Name: Solutions

Math 54: Mock Final December 11, 2015

Name: Solutions

1. Find the general solution of y - y - 2y = cos(x) - sin(2x).

The auxiliary equation for the corresponding homogeneous problem is

r2 - r - 2 = 0 (r - 2)(r + 1) = 0 r = 2, r = -1.

The general solution to the homogeneous problem is therefore yh = C1e2x + C2e-x.

For finding a particular solution, we use the method of undetermined coefficients. Assume a particular solution is of the form yp = (A1 cos(x) + A2 sin(x)) + (B1 cos(2x) + B2 sin(2x)). We then get

yp = (-A1 sin(x) + A2 cos(x)) + 2(-B1 sin(2x) + B2 cos(2x)) yp = -(A1 cos(x) + A2 sin(x)) - 4(B1 cos(2x) + B2 sin(2x))

Plugging these in the differential equations yields

yp - yp - 2yp = [-(A1 cos(x) + A2 sin(x)) - 4(B1 cos(2x) + B2 sin(2x))] -

[(-A1 sin(x) + A2 cos(x)) + 2(-B1 sin(2x) + B2 cos(2x))] - 2[(A1 cos(x) + A2 sin(x)) + (B1 cos(2x) + B2 sin(2x))] = (-3A1 - A2) cos(x) + (A1 - 3A2) sin(x) + (-6B1 - 2B2) cos(2x) + (2B1 - 6B2) sin(2x).

Since we want this to equal cos(x) - sin(2x), we require

-3A1 - A2 = 1 A1 - 3A2 = 0

,

-2B61B-1 -6B22B=2 =-10.

We

therefore

get

A1

=

-3 10

,

A1

=

-1 10

,

B1

=

-1 20

,

B2

=

3 20

.

We

conclude

that

the

general

solution of the non-homogeneous problem is

y = yh + yp

=

C1e2x

+

C2e-x

-

1 (3

10

cos(x)

+

sin(x))

+

2 (-

10

cos(2x)

+

3

sin(2x)).

1

2. Define f (x) = cos(x) on [0, ].

(a) Find the Fourier sine series of f . (You might find the formula sin(A) cos(B) = (sin(A + B) + sin(A - B))/2 useful)

The Fourier sine series of f is

n=1

bn

sin(nx)

where

2

bn

=

f (x) sin(nx) dx

0

2

=

cos(x) sin(nx) dx

0

1

=

sin((n + 1)x) + sin((n - 1)x) dx.

0

Note that

0

sin(kx)

dx

=

0

whenever

k

is

even

so

we

only

need

to

consider

even

n in the expression above (so (n ? 1) is odd). We have

1 cos((n + 1)x) cos((n - 1)x)

bn = - n + 1 - n - 1

0

1 1 - cos((n + 1)) 1 - cos((n - 1))

=

+

n+1

n-1

12

2

=

+

n+1 n-1

where we used the fact that cos(r) = -1 for odd r. To sum up, the required Fourier series is

1

2

2

b2k sin((2k)x) =

+

sin((2k)x).

2k + 1 2k - 1

k=1

k=1

(b) What does the series in (a) converge to at x = 0 and x = /4?

Note that the series in (a) is actually the Fourier series corresponding to the odd extension fo of f and therefore converges pointwise to it wherever fo is continuous. Thus, at x = /4, the series converges to fo(/4) = f (/4) = cos(/4) = 1/ 2. At x = 0, since fo is discontinuous, the series converges to (fo(0+) + fo(0-))/2 = (f (0) - f (0))/2 = 0.

2

3. (a) Show that for any a, b, c, (a cos() + b sin() + c)2 2(a2 + b2 + c2).

a

cos()

Define u = b and v = sin(). Then, u ? u = a2 + b2 + c2, v ? v =

c

1

cos2() + sin2() + 1 = 1 and u ? v = a cos() + b sin() + c so by the Cauchy-

Schwartz inequality, we have

(u ? v)2 (u ? u)(v ? v) (a cos() + b sin() + c)2 2(a2 + b2 + c2).

(b) Let a, b be non-zero vectors in Rn. Define T (v) = (v ? a)b. Find the rank and nullity of T .

Note that all the elements in the range of T are scalar multiples of b and, since a and b are non-zero, all the multiples are included in the range. Thus, rank(T ) = 1. By the rank-nullity theorem, it follows that rank(A)+nullity(T ) = n nullity(T ) = n - 1. Another way of seeing that nullity(T ) = n - 1 is that T sends precisely those elements to 0 that are orthogonal to a, i.e., the kernel of T is precisely the orthogonal complement of span{a}. Since span{a} is one-dimensional, its orthogonal complement has dimension (n - 1).

3

4. Find the solution of the wave equation

2u

2u

= 64 , 0 < x < , t > 0

t2

x2

satisfying the boundary conditions u(0, t) = 0, u(, t) = 0 (for t > 0) and initial

conditions

u(x, 0)

=

2

and

u t

(x,

0)

=

2 sin(7x)

-

4 sin(10x)

(for

0

<

x

<

).

For the given boundary conditions, the general solution of the wave equation is (note that = 64 = 8)

u(x, t) = [an cos(8nt) + bn sin(8nt)] sin(nx).

(1)

n=1

Plugging in t = 0, we get

u(x, 0) = an sin(nx).

n=1

The an's are just the Fourier sine coefficients of f (x) = 2 (on 0 < x < ) so that

2

an

=

(2) sin(nx) dx

0

4 cos(nx)

=-

n0

=

4 (1 - (-1)n) = n

0 8/n

n even n odd.

(2)

Next, differentiating (1) with respect to t and plugging in t = 0 gives

u

(x, 0) = t

(8nbn) sin(nx).

n=1

Compare

this

with

g(x)

=

2 sin(7x) - 4 sin(10x)

to

infer

that

8(7)b7

=

2

b7

=

1 28

,

8(10)b10

=

-4

b10

=

-

1 20

with

all

the

other

bn's

equal

to

zero.

We conclude that the solution is (1) with the coefficients given by (2) and the last

statement.

4

5. Solve the initial value problem x = Ax for

A=

1 -1

4 5

,

x(0) =

1 1

.

We first find the eigenvalues of A. The characteristic polynomial is 2-6+9 = (-3)2

so the only eigenvalue is = 3 with a multiplicity of two. Note that (A - 3I) =

-2 -1

4 2

is not the zero matrix so A does not have two eigenvectors corresponding to

it lone eigenvalue. We therefore need to turn to the matrix exponential method.

By the Cayley-Hamilton Theorem, we get (A - 3I)2 = 0 so that

eAt = e3It+(A-3I)t

= e3Ite(A-3I)t

= e3t I + (A - 3I)t + (A - 3I)2 t2 + (A - 3I)3 t3 + . . .

2!

3!

= e3t (I + (A - 3I)t)

=

e3t

1 - 2t -t

4t 1 + 2t

.

The general solution of the given problem is therefore x(t) = eAtc where c is a column

matrix of constants. The initial condition x(0) =

1 1

gives eA(0)c = c =

1 1

so we

conclude that the solution to the initial value problem is

x(t) = e3t

1 - 2t -t

4t 1 + 2t

1 1

= e3t

1 + 2t 1+t

.

5

6. Let V be the space of vectors in R4 such that x1 + x2 + x3 + x4 = 0 and let W be the set of vectors in V such that x1 = x4.

(a) Find an orthogonal basis {v1, v2, v3} for V with v1 = (0, 1, -1, 0) and such that {v1, v2} is a basis for W .

Observe that if we choose v2 = (1, -1, -1, 1), then v2 W as well as being orthogonal to v1. Since W is two-dimensional, {v1, v2} is a basis for it. Next, let v3 = (1, 0, 0, -1). Then, v3 is orthogonal to both v1 and v2 and also belongs to V . Since the latter has dimension 3, {v1, v2, v3} is an orthogonal basis for it.

(b) Find the orthogonal projection of v = (2, 1, 3, -6) on W .

We have

projW (v)

=

v ? v1 v1 ? v1

v1

+

v ? v2 v2 ? v2

v2

1-3 2-1-3-6 = 1 + 1 v1 + 1 + 1 + 1 + 1 v2

= -v1 - 2v2

= (-2, 1, 3, -2).

(c) Find the distance from v to W .

The distance from v to W is given by

v - projW (v)

= (4, 0, 0, -4) = 42 + 42 1/2 = 4(2)1/2.

6

7. Show that {e3x, e-x, e-4x} is a fundamental solution set for y + 2y - 11y - 12y = 0.

Let y1 = e3x, y2 = e-x, y3 = e-4x. Then, we have

y1 + 2y1 - 11y1 - 12y1 = e3x(27 + 18 - 33 - 12) = 0 y2 + 2y2 - 11y2 - 12y2 = e-x(-1 + 2 + 11 - 12) = 0 y3 + 2y3 - 11y3 - 12y3 = e-4x(-64 + 32 + 44 - 12) = 0

so y1, y2, y3 are solutions of the given differential equation. To show that {y1, y2, y3} is a fundamental solution set, we only need to prove that these functions are linearly independent. The Wronskian for these functions is

W (x) =

e3x 3e3x 9e3x

e-x -e-x e-x

e-4x -4e-4x 16e-4x

11 1 = (e3x)(e-x)(e-4x) 3 -1 -4

9 1 16

= e-2x[1(-16 + 4) - 1(48 + 36) + 1(3 + 9)] = e-2x[-12 - 84 + 12] = -84e-2x

which is never zero. Hence, {y1, y2, y3} is a linearly independent collection of functions and therefore a fundamental solution set.

7

8. Consider the matrix

A=

2 1

-1 2

.

(a) Find a fundamental matrix for z = Az.

We first find the eigenvalues of A. We have 0 = 2 - 4 + 5 so = [4 ?

16 - 20]/2 = 2 ? i. Corresponding to = 2 + i, we have A - I =

-i 1

-1 -i

which has v =

i 1

in its null space. Thus, two linearly independent solutions of

z = Az are

z1(t)

=

e2t cos(t)

0 1

- e2t sin(t)

1 0

= e2t

- sin(t) cos(t)

z2(t)

=

e2t sin(t)

0 1

+ e2t cos(t)

1 0

= e2t

cos(t) sin(t)

.

A fundamental matrix is therefore Z(t) = e2t

cos(t) sin(t)

- sin(t) cos(t)

.

e2t (b) Find a particular solution of z = Az + f(t) where f(t) = -e2t .

By variation of parameters, a particular solution is zp(t) = Z(t)v(t) where

Thus,

v(t) = = = =

Z(t)-1f(t) dt

e-2t

cos(t) - sin(t)

sin(t) cos(t)

e2t -e2t dt

cos(t) - sin(t) - sin(t) - cos(t)

dt

sin(t) + cos(t) cos(t) - sin(t)

.

zp(t)

=

e2t

cos(t) sin(t)

- sin(t) cos(t)

sin(t) + cos(t) cos(t) - sin(t)

=

e2t

cos2(t) + sin2(t) cos2(t) + sin2(t)

= e2t

1 1

.

8

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