# Elementary differential equations 7th edition - Boyce W.E

ISBN 0-471-31999-6

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The requirement that y( 1 ) = 5 is an example of an initial condition, and the combination of the ODE and an initial condition

ft=4t, 7(1) = 5 (2)

is called an initial value problem (IVP). Its solution is y = 212 + 3.

? Replace the condition y(1) = 5 in IVP (2) by y(2) = 3 and find the solution of this new initial value problem. How many solutions are there?

? Finding a Solution Formula

An ODE usually has many solutions. How can you find a solution, and how can you describe it? A solution formula provides a useful description, but graphs and tables generated by ODE Architect are also useful, especially in

the all-too-frequent case where no formula can be found. Two techniques to

find solution formulas are summarized here, and others are in your textbook.

Integration

If f (t) is a continuous function, then the general solution of the ODE

A table of integrals comes is y(t) = F (t) + C, where F (t ) is an antiderivative of f. For example, the

in handy here. general solution of dy/ dt = sin t is y = — cos t + C.

Separation of Variables

If you can write a differential equation in the form

dy

?Jt = f(ąy) then wherever g( y) = 0 you can rewrite it as

1 dy g{y) dt

so that

= f (t)

= f(t) dt

Finding a Solution Formula

29

t

Figure 2.2: Four solution curves of dy/dt = sin t/Ciy2 + 1) through the marked initial points.

Keep that table of integrals If H( y) is an antiderivative of 1/g( y) and F (t) is an antiderivative of f (t), handy! then a solution y(t) of the ODE solves the equation

for some constant C.

Here’s an example of a separable ODE:

dy sin t

(3)

dt 3 y2 + 1

Separating the variables and finding the antiderivatives, we see that

(3/ + 1)^ = sin t

y3 + y = — cos t + C (4)

We won ’ t attempt to express a solution y( t) directly in terms of t (and C), but we can check that formula (4) is correct by differentiating each side with respect to t. This gives

yF+F = siat

dt dt

which has the form of ODE (3) if we divide each side by 3y2 + 1. Figure 2.2 shows solution curves of ODE (3) through the initial points (0, -1.5), (0, -1), (0, 0), (0, 1). The curves were plotted by using ODE Architect to solve ODE (3) with the given initial data.

Solution formulas are useful, but they exist only for a small number of ODEs of special forms. That s where numerical solvers like ODE Architect come in—they don ’ t need solution formulas.

30

Chapter 2

? Modeling

The eight steps are described in Chapter 1.

? The Juggler

So the juggler's ODE for vertical motion is H' = — 32.

In the multimedia module ho = 4.5 ft.

A mathematical model is a system of mathematical equations relating specific variables that represent some aspect of a natural process. Modeling involves several steps:

1. State the problem and its context.

2. Identify and assign variables.

3. State the laws that govern the relationships between the variables.

4. Translate the laws into equations.

5. Solve the resulting equations.

6. Interpret and test the solutions in the context of the natural environment.

7. Refine the model until it predicts the empirical data.

8. Interpret the implications of the model.

The models we consider all involve ODEs.

You can observe the modeling process in the following juggler problem.

1. Find an ODE that describes the height of a ball between the time it leaves the juggler's hand, moving vertically upward, and the time it falls back into the hand.

2. Let t = time (in seconds), h = height of the ball above the floor (in feet), v = velocity (in ft/sec), and a = acceleration (in ft/sec2).

3. Apply Newton’s second law of motion to the ball: the mass m of a body times its acceleration is equal to the sum of all of the forces acting on the body. We treat the ball as a point mass encountering negligible air resistance (drag) so the only force acting on the ball is that due to gravity, which acts downward.

4. By Newton’s second law, we have that ma = —mg, where g = 32 ft/sec2 is the acceleration due to gravity near the surface of the earth, and the minus sign indicates the downward direction of the gravitational force. Since the ball’s acceleration is a = V where v is its velocity, and v = h', we can model the ball’s motion by h" = —32. The initial height h0 of the ball is that of the juggler's hand above the floor when the ball is launched upward, and that is easy to measure. The initial velocity v0 is harder to measure directly; it is simpler to solve the model first and then experiment to deduce a reasonable value for v0.

5-8. Solving and testing are up to you. See Figure 2.3 for graphs of h(t) corresponding to h0 = 4 ft and five values of V0.

The Sky Diver

31

? The Sky Diver

This kind of air resistance is called viscous damping.

Time (seconds)

Figure 2.3: Five tosses of the juggler's ball: initial velocities v0 range from 5 to 25 ft/sec. Which time-height curve corresponds to v0 = 25?

? How must you revise the process when the ball is thrown to the juggler’s other hand? (The result appears on Screen 3.4 of Module 2.)

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