Review Exam 1 MAP 2302 (Differential Equations)
Integrals and Sections 2.1, 2.2, 2.3, 3.2
Name:
Panther ID:
Show the steps of your work. Evaluate all integrals with justification
From Textbook: Intro to ODE by Shepley Ross
Assume that y is a function of the independent variable x. We’ll also assume that y is either
implicitly or explicitly defined.
(1) Solve the ODE
(2x + 1)dx + (x2 + 4)dy = 0
(2) Solve the ODE
(cos3 (2x))dx + (ln y)dy = 0
(3) Solve the ODE
(2x2 + 3)
p
1 − y 2 dx + x(x − 1)(x + 2)(x2 + 1)dy = 0
(4) The ODE M (x, y)dx + N (x, y)dy = 0 is said to be
a) an exact ODE if ... b) a separable ODE if .... c) an homogeneous ODE if ... d) a linear ODE
if... e) a Bernoulli ODE if...
(5) Solve
(2x cos y + 3x2 y)dx + (x3 − x2 sin y − y)dy = 0 with y(0) = 2
(6) Solve (y sec2 x + sec x tan x)dx + (2y + tan x)dy = 0
(7) Find the most general function N (x, y) such that (x−2 y −2 + xy −3 )dx + N (x, y)dy = 0 is exact
(8) Solve (x sin y)dx + (x2 + 1) cos ydy = 0 with y(1) = π/2
(9) Solve
(y +
p
x2 + y 2 )dx − xdy = 0 with y(1) = 0
(10) Use two different methods to solve (2x2 + 2xy + y 2 )dx + (x2 + 2xy)dy = 0
(11) Solve (y + x tan(y/x))dx − xdy = 0
p
x2 + y 2 )dy − (x3 + y 2
p
x2 + y 2 )dx = 0
(12) Solve
xy
(13) Solve
(2x − 5y)dx + (4x − y)dy = 0 with y(1) = 4
(14) Solve
(x2 + 1)dy + (4xy − x)dx = 0 with y(2) = 1
(15) Solve
(−3xy − xy 2 )dx + dy = 0
(16) Solve
xdy + (y 2 − y)dx = 0
(17) Solve
2xy 3 dx + (y 4 − 2x2 )dy = 0 with y(1) = 2
(
(18) Solve
dy
dx
+ y = f (x) where f (x) =
2 if 0 ≤ x < 1
0 if x ≥ 1
(19) Find an integrating factor and solve (4y 2 + 5xy + 1)dx + (x2 + 2xy)dy = 0
(20) Find an integrating factor and solve (2x2 + y)dx + (x2 y − x)dy = 0
(21) Find an integrating factor of the form xn y m and solve (4xy 2 + 6y)dx + (5x2 y + 8x)dy = 0
(22) a) Solve
b) Solve
(x + 2y + 3)dx + (2x + 4y − 1)dy = 0
(x + 2y + 3)dx + (4x + 8y − 1)dy = 0
(23) Solve
(5x + 2y + 1)dx + (2x + y + 1)dy = 0
(24) Solve
(x − 2y − 3)dx + (2x + y − 1)dy = 0
(25) A body weighing 8lb falls from rest toward the earth from a great height. As it falls, air resistance
acts upon it, and we shall assume that this resistance (in pounds) is numerically equal to 2v, where v
is the velocity (in feet per second). Find the velocity and distance fallen at time t seconds.
(26) A skydiver equipped with parachute and other essential equipment falls from rest toward the earth.
The total weight of the man plus the equipment is 160lb. Before the parachute opens, the air resistance
(in pounds) is numerically equal to v/2 where v is the velocity (in feet per second). The parachute
opens 5seconds after the fall begins; after it opens, the air resistance (in pounds) is numerically equal
to 5v 2 /8 Find the velocity of the skydiver (A) before the parachute opens, and (B) after the parachute
opens.
(27) A stone weighing 4lb falls from rest toward the earth from a great height. As it falls, air resistance
acts upon it, and we shall assume that this resistance (in pounds) is numerically equal to v/2, where
v is the velocity (in feet per second). (a) Find the velocity and distance fallen at time t seconds. (b)
Find the velocity and distance fallen at the end of 5 seconds.
(28) A body of mass 100grams is dropped from rest toward the earth from a height of 1000meters. As
it falls, air resistance acts upon it, and we shall assume that this resistance (in newtons) is proportional
to the velocity v (in meters per second). Suppose the limiting velocity is 245m/sec
(a) Find the velocity and distance fallen at time t seconds. (b) Find the time at which the velocity
is one-fifth of the limiting velocity.
(29) A bullet weighing 1oz is fired vertically downward from a stationary helicopter with a muzzle
velocity of 1200ft/sec.
The air resistance (in pounds) is numerically equal to 16−5 v 2 , where v is the velocity (in feet per
second). Find the velocity of the bullet as a function of time.
(30) A case of canned milk weighing 24lb is released from rest at the top of a plane metal slide which
is 30ft long and inclined 45o to the horizontal. The air resistance (in pounds) is numerically equal to
one-third the velocity (in feet per second). The coefficient of friction is 0.4
(a) Find the velocity of the moving case 1 second after it is released. (b) What is the velocity when
the case reaches the bottom of the slide?
Answers (5) 2x2 cos y + 2x3 y − y 2 = −4
(6) y tan x + sec x + y 2 = 3c
(7) N (x, y) = 2x−1 y −3 − 1.5x2 y −4 + c
(8) (x2 + 1) sin2 y = 2
(9) 2y = x2 − 1
(10) 2x3 + 3x2 y + 3xy 2 = c
(11) sin(y/x) = cx (12) (x2 + y 2 )1.5 = x3 ln(cx3 )
2
(15) y + 3 = cye−3/2x (16) y = (1 + cx−1 )−1
(14) 4(x2 + 1)2 y = x4 + 2x2 + 76
(
(17) x2 y 4 = x4 + 15
(18) y =
(19) 4x5 y + 4x4 y 2 + x4 = c
(13) (2x + y)2 = 12(y − x)
2(1 − e−x ) if 0 ≤ x < 1
2(e − 1)e−x if x ≥ 1
(20) xy 2 ex + yex = c
(21) x3 y 4 (xy + 2) = c
(22) x2 + 4xy + 4y 2 + 6x − 2y = c
(23) 5x2 + 4xy + y 2 + 2x + 2y = c
(24) ln[c(x2 + y 2 − 2x + 2y + 2)] + 4 arctan
y+1
x−1
=0
(25) v = 4(1 − e−8t ), x = 12 (8t + e−8t − 1)
(26) (A) v = 320(1 − e−t/10 ) and v ≈ 126 velocity when the parachute opens,
(B) v =
16(ce−4t + 1)
, c = 110e20 /142
1 − ce−4t
(27) (a) v = 8(1 − e−4t )
x = 2(4t − 1 + e−4t )
(28) (a) v = 245(1 − e−t/25 )
(29) v =
x = 245(t + 25(e−t/25 − 1)
256(91 + 59e−t/4 )
91 − 59e−t/4
(30) (a) 10.96ft/sec
(b) 8ft/sec;
(b) 20.46ft/sec
38feet
(b) t = 5.58s