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© 2011
Zachary S Tseng
1
MATH 251
Work sheet / Things to know
Chapter 3
1. Second order linear differential equation
Standard form:
What makes it homogeneous?
We will, for the most part, work with equations with constant coefficients only.
What is the general form of a second order linear equation with constant coefficients?
Ex. 3.1.1
Can you think of any function(s) that satisfy each equation (w/ constant
coefficients) below?
(a)
y
″  25
y
= 0
(b)
y
″ + 25
y
= 0
(c)
y
″  25
y
′ = 0
The example (c) above is an instance of a second order linear equation with the
y
±term
missing.
It is essentially a first order linear equation in disguise.
All equations of this
type can be solved by changing it into a first order equation with the substitutions
u
=
y
′
and
u
′ =
y
″, then use the integrating factor method to solve for
u
, and integrate the result
to find
y
.
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2.
The characteristic equation
Given the equation
ay
″ +
by
′ +
cy
= 0,
a
≠ 0, what is its
characteristic equation
?
Any root,
r
, of the characteristic equation has the property that
y
=
e
rt
always satisfies the
equation above.
Therefore,
y
=
e
rt
will be a particular solution for each root
r
.
Consequently, an important formula to remember for this class is (surprisingly) the
quadratic formula:
a
ac
b
b
r
2
4
2

±

=
Note that the characteristic equation method does
not
require the given differential
equation to be put into its standard form first – the quadratic formula simply doesn’t care
whether or not the leading coefficient is 1.
Suppose
r
1
and
r
2
are two distinct real roots of the characteristic equation, what is the
general solution of the differential equation?
y
=
Ex. 3.2.1
y
″ +
y
′  12
y
= 0
What is its characteristic equation?
What are the roots of the characteristic equation?
Based on the roots, what are 2 particular solutions of the equation?
The general solution is
y
=
Ex. 3.2.2
2
y
″ + 3
y
′  2
y
= 0
y
=
© 2011
Zachary S Tseng
3
3.
Initial Value Problems
What do the initial conditions of a second order differential equation look like?
A second order equation’s general solution will have 2 arbitrary constants / coefficients.
Therefore, an IVP will have 2 initial conditions in order to give 2 (algebraic) equations
needed to solve for them.
What must
the conditions look like?
Ex. 3.3.1
Take the previous example
y
″ +
y
′  12
y
= 0.
Find its solution satisfying
(a)
y
(0) = 0,
y
′(0) = 2
(b)
y
(0) = 2,
y
′(0) = 6
How to (easily) work with initial conditions where
t
0
≠ 0?
Ex. 3.3.2
y
″ +
y
′  12
y
= 0,
y
(45000) = 0,
y
′(45000) = 2
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The
Existence and Uniqueness Theorem
(for second order linear equations)
It is really the same theorem as the one we saw earlier, except this one is in the context of
second order linear equation.
What does it say?
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This note was uploaded on 07/16/2011 for the course MATH 251 taught by Professor Chezhongyuan during the Spring '08 term at Pennsylvania State University, University Park.
 Spring '08
 CHEZHONGYUAN
 Math, Differential Equations, Equations, Partial Differential Equations

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