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 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 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 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 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 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
You have .
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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 equation.
If A has two nonparallel eigenvectors x(1) and x
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 matrix with n linearly independent
eigenvectors A (k) = k (k
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 = 0:
(7.6.1)
Ax = rx
(A rI)x = 0.
The eigenvalues are so
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 and eigenvectors x = 0:
(7.5.5)
Ax = rx
(A rI)x = 0.
The
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 transformations are scalar multiplication or rotations of a v
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 = y :
x = x2
1
x = kx1
2
First order systems also show
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 writing
it is convenient to only write out the coecient m