Lecture 13: 3.6: Nonhomogeneous equation, variation of parameters.
We will now give a general method for nding particular solutions for second order
linear dierential equations that in principle works for any nonhomogeneous term.
Let us rst illustrate the

Lecture 7: 2.5 Autonomous equations and population dynamics. A differential equation is called autonomous if it has the form
(2.5.1)
dy
= f (y),
dt
y(0) = y0
The simplest model of variation of the population of a spices is exponential
growth that the rate

Lecture 5: 2.3 More models.
Model III: Mixing of chemicals. Suppose a time t = 0 a tank contains Q0 lb
of salt dissolved in 100 gal of water. Assume that water containing 1 lb of salt/gal
4
is entering the tank at a rate of r gal/min, and that the well-st

Math 302 Differential Equations
Many phenomena in physics and engineering are described by dierential equations.
Lecture 1: 1.1-2. Two Models, Direction elds, Solution curves.
Model I: A Falling object is subject to Newtons second law:
F = ma
where
a=
dv

Lecture 4: 2.2 Separable equations. We can not solve a general rst order
equation:
dy
(2.2.1)
= f (x, y)
dx
but there is another special case that we can deal with called separable equations.
First we note that there are many ways to write (2.2.1) in the

Lecture 3: 2.1 First order linear equations: Integrating Factor. In this
lecture we will learn to solve a general rst order linear dierential equation
dy
+ p(t)y = g(t).
dt
(2.1.1)
We want to nd a way to write this equation in the form
d
G(t, y(t) = f (t)

Lecture 2: Section 1.2 Analytical solution of the simple models. We can
actually solve the mice-owl model (1.1.3) analytically:
(1.2.1)
dp
= 0.5p 450
dt
The general idea is that we will try to rewrite it as a an equation of the form
(1.2.2)
d
G(t, p(t) =

Math 302 Differential Equations
Second Midterm
April 13, 2012
N AME :
1. Do not open this exam until you are told to begin.
2. This exam has 7 pages including this cover. There are 6 questions.
3. The exam is closed cover. No calculators are allowed.
4. S

Math 302 Dierential Equations Midterm 2 04/10/2009
Your PRINTED name is: Solution
Grading
Please circle your section:
(1)
T 1:30
Krieger 308
Hussey, Caleb
(2)
T 3:00
Bloomberg 274
Ariturk, Sinan
(3)
Th 3:00
Bloomberg 274
Jiang, Jin-Cheng
(4)
Th 4:30
Bloom

Solutions to Midterm 2
Problem 1.[3 15 = 45 points] Solve the following dierential equations.
(1.1) y (4) 2y + 2y y = 0.
Solution. The Character equations is
T 4 2T 3 + 2T 1 = (T 1)3 (T + 1) = 0.
It has roots 1 (repeated 3 times) and -1. So the solutions

Lecture 6: 2.4 Dierence between linear and nonlinear dierential equations. For linear equations we have the following existence theorem:
Th 1 Suppose that p and g are continuous functions on an open interval I : <
t < containing t0 . Then there is a uniqu

Lecture 8: 3.1: Second order linear dierential equations. We are now
going to study the initial value problem for second order linear dierential equations:
(3.1.1)
y + p (t) y + q(t) y = g(t),
y(t0 ) = y0 ,
y (t0 ) = y0
Such equations are likely to show u

Lecture 32: 9.6 Energy and Liapunov functions.
Energy conservation for the Pendulum. Consider the pendulum again:
+ + 2 sin = 0,
2 = g/L
With x = and y = we get the system
x = y,
y = y 2 sin x,
2 = g/L
The critical points are y = 0 and sin x = 0, i.e.

Lecture 31: 9.2-3 Trajectories. One can sometimes turn an autonomous system
(9.5)
dx/dt = F (x, y),
dy/dt = G(x, y)
into a rst order equation
dy
dy/dt
G(x, y)
=
=
dx
dx/dt
F (x, y)
(9.6)
This rst order equation can sometimes be solved at least implicitly

Lecture 30: 9.2-3. We consider a 2 2 nonlinear system
(9.1)
dx
= f (x),
dt
[
]
x1
x=
,
x2
[
]
f1 (x1 , x2 )
f (x) =
.
f2 (x1 , x2 )
We study the stability of the nonlinear system around critical points f (x0 ) = 0 by
approximating with a linear system as

Lecture 14: 3.7 Free vibrations.
Consider a mass m hanging in a spring. The mass causes an elongation L of the
spring in the downward (positive) direction. The gravitational force mg acts downwards and there is a balancing upward force Fs , due to the spr

Lecture 28: 9 Nonlinear Equations and stability. In this chapter we will
consider autonomous 2 2 nonlinear systems
(9.1)
x = f (x),
[
]
x1
where x =
,
x2
[
]
f1 (x1 , x2 )
f (x) =
.
f2 (x1 , x2 )
Autonomous mean that f does not explicitly depend on t (but

Lecture 11: 3.4: Repeated roots. In this chapter we want to solve the equation
L[y] ay + by + cy = 0
(3.4.1)
where a, b, c are real constants. We saw that y = ert is a solution to the equation
if r is a root of the characteristic equation:
ar2 + br + c =

Lecture 12: 3.5: Nonhomogeneous equation. We return to the nonhomogeneous case
(3.5.1)
L[y] y + p(t)y + q(t)y = g(t)
The complementary homogeneous equation is when g(t) 0:
(3.5.2)
L[y] y + p(t)y + q(t)y = 0
Th If Y1 and Y2 are solutions to (3.5.1) then Y1

Lecture 9: 3.2 Fundamental Solutions of linear homogeneous equations. Most
of what we will do in this chapter concerns linear second order dierential equations with constant coecients. However, the results in this section also holds for
variable coecients

Lecture 10: 3.3 Complex roots. In this chapter we want to solve the equation
ay + by + cy = 0
(3.3.1)
where a, b, c are real constants. We saw that y = ert is a solution to the equation
if r is a root of the characteristic equation:
ar2 + br + c = 0
(3.3.

Lecture 25: 7.8 Repeated eigenvalues. Recall rst that if A is a 2 2 matrix
and the characteristic polynomial have two distinct roots r1 = r2 then the corresponding eigenvectors x(1) and x(2) are linearly independent for if c1 x(1) +c2 x(2) = 0
it follows

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Lecture 17: 6.3-4 Step functions and discontinuous forcing functions. The
step function is dened to be
cfw_
0,
t < c,
1,
uc (t) =
t c.
Ex 1 The function h(t) = u3 (t) u5 (t) is given by
0,
h(t) =
1,
0,
t < 3,
3 t < 5,
t 5.
0, t < 2,
1, 2 t < 3,
f (t) =

Lecture 16: 6.1-6.2 More Laplace transforms. Recall the Laplace transform:
Lcfw_f (t) = F (s) =
(6.2.1)
est f (t) dt
0
This is dened for s > a if f is piecewise continuous and |f (t)| Keat , for t M .
We need to show that the limit as T of
T
e
st
M
f (t)

Lecture 15: 6.1-6.2 The Laplace transform.
Def The Laplace transform of the function f (t) is the function F (s) given by
Lcfw_f (t) = F (s) =
(6.1.1)
e
st
f (t) dt = lim
T
0
T
est f (t) dt,
0
if f is (piece wise) continuous and the limit exist. That f i

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