Math 3C Practice Final Problems Solutions Prepared by Vince Zaccone For Campus Learning Assistance...

Math 3CPractice Final Problems

Solutions

Prepared by Vince Zaccone

For Campus Learning Assistance Services at UCSB



1a) (find a general solution)

7y2y

First we solve the homogeneous equation:

t2h Cey

...egrateint...separate

0y2y

Now find a particular solution – it will be a constant, say yp=A.

Plug this into the equation and solve for A:

27t2

general

phgeneral

27

p

27

Cey

yyy

y

A

7A20

7)A(2)A(



1b) (find a general solution)

tyt

4y

ty4yt 2


4Ctln

h

tdt

ydy

tdt

ydy

t4

dtdy

t4

t

Cey

Ctln4yln

4

4y

0yy

Now we use variation of parameters to find a particular solution:

261

4general

261

4

661

p

661594

933458

34

44

4p

tt

Cy

tt

ty

tvtvttv

tvt4vt4tvtt

v4

t

vt4tvt

t

v

t

4

t

v

t

vy

Quotient rule

Multiply both sides by t8



1c) (solve the initial value problem)

1)0(y;e2y3y t3


t3h Cey

0y3y


t3t3general

t3p

t3t3t3t3

t3t3t3

t3p

et2Cey

et2y

t2v

2v

e2ev3ev3ev

e2ev3ev

evy

Now plug in the initial value to find C.

t3t3

0303

et2e)t(y

C1

e02Ce1

Solution to IVP



1d) (solve the initial value problem)

1)(y);tsin(y)ttan(y


)tsec(Cy

C)tcos(lnyln

dt)ttan(y

dyy)ttan(

0y)ttan(y

h

dtdy


)tsec()t(sin)tsec(Cy

)tsec()t(siny

)t(sindt)tcos()tsin(v

)tcos()tsin(v)tsin()tsec(v

)tsin()tsec(v)ttan()ttan()tsec(v)tsec(v

)tsin()tsec(v)ttan()tsec(v

)tsec(vy

221

general

221

p

221

p


)tsec()t(sin)tsec()t(y

C1101C1

)sec()(sin)sec(C1

221

21

221

Solution to IVP



2) Farmer Fred opens a bank account with an initial deposit of $10,000. The account earns a monthly interest rate of 0.5%, compounded continuously. In addition, Fred will deposit $100 each month. How many months will it take for the account value to double?

Define variables: y(t) = account value at time t (months)

The differential equation is 100y005.0dt

dy We also have an initial value y(0)=10,000

We could solve this equation by separation, but let’s use the excellent guess method instead.


t005.0h Cey0y005.0y

For the particular solution, notice that we should get a constant. Plug in and solve:

000,20Cey

000,20y

000,20A100A005.00

t005.0general

p

005.0100


000,20e000,30)t(y

000,30C000,20Ce000,10t005.0

0005.0

Finally we can use our formula to find the answer. Just set y(t)=20,000 and solve for t.

months5.57005.0

)ln(t)ln(t005.0e

000,20000,20e000,30

34

34

34t005.0

t005.0



3) A lukewarm beverage (initially at 70°F) is placed into a cold (40°F) refrigerator. Five minutes later the temperature of the still-not-cold-enough beverage is 60°F. It is replaced in the fridge to continue the chilling process. How many minutes will it take for the beverage to reach the temperature of optimal refreshment (43°F)? Assume that the temperature follows Newton’s law of cooling, i.e. the rate of change of the temperature is proportional to the difference between the temperature of the beverage and the temperature of the surroundings.

Define variables: T(t)=temperature of beverage (in°F) at time t (minutes)

)40T(kdt

dTThe differential equation is Here k is a constant of proportionality

Let’s solve this one by separation.

40CeT

Ckt40Tln

dtk40T

dT)40T(k

dt

dT

kt

We need to find the 2 constants C and k. From the given information we know y(0)=70 and y(5)=60.

081.05

)ln(ke40e3060

30C40Ce70

32

32k5k5

0k

Now we can rewrite our formula with the constants we just found, then use it to solve the problem.

utesmin4.28081.0

)ln(t

e40e3043

40e30)t(T

101

101t081.0t081.0

t081.0



4) Consider the following autonomous differential equation:

02yyy 2

a) Sketch a slope field for this equation.b) Find any equilibrium solutions and classify them as stable or unstable.c) Find an explicit formula for the solution that passes through the initial value y(0) = 1.

a) The slope field shows the equilibrium solutions, and their stability. We will calculate them algebraically as well.

b) To find equilibrium solutions, set y’=0:

1y;2y

0)1y)(2y(

02yy2

Equilibrium solutions are the horizontal lines y=2 and y=-1

To assess the stability, plug in values on either side to find if the slope is positive or negative.

stable1y

unstable2y

4)2(y;2)0(y;2)3(y



4) Consider the following autonomous differential equation:

02yyy 2

c) Find an explicit formula for the solution that passes through the initial value y(0) = 1.

We will solve this by separation:

t3

31

2

Ce1y

2y

Ct31y

2yln

Ct31yln2yln

Ctdy1y

1

2y

1

dt)1y)(2y(

dy2yy

dt

dy

We use partial fractions to simplify the integral:

1y2y)1y)(2y(

1

B1y

A2y

)2y(B)1y(A1

1y

B

2y

A

)1y)(2y(

1

31

31

31

31

Use the initial value to find C

t321

2103

e1y

2y

CCe11

21

This is an implicit solution – we can

do some algebra and solve for y to get an explicit solution

Careful with the absolute values – in this case we have y(0)=1, so we can drop the abs value bars on the bottom, but we have to switch the top of the fraction.

t3

21

t321

t321t3

21

e1

e2y1yey2e

1y

y2

This solution is valid for -1<y<2



5) Consider the following system of differential equations:

xy5.0y3dt

dy

xyx2dt

dx

a) Find and graph the nullclines of this system.b) What are the equilibria of this system?c) Classify any equilibria as stable or unstable.d) Sketch a reasonable solution to this system of differential equations.

Set x’=0 to find v-nullcline:

Set y’=0 to find h-nullcline:

0y;6x0xy5.0y3dt

dy

2y;0x0xyx2dt

dx

Equlibrium Points are

(0,0) – unstable

(6,2) - stable

h-nullclines in yellow

v-nullclines in red

Equilibria at intersections

Unstable equil. Stable equil.

Solution curve



6) Solve the following system of linear equations:

2z4yx3

5z2yx

1zyx




2z4yx3

5z2yx

1zyx

Augmented matrix form

2

5

1

413

211

111




2z4yx3

5z2yx

1zyx


2

5

1

413

211

111

5

4

1

000

120

111

2R3R*3R

1

4

1

120

120

1111R2R*2R

1R33R*3R

Row reduction:




2z4yx3

5z2yx

1zyx


2

5

1

413

211

111

5

4

1

000

120

111

2R3R*3R

1

4

1

120

120

1111R2R*2R

1R33R*3R

Row reduction:

The 3rd row represents the equation 0x+0y+0z = -5. This equation has no solution, so the original system of equations is inconsistent (not solvable).



7) Consider the following matrix:

4000

01300

4010

30226

A

a)Find the determinant of this matrix.b)Is this matrix invertible? If so, find the inverse.c)What is the determinant of the inverse of this matrix?d)Using this information, solve the following system of linear equations:

12w4

26z13

3yw4

2w3y22x6




4000

01300

4010

30226

A


12w4

26z13

3yw4

2w3y22x6

a) Since A is a triangular matrix, the determinant is the product of the diagonal elements – det(A) = -312




4000

01300

4010

30226

A


12w4

26z13

3yw4

2w3y22x6

a) Since A is a triangular matrix, the determinant is the product of the diagonal elements – det(A) = -312

b) Since det(A) is not zero, A is invertible, and we can use row reduction to find the inverse.

1A

41

131

2485

311

61

Identity

131*61*

41

485

*

41

43

*

*

41

41*

*

000

000

1010

0

1000

0100

0010

0001

3R3R

1R1R

000

0100

1010

0221

1000

01300

0010

0006

2R221R1R

000

0100

1010

001

1000

01300

0010

00226

4R42R2R

4R31R1R

000

0100

0010

0001

1000

01300

4010

30226

4R4R

2R2R

1000

0100

0010

0001

4000

01300

4010

30226




4000

01300

4010

30226

A


12w4

26z13

3yw4

2w3y22x6

c) Since A-1is a triangular matrix, the determinant is the product of the diagonal elements: det(A -1) = -1/312




4000

01300

4010

30226

A


12w4

26z13

3yw4

2w3y22x6


d) Notice that we have the inverse for the coefficient matrix, so we just need to multiply:

12

26

3

2

000

000

1010

0

w

z

y

x

41

131

2485

311

61




4000

01300

4010

30226

A


12w4

26z13

3yw4

2w3y22x6


d) Notice that we have the inverse for the coefficient matrix, so we just need to multiply:

3

2

9

w

z

y

x

12

26

3

2

000

000

1010

0

w

z

y

x

6187

41

131

2485

311

61

8) Consider the following set of vectors in R3

3

2

1

a,

2

1

1

a,

1

1

0

a,

1

0

1

a 4321

a)What is the dimension of the span of this set of vectors?b)Is the span of this set of vectors R3? c)Is this a basis for R3?d)Is each of the following vectors within the span of this set? If so, express them as a linear combination of the vectors in the set.

4

2

2

v,

3

0

2

v 21




3

2

1

a,

2

1

1

a,

1

1

0

a,

1

0

1

a 4321

4

2

2

v,

3

0

2

v 21

a) Put the column vectors in a matrix and perform elementary row operations:

0000

2110

1101

2R3R3R

2110

2110

1101

1R3R3R

3211

2110

1101**





3

2

1

a,

2

1

1

a,

1

1

0

a,

1

0

1

a 4321

4

2

2

v,

3

0

2

v 21


0000

2110

1101

2R3R3R

2110

2110

1101

1R3R3R

3211

2110

1101**

This has 2 pivot columns, so there are 2 linearly independent vectors in the set.

In other words, the span of this set is 2-dimensional.





3

2

1

a,

2

1

1

a,

1

1

0

a,

1

0

1

a 4321

4

2

2

v,

3

0

2

v 21


0000

2110

1101

2R3R3R

2110

2110

1101

1R3R3R

3211

2110

1101**



b) No, the span is 2-dimensional, and R3 is 3-dimensional.

c) No, this is not a basis because is does not span R3.





3

2

1

a,

2

1

1

a,

1

1

0

a,

1

0

1

a 4321

4

2

2

v,

3

0

2

v 21


0000

2110

1101

2R3R3R

2110

2110

1101

1R3R3R

3211

2110

1101**





d) We have several options for this part. One way is to make a good guess. If you notice that v2 is just 2 times a3 then we have our answer for that one:

2

1

1

2

4

2

2





3

2

1

a,

2

1

1

a,

1

1

0

a,

1

0

1

a 4321

4

2

2

v,

3

0

2

v 21


0000

2110

1101

2R3R3R

2110

2110

1101

1R3R3R

3211

2110

1101**






2

1

1

2

4

2

2

For v1 there is no obvious answer, so we need to try to solve for it. Since we know that our span is only 2-dimensional, we only need two basis vectors. Choose any 2 independent vectors from our set, say a1 and a2, and try to write v1 as a linear combination of them.





3

2

1

a,

2

1

1

a,

1

1

0

a,

1

0

1

a 4321

4

2

2

v,

3

0

2

v 21


0000

2110

1101

2R3R3R

2110

2110

1101

1R3R3R

3211

2110

1101**






2

1

1

2

4

2

2


ntInconsiste

cc3

c0

c2

c

c

0

c

0

c

3

0

2

acacv

21

2

1

2

2

1

1

22111





3

2

1

a,

2

1

1

a,

1

1

0

a,

1

0

1

a 4321

4

2

2

v,

3

0

2

v 21


0000

2110

1101

2R3R3R

2110

2110

1101

1R3R3R

3211

2110

1101**






2

1

1

2

4

2

2


ntInconsiste

cc3

c0

c2

c

c

0

c

0

c

3

0

2

acacv

21

2

1

2

2

1

1

22111

We get an inconsistent set of equations, so v1 is not in the span. Prepared by Vince Zaccone



9) Consider the following set of vectors in P2:

x2a,xx1a,xx1a 32

22

1

a)What is dimension of the span of this set?b)Does this set span P2?

c)Is this set a basis for P2?

d)Express each of the standard basis vectors {1, x, x2} as a linear combination of the vectors in the set.




x2a,xx1a,xx1a 32

22

1

a) These are 3 independent vectors. We can see this if we try to form a linear combination of them, and find that there is no way to cancel them out:







x2a,xx1a,xx1a 32

22

1


tindependen0ccc

0c2cc

0ccc

0cc

?0)c2cc(x)ccc(x)cc(

?0)x2(c)xx1(c)xx1(c

?0acacac

321

321

321

21

3213212

21

32

22

1

332211







x2a,xx1a,xx1a 32

22

1


tindependen0ccc

0c2cc

0ccc

0cc

?0)c2cc(x)ccc(x)cc(

?0)x2(c)xx1(c)xx1(c

?0acacac

321

321

321

21

3213212

21

32

22

1

332211

b) P2 is a 3-dimensional space, and we have 3 independent vectors in P2 – yes they span P2.







x2a,xx1a,xx1a 32

22

1


tindependen0ccc

0c2cc

0ccc

0cc

?0)c2cc(x)ccc(x)cc(

?0)x2(c)xx1(c)xx1(c

?0acacac

321

321

321

21

3213212

21

32

22

1

332211


c) P2 is a 3-dimensional space, and we have 3 independent vectors in P2 – yes they form a basis for P2.







x2a,xx1a,xx1a 32

22

1


tindependen0ccc

0c2cc

0ccc

0cc

?0)c2cc(x)ccc(x)cc(

?0)x2(c)xx1(c)xx1(c

?0acacac

321

321

321

21

3213212

21

32

22

1

332211



d) Form a linear combination of a1, a2 and a3:







x2a,xx1a,xx1a 32

22

1


tindependen0ccc

0c2cc

0ccc

0cc

?0)c2cc(x)ccc(x)cc(

?0)x2(c)xx1(c)xx1(c

?0acacac

321

321

321

21

3213212

21

32

22

1

332211




3221

121

321

221

1

321

321

21

2321321

221332211

aaa1

1c;c;c

1c2cc

0ccc

0cc

1x0x0)c2cc(x)ccc(x)cc(1acacac







x2a,xx1a,xx1a 32

22

1


tindependen0ccc

0c2cc

0ccc

0cc

?0)c2cc(x)ccc(x)cc(

?0)x2(c)xx1(c)xx1(c

?0acacac

321

321

321

21

3213212

21

32

22

1

332211




3221

121

321

221

1

321

321

21

2321321

221332211

aaa1

1c;c;c

1c2cc

0ccc

0cc

1x0x0)c2cc(x)ccc(x)cc(1acacac

Similar analysis yields the other 2 combinations.

3221

1212

321

a0aax

aaax






Math 3C Practice Final Problems Solutions Prepared by Vince Zaccone For Campus Learning Assistance...

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