Math 285 Final Part I 6-11-14
1) ( 7 points) Solve
2
x
(3 xy  )dx  ( x 2  2 )dy  0
y
y
2) ( 5 points : IQ 85 question) Prove that if n x n matrices A and B are invertible (
then their product
AB is invertible.
3) ( 7 points) Show that the set of functions satisfying the DE
space of all continuously differentiable functions
y"4 y  0 is a
A1 , B 1
exist),
subspace of the vector
C  (R)
4) ( 5 points each)
a) Find a basis for the subspace of
R4
{(1,2,3,0), (0,2,1,1), (0,0,2,1), (2,2,5,1)
x  2 y  3z  0
spanned by
b) Find two vectors that span the plane
5) ( 2 points each)
a) Find the sign of the permutation of
(1,4,3,2)
b) True/false: In a 3 dimensional vector space, any set containing 4 vectors is LD.
c) True/false: if
T : R3  R3
is onto, then it is 1-1.
6) ( 7 points) Use series to find terms up to
x 6 : y"xy  0 .
Part II
Math 285 Final Part II 6-11-14
1) ( 5 points each) Let
T : P1  P1
be defined as
T(a  bx)  (a  3b)  2bx
a) Find a matrix of T with respect to the bases B
 {2, 1  x} for the domain and C  {2,2 x} for the
range.
b) Determine if T is 1-1 or not.
2) (5 points each)
a) Compute
b) Use

t t  2
L[ f ] for f (t )  

0 t  2
CONVOLUTION to compute L1[ F ] for F ( s) 
2
3
.

( s  2) ( s  4)
3) A mass weighing 32 pounds stretches the spring 2 feet. The first mass is set aside and another
mass weighing 4 pounds is initially released from a point 1 foot above the equilibrium position
with a downward velocity of
3
ft/sec. Assume there is no damping.
a) (5points) Find the equation of the motion. In addition, find the frequency.
b) (2 points) Find

if the mass is driven by an external force
f (t )  sin t
and the system is
resonating.
4) ( 5 points each)
a) Use Gauss-Jordan to find the inverse of
b) Find the determinant of
5) ( 5 points each)
For
x1 '  3x1  2 x2
x2 '  2 x1  2 x2
1 0 1 
0 1 2 


1  1 0
A3 (2B 1 ) AT if A and B are 3 x 3 matrices, det(A)=2, det(B)=8
a) Solve the system of equations.
b) Use eigenvectors to sketch phase portrait.
6) (5 points each)
a) Find the form of the most efficient
yp :
do not solve for constants.
y"4 y  x cos 2 x  x sin 2 x  3  x
weighted inner product).


C 0 [0, ] as follows:  f , g   2 x f ( x) g ( x)dx (this is called a
0
2
Compute sin 2 x
b) An inner product is defined on
Math 285 S2014 Final Solutions
Part I
1)
M y  3 x  2 y 2 , N x  2 x  y 2 
M y  Nx
N

1
x
is a function of x only. Thus we can find an
1
 dx
 e x  x . Multiplying by x, we obtain an exact DE:
2x
x2
x2
(3x 2 y  )dx  ( x 3  2 )dy  0  f ( x, y)   Mdx  x 3 y   h( y) . Next differentiate this
y
y
y
integrating factor as I
with respect to y and compare the results with N:
f
x2
x2
 N  x 3   h' ( y)  x 3  2  h' ( y)  0 .
y
y
y
Thus
f ( x, y)  x 3 y 
x2
c
y
2) Recall that a matrix is invertible iff their determinant is not 0. Since A and B are invertible, det(A)
and det(B) are not 0. Det(AB)=det(A)det(B) is not 0 since it is a product of two nonzero numbers.
Thus AB is invertible.
3) Check the three conditions for subspace:
a)
0 is in S since 0"4(0)  0
b) If
c) If
2
y1 , y2 are solutions then ( y1  y2 )"4( y1  y )  ( y1"4 y1 )  ( y2 "4 y2 )  0  0  0
k is a scalar and y is a solution, then (ky)"4(ky)  k ( y"4 y )  k (0)  0
4)
a) It is best to find its column space. Place the vectors into columns as
1
2

3

0
0 0  2 1
2 0  2 0

1 2  5  0
 
1 1 1  0
0 0  2
1 0 1 
0 1 0

0 0 0
Taking the columns with leading 1s,
x  2 y  3z  0 has one equation with three variables:
y  t , z  s  x  2t  3s .
b)
Thus
{(1,2,3,0), (0,2,1,1), (0,0,2,1)} is a basis.
 x   2t  3s   2  3
 y  
  t  1   s 0  .
t
  
    
 z  
  0   1 
s
thus there are two free variables. Let
Thus the plane is spanned by
{( 2,1,0), (3,0,1)}
5) A ) -1 b) True C) True
6)

y   an x n
,
n 0
y" xy  0 


n 1
n2
y '   nan x n1 , y"   n(n  1)an x n2



 n(n  1)an x n2  x an x n  0   n(n  1)an x n2 
n2
n2
n 0

a x
n 0
n 1
n
0
For the first summation, add two to n, then subtract the index by 2.
For the second summation, subtract 1 from n, then add1 to the index.

 (n  2)(n  1)an2 x n 
n 0

a
n 1
n 1
xn  0
Next remove the n=0 term from the first summation to get the same starting index:

Now the summations can be combined as
 [a
n 1
We have
an2 (n  2)( n  1)  an1  0  an2 
n 2
(n  2)( n  1)  an1 ]x n  0
1
an1
(n  2)( n  1)
For
a0 , a3 , a6 : a3 
1
1
1
a0 , a6 
a3 
a0
(6)(5)
(6)(5)(3)( 2)
(2)(3)
For
a1 , a4 , a7 : a4 
1
1
1
a1 , a7 
a4 
a0
(7)(6)
(7)(6)( 4)(3)
(3)( 4)
In addition,
a2 , a5  0
y  a0  a1 x  a2 x 2  a3 x 3  a4 x 4  a5 x 5  a6 x 6  ....
Thus
 a0 (1 
1 3
1 6
1
1 7
x 
x  ...)  a1 ( x  x 4 
x  ...)
6
180
12
504
Part II
1)
a)
 a2  0
1
T (2)  2  1(2)  0(1  x)   
0 
is the first column.
 1
T (1  x)  2  2 x  1(2)  1(2 x)    is the second column.
1
Thus
1  1
A

0 1 
b) Since the matrix of A is rank 2, the kernel is trivial. Thus it is 1-1.
2)
f (t )  (0 (t )  2 (t ))t  t  2 (t )t (recall that 0 (t )  1)
1
1 2
L[ f ]  L[t ]  L[  2 (t )t ]  L[t ]  L[  2 (t )((t  2)  2)]  2  e 2 s [ 2  ]
s
s
s
t
t
2
3
L1[

]  2e 2t  3e 4t  6 e 2(t  x ) e 4 x dx  6(e 2t )  e 6 x dx  e 2t (1  e 6t )
0
0
( s  2) ( s  4)
a) Use the second shifting theorem:
b)
3)
a
1)
A)
Thus
B)
1 0 1 1 0 0
0 1 2 0 1 0  RREF


1  1 0 0 0 0
1 0 0 2  1 1 
0 1 0 2  1 2 


0 0 1  1 1  1
 2 1 1 
A   2  1 2 
 1 1  1
1
det( A3 (2 B 1 ) AT )  23 det( A)3
1
det( A)  16
det( B)
4)
a) First find the eigenvalues and eigenvectors:
For the eigenvectors,
 3
2
A
 . det( A  I )  (  4)(  1)  0    1,4
2

2


 1 
 2 
  1  t 
 ,   4  t  2 
 1 
1


 
Thus the general solution is
 1 
 2 
t 
y  c1e 4t 

c
e
 2
2


1


 1 
Since the eigenvalues are negative, all solutions approach 0.
 2 

 , the eigenvector corresponding the smaller eigenvalues.
 1 
 1 
 2  , the eigenvector corresponding the larger eigenvalues.
 1 
 
c) A)
p( D)  D 2  4  0  yc  c1 cos 2t  c2 sin 2t
For
They are initially parallel to
They end up being parallel to
Q(D )
and
annihilates
D
x cos 2x  c sin 2x
2
2
P( D)Q( D) y  ( D  4) D y  0  y  yc  y p
,
x3
2
( D 2  4) 2
.
annihilates
Thus
 c1 cos 2 x  c2 sin 2 x  Ax cos 2 x  Bx sin 2 x  Cx 2 cos 2 x  Dx 2 sin 2 x  E  Fx
B)
sin 2 x   sin 2 x, sin 2 x  

1 2

(  xdx   2 x cos 4 xdx) 
0
2 0
4




2
0
x sin 2 2 xdx
.
To
integrate
,
write
sin 2 2 x 
1
(1  cos 4 x)
2