Calculus Of One Real Variable
By Pheng Kim Ving
Chapter 16: Differential Equations
Group 16.1: Differential Equations
Section 16.1.1: Introduction To Differential Equations

 

16.1.1
Introduction To Differential Equations

 

 

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1. Differential Equations

 

In Section 5.8 Example 5.1, we were given x'' = 4, and in Part 6 of the same section, we had y'' = – g. We solved these
equations to determine the functions
x and y respectively. These are situations where a quantity isn't known but its second
derivative is known, and our task is to determine the quantity. There are situations where a quantity isn't known but its rate of
change (first derivative) is known or given, and our task is to determine the quantity. For example, we may have to determine
y if dy/dx = cos x. There are also situations where a quantity isn't known but it and its derivatives of order 1 or higher are
related by an equation, and our task is to determine the quantity. For example, we may have to determine
y if y'' + 3y' – 2y =
0. Recall that a function is considered as its own zeroth (0th) derivative:
y(0) = y. An equation involving derivatives of order 1 or
higher of a function is called a differential equation. To solve a differential equation is to determine the function that solves or
satisfies the equation (it and its derivatives solve or satisfy the equation).

 

Let y = x2. Then y' = 2x. The equation y' = 2x involves the derivative of a function, and for this reason is called a differential
equation
. Now, to work in reverse, suppose a function
y of x is unknown but its derivative is known to be y' = 2x. This is a
differential equation. Clearly an antiderivative of 2
x is x2. If y = x2, then y' = 2x. The function y = x2 is a function that satisfies
or solves the differential equation
y' = 2x, and for this reason is called a solution  of that differential equation. The equation y''
=
x – 1 is also a differential equation. Note that the equations y' = 2x and y'' = x – 1 can also be written as y' – 2x = 0 and
y'' – x + 1 = 0 respectively.

 

Now suppose we have a function y that's unknown and that has derivatives y' and y'' that are also unknown, but all three
satisfy the equation
x2y'' – xy' – 3y = 0. This is a differential equation too. Any function that satisfies or solves it is called a
solution of it.

 

The function y = x2 is a solution of the differential equation y' = 2x, and so are the functions y = x2 + 1,  y = x2 – 200, or y =
x2 + C for any constant C, because the derivative of a constant is 0. Any solution can be obtained from the solution y = x2 + C
by giving a particular value to
C. For this reason, y = x2 + C is called the general solution. It represents all the solutions. A
solution with a particular value of
C is called, well, a particular solution. Remember that a function y = f(x) can be considered
as its own 0th (zeroth) derivative.

 

Consider the differential equation y'' = 2x. Then y' = x2 + C1 and y = (x3/3) + C1x + C2. The function y = (x3/3) + C1x + C2,
where
C1 and C2 are arbitrary constants, is the general solution of the differential equation y'' = 2x.

 

Definition 1.1 – Differential Equations

 

An equation that involves derivative(s) of order(s) 1 and/or higher of an unknown function is called a differential equation.
Any function
f(x) that is defined on an interval and satisfies the differential equation for all x in that interval is called a
solution of the differential equation. A solution containing one ore more arbitrary constants and such that every solution can
be obtained from it by giving particular value(s) to the constant(s) is called the general solution. A solution obtained from
the general solution by giving particular value(s) to the constant(s) is called a particular solution.

 

 

For the derivatives in a differential equation, notations other than the prime notations ( ', '', etc ) are also used. For example,
the following three equations are the same equation:

 

x2y'' – xy' – 3y = 0,

 

 

In this section we investigate two types of differential equations: y' = f(x) and y'' = f(x), where f is a known function (not the
unknown
 function y). For the remainder of this section, the abbreviation “ DE” stands for “ differential equation”.

 

Ordinary And Partial Differential Equations

 

Equations that involve only derivatives of functions of one variable are called ordinary differential equations. Equations that
involve partial derivatives of functions of several (two or more) variables are called partial differential equations. In this
chapter we of course touch on the surface of only ordinary DEs.

 

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2. The Differential Equation y' = f (x)

 

Suppose the function y = F(x) is a solution of the DE y' = f(x), ie F '(x) = f(x). Then so are y = F(x) + 2 and y =
F(x) – 100/7, as the derivative of a constant is 0. In fact, for any constant C, the function y = F(x) + C is a solution. We
see that there are infinitely many solutions. Any two solutions
F(x) + C1 and F(x) + C2 of the DE y' = f(x) differ by a
constant, which here is
C1 – C2. We use the function y = F(x) + C, where C is an arbitrary constant, to represent the
collection of all the solutions, and call it the general solution of the DE
y' = f(x). To solve a differential equation means
to find its general solution.

 

Solving  y' = f (x)

 

If y' = f(x) then f is the derivative of y so y is an antiderivative of f. Thus solving the DE y' = f(x) clearly amounts to
finding the general antiderivative of
f(x). See Section 5.7 Definition 3.1. If f has an antiderivative, then the DE y' = f(x)
has that antiderivative as a solution of it and the general antiderivative of
f as its general solution.

 

Example 2.1

 

Solve y' = x3 – 5x2 + 2x – 4.

 

Solution

EOS

 

Graphs Of Solutions

 

Any two solutions F(x) + C1 and F(x) + C2 of the DE y' = f(x) differ by a constant, which here is C1 – C2. So, geometrically, the set of the graphs of F(x) + C obtained by running the constant C thru the set R of real numbers can be visualized as a collection of graphs congruent to each other. Some members of this collection are sketched in Fig. 2.1. Also see Section 5.7 Graphs Of Antiderivatives. Let x1 be any point of dom( f ) and F1 and F2 any 2 solutions. We have F1'(x1) = f(x1) = F2'(x1). This shows that the graphs of F1 and F2 have the same slope, ie, have parallel tangent lines, at x1. At any point x the graphs of all the solutions of the DE y' = f(x) have the same slope or parallel tangent lines.

 

Fig. 2.1

 

Graphs Of Some Solutions Of y' = f(x).

 

It's worth repeating that any 2 solutions of y' = f(x) differ by a constant. If F(x) is a solution then any other solution can
be obtained by adding a constant to F(x).

 

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3. The Differential Equation y'' = f (x)

 

Example 3.1

 

Solve the differential equation y'' = 5x.

 

Solution

EOS

 

Remarks 3.1

 

a. C1 and C2 are arbitrary constants. The general solution of y'' = 5x is y = (5/6)x3 + C1x + C2. To verify the answer y
    = (5/6)
x3 + C1x + C2 we just differentiate it, twice: y' = (5/2)x2 + C1, y'' = 5x. The general solution of the DE y'' =
    
f(x) is of the form y = G(x) + C1x + C2, where G''(x) = f(x) and C1 and C2 are arbitrary constants.

 

b. We saw above that any 2 solutions of the DE y' = f(x) differ by a constant. Now let's check to see if the same is true
    for the solutions of the DE
y'' = f(x). Let F1(x) = G(x) + C1x + C2 and F2(x) = G(x) + D1x + D2 be any 2 of its
    solutions. Clearly they differ by a constant if
C1 = D1. Now let's check:

 

    F1(x) – F2(x) = (G(x) + C1x + C2) – (G(x) + D1x + D2) = (C1 – D1)x + (C2 – D2).

 

    Thus indeed they differ by a constant if C1 – D1 = 0 or C1 = D1, ie if they are of the form F1(x) = G(x) + Cx + K1
    and
F2(x) = G(x) + Cx + K2; otherwise they don't.

 

    We see that any 2 solutions of y'' = f(x) differ by a linear function including a constant (a constant function is a linear
    function). If
F(x) is a solution of y'' = f(x) then any other solution can be obtained by adding a linear function including
    a constant to
F(x).

 

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4. Initial-Value Problems

 

Note 4.1

 

In Section 5.8 Example 5.1, we solved the DE x'' = 4 subject to two initial conditions x'(0) = 0 and x(0) = 0, and we got
x = 2t2. In Part 6 of the same section, we solved the DE y'' = – g subject to two initial conditions y'(0) (= v(0)) = v0 and
 
y(0) = y0, and we got  y = – (1/2)gt2 + v0t + y0. The problem of solving a DE subject to one or more prescribed values
for the solution function or its derivatives is referred to as an initial-value problem.

 

The Case Of y' = f (x)

 

As we saw above, the DE y' = f(x) has infinitely many solutions, the graphs of some of which are illustrated in Fig. 2.1.
However, as seen there, given a point (
x0, y0) [ y0 is not  f(x0)], there's only one graph that passes thru the point (x0, y0).
If
y = F1(x) is the solution whose graph is that one that passes thru the point (x0, y0), then we must have y0 = F1(x0).

 

Now suppose we're given the DE y' = f(x) and asked to find the particular solution whose graph passes thru a given point
(
x0, y0), ie, the solution y = y(x) such that y(x0) = y0. This type of problem is called an initial-value problem, because
the point (
x0, y0) is considered as the initial point of the solution. For the reason for the use of the adjective initial,  see
Note 4.1 above and Remarks 4.2 c below.

 

Example 4.1

 

Solve the initial-value problem:

 

 

Solution

EOS

 

Remarks 4.1

 

a. For clarity, an initial-value problem is stated in the following format:

 

    Solve the initial-value problem:

 

  

 

b. From the equation y' = x – 2 we get y = x2/2 – 2x + C, which represents a collection of infinitely many antiderivatives y of
    
y'; the graphs of y are congruent to each other. The condition y(1) = 3 tells us to select the particular antiderivative y
    whose graph passes thru the point (1, 3). That antiderivative is
y = x2/2 – 2x + 9/2.

 

c. In the equation y' = f(x), the highest derivative order is 1, and we need 1 initial condition to solve the initial-value problem.

 

The Case Of y'' = f (x)

 

Again consider the DE y'' = 5x. As seen in Example 3.1, its general solution is y = (5/6)x3 + C1x + C2, where there are two
arbitrary constants,
C1 and C2. One way to obtain a particular solution is to specify two initial conditions to determine the
two constants, as shown in the following example.

 

Example 4.2

 

Solve the initial-value problem:

 

 

Solution

EOS

 

Remarks 4.2

 

a.  The initial conditions y'(1) = 2 and y(1) = –1 describe the behavior of the solution at the same  point, x = 1, which is     considered as the initial point of the solution.

 

b.  First, from the equation y'' = 5x we get y' = (5/2)x2 + C1, which represents a collection of infinitely many antiderivatives y'
of
y''; the graphs of y' are congruent to each other. The condition y'(1) = 2 tells us to select the particular antiderivative y'
whose graph passes thru the point (1, 2). That antiderivative is
y' = (5/2)x2 – 1/2.

 

Next, from that antiderivative y' = (5/2)x2 – 1/2 we get its own general antiderivative y = (5/6)x3 – (1/2)x + C2, which
represents a collection of infinitely many antiderivatives
y of y'; the graphs of y are congruent to each other. The condition
y(1) = –1 tells us to select the particular antiderivative y whose graph passes thru the point (1, –1). That antiderivative is y
= (5/6)
x3 – (1/2)x – 4/3. It's the desired solution.

 

c.  The adjective initial  undoubtedly first appeared in the study of motion where it was used to describe initial conditions, ie conditions at initial time t0 = 0. See Section 5.8 Remarks 5.1 iii and Remarks 6.1 iv of the same section. However, presently the phrase initial conditions  is utilized to refer to conditions of the unknown function or its derivatives at any
single value of the independent variable.

 

d. In the equation y'' = f(x), the highest derivative order is 2, and we need 2 initial conditions to solve the initial-value problem.

 

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5. Boundary-Value Problems

 

Since the general solution of the DE y'' = f(x) is of the form y = G(x) + C1x + C2 (see Remarks 3.1), another way to obtain a particular solution is to generate from this equation a system of two equations in the unknowns C1 and C2 that

 

Example 5.1

 

Solve the boundary-value problem:

 

 

Multiplying the first equation by –1 and adding the resulting equation to the second we obtain 2C1 = –  82, so that C1 =
–
41 . Then C2 = 1 – C1 = 1 – (– 41) = 42. Thus the solution is y = x4 – 41x + 42.

EOS

 

Remark 5.1

 

The boundary conditions y(1) = 2 and y(3) = 0 describe the behavior of the solution at different  points, x1 = 1 and x2 =
3, which are viewed as the boundary points of the solution.

 

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Problems & Solutions

 

1.  Solve the following differential equations.

 

   

 

Solution

 

 

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2.  Solve the following differential equations.

 

     a.  y'' = 0.

 

    

 

Solution

 

 

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3.  Solve the following initial-value problems.

 

    

 

Solution

 

 

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4.  Solve the following initial-value problems.

 

    

 

Solution

 

 

 

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5.  Suppose the differential equation y'' = f (x) has a solution y = F(x) on [a, b], where a < b. Show that for any real
    

 

Solution

 

 

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