THE PETROLEUM INSTITUTE
MATH 261, Differential Equations, Spring 2015
Course Syllabus
Prerequisites: Students should have exited the MATH 212 course with a grade D or higher
Instructor: Dr. Alip Mohammed
Office: 8-292
E-mail: [email protected]
Phone: ext.75

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Differential Equations Worksheet 18 (Section 4.
Variable-coefficient equations . ‘
In this section, we consider the non—homogeneous linear second-order equation of the form
(I) (12(t)y"+a](r)y'+an(t)y = g(r)
The following result

Solving lnltial Value Problems
We now use Laplace and Inverse Laplace transforms to solve initial value proble
D.E. This can be done following two steps:
1) Take Laplace transform of both sides, to obtain J {y}
2) Take the inverse Laplace transform to obt

Differentlal Equatlons Worksheet 22 (Sectlon
Inverse Laplace Transform
In Section 7.2, we turned a function f (r) into F (s). In this section, we want to do the other
way round, that is, turn a function F (s) into f (I) using the inverse mapping or invers

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orentlal Equations Worksheet 26 (Sectlon 7.9)
Solvlng Ilnear systems with Laplace transform
We can use Laplace transform to reduce certain systems of linear differential equations with
initial conditions to a system of li

Differential Equations Worksheet 27 (Section 8.1)
Introduction: Taylor polynomlal approximatlon
One of the best tools to approximate a ﬁmction f (x) near a particular point x0 is the Taylor
polynomial. The formula for the Taylor polynomial of degree n cen

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thhed 25 (section 7.7)
Convolutlon
Let’s consider the initial value problem
y"+y = gm, y(0) = y'(0) = 0-
Take Laplace transform by sides, we have
s2Y(s) + Y (s) = 6(3) , where Y(s) and 0(3) are Laplace transform of y(t) and g(t).
And he

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leferentlal Equatlons Worksheet 24 (Sectlon 7.6)
Transforms of Dlseontlnuous functlons
One of the advantages of Laplace transform is that it can handle ﬁmctions that have jump
discontinuity such as step functio

leferentlal Equations Worksheet 28 (Section 8.2)
Power serles and Analytic functlons
Power Series:
A power series about the point x0 is an expression of the form
(1) Zane—x0)" =a0 +a](x—xo)+a2(x—x0)2 + .
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where x is a variable and the an ’s are const

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_ . . .7 7 _ r. a . . Differential Equations Worksheet 29 (Section 8.3)
Power series solution to linear differential equation -
In this section, we study a method for obtaining a power series solution to a linear diff

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Power series expansions about x0 : 0 are easier to manipulate than expansions about nonzero
points. As the next example shows, a simple shift in variable enables us always to expand
about the origin.
leferentlal Equatlons Worksheet 30 (Soctlon 8.4
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Differential Equations Workshee 31 (Section 8.6)
Method of Frobenlus
Consider the equation of the form
(1) y"(I)+-P(leTI)t-q(I)y(X)=0-
Let’s assume xp(.r) and x2q(.r) are analytic functions. That is in some open interval about
x

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leferentlal Equatlons Worksheet 19 (Sectlon 4. '
A closer look at free mechanical vlhratlons
CASE A No friction:I no damping, free motion — no external force
rash-MAX
Recall the mass-spring system is governed by the eq

Deflnltlon of Laplace Transform
Laplace Transform
Let f (t) be a function on [0,00). The Laplace transform of f is the function F deﬁned by
the integral
(1) F(s) = j e*5‘f(t)d;.
0
The domain of F is all values of s for which the integral in (1) exists. Th

Differentlal Equations Worksheet 21 (Sectlon 7.3)
Properties of the Laplace Transform
In the previous section, we deﬁned the Laplace transform of a function f (t) as
:0
. 't
F(s)=£e"‘f(t)dt. ‘\ [and‘h‘m h S.
In this section, we discuss some properties of

Section 11 Background 5
Although the majority of equations one is likeiy to encounter in practice. fail into the nonlin-
ear category. knowing how to deal with the simpler linear equations is an important first step (just
as tangent lines heip our underst

1.4
The Approximation
Method of Euler
p.24
Eulers method (or the tangentline method) is a procedure for
constructing
approximate
solutions to an initial value
problem for a first order
differential equation.
y f ( x, y ), y ( x0 ) y0
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Le

Differential Equatlons Worksheet 4 (Section 2.2)
Separahie Equatlons
A simple class of ﬁrst order differentiai equation that can be solved using integration is the
class of separable equations. These are equations % m f (x, y) , that can be rewritten

Differential Equations Worksheet 2 (Section 1.2)
Solutions and Initial Value Problems
Deﬁnition 1: Explicit Solution
A function f(x) is called an explicit solution of a DB. if it satisﬁes the DE. (Replacing
f (x) for y makes the equation true for al

Differential Equations Worksheet 1 (Section 1.1)
Introduction: Background
When we develop models to try to understand physical phenomena, often our models yield
equations that contain some derivatives. These equations are called Differential Equations. Fo

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leferentlal Equations Worksheet 5 (Sectlon 2.3)
Linear Equations
A type of ﬁrst order D.E that occurs frequently in applications is the linear DB of the form:
al (ﬁg + ao(x) y = b(x) , where no (x), a] (x) and b(x) only depend on

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" " " " leferentlal Equations Worksheets (Sectlon 2.4)
Exact Equatlons
In Calculus III, to apply the Fundamental Theorem for Line Integral, we used the test: if
6%! = Z—N, then there exists a function F (x, y), such that 6F (x, y) =

Dlﬁerentlal Equatlons Worksheet 7 (Sectlon 2.5)
Speclal Integratlng Factor
In the previous lesson, if the equation M (x, y)dx+ N (x, y)dy = 0 satisﬁes the compatibility
. . 6M 6N . . .
condition ——- (x, y) = — (x, y), then the equatlon IS exact and we kno

Differentlal Equatlons Worksheet 3 (Sectlon 1.4)
The Approxlmatlon Method of Euler (Euler’s Method or The Tangent Llne Method)
The Euler’s method is a procedure for constructing approximate solutions to an initial value
problem for a ﬁrst order D.E.
Quest

Dlﬁerentlal Equatlons Worksheet 8 (Sectlon 2.6)
Solvlng by Substltutlons
Substitutions can be used to transform a DB into the forms we have learnt. This section
introduces three types of substitutions.
Type 1: Homogeneous Equations (substitute v =
If th

Chapter 1
Introduction
1.1
Background
p.1
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const ' 0
x ' nx
n
n-1
sin x ' cos x
sin ax ' a cos ax
sin x/b ' cos bx
b
cos x ' sin x
cos ax ' a sin ax
cos bx ' sin bx
b
e ' e
x
x
e ' a e
ax
ax
1
e ' b e
x/b
u(x) v(x)' u' (x) v' (x)
cu(x)' cu