Solutions to Homework #1:
Section 1.2, #4: The order is three.
9: a) With y = C exp(t2 ), we nd y = 2tC exp(t2 ), hence y 2ty = 0.
b) We have y (1) = Ce. To satisfy y (1) = 2, we need C = 2/e.
15: With y = tr , we nd y = rtr1 , y = r(r 1)tr2 , and
t2 y 2t
Solutions to Test 7% 1
MATH 2214H, Fall 2013
Show all your work. No credit for unsupported answers. No calculators allowed.
1. (3 pts.) For the differential equation
yIZylt'l'l): tZO:
plot the isocline corresponding to slope 2, together with a few line
MATH 2214 ASSIGNMENT SHEET
TEXTBOOK: Elementary Differential Equations, by W. Kohler and L. Johnson
It is recommended that graphical and numerical problems are solved using "Matlab for 2214,"
available from the course web site. Note that all problem numbe
Homework 5
Find the general solution to these differential equations using the method of undetermined
coefficients.
1. y00 + y = et sin t
2. y00 2y0 + y = 12et
Find the particular solution to this initial value problem using the method of undetermined
coe
Practice Test 3
1. This question appeared on the Fall 2002 final exam.
Let A be a 2 2 matrix
with real entries. Suppose that 1 + i is an eigenvalue of A with
!
1
an eigenvector
. Find the solution of
i
dx
= Ax !
dt
1
x(0) = 2
!
(a)
et sin t + et cos t
CEE 2814
Homework 4
Fall 2016
1. A map is drawn to a scale of 1 inch = 60 feet and with a contour interval of 2 feet. If adjacent
contour lines are 0.5 inch apart in a particular location what is the ground slope in percent?
2. Sketch, at a scale of 1 inc
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Quadratic (air)resistance:
The dierential equation to be solved is:
dv
= mg + kv 2 ,
v (0) = 0.
(1)
dt
The terminal velocity is obtained by settin the right-hand side of (1) equal
to 0 and solving for v :
mg
vterm =
.
k
We choose the minus sign because t
Solutions to homework #2
Section 2.2
26. The general solution of the dierential equation is y = Cet and the
initial condition leads to y = y0 et . To pass through the two given points,
we must satisfy
4 = y0 e , 1 = y0 e3 .
By dividing the two equations,
Solutions to homework #5
Section 3.1
3. The singular points are at 1, 0 and 1. The solution is guaranteed to
exist on (, 1). Note that 1 is the singularity nearest to 2, where the
initial condition is given.
11. a) For t = 0, y = 1 and y = 1. Only Graph B
Additional problem to homework #3
MATH 2214H, Fall 2013
1. A pursuit problem:
At t = 0, a pig, initially at the origin, runs along the x-axis with constant
speed v. At t = 0, a farmer, initially 20 yards north of the origin, also
runs with constant speed
Additional problem to homework #4
MATH 2214H, Fall 2013
1. In class we solved the dierential equation
2
y + y = 4t
t
for various initial conditions (see your notes).
Use Eulers method on this equation but with a negative step size of say
h = 0.1. This all
Practice Problems for Test #1
MATH 2214H, Fall 2013
1. Find the general solution to
t y + y = tet ,
t > 0.
2. Find the general solution to
t y + 2y =
1
.
1 + 4t2
3. Solve
ey
,
t > 0,
ty
with initial condition y (1) = 0. Leave the solution in implicit form
Solutions to Practice Problems for Test #1
MATH 2214H, Fall 2013
1. Find the general solution to
t y + y = tet ,
t > 0.
Solution: y + y = et = (t) = eln t = t = (ty ) = tet = ty =
t
tet et + C =
y (t) = et t1 et + Ct1 .
To integrate tet use integration by