4.2E: Exercises for Section 8.1
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- Jun 25, 2021
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In exercises 1 - 7, determine the order of each differential equation.
1) y′+y=3y2
- Answer
- 1st-order
2) (y′)2=y′+2y
3) y‴+y″y′=3x2
- Answer
- 3rd-order
4) y′=y″+3t2
5) dydt=t
- Answer
- 1st-order
6) dydx+d2ydx2=3x4
7) (dydt)2+8dydt+3y=4t
- Answer
- 1st-order
In exercises 8 - 17, verify that the given function is a solution to the given differential equation.
8) y=x33 solves y′=x2
9) y=2e−x+x−1 solves y′=x−y
10) y=e3x−ex2 solves y′=3y+ex
11) y=11−x solves y′=y2
12) y=ex22 solves y′=2xy
13) y=4+lnx solves xy′=1
14) y=3−x+xlnx solves y′=lnx
15) y=2ex−x−1 solves y′=y+x
16) y=ex+sinx2−cosx2 solves y′=cosx+y
17) y=πe−cosx solves y′=ysinx
In exercises 18 - 27, verify the given general solution and find the particular solution.
18) Find the particular solution to the differential equation y′=4x2 that passes through (−3,−30), given that y=C+4x33 is a general solution.
19) Find the particular solution to the differential equation y′=3x3 that passes through (1,4.75), given that y=C+3x44 is a general solution.
- Answer
- y=4+3x44
20) Find the particular solution to the differential equation y′=3x2y that passes through (0,12), given that y=Cex3 is a general solution.
21) Find the particular solution to the differential equation y′=2xy that passes through (0,12), given that y=Cex2 is a general solution.
- Answer
- y=12ex2
22) Find the particular solution to the differential equation y′=(2xy)2 that passes through (1,−12), given that y=−3C+4x3 is a general solution.
23) Find the particular solution to the differential equation y′x2=y that passes through (1,2e), given that y=Ce−1/x is a general solution.
- Answer
- y=2e−1/x
24) Find the particular solution to the differential equation 8dxdt=−2cos(2t)−cos(4t) that passes through (π,π), given that x=C−18sin(2t)−132sin(4t) is a general solution.
25) Find the particular solution to the differential equation dudt=tanu that passes through (1,π2), given that u=sin−1(eC+t) is a general solution.
- Answer
- u=sin−1(e−1+t)
26) Find the particular solution to the differential equation dydt=et+y that passes through (1,0), given that y=−ln(C−et) is a general solution.
27) Find the particular solution to the differential equation y′(1−x2)=1+y that passes through (0,−2), given that y=C√x+1√1−x−1 is a general solution.
- Answer
- y=−√x+1√1−x−1
In exercises 28 - 37, find the general solution to the differential equation.
28) y′=3x+ex
29) y′=lnx+tanx
- Answer
- y=C−x+xlnx−ln(cosx)
30) y′=sinxecosx
31) y′=4x
- Answer
- y=C+4xln4
32) y′=sin−1(2x)
33) y′=2t√t2+16
- Answer
- y=23√t2+16(t2+16)+C
34) x′=cotht+lnt+3t2
35) x′=t√4+t
- Answer
- x=215√4+t(3t2+4t−32)+C
36) y′=y
37) y′=yx
- Answer
- y=Cx
In exercises 38 - 42, solve the initial-value problems starting from y(t=0)=1 and y(t=0)=−1. Draw both solutions on the same graph.
38) dydt=2t
39) dydt=−t
- Answer
- y=1−t22, and y=−t22−1
40) dydt=2y
41) dydt=−y
- Answer
- y=e−t and y=−e−t
42) dydt=2
In exercises 43 - 47, solve the initial-value problems starting from y0=10. At what time does y increase to 100 or drop to 1?
43) dydt=4t
- Answer
- y=2(t2+5), When t=3√5, y will increase to 100.
44) dydt=4y
45) dydt=−2y
- Answer
- y=10e−2t, When t=−12ln(110), y will decrease to 1.
46) dydt=e4t
47) dydt=e−4t
- Answer
- y=14(41−e−4t), Neither condition will ever happen.
Recall that a family of solutions includes solutions to a differential equation that differ by a constant. For exercises 48 - 52, 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?
48) [T] y′=y(x)
49) [T] xy′=y
- Answer
- Solution changes from increasing to decreasing at y(0)=0.
50) [T] y′=t3
51) [T] y′=x+y (Hint: y=Cex−x−1 is the general solution)
- Answer
- Solution changes from increasing to decreasing at y(0)=0.
52) [T] y′=xlnx+sinx
53) Find the general solution to describe the velocity of a ball of mass 1 lb that is thrown upward at a rate of a ft/sec.
- Answer
- v(t)=−32t+a
54) 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.
55) You throw two objects with differing masses m1 and m2 upward into the air with the same initial velocity of a ft/s. What is the difference in their velocity after 1 second?
- Answer
- 0 ft/s
56) [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.
57) [T] For the previous problem, use your calculator to approximate how much higher the ball went on Mars.
- Answer
- 4.86 meters
58) [T] A car on the freeway accelerates according to a=15cos(πt), 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?
59) [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.
- Answer
- x=50t−15π2cos(πt)+3π2,2 hours 1 minute
60) [T] For the previous problem, find the total distance traveled in the first hour.
61) Substitute y=Be3t into y′−y=8e3t to find a particular solution.
- Answer
- y=4e3t
62) Substitute y=acos(2t)+bsin(2t) into y′+y=4sin(2t) to find a particular solution.
63) Substitute y=a+bt+ct2 into y′+y=1+t2 to find a particular solution.
- Answer
- y=1−2t+t2
64) Substitute y=aetcost+betsint into y′=2etcost to find a particular solution.
65) Solve y′=ekt with the initial condition y(0)=0 and solve y′=1 with the same initial condition. As k approaches 0, what do you notice?
- Answer
- y=1k(ekt−1) and y=t