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
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 =
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 orde
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 point
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 an
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 )
Auto
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
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
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 re
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
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 )
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 i
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 pop
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 enterin
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 objec
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
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
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
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
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)
T
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 h
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 linea
You have .
No calculators, books or notes allowed.
Academic Honesty Certicate. I agree to complete this exam without unauthorized
assistance from any person, materials or device.
Signature:
Dat
Lecture 26: 7.7 The exponential matrix.
We will now nd a nice way to the express the solution to the system
x = Ax,
(7.7.1)
where A is a 22 matrix, analogous to the formula for the solution of one equ
Lecture 27: 7.9 Nonhomogeneous equations. There are several methods in
the book but we will only go over using diagonalization and the exponential matrix.
Diagonalization Suppose that A is an nn matri
Lecture 24: 7.6 Complex eigenvalues.
Ex Find the solution to the system
]
[ ]
[
[ ]
x1
1 2
a
x = Ax,
where x =
, A=
,
x(0) =
,
x2
2 1
b
Sol 1 First we want to nd the eigenvalues r and eigenvectors x =
Lecture 23: 7.5 Linear systems of dierential equations.
Ex 2 Find the solution to the system
[ ]
[
]
[ ]
x1
6 2
a
x = Ax,
where x =
, A=
,
x(0) =
,
x2
2 9
b
Sol First we want to nd the eigenvalues r a
Lecture 22 7.3 Eigenvectors. The equation
(7.3.1)
Ax = y
or
a11 x1 + a12 x2 = y1
a21 x1 + a22 x2 = y2
can be viewed as a map transforming the vector x R2 into the vector y R2 .
Examples of such transf
Lecture 20: 7.1 Systems of rst order dierential equations. A second
order equation can always be written as a rst order system:
Ex The equation of a spring y + ky = 0 can be written with x1 = y and x2
Lecture 21 7.2-3 Algebraic Systems.
Ex. 1 We want to solve the 3 3 system
x1 2x2 + x3 = 0
2x2 8x3 = 8
4x1 + 5x2 + 9x3 = 9
Geometrically this represents the intersection of 3 planes. To minimize the wr