how to solve this equation.) Also, it is easy to see that the constant function y ≡ 1 is a solution of the equation: y ≡ 1 implies y′^ ≡ 0.
and 0 = 4x(1 − 1)^1 /^2 = 0;
the equation is satisfied. This solution is not included in the general solution because there is no value that you can assign to C that will produce the solution y ≡ 1; y ≡ 1 is a singular solution.
Additional examples of differential equations having singular solutions are given in the Exercises 1.3.
PARTICULAR SOLUTION If specific values are assigned to the arbitrary constants in the general solution of a differential equation, then the resulting solution is called a particular solution of the equation.
Example 9. Find the differential equation of the two-parameter family
y = C 1 cos 2x + C 2 sin 2x
SOLUTION We differentiate twice:
y = C 1 cos 2x + C 2 sin 2x y′^ = − 2 C 1 sin 2x + 2C 2 cos 2x y′′^ = − 4 C 1 cos 2x − 4 C 2 sin 2x
Multiplying the first equation by 4 and adding it to the third equation, we get
y′′^ + 4y = 0.
This is the differential equation of the given family.
Remark These examples illustrate that there is no ”general method” for finding the differential equation of a given n-parameter family of functions. You can only follow the strategy and try to find some way to eliminate the parameters from the system of equations.
1.3 Initial Conditions; Initial-Value Problems
As we noted in the preceding section, we can obtain a particular solution of an nth order differential equation simply by assigning specific values to the n constants in the general solution. However, in typical applications of differential equations you will be asked to find a solution of a given equation that satisfies certain preassigned conditions.
Example 1. Find a solution of y′^ = 3x^2 − 2 x
that passes through the point (1, 3).
SOLUTION In this case, we can find the general solution by integrating:
y =
3 x^2 − 2 x
dx = x^3 − x^2 + C.
The general solution is y = x^3 − x^2 + C.
To find a solution that passes through the point (2, 6), we set x = 2 and y = 6 in the general solution and solve for C:
6 = 2^3 − 22 + C = 8 − 4 + C which implies C = 2.
Thus, y = x^3 − x^2 + 2 is a solution of the differential equation that satisfies the given condition. In fact, it is the only solution that satisfies the condition since the general solution represented all solutions of the equation and the constant C was uniquely determined.
Example 2. Find a solution of x^2 y′′^ − 2 xy′^ + 2y = 4x^3
which passes through the point (1, 4) with slope 2.
SOLUTION As shown in Example 4 in the preceding section, the general solution of the differential equation is y = C 1 x^2 + C 2 x + 2x^3.
Setting x = 1 and y = 4 in the general solution yields the equation
C 1 + C 2 + 2 = 4 which implies C 1 + C 2 = 2.
The second condition, slope 2 at x = 1, is a condition on y′. We want y′^ (1) = 2. We calculate y′: y′^ = 2C 1 x + C 2 + 6x^2 ,
and then set x = 1 and y′^ = 2. This yields the equation
2 C 1 + C 2 + 6 = 2 which implies 2 C 1 + C 2 = − 4.
Now we solve the two equations simultaneously:
by moving all the non-zero terms to the left-hand side. Since we are talking about an n-th order equation, y(n)^ must appear explicitly in the expression F. Each of the other arguments may or may not appear explicitly. For example, the third-order differential equation
x^2 y′′′^ − 2 xy′′^ = y^2 exy
written in the form of equation (1) is
x^2 y′′′^ − 2 xy′′^ − y^2 exy^ = 0
and F (x, y, y′, y′′, y′′′) = x^2 y′′′^ − 2 xy′′^ − y^2 exy^. Note that y′^ does not appear explicitly in the equation. However, it is there “implicitly.” For example, y′′^ = (y′^ )′, y′′′^ = (y′)′′.
n-th ORDER INITIAL-VALUE PROBLEM An n-th order initial-value problem consists of an n-th order differential equation
F (x, y, y′, y′′,.. ., y(n)) = 0
together with n (initial) conditions of the form
y(c) = k 0 , y′^ (c) = k 1 , y′′(c) = k 2 ,... , y(n−1)(c) = kn− 1.
where c and k 0 , k 1 ,... , kn− 1 are given numbers.
It is important to understand that to be an n-th order initial-value problem there must be n conditions (same n) of exactly the form indicated in the definition. For example, the problem:
- Find a solution of the differential equation
y′′^ + 9y = 0
satisfying the conditions y(0) = 0, y(π) = 0 is not an initial-value problem; the two conditions are not of the form in the definition.
Similarly, the problem:
- Find a solution of the differential equation
y′′′^ − 3 y′′^ + 3y′^ − y = 0
satisfying the conditions y(0) = 1, y′^ (0) = 2 is not an initial-value problem; a third order equation requires three conditions: y(c) = k 0 , y′^ (c) = k 1 , y′′(c) = k 2.
EXISTENCE AND UNIQUENESS The fundamental questions in any course on differential equations are:
(1) Does a given initial-value problem have a solution? That is, do solutions to the problem exist? (2) If a solution does exist, is it unique? That is, is there exactly one solution to the problem or is there more than one solution.
The initial-value problems in Examples 1, 2, and 3 each had a unique solution; values for the arbitrary constants in the general solution were uniquely determined.
Example 4. The function y = x^2 is a solution of the differential equation y′^ = 2√y and y(0) = 0. Thus the initial-value problem y′^ = 2
y; y(0) = 0.
has a solution. However, y ≡ 0 also satisfies the differential equation and y(0) = 0. Thus, the initial-value problem does not have a unique solution. In fact, for any positive number a, the function
ya(x) =
0 , x ≤ a (x − a)^2 , x > a
is a solution of the initial-value problem.
a
x
y
Example 5. The one-parameter family of functions y = Cx is the general solution of
y′^ = y x
There is no solution that satisfies y(0) = 1; the initial-value problem
y′^ = y x , y(0) = 1
does not have a solution.
The questions of existence and uniqueness of solutions will be addressed in the specific cases of interest to us. A general treatment of existence and uniqueness of solutions of initial-value problems is beyond the scope of this course.
(d) Is there a member of the two-parameter family which satisfies the initial condition y(0) = 0 , y′(0) = 1? If not, why not?
- (a) Show that each member of the two-parameter family of functions
y = C 1 x + C 2 x^1 /^2
is a solution of the differential equation 2 x^2 y′′^ − xy′^ + y = 0.
(b) Find a solution of the initial value problem 2 x^2 y′′^ − xy′^ + y = 0; y(4) = 1, y′(4) = −2. (c) Is there a member of the two-parameter family which satisfies the initial condition y(0) = 1 , y′^ (0) = 2? If not, why not?
- Each member of the two-parameter family of functions
y = C 1 sin x + C 2 cos x
is a solution of the differential equation y′′^ + y = 0. (a) Determine whether there are one or more members of this family that satisfy the condi- tions y(0) = 0, y(π) = 0. (b) Show that the zero function, y ≡ 0, is the only member of the family that satisfies the conditions y(0) = 0, y(π/2) = 0.
- Given the differential equation (y′^ )^2 − xy′^ + y = 0. (a) Show that the family of straight lines y = Cx−C^2 is the general solution of the equation (b) Show that y = 14 x^2 is a solution of the equation. Note that this function is not included in the general solution of the equation; it is a singular solution of the equation.
- Given the differential equation x(y′^ )^2 − 2 yy′^ + 4x = 0.
(a) Show that the one-parameter family y = x^2 + C^2 C is the general solution of the equation (b) Show that each of y = 2x and y = − 2 x is a solution of the equation. Note that these functions are not included in the general solution of the equation; they are singular solutions of the equation.
Find the differential equation of the given family.
- y = Cx^3 + 1.
- y = Cx^2 + 3.
- y = Ce^2 x^ + e−^2 x
- y = Cex^ + sin x.
- y = Cx + C^2.
- y = C 1 ex^ + C 2 e−^2 x.
- y = (C 1 + C 2 x)e^2 x
- y = C 1 x + C 2 x^2.
- y = C 1 x + C 2 x−^1.
- y = C 1 cos 3x + C 2 sin 3x.
- y = C 1 sin (3x + C 2 ).
- y = C 1 e^2 x^ cos (3x + C 2 ).
- y = C 1 + C 2 x + C 3 x^2.
- y = C 1 x + C 2 x^2 + C 3 x^3.
- Verify that the function ya(x) is a solution of the initial-value problem y′^ = 2
y, y(0) = 0.
