CalculusVolume2-OP-86

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Solution
We have a resistor and an inductor in the circuit, so we use Equation 4.24. The voltage drop across the resistor is
given by ER = Ri = 5i. The voltage drop across the inductor is given by EL = Li′ = 0.4i′. The electromotive
force becomes the right-hand side of Equation 4.24. Therefore Equation 4.24 becomes
0.4i′ + 5i = 50sin20t.
Dividing both sides by 0.4 gives the equation
i′ + 12.5i = 125sin20t.
Since the initial current is 0, this result gives an initial condition of i(0) = 0. We can solve this initial-value
problem using the five-step strategy for solving first-order differential equations.
Step 1. Rewrite the differential equation as i′ + 12.5i = 125sin20t. This gives p(t) = 12.5 and
q(t) = 125sin20t.
Step 2. The integrating factor is µ(t) = e
∫ 12.5dt
= e12.5t.
Step 3. Multiply the differential equation by µ(t):
e12.5t i′ + 12.5e12.5t i = 125e12.5t sin20t
d
dt
⎡
⎣ie12.5t⎤
⎦ = 125e12.5t sin20t.
Step 4. Integrate both sides:
⌠
⌡
d
dt
⎡
⎣ie12.5t⎤
⎦dt = ∫ 125e12.5t sin20t dt
ie12.5t = ⎛
⎝
250sin20t − 400cos20t
89
⎞
⎠e12.5t + C
i(t) = 250sin20t − 400cos20t
89 + Ce−12.5t.
Step 5. Solve for C using the initial condition v(0) = 2:
i(t) = 250sin20t − 400cos20t
89 + Ce−12.5t
i(0) = 250sin20(0) − 400cos20(0)
89 + Ce−12.5(0)
0 = −400
89 + C
C = 400
89 .
Therefore the solution to the initial-value problem is
i(t) = 250sin20t − 400cos20t + 400e−12.5t
89 = 250sin20t − 400cos20t
89 + 400e−12.5t
89 .
The first term can be rewritten as a single cosine function. First, multiply and divide by 2502 + 4002 = 50 89:
250sin20t − 400cos20t
89 = 50 89
89
⎛
⎝
250sin20t − 400cos20t
50 89
⎞
⎠
= − 50 89
89
⎛
⎝
8cos20t
89
− 5sin20t
89
⎞
⎠.
Next, define φ to be an acute angle such that cosφ = 8
89
. Then sinφ = 5
89
and
418 Chapter 4 | Introduction to Differential Equations
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4.19
−50 89
89
⎛
⎝
8cos20t
89
− 5sin20t
89
⎞
⎠ = − 50 89
89
⎛
⎝cosφcos20t − sinφsin20t⎞
⎠
= − 50 89
89 cos⎛
⎝20t + φ⎞
⎠.
Therefore the solution can be written as
i(t) = − 50 89
89 cos ⎛
⎝20t + φ⎞
⎠ + 400e−12.5t
89 .
The second term is called the attenuation term, because it disappears rapidly as t grows larger. The phase shift is
given by φ, and the amplitude of the steady-state current is given by 50 89
89 . The graph of this solution appears
in Figure 4.26:
Figure 4.26
A circuit has in series an electromotive force given by E = 20sin5t V, a capacitor with capacitance
0.02 F, and a resistor of 8 Ω. If the initial charge is 4 C, find the charge at time t > 0.
Chapter 4 | Introduction to Differential Equations 419
4.5 EXERCISES
Are the following differential equations linear? Explain
your reasoning.
208.
dy
dx = x2 y + sinx
209.
dy
dt = ty
210.
dy
dt + y2 = x
211. y′ = x3 + ex
212. y′ = y + ey
Write the following first-order differential equations in
standard form.
213. y′ = x3 y + sinx
214. y′ + 3y − lnx = 0
215. −xy′ = (3x + 2)y + xex
216.
dy
dt = 4y + ty + tan t
217.
dy
dt = yx(x + 1)
What are the integrating factors for the following
differential equations?
218. y′ = xy + 3
219. y′ + ex y = sinx
220. y′ = x ln(x)y + 3x
221.
dy
dx = tanh(x)y + 1
222.
dy
dt + 3ty = et y
Solve the following differential equations by using
integrating factors.
223. y′ = 3y + 2
224. y′ = 2y − x2
225. xy′ = 3y − 6x2
226. (x + 2)y′ = 3x + y
227. y′ = 3x + xy
228. xy′ = x + y
229. sin(x)y′ = y + 2x
230. y′ = y + ex
231. xy′ = 3y + x2
232. y′ + lnx = y
x
Solve the following differential equations. Use your
calculator to draw a family of solutions. Are there certain
initial conditions that change the behavior of the solution?
233. [T] (x + 2)y′ = 2y − 1
234. [T] y′ = 3et/3 − 2y
235. [T] xy′ + y
2 = sin(3t)
236. [T] xy′ = 2cosx
x − 3y
237. [T] (x + 1)y′ = 3y + x2 + 2x + 1
238. [T] sin(x)y′ + cos(x)y = 2x
239. [T] x2 + 1y′ = y + 2
240. [T] x3 y′ + 2x2 y = x + 1
Solve the following initial-value problems by using
integrating factors.
241. y′ + y = x, y(0) = 3
242. y′ = y + 2x2, y(0) = 0
243. xy′ = y − 3x3, y(1) = 0
244. x2 y′ = xy − lnx, y(1) = 1
245. ⎛
⎝1 + x2⎞
⎠y′ = y − 1, y(0) = 0
246. xy′ = y + 2x lnx, y(1) = 5
420 Chapter 4 | Introduction to Differential Equations
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247. (2 + x)y′ = y + 2 + x, y(0) = 0
248. y′ = xy + 2xex, y(0) = 2
249. xy′ = y + 2x, y(0) = 1
250. y′ = 2y + xex, y(0) = −1
251. A falling object of mass m can reach terminal
velocity when the drag force is proportional to its velocity,
with proportionality constant k. Set up the differential
equation and solve for the velocity given an initial velocity
of 0.
252. Using your expression from the preceding problem,
what is the terminal velocity? (Hint: Examine the limiting
behavior; does the velocity approach a value?)
253. [T] Using your equation for terminal velocity, solve
for the distance fallen. How long does it take to fall 5000
meters if the mass is 100 kilograms, the acceleration due
to gravity is 9.8 m/s2 and the proportionality constant is
4?
254. A more accurate way to describe terminal velocity is
that the drag force is proportional to the square of velocity,
with a proportionality constant k. Set up the differential
equation and solve for the velocity.
255. Using your expression from the preceding problem,
what is the terminal velocity? (Hint: Examine the limiting
behavior: Does the velocity approach a value?)
256. [T] Using your equation for terminal velocity, solve
for the distance fallen. How long does it take to fall 5000
meters if the mass is 100 kilograms, the acceleration due
to gravity is 9.8 m/s2 and the proportionality constant
is 4? Does it take more or less time than your initial
estimate?
For the following problems, determine how parameter a
affects the solution.
257. Solve the generic equation y′ = ax + y. How does
varying a change the behavior?
258. Solve the generic equation y′ = ax + y. How does
varying a change the behavior?
259. Solve the generic equation y′ = ax + xy. How does
varying a change the behavior?
260. Solve the generic equation y′ = x + axy. How does
varying a change the behavior?
261. Solve y′ − y = ekt with the initial condition
y(0) = 0. As k approaches 1, what happens to your
formula?
Chapter 4 | Introduction to Differential Equations 421
asymptotically semi-stable solution
asymptotically stable solution
asymptotically unstable solution
autonomous differential equation
carrying capacity
differential equation
direction field (slope field)
equilibrium solution
Euler’s Method
general solution (or family of solutions)
growth rate
initial population
initial value(s)
initial velocity
initial-value problem
integrating factor
linear
logistic differential equation
order of a differential equation
particular solution
phase line
separable differential equation
separation of variables
solution curve
solution to a differential equation
CHAPTER 4 REVIEW
KEY TERMS
y = k if it is neither asymptotically stable nor asymptotically unstable
y = k if there exists ε > 0 such that for any value c ∈ (k − ε, k + ε) the solution
to the initial-value problem y′ = f (x, y), y(x0) = c approaches k as x approaches infinity
y = k if there exists ε > 0 such that for any value c ∈ (k − ε, k + ε) the
solution to the initial-value problem y′ = f (x, y), y(x0) = c never approaches k as x approaches infinity
an equation in which the right-hand side is a function of y alone
the maximum population of an organism that the environment can sustain indefinitely
an equation involving a function y = y(x) and one or more of its derivatives
a mathematical object used to graphically represent solutions to a first-order differential
equation; at each point in a direction field, a line segment appears whose slope is equal to the slope of a solution to
the differential equation passing through that point
any solution to the differential equation of the form y = c, wherec is a constant
a numerical technique used to approximate solutions to an initial-value problem
the entire set of solutions to a given differential equation
the constant r > 0 in the exponential growth function P(t) = P0 ert
the population at time t = 0
a value or set of values that a solution of a differential equation satisfies for a fixed value of the
independent variable
the velocity at time t = 0
a differential equation together with an initial value or values
any function f (x) that is multiplied on both sides of a differential equation to make the side
involving the unknown function equal to the derivative of a product of two functions
description of a first-order differential equation that can be written in the form a(x)y′ + b(x)y = c(x)
a differential equation that incorporates the carrying capacity K and growth rate r into
a population model
the highest order of any derivative of the unknown function that appears in the
equation
member of a family of solutions to a differential equation that satisfies a particular initial condition
a visual representation of the behavior of solutions to an autonomous differential equation subject to various
initial conditions
any equation that can be written in the form y′ = f (x)g(y)
a method used to solve a separable differential equation
a curve graphed in a direction field that corresponds to the solution to the initial-value problem passing
through a given point in the direction field
a function y = f (x) that satisfies a given differential equation
422 Chapter 4 | Introduction to Differential Equations
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	Chapter 4. Introduction to Differential Equations
	Glossary