Differential equations are among the most powerful tools we have for analyzing the world mathematically. They are used to formulate the fundamental laws of nature (from Newton’s Laws to Maxwell’s equations and the laws of quantum mechanics) and to model the most diverse physical phenomena. This chapter provides an introduction to some elementary techniques and applications of this important subject.

A differential equation is an equation that involves an unknown function y and its first or higher derivatives. A solution is a function y = f (x) satisfying the given equation. As we have seen in previous chapters, solutions usually depend on one or more arbitrary constants (denoted A, B, and C in the following examples):

The first step in any study of differential equations is to classify the equations according to various properties. The most important attributes of a differential equation are its order and whether or not it is linear.The order of a differential equation is the order of the highest derivative appearing in the equation. The general solution of an equation of order n usually involves n arbitrary constants. For example,

has order 2 and its general solution has two arbitrary constants A and B.

" 0y y

A differential equation is called linear if it can be written in the form

11 1 0'n n

n na x y a x y a xy a x y b x

The coefficients aj(x) and b(x) can be arbitrary functions of x, but

a linear equation cannot have terms such as y3, yy , or siny. '

ththe derivative of ny n y

is separable but not linear. Then find the

general solution and plot the family of solutions.

Separation of Variables

sin is separable.

is not separable because is not the product of & .

dyx y

dxdy

x y x y f x g ydx

Separable Equations have the form .dy

f x g ydx

Show that 0dy

y xdx

A differential equation is called linear if it can be written in the form

11 1 0'n n

n na x y a x y a xy a x y b x

'yy NL2

2 2

2 2

2 2

y x C

y xydy xdx C

y x C

2Solutions to 0. dy

y x C y xdx

Although it is useful to find general solutions, in applications we are usually interested in the solution that describes a particular physical situation. The general solution to a first-order equation generally depends on one arbitrary constant, so we can pick out a particular solution y(x) by specifying the value y(x0) for some fixed x0. This specification is called an initial condition. A differential equation together with an initial condition is called an initial value problem.

Family of Solutions

Initial Value Problem Solve the initial value problem

2 2

2

2

2

2 2

02

/2

ln2

, 0 3 3 3

3

Ct t

C

t

t

dy dyty tdt

dt y

dy ttdt y C

y

y e y e

y Ce y Ce C

e

y e

Since C is arbitrary, eC represents an arbitrary positive number, and ±eC is an arbitrary nonzero number. We replace ±eC by C and write the general solution as

' , 0 3y ty y

Law of Exponents

In the context of differential equations, the term “modeling” means finding a differential equation that describes a given physical situation. As an example, consider water leaking through a hole at the bottom of a tank. The problem is to find the water level y (t) at time t. We solve it by showing that y (t) satisfies a differential equation

The key observation is that the water lost during the interval from t to t + Δt can be computed in two ways. Let

First, we observe that the water exiting through the hole during a time interval Δt forms a cylinder of base B and height υ(y)Δt (because the water travels a distance υ(y)Δt). The volume of this cylinder is approximately Bυ(y)Δt [approximately but not exactly, because υ(y) may not be constant]. Thus,

velocity of water flowing through the hole

when the tank is filled to height

area of the hole

area of horizontal cross-section of the tank

at height

v y

y

B

A y

y

Water lost between and t t t Bv y t Water leaks out of a tank through a hole of area B at the bottom.

V Bh

Second, we note that if the water level drops by an amount Δy during the interval Δt, then the volume of water lost is approximately A(y)Δy. Therefore,

In the context of differential equations, the term “modeling” means finding a differential equation that describes a given physical situation. As an example, consider water leaking through a hole at the bottom of a tank. The problem is to find the water level y (t) at time t. We solve it by showing that y (t) satisfies a differential equation

The key observation is that the water lost during the interval from t to t + Δt can be computed in two ways. Let

velocity of water flowing through the hole

when the tank is filled to height

area of the hole

area of horizontal cross-section of the tank

at height

v y

y

B

A y

y

Water lost between and t t t A y y

Bv y Bv yy dy

t A y dt A y

Water leaks out of a tank through a hole of area B at the bottom.

A y y Bv y t

0lim

t

0

' limt

f t t f tf t

t

To use our differential equation, we need to know the velocity of the water leaving the hole. This is given by Torricelli’s Law with (g = 9.8 m/s2):

Bv ydy

dt A y

Given 2

2 9.8 4.43 m/s

v y gy

y y

Bv ydy

dt A y 2 2 9.8 4.43 m/sv y gy y y

Application of Torricelli’s Law A cylindrical tank of height 4 m and radius 1 m is filled with water. Water drains through a square hole of side 2 cm in the bottom. Determine the water level y(t) at time t (seconds). How long does it take for the tank to go from full to empty?Solution We use units of centimeters .

2 210,000 cmA y r 24 cmB 2 29.8 m/s 980 cm/sg

2 980 44.3 m/sv y y y

4 44.30.0056

10,000

yBv ydyy

dt A y

1 m 100 cmis constant...

1/ 2

2

0.0056 0.0056

0.0056 2 0.0056

0.0028 0.0028

dy dyy dt

dt y

dydt y t C

y

y t C y C t

Step 2. Use the initial condition (the tank was full).

20 400 cm 400 20y C C Which sign is correct?

220 0.0028y t t

7142 set

0y

has to be decreasing and a 20 makes an increasing function. y y

CONCEPTUAL INSIGHT The previous example highlights the need to analyze solutions to differential equations rather than relying on algebra alone. The algebra seemed to suggest that C = ±20, but further analysis showed that C = −20 does not yield a solution for t ≥ 0. Note also that the function

y (t) = (20 − 0.0028t)2

is a solution only for t ≤ te—that is, until the tank is empty. This function cannot satisfy our original differential equation for t > te because its derivative is positive for t > te, and solutions of the given differential must have nonpositive derivatives.

Only Separation of Variables is tested on the BC Exam.