Math 3F — Exam #1 Sample

Math 3F — Exam #1 Sample
Laney College, Fall 2010
Fred Bourgoin
1. Solve the initial-value problems.
dy
= (1 − y) dx, y(π) = 2
cos x
(b) y 0 = 2xy 2 + 3x2 y 2 , y(1) = −1
(a)
2. Brine containing 0.2 kg of salt per liter runs into a tank initially filled with 500 L of
water containing 5 kg of salt. The brine enters the tank at a rate of 5 L/min. The
mixture is continually stirred and flows out at a rate of 5 L/min.
(a) Find the amount of salt in the tank after 10 minutes.
(b) After 10 minutes, the tank develops a leak and an additional liter per minute
flows out of the tank. How much salt is in the tank 20 minutes after the leak
develops?
3. Is y = 3 sin 2x + e−x a solution to the differential equation y 00 + 4y = 5e−x ?
4. Recall that Newton’s law of cooling states that the rate at which the temperature of
an object changes is proportional to the difference between the temperature of the
object and the temperature of its surroundings. A body is discovered at noon on a
cold Southern California day (16◦ C); the temperature of the body is then recorded as
34.5◦ C. When the homicide detectives show up an hour later, the body’s temperature
is 33.7◦ C. When did the murder occur? (The normal temperature of a living human
being is 37◦ C.)
5. Without solving the problem, determine the largest interval in which the solution to
the initial-value problem is certain to exist.
3
(4 − t2 )y 0 + 2ty = ,
t
y(−1) = −3
6. For each differential equation, determine its order and decide whether it is linear or
nonlinear.
d2 y
+ sin(t + y) = sin t
dt2
d3 y
dy
(b)
+ t + (cos2 t)y = t3
3
dt
dt
(a)
7. Solve the linear equation. (12 pts)
y 0 + (cos x)y = cos x
1
8. Consider the differential equation
dy
= (y + 2)(y − 2)2 .
dt
(a) What are the equilibrium solutions?
(b) Draw a phase line.
(c) For each equilibrium solution, determine its stability.
(d) Sketch the equilibrium solutions a few other solutions.
XC. A 30-year-old woman accepts an engineering position with a starting salary of $30,000
a year. Her salary S(t) increases exponentially, with S(t) = 30et/20 (in thousands of
dollars) after t years. Meanwhile, 12% of her salary is deposited continuously in a
retirement account, which accumulates interest at a continuous annual rate of 6%.
How much money will she have available in her retirement account when she retires
at age 70?
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Math 3F — Exam #1 Sample Solutions
Laney College, Fall 2010
Fred Bourgoin
1. Solve the initial-value problems.
(a)
dy
= (1 − y) dx, y(π) = 2
cos x
Solution. The equation is separable. Just be careful when you integrate.
Z
Z
dy
= cos x dx
=⇒
− ln |1 − y| = sin x + C
1−y
=⇒
1 − y = Ce− sin x
y = 1 + Ce− sin x .
=⇒
Plug in the initial condition to find the value of C:
y(π) = 1 + C = 2
=⇒
C = 1.
The solution is y(x) = 1 + e− sin x .
(b) y 0 = 2xy 2 + 3x2 y 2 ,
y(1) = −1
Solution. This equation is also separable.
Z
Z
dy
= (2x + 3x2 ) dx
=⇒
y2
=⇒
1
= x2 + x3 + C
y
1
.
y=
C − x2 − x3
−
Again, use the initial condition to determine the value of C:
y(1) =
The solution is y(x) =
1
= −1
C −2
=⇒
C = 1.
1
.
1 − x2 − x3
2. Brine containing 0.2 kg of salt per liter runs into a tank initially filled with 500 L of
water containing 5 kg of salt. The brine enters the tank at a rate of 5 L/min. The
mixture is continually stirred and flows out at a rate of 5 L/min.
(a) Find the amount of salt in the tank after 10 minutes.
Solution. If we let y(t) be the amount of salt in the tank at time t, then the
IVP we need to solve is
y
y
dy
= (0.2)(5) −
(5) = 1 −
,
dt
500
100
3
y(0) = 5.
Let us solve the DE as a separable equation.
dy
1
=−
(y − 100)
dt
100
=⇒
Z
=⇒
ln |y − 100| = −
dy
=
y − 100
Z
−
1
dt
100
t
+C
100
y = 100 + Ce−t/100 .
=⇒
Now, plug in the initial condition:
y(0) = 100 + C = 5
=⇒
C = −95.
The solution to the IVP is y(t) = 100 − 95e−t/100 . So, after 10 minutes the
amount of salt in the tank is y(10) = 100 − 95e−1/10 ≈ 14.04 kg.
(b) After 10 minutes, the tank develops a leak and an additional liter per minute
flows out of the tank. How much salt is in the tank 20 minutes after the leak
develops?
Solution. This is a little trickier because the volume changes over time. The
IVP to solve is
6y
dy
=1−
,
y(0) = 100 − 95e−1/10 .
dt
500 − t
The equation is not separable, but it is linear:
y0 +
6
t = 1.
500 − t
The integrating factor is
µ=e
R
6
500−t
dt
= e−6 ln |500−t| =
1
.
(500 − t)6
Multiply the equation by the integrating factor to get:
0
1
1
y =
.
6
(500 − t)
(500 − t)6
Integrate both sides, then solve for y;
1
1
y=
+C
6
(500 − t)
5(500 − t)5
=⇒
y=
1
(500 − t) + C(500 − t)6 .
5
Find C using the initial condition:
y(0) = 100 + 5006 C = 100 − 95e−1/10
=⇒
C=
Therefore, the solution to the IVP is
y(t) =
and y(20) = 96 −
95e−1/10
1
(500 − t) −
(500 − t)6
5
5006
95e−1/10
4806 ≈ 28.72 kg.
5006
4
−95e−1/10
.
5006
3. Is y = 3 sin 2x + e−x a solution to the differential equation y 00 + 4y = 5e−x ?
Solution. Take a couple of derivatives, plug in, and see if you get the right result.
y 0 = 6 cos 2x − e−x
and
y 00 = −12 sin 2x + e−x .
So,
y 00 + 4y = −12 sin 2x + e−x + 4(3 sin 2x + e−x ) = 5e−x .
Therefore, the given function is a solution to the DE.
4. Recall that Newton’s law of cooling states that the rate at which the temperature of
an object changes is proportional to the difference between the temperature of the
object and the temperature of its surroundings. A body is discovered at noon on a
cold Southern California day (16◦ C); the temperature of the body is then recorded as
34.5◦ C. When the homicide detectives show up an hour later, the body’s temperature
is 33.7◦ C. When did the murder occur? (The normal temperature of a living human
being is 37◦ C.)
Solution. The information we are given can be translated as
du
= k(u − 16),
dt
u(0) = 34.5,
u(1) = 33.7.
Solving the equation (which is separable) yields u(t) = 16 + Cekt . Furthermore,
u(0) = 16 + C = 34.5 so that C = 18.5; and u(1) = 16 + 18.5ek = 33.7 so that
k = ln 17.7
18.5 . Hence, the solution to the IVP is
17.7
u(t) = 16 + 18.5e(ln 18.5 )t .
Setting u(t) = 37, we get
17.7
16 + 18.5e(ln 18.5 )t = 37
=⇒
t=
21
ln 18.5
17.7
18.5
≈ −2.87.
That is about 2 hours and 52 minutes before t = 0 (noon). So, the murder occurred
at approximately 9:08AM.
5. Without solving the problem, determine the largest interval in which the solution to
the initial-value problem is certain to exist.
3
(4 − t2 )y 0 + 2ty = ,
t
y(−1) = −3
Solution. First rewrite the equation:
y0 +
3
2t
y=
.
4 − t2
t(4 − t2 )
The functions p(t) and g(t) have discontinuities at −2, 0, and 2. The largest interval
containing −1 (where the initial condition is given) is (−2, 0).
5
6. For each differential equation, determine its order and decide whether it is linear or
nonlinear.
d2 y
+ sin(t + y) = sin t
dt2
Solution. This is a nonlinear, second-order DE.
d3 y
dy
(b)
+ t + (cos2 t)y = t3
dt3
dt
Solution. This is a linear, third-order DE.
(a)
7. Solve the linear equation.
y 0 + (cos x)y = cos x
R
Solution. The integrating factor is µ = e cos x dx = esin x . Multiplying the equation
by µ we get
(esin x y)0 = (cos x) esin x .
Now integrate both sides and solve for y:
esin x y = esin x + C
8. Consider the differential equation
=⇒
y = 1 + Ce− sin x .
dy
= (y + 2)(y − 2)2 .
dt
(a) What are the equilibrium solutions?
Solution. Set
dy
dt
= 0 to find the equilibrium solutions: y = ±2.
(b) Draw a phase line.
Solution. The graph of y vs. f (y) and the phase line are as follows.
−2
2
(c) For each equilibrium solution, determine its stability.
Solution. The equilibrium solution y = −2 is unstable, while y = 2 is semistable.
6
(d) Sketch the equilibrium solutions a few other solutions.
Solution.
XC. A 30-year-old woman accepts an engineering position with a starting salary of $30,000
a year. Her salary S(t) increases exponentially, with S(t) = 30et/20 (in thousands of
dollars) after t years. Meanwhile, 12% of her salary is deposited continuously in a
retirement account, which accumulates interest at a continuous annual rate of 6%.
How much money will she have available in her retirement account when she retires
at age 70?
Solution. If we let y(t) be the amount in her retirement account at time t, then the
IVP is
dy
= 0.12S(t) + 0.06y = 3.6et/20 + 0.06y,
y(0) = 0.
dt
This is a linear equation: y 0 − 0.06y = 3.6et/20 . The integrating factor is
µ=e
R
(−0.06) dt
= e−0.06t .
As usual, multiply the equation by µ, integrate, then solve for y:
(e−0.06t y)0 = 3.6e−0.01t
=⇒
e−0.06t y = −360e−0.01t + C
=⇒
y = Ce0.06t − 360e0.05t .
Apply the initial condition:
y(0) = C − 360 = 0
=⇒
C = 360.
Hence, y(t) = 360(e0.06t − e0.05t ) and y(40) = 360(e2.4 − e2 ) ≈ 1,308,283.30.
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