BC Exam

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):

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

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,

" 0y y

Just like the order of a Taylor Polynomial!

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 x y 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. '

2

2 2

2 2

2 2

y x C

y xydy xdx C

y x C

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 hav e the form .dy

f x g ydx

Show that 0dyy 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 x y a x y b x

'yy NL

Here we go ;-)

Solve for y

First-Order

This is a conic... A family of Hyperbolas

2 2

2 21

y x

a b

2Solutions to 0. dy

y x C y xdx

2y x C A Degenerate Hyperbola

Let 0c

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

to a Particular

Differential Equation

2 2

2

2

2

2 2

02

/2

ln2

, 0 3 3 3

3

Ct tC

t

t

dy dyty tdt

dt y

dy ttdt y C

y

y e y e

y e y Ce C

e

C

y e

Initial Value Problem Solve the initial value problem

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

or

c

c c

y e f t

y e f t e f t

Family of Solutions

2

2

t

y Ce

2

23t

y e

Laws 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.

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

V BV y t d rt

Differential Equations UP

Not constant,

but close

Bv y Bv yy dy

t A y dt A y

A y By y tv

Second, we note that the water level drops by an amount Δy during the interval Δt.

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

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

Now we can set up our

differential equation!

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 4.43 m/sv y gy y

Velocity of the water passing through the hole is...

Now we simply plug in our known

values and solve the differential equation

using separation of variables.

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?

Bdy

dt y

v y

A

Solution We can use units of centimeters .

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

2 980 44.3 cm/sv y y y

4 44.30.0056

10,000

yBv ydyy

dt A y

1 m 100 cmis constant...

2 4.43 m/sv y gy y

20 400 cm 400 20y C C

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).

Which sign is correct?

7143 set

0y

Separation of VariablesThis is actually an .IVP

10,000

200

400

220 0.0028y t t

400

200

10,000

220 0.0028y t t

Why can't we just find ?dy

dt

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.