Basics of Differential Equations

Calculus is the mathematics of change, and rates of change are expressed by derivatives. Thus, one of the most common ways to use calculus is to set up an equation containing an unknown function

y = f (x)

and its derivative, known as a differential equation. Solving such equations often provides information about how quantities change and frequently provides insight into how and why the changes occur.

Techniques for solving differential equations can take many different forms, including direct solution, use of graphs, or computer calculations. We introduce the main ideas in this chapter and describe them in a little more detail later in the course. In this section we study what differential equations are, how to verify their solutions, some methods that are used for solving them, and some examples of common and useful equations.

General Differential Equations

which is an example of a differential equation because it includes a derivative. There is a relationship between the variables

x

Therefore we can interpret this equation as follows: Start with some function

y = f (x)

Examples of Differential Equations and Their Solutions
Equation	Solution
$y' = 2 x$	$y = x^{2}$
$y' + 3 y = 6 x + 11$	$y = e^{−3 x} + 2 x + 3$
$y'' - 3 y' + 2 y = 24 e^{−2 x}$	$y = 3 e^{x} - 4 e^{2 x} + 2 e^{−2 x}$

Note that a solution to a differential equation is not necessarily unique, primarily because the derivative of a constant is zero. For example,

y = x^{2} + 4

is also a solution to the first differential equation in [link]. We will return to this idea a little bit later in this section. For now, let’s focus on what it means for a function to be a solution to a differential equation.

It is convenient to define characteristics of differential equations that make it easier to talk about them and categorize them. The most basic characteristic of a differential equation is its order.

General and Particular Solutions

The only difference between these two solutions is the last term, which is a constant. What if the last term is a different constant? Will this expression still be a solution to the differential equation? In fact, any function of the form

y = x^{2} + C,

represents any constant, is a solution as well. The reason is that the derivative of

x^{2} + C

It can be shown that any solution of this differential equation must be of the form

y = x^{2} + C .

This is an example of a general solution to a differential equation. A graph of some of these solutions is given in [link]. (Note: in this graph we used even integer values for

C

In this example, we are free to choose any solution we wish; for example,

y = x^{2} - 3

is a member of the family of solutions to this differential equation. This is called a particular solution to the differential equation. A particular solution can often be uniquely identified if we are given additional information about the problem.

Initial-Value Problems

Usually a given differential equation has an infinite number of solutions, so it is natural to ask which one we want to use. To choose one solution, more information is needed. Some specific information that can be useful is an initial value, which is an ordered pair that is used to find a particular solution.

A differential equation together with one or more initial values is called an initial-value problem. The general rule is that the number of initial values needed for an initial-value problem is equal to the order of the differential equation. For example, if we have the differential equation

y^{'} = 2 x,

is an initial value, and when taken together, these equations form an initial-value problem. The differential equation

y ″ - 3 y^{'} + 2 y = 4 e^{x}

is second order, so we need two initial values. With initial-value problems of order greater than one, the same value should be used for the independent variable. An example of initial values for this second-order equation would be

y (0) = 2

These two initial values together with the differential equation form an initial-value problem. These problems are so named because often the independent variable in the unknown function is

t,

In [link], the initial-value problem consisted of two parts. The first part was the differential equation

y^{'} + 2 y = 3 e^{x},

The same is true in general. An initial-value problem will consists of two parts: the differential equation and the initial condition. The differential equation has a family of solutions, and the initial condition determines the value of

C .

The family of solutions to the differential equation in [link] is given by

y = 2 e^{−2 t} + C e^{t} .

This family of solutions is shown in [link], with the particular solution

y = 2 e^{−2 t} + e^{t}

In physics and engineering applications, we often consider the forces acting upon an object, and use this information to understand the resulting motion that may occur. For example, if we start with an object at Earth’s surface, the primary force acting upon that object is gravity. Physicists and engineers can use this information, along with Newton’s second law of motion (in equation form

F = m a,

In [link] we assume that the only force acting on a baseball is the force of gravity. This assumption ignores air resistance. (The force due to air resistance is considered in a later discussion.) The acceleration due to gravity at Earth’s surface,

g,

We introduce a frame of reference, where Earth’s surface is at a height of 0 meters. Let

v (t)

To do this, we set up an initial-value problem. Suppose the mass of the ball is

m,

is measured in kilograms. We use Newton’s second law, which states that the force acting on an object is equal to its mass times its acceleration

(F = m a) .

However, this force must be equal to the force of gravity acting on the object, which (again using Newton’s second law) is given by

F_{g} = − m g,

since this force acts in a downward direction. Therefore we obtain the equation

F = F_{g},

Notice that this differential equation remains the same regardless of the mass of the object.

We now need an initial value. Because we are solving for velocity, it makes sense in the context of the problem to assume that we know the initial velocity, or the velocity at time

t = 0 .

A natural question to ask after solving this type of problem is how high the object will be above Earth’s surface at a given point in time. Let

s (t)

denote the height above Earth’s surface of the object, measured in meters. Because velocity is the derivative of position (in this case height), this assumption gives the equation

s^{'} (t) = v (t) .

An initial value is necessary; in this case the initial height of the object works well. Let the initial height be given by the equation

s (0) = s_{0} .

If the velocity function is known, then it is possible to solve for the position function as well.

Key Concepts

Determine the order of the following differential equations.

y^{'} + y = 3 y^{2}

1

{(y^{'})}^{2} = y^{'} + 2 y

y ‴ + y ″ y^{'} = 3 x^{2}

3

y^{'} = y ″ + 3 t^{2}

\frac{d y}{d t} = t

1

\frac{d y}{d x} + \frac{d^{2} y}{d x^{2}} = 3 x^{4}

{(\frac{d y}{d t})}^{2} + 8 \frac{d y}{d t} + 3 y = 4 t

1

Verify that the following functions are solutions to the given differential equation.

y = \frac{x^{3}}{3}

solves $y^{'} = x^{2}$

y = 2 e^{− x} + x - 1

solves $y^{'} = x - y$

y = e^{3 x} - \frac{e^{x}}{2}

solves $y^{'} = 3 y + e^{x}$

y = \frac{1}{1 - x}

solves $y^{'} = y^{2}$

y = e^{x^{2} / 2}

solves $y^{'} = x y$

y = 4 + ln x

solves $x y^{'} = 1$

y = 3 - x + x ln x

solves $y^{'} = ln x$

y = 2 e^{x} - x - 1

solves $y^{'} = y + x$

y = e^{x} + \frac{sin x}{2} - \frac{cos x}{2}

solves $y^{'} = cos x + y$

y = π e^{− cos x}

solves $y^{'} = y sin x$

Verify the following general solutions and find the particular solution.

Find the particular solution to the differential equation $y^{'} = 4 x^{2}$

that passes through $(−3, −30),$

given that $y = C + \frac{4 x^{3}}{3}$

is a general solution.

Find the particular solution to the differential equation $y^{'} = 3 x^{3}$

that passes through $(1, 4.75),$

given that $y = C + \frac{3 x^{4}}{4}$

is a general solution.

y = 4 + \frac{3 x^{4}}{4}

Find the particular solution to the differential equation $y^{'} = 3 x^{2} y$

that passes through $(0, 12),$

given that $y = C e^{x^{3}}$

is a general solution.

Find the particular solution to the differential equation $y^{'} = 2 x y$

that passes through $(0, \frac{1}{2}),$

given that $y = C e^{x^{2}}$

is a general solution.

y = \frac{1}{2} e^{x^{2}}

Find the particular solution to the differential equation $y^{'} = {(2 x y)}^{2}$

that passes through $(1, - \frac{1}{2}),$

given that $y = - \frac{3}{C + 4 x^{3}}$

is a general solution.

Find the particular solution to the differential equation $y^{'} x^{2} = y$

that passes through $(1, \frac{2}{e}),$

given that $y = C e^{− 1 / x}$

is a general solution.

y = 2 e^{− 1 / x}

Find the particular solution to the differential equation $8 \frac{d x}{d t} = −2 cos (2 t) - cos (4 t)$

that passes through $(π, π),$

given that $x = C - \frac{1}{8} sin (2 t) - \frac{1}{32} sin (4 t)$

is a general solution.

Find the particular solution to the differential equation $\frac{d u}{d t} = tan u$

that passes through $(1, \frac{π}{2}),$

given that $u = {sin}^{−1} (e^{C + t})$

is a general solution.

u = {sin}^{−1} (e^{−1 + t})

Find the particular solution to the differential equation $\frac{d y}{d t} = e^{(t + y)}$

that passes through $(1, 0),$

given that $y = − ln (C - e^{t})$

is a general solution.

Find the particular solution to the differential equation $y^{'} (1 - x^{2}) = 1 + y$

that passes through $(0, −2),$

given that $y = C \frac{\sqrt{x + 1}}{\sqrt{1 - x}} - 1$

is a general solution.

y = - \frac{\sqrt{x + 1}}{\sqrt{1 - x}} - 1

For the following problems, find the general solution to the differential equation.

y^{'} = 3 x + e^{x}

y^{'} = ln x + tan x

y = C - x + x ln x - ln (cos x)

y^{'} = sin x e^{cos x}

y^{'} = 4^{x}

y = C + \frac{4^{x}}{ln (4)}

y^{'} = {sin}^{−1} (2 x)

y^{'} = 2 t \sqrt{t^{2} + 16}

y = \frac{2}{3} \sqrt{t^{2} + 16} (t^{2} + 16) + C

x^{'} = coth t + ln t + 3 t^{2}

x^{'} = t \sqrt{4 + t}

x = \frac{2}{15} \sqrt{4 + t} (3 t^{2} + 4 t - 32) + C

y^{'} = y

y^{'} = \frac{y}{x}

y = C x

Solve the following initial-value problems starting from $y (t = 0) = 1$

and $y (t = 0) = −1 .$

Draw both solutions on the same graph.

\frac{d y}{d t} = 2 t

\frac{d y}{d t} = − t

y = 1 - \frac{t^{2}}{2}, y = - \frac{t^{2}}{2} - 1

\frac{d y}{d t} = 2 y

\frac{d y}{d t} = − y

y = e^{− t}, y = − e^{− t}

\frac{d y}{d t} = 2

Solve the following initial-value problems starting from $y_{0} = 10 .$

At what time does $y$

increase to $100$

or drop to $1 ?$

\frac{d y}{d t} = 4 t

y = 2 (t^{2} + 5), t = 3 \sqrt{5}

\frac{d y}{d t} = 4 y

\frac{d y}{d t} = −2 y

y = 10 e^{−2 t}, t = - \frac{1}{2} ln (\frac{1}{10})

\frac{d y}{d t} = e^{4 t}

\frac{d y}{d t} = e^{−4 t}

y = \frac{1}{4} (41 - e^{−4 t}),

never

Recall that a family of solutions includes solutions to a differential equation that differ by a constant. For the following problems, use your calculator to graph a family of solutions to the given differential equation. Use initial conditions from $y (t = 0) = −10$

to $y (t = 0) = 10$

increasing by $2 .$

Is there some critical point where the behavior of the solution begins to change?

[T] $y^{'} = y (x)$

[T] $x y^{'} = y$

Solution changes from increasing to decreasing at $y (0) = 0$

[T] $y^{'} = t^{3}$

[T] $y^{'} = x + y$

(Hint: $y = C e^{x} - x - 1$

is the general solution)

Solution changes from increasing to decreasing at $y (0) = 0$

[T] $y^{'} = x ln x + sin x$

Find the general solution to describe the velocity of a ball of mass $1 lb$

that is thrown upward at a rate $a$

ft/sec.

v (t) = −32 t + a

In the preceding problem, if the initial velocity of the ball thrown into the air is $a = 25$

ft/s, write the particular solution to the velocity of the ball. Solve to find the time when the ball hits the ground.

You throw two objects with differing masses $m_{1}$

and $m_{2}$

upward into the air with the same initial velocity $a$

ft/s. What is the difference in their velocity after $1$

second?

0

ft/s

[T] You throw a ball of mass $1$

kilogram upward with a velocity of $a = 25$

m/s on Mars, where the force of gravity is $g = −3.711$

m/s². Use your calculator to approximate how much longer the ball is in the air on Mars.

[T] For the previous problem, use your calculator to approximate how much higher the ball went on Mars.

4.86

meters

[T] A car on the freeway accelerates according to $a = 15 cos (π t),$

where $t$

is measured in hours. Set up and solve the differential equation to determine the velocity of the car if it has an initial speed of $51$

mph. After $40$

minutes of driving, what is the driver’s velocity?

[T] For the car in the preceding problem, find the expression for the distance the car has traveled in time $t,$

assuming an initial distance of $0 .$

How long does it take the car to travel $100$

miles? Round your answer to hours and minutes.

x = 50 t - \frac{15}{π^{2}} cos (π t) + \frac{3}{π^{2}}, 2

hours $1$

minute

[T] For the previous problem, find the total distance traveled in the first hour.

Substitute $y = B e^{3 t}$

into $y^{'} - y = 8 e^{3 t}$

to find a particular solution.

y = 4 e^{3 t}

Substitute $y = a cos (2 t) + b sin (2 t)$

into $y^{'} + y = 4 sin (2 t)$

to find a particular solution.

Substitute $y = a + b t + c t^{2}$

into $y^{'} + y = 1 + t^{2}$

to find a particular solution.

y = 1 - 2 t + t^{2}

Substitute $y = a e^{t} cos t + b e^{t} sin t$

into $y^{'} = 2 e^{t} cos t$

to find a particular solution.

Solve $y^{'} = e^{k t}$

with the initial condition $y (0) = 0$

and solve $y^{'} = 1$

with the same initial condition. As $k$

approaches $0,$

what do you notice?

y = \frac{1}{k} (e^{k t} - 1)

and $y = x$

Basics of Differential Equations

General Differential Equations

General and Particular Solutions

Initial-Value Problems

Key Concepts

Glossary