Solutions #1
Page 1 of 4
Solutions for Homework 1
1. Calculate the integral
e3x cos x dx.
solution:Applying integration by parts twice, we see that:
e3x cos x dx = e3x sin x 3
e3x sin x dx
= e3x sin x 3 e3x cos x + 3
= e3x sin x + 3e3x cos x 9
Therefore,

Week 2, Section 1-6
2. 2xyy = x2 + 2y 2 .
We can write it in the form
y
=
=
1x y
+
2y x
11
y
y + .
2x
x
It is of the form
y
y = F ( ).
x
Hence it is homogeneous and we make a substitution
v=
y
dy
dv
y = vx
=v+x ,
x
dx
dx
We get
v+x
dv
1
dv
1
dx
=
+v x
=

Week 1 - DE Intro, First-Order DEs
Section 1.1
1. DE is y = 3x2 , proposed solution is y = x3 + 7. Checking solution:
d
(x3 + 7) = 3x2
dx
LHS = y =
RHS = 3x2
Since RHS = LHS, y = x3 + 7 is a solution to the given DE.
3. DE is y + 4y = 0. There are two pro

Week 4, 4.1 Solutions
1.
x + 3x + 7x = t2
or, isolating x which we will need soon,
x = 3x 7x + t2
To turn this single second-order DE into a system of 2 rst-order DEs, we begin by dening
the new variables,
y1 = x
y2 = x
If we dierentiate each of these wil

Section 5.1
11. x = 3y, y = 3x.
x
y
=
0 3
3 0
x
y
x
y
=
3 2
2 1
x
y
12. x = 3x 2y, y = 2x + y
13. x = 2x + 4y + 3et , y = 5x y t2
x
y
2 4
5 y
=
x
3e5
+
y
t2
14. x = tx et y + cos(t), y = et x + t2 y sin(t)
x
y
=
t
et
et
t2
cos(t)
x
+
sin(t)
y
15. x = y +

Week 4- Non-Homogeneous Linear DEs
Section 3.5
1.
y + 16y = e3x
Given the RHS of e3x , we would prefer to use yp = Ae3x .
For yc , we solve r2 + 16 = 0 which gives r = 4i and yc = c1 cos(4x) + c2 sin(4x). There
is no overlap between yc then and the prefer

Week 4, 3.7 Solutions
11. R = 30 , L = 10 H, C = 0.02 F ; E(t) = 50 sin(2t) V
The two DEs we can derive are DE is
1
1
Q = RQ + LQ + Q
C
C
1
E = RI + LI + I
C
E = RI + LI +
or, dierentiating
We are asked to solve for current, so we use the second equation.

Week 1 - DE Intro, First-Order DEs
Section 1.4
1.
dy
+ 2xy
dx
dy
dx
1
Separate:
dy
y
1
dy
Integrate both sides:
y
ln y
=0
= 2xy
= 2x dx
=
2x dx
= x2 + C
Exponentiate both sides: eln y = ex
2 +C
y = ex
2 +C
Most commonly, this type of solution is rewritten

MTHE 235 - Test 2 - November 14, 2012
Last name:(blockletters)
First/Given Name:
Student Number:
This test consists of 7 pages, with questions to be answered in the space provided.
Show all work and give explanations when needed.
1
Last name:(blockletters

Last name:(blockletters)
Student Number:
First/Given Name:
MATH 235 - TEST 1 (Based on Assignments 1, 2, 3 and 4)
Version 1B Fall 2012
This test consists of 7 pages, with questions to be answered in the space provided.
Show all work and give explanations

Week 8- Graphical Interpretation; Non-Homogeneous Systems
Section 5.6
1.
x
y
=
1 2
2 1
x
3
+
y
2
x
a
=
for the particular solution.
y
b
There is no overlap between this solution and the eigenvalue/eigenvector solutions to the
homogeneous form, so we can u

Last name:(blockletters)
Student Number:
First/Given Name:
MATH 235 - TEST 1 (Based on Assignments 1, 2, 3 and 4)
Version 1A Fall 2010
This test consists of 4 questions to be answered in the space provided.
Show all work and give explanations when needed.

For each of the following piece-wise functions,
write them in on line using step functions,
nd F (s) = L(f (t).
1. f (t) =
0
0<t<2
2 4) 2 < t
(t
In step-function form,
f (t) = u2 (t2 4)
L(u2 (t2 4) = e2s L(t + 2)2 4)
= e2s L(t2 + 4t + 4 4)
= e2s L(t2 +

Week 2 - First-order Linear DEs
Section 1.5
1.
y + 1 y=2
p(t)
Find integrating factor: w(t) = e
1 dt
= et
Multiplying DE by factor: et y + et y = 2et
d
et y = 2et
dt
Integrate both sides: et y =
2et dt = 2et + C
y =2+
C
= 2 + Cet
et
This is the general so

Solutions #2
Page 1 of 6
Solutions for Homework 2
1. Solve the initial value problem y = (1 y) cos() with y() = 2.
solution: This is a separable equation:
1
dy =
1y
cos()d ln |1 y| = sin + C1 (1 y)1 = Cesin
1
1
y =1
sin
Ce
Cesin
1y =
Applying the initi