2 Problem 2. - UMass

Problem

1.

Solve

the initial value

problem

dy dt

= y2[t + cos(t)] with

y(0) = 6.

Solution: Noting the ODE is separable, we get

dy = (t + cos(t))dt

y2

dy

= t + cos(t)dt

y2

1 t2 - = + sin(t) + C

y2

-1

y

=

1 2

t2

+

sin(t)

+

C

then plugging in the initial condition y(0) = 6,

-1

6=

1 2

02

+

sin(0)

+

C

1 =-

C

1 C=-

6

-1

y

=

1 2

t2

+

sin(t)

-

1 6

Problem

2.

Solve the initial value problem y

=

3x sin(x2)-1 3+2y

with y(0) = 2.

Solution: Noting the ODE is separable, we get

(3 + 2y)dy = (3x sin(x2) - 1)dx

(3 + 2y)dy = 3x sin(x2) - 1dx

3y

+

y2

=

3 -

cos(x2)

-

x

+

C

2

then plugging in the initial condition y(0) = 2,

3(2)

+

(2)2

=

3 -

cos(02)

-

0

+

C

2

3 10 = - + C

2

23 C=

2

Thus, the solution y(x) is given implicitly by

3y

+

y2

=

3 -

cos(x2)

-

x

+

23

2

2

Problem 3. Mary initially deposits $1000 in a savings account that pays interest at the rate of 5% per year (compounded continuously). She also arranges for $25 per week to be deposited automatically into her account.

(a) Assume that weekly deposits can be approximated by continuous deposits. Write down an initial value problem for her account balance S(t) over time (t measured in years).

(b) How long does she needs to save to buy a $5000 car?

Solution:

(a) Let M (t) be the amount of money in the account at time t. Then

dM = interest rate (in years) + deposit rate (in years)

dt = 0.05M + (25)(52) = 0.05M + 1300

Since we are also told that $ is initially deposited, our IVP is

dM - 0.05M = 1300,

dt

M (0) = 1000

(b) Let ?(t) = e-0.05t, then

?M - 0.05?M = 1300? e-0.05tM - 0.05e-0.05tM = 1300e-0.05t d [e-0.05tM ] = 1300e-0.05t

dt

e-0.05tM = 1300 e-0.05tdt

e-0.05tM = - 1300 e-0.05t + C 0.05

M

=

1300 -

+

C e0.05t

0.05

then plugging in the initial condition M (0) = 1000,

thus,

1300 1000 = - + C

0.05 1300

C = 1000 + 0.05

1350 =

0.05

M (t)

=

1000

=

1300 -

+

1350 e0.05t

0.05 0.05

= -26000 + 27000e0.05t

Therefore, to buy a car worth $ 5000, she has to wait:

5000 = -26000 + 27000e0.05t

31000 = e-0.05t 27000

31000

0.05t = ln(

)

27000

1 31000

t = ln(

)

0.05 27000

2.76 years

Problem 4. The half-life of a radioactive substance is 2 days. Find the time required for a given amount of the material to decay to 1/10 of its original mass.

Solution: Let M = mass of the substance, and suppose its half-life is 2 days. Then since

M is radioactive, it satisfies

dM = kM

dt

for some constant k. Since this ODE is separable, we get

dM = kdt

M

dM

= kdt

M

ln M = kt + C

M = Cekt

Now, a 2 days half-life gives So,

1 M (t + 2) = M (t)

2 Cek(t+2) = 1 Cekt

2 Cekte2k = 1 Cekt

2 e2k = 1

2 11 k = ln( ) 22

M (t)

=

1

Ce2

ln(

1 2

)t

Finally, for a given mass to decay to 1/10 its mass, it must satisfy

1 M (t) = M (0)

10

1

Ce2

ln(

1 2

)t

=

1

1

Ce2

ln(

1 2

)(0)

10

1

Ce2

ln(

1 2

)t

=

1 C

10

e = 1 2

ln(

1 2

)t

1

10

11

1

ln( )t = ln( )

22

10

t

=

2

ln(

1 10

)

ln(

1 2

)

=

2 log2(10)

Problem 5. A radioactive material loses 25% of its mass in 10 minutes. What is its halflife?

Solution: Same the the previous question, M (t) has the form M (t) = Cekt

To lose 25 % of its mass in 10 minutes, it must satisfy

3 M (t + 10) = M (t)

4 Cek(t+10) = 3 Cekt

4 Cekte10k = 3 Cekt

4 e10k = 3

4 13 k = ln( ) 10 4

So its half-life th is given by

1

M (th)

=

M (0) 2

Ce 1 10

ln(

3 4

)th

=

1

Ce

1 10

ln(

3 4

)(0)

2

Ce 1 10

ln(

3 4

)th

=

1 C

2

1 1 e = 10

ln(

3 4

)th

2

13

1

10

ln( 4 )th

=

ln( ) 2

th

=

10

ln(

1 2

)

ln(

3 4

)

=

10 ln(2) ln(4) - ln(3)

Problem 6. At what yearly rate of interest, compounded continuously, will a bank deposit double in value in 8 years?

Solution: Assuming the bank makes no deposit or withdrawal, the amount of money, M (t), in the account satisfies

dM = rM

dt

dM

= rdt

M

ln M = rt + C

M = Cert

For the initial deposit to double in 8 years, we must have

M (8) = 2M (0)

Ce8r = 2Cer(0)

e8r = 2 1

r = ln(2) 8

0.086

Problem 7. Newton's law of cooling states that if an object with temperature T (t) at time t is in a medium with temperature Tm the rate of change of T at time t is proportional to T (t) - Tm, thus T satisfies a differential equation of the form

T = -k(T - Tm).

Here, k > 0 since the temperature of the object must decrease if T > Tm or increase if T < Tm. We call k the temperature decay constant of the medium.

(a) A thermometer is moved from a room where the temperature is 70F to a freezer where the temperature is 12F. After 30 seconds, the thermometer reads 40F. What does it read after 2 minutes?

(b) An object is placed in a room where the temperature is 20C. The temperature of the object drops by 5C in 4 minutes and by 7C in 8 minutes. What is the temperature of the object when it was initially placed in the room?

Solution:

(a) Here, denoting the temperature in units of F and time in units of minutes, we have

T (0) = 70, T (.5) = 40, Tm = 12, T (2) =?

So, solving the differential equation, we can use either separation of variables or integrating factors.

Separation of Variables

dT dt

= -k(T

- Tm)

dT T -Tm

=

-k

dt

dT T -Tm

=

-

k dt

ln |T - Tm| = -k t + C

T (t) = Tm + Ce-kt

Integrating Factors

Let ?(t) = ekt

?(t)T (t) + k?(t)T (t) = k?(t) Tm

d dt

(?(t)T (t))

=

k?(t)Tm

T (t)

=

1 ?(t)

-k Tm

?(t)dt + C

T (t) = Tm + Ce-kt

Thus, for Tm = 12, either method gives T (t) = 12 + Ce-kt. Note that there are two unknowns here, C and k. To solve for C, we use T (0) = 70. Thus,

T (0) = 70 = 12 + C C = 58

To solve for k, we use T (.5) = 40. Thus

T (.5) = 40 = 12 + 58e-k/2

28 = e-k/2

29 k = 2 ln

58

14

Finally,

T (2)

=

12

+

58e-4

ln(

29 14

)

=

12

+

58

14

4

15.15F

29

(b) From, part a), the solution to the differential equation is still T (t) = Tm + Ce-kt, but now we have temperatures measured in C and

Tm = 20, T (4) = T (0) - 5, T (8) = T (0) - 7, T (0) =? Since T (t) = 20 + Ce-kt, we can use the above to get T (0) = 20 + C, T (4) = T (0) - 5 = 20 + Ce-4k, T (8) = T (0) - 7 = 20 + Ce-8k

Thus giving a system of three equations with three unknowns, C, k, and T (0). Subtracting the third from the second gives

2 = C(e-4k - e-8k), C =

2

e-4k - e-8k

Subtracting the first from the second and substituting C from above gives

0 = 5+C(e-4k-1), 0 = 5+

2

(e-4k-1)

e-2k (e-2k - e2k) 0 = 5+2

e-4k - e-8k

e-6k (e2k - e-2k)

So, we get

e4k = 5 2

15 k = ln 0.229

42

2

2

25

C = e-4k - e-8k = 4/25 - 2/5 = 3 8.33

And finally, we get from the first equation, 25 85

T (0) = 20 + C = 20 + = 28.33 33

Problem 8. Consider the following equation for a certain population of squirrels given by P (t) (t is measured in years).

dP

P

= 2P 1 - (P - 1)

dt

2

(a) Find all thee equilibrium points of the equations. Draw the phase line and determine the stability of each equilibrium point.

(b) Make a graph of the solutions with initial conditions P (0) = 1/4, P (0) = 3/2 and P (0) = 3.

(c) At a certain time the hunting of squirrels become permitted and the law allows that a certain percentage of the squirrel population be eliminated every year. A new equation for the squirrel population is then

dP

P

= 2P 1 - (P - 1) - P

dt

2

The IALS asserts that no more than 10% of squirrels should be eliminated every year (i.e. = 0.1), otherwise the population would go extinct. On the contrary the UHA asserts that it is safe to hunt half the squirrel population every year (i.e. = 0.5). Analyze the systems as varies and determine who is right.

Solution:

(a)

Equilibrium points

correspond

to

dP dt

= 0.

So

we

solve,

dP

P

= 0 = 2P 1 - (P - 1)

dt

2

which has solutions for P = 0, 1 or 2.

Figure 1: Phase Line for Problem 8

From

the

figure,

we

can

see

that

dP dt

>

0

when

P

................
................

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