Intro to Differential Equations and Linear Algebra
MATH 250B

Spring 2013
1
Solutions to Section 1.1
TrueFalse Review:
1. FALSE. A derivative must involve some derivative of the function y = f (x), not necessarily the rst
derivative.
2. TRUE. The initial conditions accompanying a dierential equation consist of the values of y,
Intro to Differential Equations and Linear Algebra
MATH 250B

Spring 2010
Math 250B
Exam #2 Solutions
Spring 2010
2x1 + x2 = b
Determine conditions on a and b such that there exists (i)
x1 + ax2 = 1
exactly one solution, (ii) no solution, and (iii) innitely many solutions.
1. Consider the system of equations
21
= 2a1. Thus, as
Intro to Differential Equations and Linear Algebra
MATH 250B

Spring 2010
Math 250B
Test #3
Spring 2010
1. (a) Circle TRUE or FALSE corresponding to the truth value of the statement.
i.
TRUE FALSE If A is a 2 3 matrix, then nullity(A) > 0.
ii. TRUE
FALSE
Any set of three vectors in R4 is linearly independent.
iii.
TRUE FALSE
An
Intro to Differential Equations and Linear Algebra
MATH 250B

Spring 2010
Math 250B
Test #4
Spring 2010
1. Consider the dierential equation
(D2 + 4D + 5)(D2 + 4)y = 5 cos 3x.
(a) Determine the general solution to the homogeneous problem.
SOLUTION: The Auxilliary equation is (r2 + 4r + 5)(r2 + 4) = 0, which has the solutions r =
Intro to Differential Equations and Linear Algebra
MATH 250B

Spring 2010
Math 250B
Solutions to Problem Set #1
Spring 2010
Solutions to 1.1
d2 y
dy
=g
= gt + C1
2
dt
dt
Now impose the initial conditions
1.
y (t) =
y (0) = 0 C2 = 0
g2
t + C1 t + C2 .
2
and
y (0) = 0 C1 = 0.
Hence the solution to the IVP is
g2
t.
2
The object h
Intro to Differential Equations and Linear Algebra
MATH 250B

Spring 2010
Solutions to 1.9
Math 250B
Spring 2010
1. (y + 3x2 )/y = 1 and x/x = 1. So, My = Nx , hence the equation is exact.
3. (yexy )/y = exy (1 + xy ) and (2y xexy )/x = exy (1 xy ) so the equation is NOT exact.
5. (y 2 + cos x)dx + (2xy + sin y )dy = 0. A poten
Intro to Differential Equations and Linear Algebra
MATH 250B

Spring 2010
Solutions to 5.7
3.
det(A I ) = 0
1
2
2 5
= 0 ( 3)2 = 0 1 = 3 of multiplicity two.
0
2 2 v1
=
v1 v2 = 0. The
0
2 2 v2
solution set of this system is cfw_(r, r) : r R so the eigenspace corresponding to 1 = 3 is E1 =
y R2 : v = r(1, 1), r R . A basis for
Intro to Differential Equations and Linear Algebra
MATH 250B

Spring 2010
Solutions to 4.9
Math 250B
Spring 2010
1. It is easy to show that
nullspace(A)
= cfw_(6s + t, r, s, t) : r, s, t R
= cfw_r(0, 1, 0, 0) + s(6, 0, 1, 0) + t(1, 0, 0, 1) : r, s, t R
Therefore a basis for nullspace(A) is cfw_(0, 1, 0, 0), (6, 0, 1, 0), (1, 0,
Intro to Differential Equations and Linear Algebra
MATH 250B

Spring 2010
Solutions to 6.3
17. The auxiliary polynomial is P (r) = r2 + 4r + 4 = (r + 2)2 . Therefore, the complementary function is
yc (x) = c1 e2x + c2 xe2x .
The given dierential equation can be written in operator form as
(D + 2)2 y = 5xe2x .
An annihilator for
Intro to Differential Equations and Linear Algebra
MATH 250B

Spring 2010
Solutions to 1.8
9.
15.
dy
y dy
y
dv
= 3y (3 2 )
=3
(3 2v)(v + x
) = 3v
dx
x dx
x
dx
dv
3v
3 2v
dx
3
x dx = 3 2v v
2v2 dv = x 2v lnv = lnx + c1
3x
y
3x
2y lnx  = lnx + c1 lny = 2y + c2 y2 = ce3x/y.
(3x 2y)
dy
dy
y
dv
x dx + yln x = yln y dx
Intro to Differential Equations and Linear Algebra
MATH 250B

Spring 2010
Solutions to 1.6
1.
dx
d (e x y )
yy = e2x. I ( x) = e = e x
= e x e x y = e x + c y = e 2 x + ce x
dx
2x
5.
2x
4
2 dx
y=
.
I ( x) = e 1+ x = 1 + x 2
2
2
1+ x
1+ x2
d
4
(1 + x 2 ) y =
dx
1+ x2
dx
(1 + x 2 ) y = 4
= 4 arctan( x) + c
1+ x2
1
y ( x) =
Intro to Differential Equations and Linear Algebra
MATH 250B

Spring 2010
Solutions to 7.3
1.
4t
4t
2 3 e
= 4e = x .
x1 is a solution: Ax1 =
1
4t
2 5 2e4t
8e
t
t
2 3 3e
= 3e = x .
x2 is a solution: Ax2 =
2
2 5 et
et
e4t 3et
and det[x(0)] = 1 3 = 5 0.
x(t) = [x1, x2] =
4t
2 1
2e
et
Consequently, x1 and
Intro to Differential Equations and Linear Algebra
MATH 250B

Spring 2010
Math 250B
Exam 1 solutions
Spring 2010
1. For the following equations sketch the slope eld and some representative solution curves.
(a)
y = y x
The isoclines are y = x + k , that is to say the lines of slope 1 and yintercept k. On each such
line the slop