MATH 225: DIFFERENTIAL EQUATIONS AND LINEAR ALGEBRA ...

MATH 225: DIFFERENTIAL EQUATIONS AND LINEAR ALGEBRA
Solutions of First Midterm Exam
March 13, 2006
1.a) Show that the function y(x) = A + Be x + Ce 2x satisfies the differential
equation y ′′′ − 3y ′′ + 2y ′ = 0, where A, B, and C are arbitrary constants.
Solution:
Hence
y(x) = A + Be x + Ce 2x ,
y ′ (x) = Be x + 2Ce 2x ,
y ′′ (x) = Be x + 4Ce 2x
y ′′′ (x) = Be x + 8Ce 2x
y ′′′ − 3y ′′ + 2y ′ = Be x + 8Ce 2x − 3(Be x + 4Ce 2x ) + 2(Be x + 2Ce 2x ) = 0
1.b) Find the constants A, B, and C so that y(0) = 1, y ′ (0) = −1, y ′′ (0) = 0.
Solution:
y(0) = A + B + C = 1,
y ′ (0) = B + 2C = −1,
y ′′ (0) = B + 4C = 0
There is unique solution A = 5/2, B = −2, C = 1/2.
2.a) Solve the following system by Gauss-Jordan Elimination
Solution :
x1 − 2x2 + x3 − 4x4 = 1,
x1 + 3x2 + 7x3 + 2x4 = 2,
x1 − 12x2 − 11x3 − 16x4 = −1
⎡
1 −2 1 −4
⎤
1
⎡
1 −2 1 −4
⎤
1
⎡
1 −2 1 −4
⎤
1
⎣1
3 7 2 2 ⎦ → ⎣0
5 6 6 1 ⎦ → ⎣0
5 6 6 1⎦
→
1 −12 −11 −16 −1 1 −12 −11 −16 −1 0 0 0 0 0
⎡
1 −2 1 −4
⎤
1
⎡
1 0 17/5 −8/5
⎤
7/5
⎣0
1 6/5 6/5 1/5⎦
→ ⎣0
1 6/5 6/5 1/5⎦
0 0 0 0 0 0 0 0 0 0
Hence the there are infinitely many solutions with two free parameters
x1 = 7 17t 8s
− +
5 5 5 ,
x2 = 1 6t 6s
− −
5 5 5 ,
x3 = t, x4 = s

2.b) What conditions must b1,b2, and b3 satisfy in order for the system
x1 + x2 + 2x3 = b1,
x1
to be consistent. Find the solutions.
+ x3 = b2,
2x1 + x2 + 3x3 = b3
Solution :
⎡
1
⎣1
1
0
2
1
⎤
b1
b2⎦
⎡
1
→ ⎣0
1
−1
2
−1
⎤
b1
b2 − b1⎦
⎡
1
→ ⎣0
1
−1
2
−1
⎤
b1
b2 − b1 ⎦
2 1 3 b3 2 1 3 b3 0 −1 −1 b3 − 2b1
⎡
1
→ ⎣0
1
1
2
1
b1
b1 − b2
⎤
⎦
0 0 0 b1 − b3 + b2
Hence for consistency b3 = b1 + b2. In this case there are infinitely many solutions
x1 = b2 − t, x2 = b1 − b2 + t, x3 = t where t ∈ R is an arbitrary real variable.
3.a) Solve the differential equation x 2 y ′ + 2xy = 5y 4 with y(1) = 1. Discus the
existence and uniqueness of the solution of the differential equation with the initial
condition y(0) = 0.
Solution : Let v(x) = y −3 then
v ′ − 6
x
15
v + = 0
x2 This is a linear equation with the integrating factor ρ(x) = 1
x 6 . Hence the solution
is
v(x) = C x 6 + 15
7
1
x
�
5x
→ y(x) =
7Cx7 + 15
where C is an arbitrary constant. If y(1) = 1 then C = − 8
7 .
The singular solution of this equation is y = 0. Furthermore at x = 0 the existence
and uniqueness is not guaranteed. It is clear from the solution that y(0) = 0 is
satisfied for all C, hence there are infinitely many solutions.
3.b) Solve the differential equation (x + ln y)dx + ( x
y + ey )dy = 0 with y(0) = 1
Solution: M(x, y) = x + ln y and N(x, y) = x
y + ey . Then
but
Fx = M(x, y) = x + ln y → F (x, y) = x2
2
� 1/3
+ x ln y + g(y),
Fy = x
y + g′ = x
y + ey , → F (x, y) = x2
+ x ln y + ey
2

Hence the solution is
x 2
2 + x ln y + ey = C
where C is an arbitrary constant. Since y(0) = 1 then C = e. Therefore the solution
is
x 2
2 + x ln y + ey = e
4.a) Find the Reduced Row-Echelon matrix of the augmented coefficient matrix of
the following system. Then find the solution of the system.
Solution:
x1 + 2x2 + 3x3 = 4,
2x1 + 5x2 + 3x3 = 5,
x1
+ 8x3 = 9
⎡
1 2 3
⎤
4
⎡
1 2 3
⎤
4
⎡
1 2 3
⎤
4
⎡
1 2 3
⎤
4
⎣2
5 3 5⎦
→ ⎣0
1 −3 −3⎦
→ ⎣0
1 −3 −3⎦
→ ⎣0
1 −3 −3⎦
1 0 8 9 1 0 8 9 0 −2 5 5 0 0 −1 −1
⎡
1 0 9
⎤
10
⎡
1 0 0
⎤
1
⎡
1 0 0
⎤
1
→ ⎣0
1 −3 −3⎦
→ ⎣0
1 −3 −3⎦
→ ⎣0
1 0 0⎦
0 0 1 1 0 0 1 1 0 0 1 1
Hence the solution is x1 = 1, x2 = 0, x3 = 1.
4.b) Let XA = Z where A, Z are matrices given by
⎡
0
A = ⎣1
2
−2
2
0
⎤
3
−3⎦
,
4
�
4
Z =
−1
−2
2
�
3
.
0
Find A −1 and the matrix X.
Solution:
⎡
0 −2 3 1 0
⎤
0
⎡
1 2 −3 0 1
⎤
0
⎡
1 2 −3 0 1
⎤
0
⎣1
2 −3 0 1 0⎦
→ ⎣0
−2 3 1 0 0⎦
→ ⎣0
−2 3 1 0 0⎦
2 0 4 0 0 1 2 0 4 0 0 1 0 −4 10 0 −2 1
⎡
1 2 −3 0 1
⎤
0
⎡
1 2 −3 0 1
⎤
0
→ ⎣0
1 −3/2 −1/2 0 0⎦
→ ⎣0
1 −3/2 −1/2 0 0⎦
0 −4 10 0 −2 1 0 0 4 −2 −2 1
⎡
1 0 0 1 1
⎤
0
⎡
1 0 0 1 1
⎤
0
→ ⎣0
1 −3/2 −1/2 0 0 ⎦ → ⎣0
1 0 −5/4 −3/4 3/8⎦
0 0 1 −1/2 −1/2 1/4 0 0 1 −1/2 −1/2 1/4

Hence
A −1 ⎡
1 1
⎤
0
= ⎣−5/4
−3/4 3/8⎦
, then X = ZA
−1/2 −1/2 1/4
−1 =
� �
5 4 0
.
−7/2 −5/2 3/4