Math 2142 Homework 4, Linear ODE Solutions

Math 2142 Homework 4, Linear ODE Solutions

Problem 1. Consider the first initial value problem.

dy = (1 - 2x)y2 with y(0) = -1/6 dx Rewriting this separable equation, we get

1 y2 dy = (1 - 2x) dx

-1 = x - x2 + c y

Plugging in y = -1/6 and x = 0 from the initial condition gives 6 = c. Solving for y gives

-1 = x - x2 + 6 y 1 = -x + x2 - 6 y

1 y = x2 - x - 6

Now consider the second initial value problem.

dy = xy3(1 + x2)-1/2 dx

with

y(0) = 1

Rewriting this separable equation, we get

1 dy =

y3 -1 = 2y2

x

dx

1 + x2

x

dx

1 + x2

To do the remaining integral, use the substitution u = 1 + x2, so du = 2x dx. This gives

x

1 dx =

1

du = u + c = 1 + x2 + c

1 + x2

2u

Therefore, altogether we have

-1

=

1 + x2

+c

2y2

Plugging in y = 1 and x = 0 from the initial condition gives -1/2 = 1 + c, so c = -3/2. To solve for y

-1

3

2y2 =

1 + x2 - 2

1

y2

=

-2

1 + x2 + 3

y = (3 - 2 1 + x2)-1/2

Problem 2. For the first differential equation

dy

x2

=

dx y(1 + x3)

we separate the variables to get

x2

y dy =

dx

1 + x3

The y integral is just y2/2. To do the x integral, use the substitution u = 1+x3, so du = 3x2 dx.

x2

1

dx =

1

du

=

1

ln |u|

+

c

=

1

ln |1

+

x3|

+

c

=

ln( 3 1

+

x2)

+

c

1 + x3

3u

3

3

We can remove the absolute value signs because x > 0. Altogether, we have

y2/2 = ln( 3 1 + x2) + c

y = ? 2 ln( 3 1 + x2) + c

where the c in the second line is really twice the constant from the first line. For the second differential equation

dy

2x

=

dx yey - x2yey

we separate the variables to get

yey dy =

2x dx

1 - x2

To calculate the y integral, we use integration by parts. Let u = y and dv = ey dy. Then du = dy and v = ey, giving us

yey dy = yey - ey dy = yey - ey

(and I'll save the integration constant for the end). To do the x integral, use the substitution u = 1 - x2, so du = -2x dx.

2x dx = -

1 du = - ln |1 - x2| + c = - ln(1 - x2) + c

1 - x2

u

Note that we can remove the absolute value signs because -1 < x < 1. Altogether, we get

yey - ey = - ln(1 - x2) + c

which we will not attempt to solve for y!

Problem 3(a). Solve the following algebraic equation explicitly for y. y2 - 2y + 4 = x4 + x2 + 6

Solution. As indicated on the homework, we completing the square on the left side and solve for y.

(y - 1)2 - 1 + 4 = x4 + x2 + 6 (y - 1)2 + 3 = x4 + x2 + 6 (y - 1)2 = x4+ x2 + 3 y = ? x4 + x2 + 3 + 1

3(b). Solve the following separable initial value problem and give the solution explicitly as a

function y(x).

dy 3x2 + ex

=

with y(0) = 1

dx 2y - 4

Solution. Separating the variables, integrating and completing the square gives

2y - 4 dy = 3x2 + ex dx

y2 - 4y = x3 + ex + c (y - 2)2 - 4 = x3 + ex + c

(y - 2)2 = x3 +ex + c y = 2 ? x3 + ex + c

where c = c + 4 is an altered constant of integration. To solve for the constant c, we plug in

y = 1 and x = 0 to get

1=2? 0+1+c

It follows that c = 0 and we need to use the negative square root. So, the specific solution is

y = 2 - x3 + ex

Problem 4. A tank contains 1000 liters of water with 15 kg of dissolved salt. Pure water enters the tank at a rate of 10 liters per minutes and the mixture is drained from the bottom at the same rate. Assume that the solution is kept thoroughly mixed, how many kilograms of salt are in the tank after t minutes?

Solution. Let y(t) be the amount of salt (in kilograms) in the tank after t minutes. We are given y(0) = 15. To express dy/dt, we need to determine the rates at which salt is entering

and leaving the tank. Since the water entering the tank contains no salt, the rate at which salt enters is 0. The rate at which salt is leaving the tank is

y kg

Ly

? 10 =

1000 L min 100

measured in kilograms per minute. Therefore, our initial value problem for this situation is

dy

y

=-

with y(0) = 15

dt 100

We know that the solution to a differential equation of the form dy/dt = ky is y = Aekt where A = y(0). Therefore, the solution here is

y(t) = 15e-t/100

Problem 5. Newton's Law of Cooling states that the rate of cooling (or heating) of an object is proportional to the temperature difference between the object and its surroundings. In other words, if T (t) is the temperature of the object at time t and Ts is the constant temperature of its surroundings, then Newton's Law of Cooling says that T (t) satisfies

dT dt = k(T (t) - Ts)

for some constant k.

5(a). A can of soda at 72 degrees is placed in a refrigerator where the temperature is 44 degrees. Use Newton's Law of Cooling to set-up and solve a differential equation for T (t), the temperature of soda after t minutes. (The constant k will appear in your answer but you can solve for the constant of integration using the initial temperature of the can.)

Solution. Let T (t) be the temperature of the can t minutes after it is placed in the refrigerator. Since the temperature of the refrigerator is 44 degrees, we have Ts = 44. Since the initial temperature of the can is 72 degrees, we have T (0) = 72. Therefore, our initial value problem is

dT = k(T - 44) with T (0) = 72

dt Separating the variables gives

1 dT = k dt

T - 44 ln(T - 44) = kt + c

T - 44 = Aekt T (t) = Aekt + 44

where A = ec is the altered constant of integration. Note that we do not need absolute value signs for ln |T - 44| because T (0) = 72, so we know T (t) - 44 is positive. To find the constant

A of integration, we plug in T = 72 and t = 0 to get 72 = A + 44, or in other words, A = 28. Altogether, our solution looks like

T (t) = 28ekt + 44

5(b). After 20 minutes, the temperature of the can is 61 degrees. Use this information to find the constant k.

Solution. We are given T (20) = 61, so we can find k by plugging in t = 20 and T = 61. 61 = 28e20k + 44

which means k = -0.025 (approximately). 5(c). How long does it take for the soda to cool to 50 degrees?

Solution. We need to solve

50 = 28e-0.025t + 44

which gives t = 62 (approximately).

Problem 6. One model for the spread of a rumor is that the rate of spread is proportional to the product of the fraction of people who have heard the rumor and the fraction of the people who have not heard the rumor. If y(t) denotes the fraction of people who have heard the rumor after t days, then the modeling differential equation is

dy = ky(1 - y)

dt

where k is the proportionality constant.

6(a). Find the general solution to this separable differential equation. (You can assume that 0 < y < 1, and hence 0 < 1 - y < 1 as well. The constant k will appear in your solution in addition to the constant of integration.)

Solution. Separating the variables gives

1 dy = k dt

y(1 - y)

The t integral is just kt + c, but to do the y integral, we need to use partial fractions.

1

A B A(1 - y) + By

=+

=

y(1 - y) y 1 - y

y(1 - y)

and therefore, 1 = A(1 - y) + By. Since this equality has to hold for all y, we can plug in y values to help us determine the constants A and B. Plugging in y = 1 gives us 1 = B and plugging in y = 0 gives us 1 = A. So, our partial fraction decomposition is

1

11

=+

y(1 - y) y 1 - y

Returning to our original integrals, we have

1 dy = k dt

y(1 - y)

11

+

dy = kt + c

y 1-y

ln(y) - ln(1 - y) = kt + c

y

ln

= kt + c

1-y

y = Aekt where A = ec 1-y

1-y

1

y = Aekt

1

1

-1 =

y

Aekt

1

1

=

+1

y

Aekt

1

1 + Aekt

=

y

Aekt

Aekt y=

1 + Aekt

6(b). Suppose at t = 0, one person is a class of 25 has heard a rumor and at t = 1, two people in the class have heard the rumor. Use this information to find the value of the constant of integration and the constant k.

Solution. We are given y(0) = 1/25 and y(1) = 2/25. Plugging in y = 1/25 and t = 0,

1

A

=

25 1 + A

1 + A = 25A

1 =A

24

To find k, we plug in y = 2/25 and t = 1 to get

2

ek/24

25 = 1 + ek/24

Multiplying the right side by 24/24, the right side becomes ek/(24 + ek). From here, we

simplify to find k.

2

ek

=

25

24 + ek

25

24 + ek

=

2

ek

25 24 2 = ek + 1

23 24 2 = ek

48 = ek 23

and so k = ln 48/23. Plugging A and k into our original solution gives

1/24 ? et ln(48/23) y(t) =

1 + 1/24 ? et ln 48/23

Multiplying the right side by 24/24, move the t into the logarithm as an exponent and cancel the exponential and logarithm to get

(48/23)t y(t) =

24 + (48/23)t

6(c). When will 90% of the class have heard the rumor?

Solution. To find when 90 percent of the class has heard the rumor, we need to solve

9

(48/23)t

10 = 24 + (48/23)t

for t. I don't have a calculator handy, but you can simplify this expression as follows.

9

(48/23)t

10 = 24 + (48/23)t

10

24 + (48/23)t

=

9

(48/23)t

10

24

9 = (48/23)t + 1

1

24

9 = (48/23)t

9 ? 24 = (48/23)t

ln 216 = t ln 48/23 ln 216

=t ln 48/23

Problem 7. For the first initial value problem

dy + 2xy = 4e-x2 dx

with

y(0) = 3

our integrating factor is

I(x) = e 2x dx = ex2

The next integral we need to calculate is

I(x) ? Q(x) dx = ex2 ? 4e-x2 dx = 4 dx = 4x + c

Since 1/I(x) = e-x2, the general solution is y(x) = e-x2(4x + c) = 4xe-x2 + ce-x2

To find the constant c, we plug in y = 3 and x = 0 to get 3 = 4(0)e0 + ce0

Therefore, c = 3 and the specific solution is y(x) = 4xe-x2 + 3e-x2

Next, consider the second differential equation. We rewrite it in the form of a linear

equation as

dy - y = 2t2 dt t

The integrating factor is

I(t) = e -1/t dt = e- ln t = eln t-1 = t-1

The next integral to calculate is

I(t) ? Q(t) dt = t-1 ? 2t2 dt = 2t dt = t2 + c

Since 1/I(t) = t, the general solution is

y(t) = t ? (t2 + c) = t3 + ct

Consider the third integral The integrating factor is

dy + y = 3x

dx I(x) = e 1 dx = ex

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