Exam2 Sample Key

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Sample Exam 2 - SolutionsMAT 2150 - Winter 2016

February 19, 2016 Friday

(1) [Sec. 3.2: 21] A snowball melts in such a way that the rate of change in its volume is proportionalto its surface area. If the snowball was initially 4 in. in diameter, and after a half hour itsdiameter is 3 in.

(a) When will its diameter be 2 in.?

(b) Mathematically speaking, when will the snowball disappear?

Solution: The governing equation is

dV

dt = −kS

where

V = V (t) = 4

3πr3, S = S (t) = 4πr2

with the understanding that r = r(t) is also a function of time. Let us measure time in hours.The minus sign in the model comes from the fact that the volume is decreasing and hence weexpect k to be a positive constant which we will compute using the given data in the problem.

Inserting the expressions for V and S into the governing equation we obtain

4π

3

d(r3)

dt = −k · 4πr2 =⇒

1

3 · 3r2

dr

dt = −kr2 =⇒

dr

dt = −k.

Solving this simple first order differential equations we get

r(t) = −kt + C.

Initially (i.e. at t = 0) the diameter of the snowball is 4 in. which means that the radius is 2in. Hencer(0) = 2 =⇒ C = 0 =⇒ r(t) = 2 − kt.

It is also given that the diameter after half an hour (i.e. at time t = 1/2) is 3 in. and hencethe radius is 3/2 in. when t = 1/2. Thus

r(1/2) = 3

2 =⇒ 2 − k

1

2 =

3

2 =⇒ k = 1 =⇒ r(t) = 2 − t.

We are now ready to answer the questions.

(a) The diameter will be 2 in. when the radius is 1 in. Thus

r(T ) = 1 =⇒ 2 − T = 1 =⇒ T = 1.

Hence the diameter will be 2 in. in one hour.


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(b) The snowball will disappear when r = 0 i.e.

r(T ) = 2 − T = 0 =⇒ T = 2.

That is, the snowball will disappear in two hours.

(2) [Sec. 4.2: 19] Solve the initial value problem y + 2y + y = 0, y(0) = 1, y(0) = −3.

Solution: The associated auxiliary equation is

r2 + 2r + 1 = 0

which has the repeated root r1,2 = −1. Hence two linearly independent solutions are

y1(t) = e−t and y2(t) = te

−t.

A general solution is given byy(t) = c1e

−t + c2te−t.

Upon differentiation y (t) = −c1e−t + c2(1 − t)e

−t. The initial value

y(0) = 1 =⇒ c1 = 1.

Hencey(0) = −3 =⇒ −1 + c2 = −3 =⇒ c2 = −2.

Consequently, the solution of the IVP is

y(t) = (1 − 2t)e−t.

(3) [Sec. 4.3: 21] Solve the IVP y + 2y + 2y = 0, y(0) = 2, y(0) = 1.

Solution: The auxiliary equation is r2 + 2r + 2 = 0 whose roots are

r1,2 = −2 ±

22 − 4(1)(2)

2 =

−2 ± 2i

2 = −1 ± i.

Hence a general solution to the DE is

y(t) = c1e−t cos t + c2e

−t sin t

and hencey(0) = 2 =⇒ c1 = 2.

The derivative of y(t) is

y(t) = c1[−e−t cos t − e−t sin t] + c2[−e

−t sin t + e−t cos t]

and hence

y

(0) = 1 =⇒ −

c1 + c2 = 1 =⇒

c2 = 3.Thus the desired solution is

y(t) = 2e−t cos t + 3e−t sin t.


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(4) [Sec. 4.5: 29] Solve the IVP y − y = sin t − e2t, y(0) = 1, y(0) = −1.

Solution: The auxiliary equation

r2 − 1 = 0 =⇒ r1,2 = ±1 =⇒ yh(t) = c1et + c2e

−t.

We need to find two particular solutions, one for the sin t part and one for the e2t part of theright hand side function. For the sin t part we set

y p(t) = A cos t + B sin t =⇒ y

p(t) = −A sin t + B cos t =⇒ y

p(t) = −A cos t − B sin t.

Inserting into the DE y p − y p = sin t yields

−A cos t − B sin t − (A cos t + B sin t) = sin t =⇒ −2A cos t − 2B sin t = sin t

=⇒ 2A = 0, 2B = −1

=⇒ A = 0, B = −1

2.

Hence,

y p(t) = −1

2

sin t.

For the e2t part,y p(t) = Ae

2t =⇒ y p(t) = 2Ae2t =⇒ y p(t) = 4Ae

2t.

Inserting into the DE y p − y p = −e2t yields

4Ae2t − Ae2t = −e2t =⇒ 3Ae2t = −e2t =⇒ 3A = −1 =⇒ A = −1

3.

Hence,

y p(t) = −1

3e2t.

Hence a general solution to the DE given in the problem is

y(t) = c1et + c2e

−t−

1

2 sin t −

1

3e2t.

The initial condition

y(0) = 1 =⇒ c1 + c2 −1

3 = 1 =⇒ c1 + c2 =

4

3.

Also,

y(t) = c1et

− c2e−t

−1

2 cos t −

2

3e2t.

Hence the other initial condition

y(0) = −1 =⇒ c1 − c2 −1

2 −

2

3 = −1 =⇒ c1 − c2 =

1

6.


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Thus,

c1 = 3

4, c2 =

7

12.

Thus, the desired solution to the IVP is

y(t) = 3

4et +

7

12e−t −

1

2 sin t −

1

3e2t.

(5) Find a particular solution to y

+ 2y

+ y = e−t

(a) by the method of undetermined coefficients.

(b) by the method of variation of parameters.

Solution:

(a) The auxiliary equation is

r2 + 2r + 1 = 0 =⇒ (r + 1)2 = 0 =⇒ r1,2 = −1.

Hence, the trial solution is

y p(t) = At2e−t =⇒ y p = Ae

−t(2t − t2) =⇒ y p = Ae−t(t2 − 4t + 2).

Inserting these into the DE gives

y p + 2y

p + y + p = Ae−t[t2 − 4t + 2 + 2(2t − t2) + t2] = e−t

=⇒ 2Ae−t = e−t =⇒ 2A = 1 =⇒ A = 1

2.

Hence, a particular solution is

y p(t) = 1

2t2e−t.

(b) Since r1,2 = −1, the complimentary solution is

yh(t) = c1e−t + c2te

−t.

Hence, a variation of parameter solution is

y p(t) = v1(t)e−t + v2(t)te

−t.

Differentiating y p gives,

y p

= v 1e−t − v1e

−t + v2te−t + v2(te

−t)

= v 1e−t + v

2te−t − v1e

−t + v2(1 − t)e−t

= −v1e−t + v2(1 − t)e

−t


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where we assumed thatv1e−t + v

2te−t = 0

to avoid second order derivatives of v1 and v2. Multiplying both sides of this equality bye−t gives

v1 + v

2t = 0.

Differentiating once more we obtain

y p = −v1e−t + v1e−t + v2(1 − t)e−t + v2[(1 − t)e−t]

= −v1e−t + v1e

−t + v2(1 − t)e−t + v2[−e

−t− (1 − t)e−t]

= e−t[−v1 + v1 + v

2(1 − t) + v2(t − 2)].

Inserting these into the DE yields

y p + 2y

p + y p = e−t[−v

1 + v1 + v

2(1 − t) + v2(t − 2) + 2v2(1 − t) − 2v1 + v1 + tv2]

= e−t[−v1 + v

2(1 − t)] = e−t

after cancelations. Comparing two sides of this equality gives

−v1 + v

2(1 − t) = 1.

Hence we have the following system of equations for v 1

and v 2:

v1

+ t v2

= 0−v

1 + (1 − t)v

2 = 1

The first equation gives −v1

= tv 2. Inserting this into the second equations we obtain

tv2 + (1 − t)v

2 = 1 =⇒ v

2 = 1 =⇒ v2 = t

where we set the constant of integration to be zero. Then,

v1

= −tv2

=⇒ v1

= −t =⇒ v1 = −1

2t2

where we again set the constant of integration to zero. Thus, a particular solution is

y p(t) = v1(t)e−t + v2(t)te

−t = −1

2t2 + t · te−t = (t2 −

1

2t2)e−t =

1

2t2e−t.

The same as the one we found in (a).

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