Math 2250-1
Summer 2010
Final Exam
You will have two hours to complete this test. Please show all your work, since your score on a problem is determined from that and not necessarily from your answer. All solutions should be real-valued unless specifically instructed otherwise. No calculators or other electronic devices are allowed. Good Luck!
1. Find the general solution to the equation y
0
= x 2 + 1 x + 1
The easiest way to evaluate the integral is to rewrite it. To do this, we perform long division to find that x
2
+ 1 x + 1
= x − 1 +
2 x + 1
.
We then have that y ( x ) =
Z x − 1 +
2 x + 1 dx = x
2
2
− x + 2 ln | x + 1 | + C.
2. Solve the initial value problem dy dx
= 2 xy
2
+ 3 x
2 y
2
, y (1) = − 1
We first rewrite the right hand side of the equation as dy dx
= y
2
(2 x + 3 x
2
) , so that upon separating variables we have
Z dy y 2
=
Z
(2 x + 3 x
2
) dx ⇒ − y
− 1
=
− 1 x 2 + x 3 + C
⇒ y ( x ) =
− 1 x 2 + x 3 − 1
.
Applying the initial condition, we find that
− 1 =
− 1
2 + C
⇒ 2 + C = 1 ⇒ C = − 1 , so y ( x ) =
− 1 x 2 + x 3 − 1
.
3. Consider a tank initially containing 2 lbs of salt dissolved in 20 gal of salt water. Salt water containing 1 lb of salt per gallon flows into the tank at a rate of 3 gal/min. The well-mixed solution flows out of the tank at 2 gal/min. Find the amount of salt in the tank as a function of time.
1

If x ( t ) is the amount of salt in the tank and V ( t ) is the volume in the tank, then the initial conditions are x (0) = 2 and V (0) = 20. We also have that V ( t ) = ( r i
− r o
) t + V
0
= (3 − 2) t + 20 = t + 20.
The equation for x is x
0
= 3 −
2
20 + t x ⇒ x
0
+
2
20 + t x = 3 .
This is a first-order linear equation with integrating factor
ρ ( t ) = exp
Z 2
20 + t dt = exp(2 ln(20 + t )) = (20 + t )
2
.
Multiplying both sides of the equation by ρ ( t ) yields
(20 + t )
2 x
0
+ 2(20 + t ) x = 3(20 + t )
2
.
At this point we recognize that the left hand side can be expressed as
D t x (20 + t )
2
= 3(20 + t )
2
, and when we integrate, we have x (20 + t )
2
=
Z
3(20 + t )
2 dt = ( t + 20)
3
+ C.
Solving this expression for x ( t ), we find x ( t ) = (20 + t ) +
C
(20 + t ) 2
.
The initial condition implies that
C
2 = 20 +
400
⇒ C = − 7200 .
The solution is x ( t ) = (20 + t ) −
7200
(20 + t ) 2
.
4. Use Euler’s method to find approximations to y (1 / 4) and y (1 / 2) using stepsize 1 / 4, where y is the solution to the initial value problem y
0
= 2 x + 4 y
2
, y (0) = 1 .
The initial conditions imply that x
0
= 0 , y
0
= 1 .
Using the relation y k +1
= y k
+ hf ( x k
, y k
) with f ( x, y ) = 2 x + 4 y 2 , we have that x
1
= x
0
+ h =
1
4
, y
1
= 1 +
1
4
(4) = 2 .
2

Taking one more step, we find that x
2
=
1
2
, y
2
= 2 +
1
4
2
1
4
+ 4 (2)
2
= 2 +
1
4
1
2
+ 16 = 2 +
33
8
=
49
.
8
5. Find the inverse of the matrix

A =

1 4 1
2 8 3
2 7 4

 and use it to solve the equation AX = B , where

B =

1 0
0 2
− 1 1


.
To find A
− 1
, we augment A with the identity matrix and row reduce:

1 4 1




2 8 3
2 7 4
..
.
1 0 0
..
.
0 1 0
..
.
0 0 1





∼





1 4 1
0 0 1
0 − 1 2
..
.
1 0 0
..
.
− 2 1 0
..
.
− 2 0 1





∼





1
0
0
0
1
0
9
−
1
2
..
.
− 7 0 4
..
.
2 0 − 1
..
.
− 2 1 0






1 0 0
∼




0 1 0
0 0 1
..
.
11 − 9 4
..
.
− 2 2 − 1
..
.
− 2 1 0


 .


This implies that

A
− 1
=

11 − 9 4
− 2 2 − 1
− 2 1 0


.
If AX = B , then

X = A
− 1
B =

11 − 9 4
− 2 2 − 1
− 2 1 0
 
 
1 0
0 2
− 1 1
 

=

7 − 14
− 1 3
− 2 2


.
6. Let

A =

3 − 1 2
− 1 1 0
4 3 2


.
Find det( A 3 ).
Expanding along the third column, we have that det( A ) = 2
− 1 1
4 3
+ 2
3 − 1
− 1 1
= 2( − 3 − 4) + 2(3 − 1) = − 14 + 4 = − 10 .
3

For any two square matrices A and B of the same size, we have that det( AB ) = det( A )det( B ).
Therefore, det( A
3
) = det( A )det( A )det( A ) = (det( A ))
3
= − 1000 .
7. Find a basis for the subspace of all vectors ~ ∈
R
3 such that ~ · ~
1
= ~ · ~
2
= 0, where

1
=

1
0
− 1


 and ~
2
=

− 3
1
− 2


.
If ~ · ~
1
= 0, then x
1
− x
3 this system of equations is
= 0. If ~ · ~
2
= 0, then − 3 x
1
+ x
2
− 2 x
3
= 0. The coefficient matrix for
1 0 − 1
− 3 1 2
∼
1 0 − 1
0 1 − 5
.
Since there is no leading one in the third column of the reduced row echelon form of the coefficient matrix, the variable x
3 is arbitrary. If we let x
3
= r , then x
1
= r, x
2
= 5 r, x
3
= r.
Taking r = 1, we have that the set of vectors in question is the set of all linear combinations of the vector
 1 

5
1

, so this is our basis.
8. Solve the initial value problem y
00
− 6 y
0
+ 13 y = 0 , y (0) = 2 , y
0
(0) = − 1 .
The characteristic equation for this differential equation is r
2 − 6 r + 13 = 0 ⇒ r
2 − 6 r + 9 = − 13 + 9 ⇒ ( r − 3)
2
= − 4 ⇒ r = 3 ± 2 i.
Therefore, the general solution is y ( x ) = c
1 e
3 x cos 2 x + c
2 e
3 x sin 2 x.
Taking the derivative, we have that y
0
( x ) = 3 c
1 e
3 x cos 2 x − 2 c
1 e
3 x sin 2 x + 3 c
2 e
3 x sin 2 x + 2 c
2 e
3 x cos 2 x.
Applying the initial conditions, we arrive at the equations c
1
= 2
3 c
1
+ 2 c
2
= − 1
⇒ 2 c
2
= − 1 − 6 = − 7 ⇒ c
2
= −
7
2
.
4

Therefore, the solution is y ( x ) = 2 e
3 x cos 2 x −
7
2 e
3 x sin 2 x.
9. Find the general solution to the equation y
(4) − 6 y
(3)
+ 10 y
00 − 6 y
0
+ 9 y = 0 .
[Hint: i is a root of the characteristic polynomial.]
Since i is a root, we know that − i is too, which implies that ( r 2 + 1) is a factor of the characteristic polynomial. Performing the corresponding long division, we have that r 2 − 6 r + 9 r 2 + 1 ) r 4
− r 4
− 6 r 3
− r 2
+ 10
− 6 r 3
6 r 3
+ 9
+ 6 r r 2 − r
6
2 r
9 r 2 + 9
− 9 r 2 − 9
− 6 r + 9
0
.
The other two roots must satisfy
0 = r
2 − 6 r + 9 = ( r − 3)
2
, so the four roots of the characteristic polynomial are r = i, − i, 3 , 3. This means that the general solution is y ( x ) = c
1 cos x + c
2 sin x + c
3 e
3 x
+ c
4 xe
3 x
.
10. Find the general solution to the equation y
00
+ 2 y
0
− 3 y = sin 3 x + 4 e x
+ x.
The characteristic equation for the associated homogeneous equation is r
2
+ 2 r − 3 = 0 ⇒ r
2
+ 2 r + 1 = 3 + 1 ⇒ ( r + 1)
2
= 4 ⇒ r = − 1 ± 2 = 1 or − 3 .
Therefore, the general solution to the homogeneous problem is y c
( x ) = c
1 e x
+ c
2 e
− 3 x
.
Our first guess for y p
( x ), based on the form of the right-hand side of the equation is y p
( x ) = A sin 3 x + B cos 3 x + Ce x
+ Dx + E.
However, we notice that the term e x is repeated in y c
, so we multiply this term by an x to get y p
( x ) = A sin 3 x + B cos 3 x + Cxe x
+ Dx + E ⇒ y
0 p
( x ) = 3 A cos 3 x − 3 B sin 3 x + Ce x
+ Cxe x
+ D,
5

y
00 p
( x ) = − 9 A sin 3 x − 9 B cos 3 x + 2 Ce x
+ Cxe x
.
Substituting these expressions into the left-hand side of the equation, we find that y
00 p
+ 2 y
0 p
− y p
=
− 9 A sin 3 x − 9 B cos 3 x + 2 Ce x
+ Cxe x
+ 2(3 A cos 3 x − 3 B sin 3 x + Ce x
+ Cxe x
+ D )
− 3( A sin 3 x + B cos 3 x + Cxe x
+ Dx + E ) = ( − 12 A − 6 B ) sin 3 x +(6 A − 12) B cos 3 x +4 Ce x − 3 Dx +2 D − 3 E.
Equating coefficients, we find that
− 12 A − 6 B = 1
6 A − 12 B = 0
4 C = 4
− 3 D = 1
2 D − 3 E = 0
⇒
C = 1
D = − 1
E = −
3
2
9
.
As for A and B , we have


− 12 − 6
6 − 12
..
.
1
..
.
0



∼

1
1
2
0 − 15 .
..
..
.
− 1
12
1
2
 

∼

1 0
0 1
Our general solution is therefore
..
.
− 2
30
..
.
− 1
30


.
y ( x ) = c
1 e x
+ c
2 e
− 3 x
−
2
30 sin 3 x −
1
30 cos 3 x + xe x
−
1
3 x −
2
9
.
11. Find an invertible matrix P and a diagonal matrix D such that A = P DP
− 1 , where

A =

6 − 5 2
4 − 3 2
2 − 2 3


.
The eigenvalues must satisfy
0 = det( A − λI ) =
6 − λ
4
2
− 5
− 3 − λ
− 2
2
2
3 − λ
= (6 − λ )
− 3 − λ
− 2
2
3 − λ
+5
4 2
2 3 − λ
+2
4 − 3 − λ
2 − 2
= (6 − λ )[( − 3 − λ )(3 − λ ) + 4] + 5[4(3 − λ ) − 4] + 2[ − 8 + 2(3 + λ )]
= (6 − λ )[ − 9 + 3 λ − 3 λ + λ
2
+ 4] + 5[8 − 4 λ ] + 2[ − 2 + 2 λ ] = (6 − λ )[ λ
2 − 5] + 20 λ + 40 λ − 4 + 4 λ
= 6 λ
2 − 30 − λ
3
+ 5 λ − 16 λ + 36 = − λ
3
+ 6 λ
2 − 11 λ + 6 .
We first seek to find a rational root of the characteristic polynomial by trying λ = ± 1 , ± 2 , ± 3 , ± 6.
When we try λ = 1, we get
− 1 + 6 − 11 + 6 = 0 ,
6

so λ = 1 is one eigenvalue of A . This means that ( λ − 1) should factor out of the characteristic polynomial, and upon performing long division, we find
− λ 2 + 5 λ − 6
λ − 1 ) − λ 3
λ 3 − λ
+ 6
2
λ 2
5 λ 2 − 11 λ
− 5 λ 2 + 5 λ
− 6 λ + 6
6 λ − 6
0
− 11 λ + 6
The remaining two eigenvalues must satisfy
− λ
2
+ 5 λ − 6 = 0 ⇒ λ
2 − 5 λ +
25
4
= − 6 +
25
4
⇒ λ −
5
2
2
=
1
4
⇒ λ =
5
2
±
1
2
= 3 or 2 .
Thus, the complete list of eigenvalues is λ = 1 , 2 , 3. For λ = 1, we have
A − I =


5 − 5 2
4 − 4 2
2 − 2 2



∼

1 − 1
0 0
0 0
5
6
5
2
5
2
 

∼

1 − 1 0
0 0 1
0 0 0


⇒ a = r b = r c = 0
For λ = 3, we have

A − 3 I =

3 − 5 2
4 − 6 2
2 − 2 0


∼


1 − 1 0
0 − 2 2
0 − 2 2


∼


1 0 − 1
0 1 − 1
0 0 0


⇒ a b c
=
=
= r r r

⇒ ~
1
=

1
1
0


.
⇒ ~
2

=

1
1
1


.
And finally, for λ = 2 we have

A − 2 I =

4 − 5 2
4 − 5 2
2 − 2 1


∼


1 − 1
1
2
0 − 1 0
0 − 1 0


∼


1 0
1
2
0 1 0
0 0 0


⇒ a = − b = 0
1
2 r c = r

⇒ ~
3
=

− 1
0
2


.
This implies that our matrices P and D are

P =

1 1 − 1
1 1 0
0 1 2

 and D =


1 0 0
0 3 0
0 0 2


.
Also note that any permutation of the columns of these two matrices would work just as well.
12. Use the definition of the Laplace transform to find the Laplace transform of the function defined by t 0 ≤ t ≤ 1 f ( t ) = .
0 otherwise
By definition,
L{ f ( t ) } =
Z
∞ e
− st f ( t ) dt =
0
Z
1 e
− st t dt.
0
7

Making the substitution u = t, du = dt, dv = e
− st dt, and v = − (1 /s ) e
− st yields and integrating by parts
L{ f ( t ) } = −
1 s te
− st
1
0
+
1 s
Z
1
0 e
− st dt = −
1 s e
− s
+
1 s
−
1 s e
− st
1
0
= − e
− s s
+
1 s
= − e
− s s
− e
− s s 2
1
+ s 2
− e
− s s
+
1 s for s > 0.
13. Use Laplace transforms to solve the initial value problem x
00
+ 3 x
0
+ 2 x = sin t, x (0) = 0 , x
0
(0) = − 1 .
Taking the transform of both sides and using the theorem about transforms of derivatives, we have s
2
X ( s ) + 1 + 3 sX ( s ) + 2 X ( s ) =
1 1
− 1 .
s 2 + 1
⇒ ( s
2
+ 3 s + 2) X ( s ) = s 2 + 1
Therefore,
X ( s ) = s 2 + 2 s 2 + 1
1
·
( s + 1)( s + 2)
=
( s 2 s 2 + 2
+ 1)( s + 1)( s + 2)
.
If we write
− s
2
( s 2 + 1)( s + 1)( s + 2)
=
As + B s 2 + 1
C
+ s + 1
D
+ s + 2 and multiply this expression through by the denominator on the left, we find that
− s
2
= ( As + B )( s + 1)( s + 2) + C ( s
2
+ 1)( s + 2) + D ( s
2
+ 1)( s + 1) .
Evaluating this equation at s = − 1 yields
− 1 = 2 C ⇒ C = −
1
2
.
Evaluating at s = − 2 yields
− 4 = − 5 D ⇒ D =
4
5
.
Equating the constant terms on each side of the equation, we have
0 = 2 B + 2 C + D ⇒ B =
1
2
( − 2 C − D ) =
1
2
1 −
4
5
Equating the coefficients of s 3 shows that
A = − C − D = −
1
2
−
4
5
3
= −
10
.
We now have that
1 1
X ( s ) =
10 s 2 + 1
3
−
10 s s 2 + 1
−
1 1
2 s + 1
+
4
5
1 s + 2
.
=
1
10
.
8

These terms can be immediately inverse transformed to find that
1 x ( t ) =
10 sin t −
3
10 cos t −
1
2 e
− t
+
4
5 e
− 2 t
.
14. Find the general solution to the system of equations x
0
=
3 − 1
5 − 3
First, we must find the eigenvalues of the coefficient matrix, which satisfy
0 =
3 − λ
5
− 1
− 3 − λ
= (3 − λ )( − 3 − λ ) + 5 = − 9 − 3 λ + 3 λ + λ
2
+ 5 = λ
2 − 4 .
Therefore the eigenvalues are λ = 2 , − 2. To find an eigenvalue corresponding to λ = 2, we calculate
A − 2 I =
1 − 1
5 − 5
∼
1 − 1
0 0
⇒ a b
=
= r r
⇒ ~
1
=
1
1
.
For λ = 2, we have
A + 2 I =
5
5
−
−
1
1
∼
1 −
0 0
1
5 ⇒ a = 1
5 b = r r
⇒ ~
2
=
1
5
.
The general solution is therefore
~ ( t ) = c
1
1
1 e
2 t
+ c
2
1
5 e
− 2 t
.
15. Find the solution to the initial value problem x
0
=
0 − 1
1 0 x, ~ (0) =
0
1
.
The eigenvalues of A must satisfy
0 =
− λ − 1
1 − λ
= λ
2
+ 1 ⇒ λ = ± i.
Now we need to find an eigenvector corresponding to one of these complex conjugate eigenvectors, say λ = i . We have
A − iI =
− i − 1
1 i
∼
1
0
−
0 i
⇒ a = ir b = r
⇒ ~
1
= i
1
.
Recall that e
( p + qi ) t
( ~a +
~bi
) = e pt
(cos qt + i sin qt )( ~a +
~bi
) = e pt
[cos qt~a − sin qt~b + i (cos qt~b + sin qt~a )] ,
9

so using the real and imaginary parts of e it v
1 have
~ ( t ) = c
1 cos t
0
1
− sin t
1
0 as our two linearly independent solutions, we
+ c
2 sin t
0
1
+ cos t
1
0
.
Applying the initial condition, we find that c
1
0
1
+ c
2
1
0
=
1
0
⇒ c
2
= 0 , c
1
= 1 .
Thus,
~ ( t ) =
− sin t cos t
.
Laplace Transforms t n e at cos kt sin kt n !
s n +1
1 s − a s s 2 + k 2 k s 2 + k 2
10