- •Lecture 1.
- •§1. Determinants and their properties
- •Is the minor of .
- •§2. Matrices and operations on them. Inverse matrix
- •Definition 4. (Definition of a matrix)
- •§3. Systems of linear equations
- •Vector Addition and Multiplication of a vector by a scalar
- •2 The inner, vector and mixed product of vectors
- •Inner Product and its Properties
- •Lecture 2.
- •Vectors. The elementary operations over vectors. The scalar, vector and mixed product of vectors.
- •§1. The elementary operations over vectors
- •Vector Product and its Properties
- •The vector product in coordinates. Consider vectors
- •Triple Product in Coordinates. Given three vectors , , and , let us express the triple product of these vectors in terms of their coordinates. Consider the triple product
- •The simplest problem of analytic geometry
- •Division of an interval in a given ratio. Suppose given an interval м1м2. Let us find the coordinates a of point м on the interval for which .
- •3. Second-Order Curves in the Plane Ellipse
- •Parabola
- •Lecture 4 Function. The function limit. Fundamental theorems on limits. Infinitely small and infinitely large quantities
- •1. Functions
- •2. The Theory of Limits
- •Infinitesimals and bounded functions.
- •3. Fundamental Theorems on Limits
- •4. Continuity of Functions
- •Lecture 5 The derivative of the function. Geometric and mechanical meaning of derivative. Table of derivatives. The differential of a function
- •1. The derivative of a function
- •Implicit Function Derivative
- •2. Differential of a function
- •3. Higher Derivatives
- •1. Properties of differentiable functions
- •L’Hospital’s Rule for Form
- •L’Hospital’s Rule for Form
- •Taylor’s formula
- •2. Monotonic conditions. Extremum of function
- •3. Convexity and concavity. Point of inflection
- •4. Asymptotes
- •5. General Scheme for the Investigation of the Graph of a Function
- •1. Concept of functions of several variables
- •2. Partial derivatives. Total differential
- •3. Differentiation of composite and implicit functions. Tangent and surface normal
- •Implicit function of two variables
- •4. Partial derivatives and higher order differentials
- •5. Extrema of functions of two variables
- •Lecture 8 Antiderivative. Indefinite integral and its properties. Table of integrals. Main methods of integration
- •1. Antiderivative and indefinite integral
- •2. Main methods of integration
- •Integration by substitution ( or change of variable )
- •Integration by parts
- •3. Integration of fractional rational functions
- •I. Integrating Proper Rational Functions
- •II. Integrating Improper Rational Functions
- •4. Integration of irrational functions
- •5. Integration of trigonometric functions
- •1. Concept of definite integral
- •2. Main properties of definite integrals
- •Integration by substitution
- •Integration by Parts
- •3. Applications of definite integrals
- •The areas of plane figures
- •Lecture 11 Differential equations of the first and second order. Homogeneous and non-homogeneous linear differential equations. Linear differential equations with constant coefficients
- •1. Problems that lead to differential equations
- •2. Equations with separable variables
- •3. Bernoulli’s equation
- •4. The equations that allow lower-order
- •5. Homogeneous equation
- •1. Numerical series
- •6. Types of series:
- •2. The convergence of the sum of the series
- •Integral
- •3. Alternating series
- •4. Power series
- •Main concept. Definition of probability
- •2. Properties of probability
- •Lecture 15. Elements of Mathematical Statistics. Random variables, their types. Distribution laws of random variables
- •1. Random variables, their types
- •2. Distribution laws of random variables
5. Homogeneous equation
Homogeneous equations can be written in the form:
,
and also in the form
М(x, y)dx+N(x, y)dy=0,
where М(x, y) and N(x, y) - homogeneous functions of the same degree.
To solve the homogeneous equation, we must make the substitution y = tx, then we obtain an equation with separable variables. At this change dy=tdx+xdt.
Problem Statement: Given a first order linear differential equation and (possibly) an initial condition of the form
,
.
Find the general solution for the given equation. That is, find the one-parameter family of functions that are solutions of the given equation.
Notice that the terms involving
and
y are on the left-hand side of the equation and the only variable on
the right-hand side is the independent variable t.
Step 0: Put the problem into standard form.
If the problem is presented in any form other than
the one described above, re-write the problem in the form above
first. Then divide the equation through by
making the coefficient of
exactly one. This means that the problem has the (standard) form
,
.
Step 1: Determine
the integrating
factor, i.e.,
.
You will need to evaluate the integral by techniques from the calculus.
Step 2: Multiply the equation through by the integrating factor found in the previous step.
This step is straight forward algebra. Just
remember to multiply both sides of the equation by
.
Step 3: Integrate both sides of the equation.
You should check (by mentally differentiating)
that the left hand side is exactly the derivative of the product
.
If it isn’t, you have made a mistake in Steps 1 – 3. Don’t
forget the constant of integration.
Step 4: Solve (algebraically) the result in Step 3 for y explicitly.
The result is the general solution for the equation.
Step 5: (only if an IVP is given) Apply the initial conditions and solve (algebraically) for the value of the constant of integration.
This is accomplished by substituting
for
t and
for
y. The
constant of integration is the unknown in the resulting equation.
A second order linear differential equation with constant coefficients is a differential equation of the form
,
where
and
are constants.
The characteristic equation associated with the differential equation above is
.
This
quadratic has two (real or imaginary) roots:
and
.
If
,
then the general solution to the second order differential equation
above is of the form
.
If
,
then the general solution is of the form
.
Example 9. Solve the following initial value problem:
The characteristic equation is
,
so
that the characteristic equation has two roots:
and
The general solution to the second order linear differential equation is therefore
In
order to solve the initial value problem, we need to find the
constants
and
:
The solution to the initial value problem is therefore
6. Non-homogeneous second-order equations
Special case of the general linear equation of order n
(1)
where
and
are
given functions of
or are constants.
Consider equation (1). If
,
it is called homogeneous equation
with variable or constant coefficients, depending on whether
are
functions of
or are constants. It is also called the reduced
linear differential equation.
Example 10.
,is
a second order homogeneous equation with variable coefficient.
If
in equation (1), it is called non- homogeneous equation with variable
or constant coefficients, depending on whether
are
functions of
or are constants. Example 11.
is a second order non-homogeneous equation with variable coefficient.
Suppose
and
are
two independent solutions of the reduced equation of (1), namely
(2)
then the linear combination
where
are
arbitrary constants, is also a solution.
Theorem 1. If
are
n linearly
independent functions of
which satisfy a homogeneous equation (3), then the linear combination
(3)
where
are
arbitrary constants, is its solution. Equation (3), which provides
the solution for the homogeneous equation is called the complementary
function.
Theorem 2. The
general solution of a complete non-homogeneous differential equation
is equal to the sum of its complementary
function and any particular
integral. If
is a particular solution of (1), then the general solution is
+
(4)
Hence, for non-homogeneous equations:
General solution = Complementary function + Particular integral
The general linear equation (4) is usually difficult to solve and requires special techniques. However an important and special case occurs when the coefficients are constants, the equation being called a constant coefficient equation. Consider a constant coefficient homogeneous equation
(5)
Denoting,
,
in equation (5) we have
.
(6)
If we make a formal substitution
in (6), we obtain a polynomial in
of degree
given:
,
(7)
and if we equate equation (7) to zero we have an
algebraic equation of degree
which
must have
roots. The equation
is called the auxiliary equation
of the differential equation of the differential equation (7).
Theorem 3. If
is a root of the auxiliary equation
,
then
is a solution of the homogeneous linear differential equation
where
is a constant.
Summary to solving a homogeneous differential equation
The solution of the differential equation then
reduces to solving the algebraic auxiliary equation for its
roots and forming a linearly independent combination
as the general solution, if the roots
were all distinct.
Theorem 4. If the
auxiliary equation of a homogeneous linear differential equation
contains
as
an s-fold root,
then
is a solution of the differential equation.
Example 12.
If the
(twice),
the solution is
Auxiliary equation with complex roots
If the auxiliary equation with real coefficients
contains two complex roots
and
then
is
a solution of the differential equation,
are constants.
Generalization If
the complex roots occur as multiple roots, that is if (
is an s-fold
pair of
complex roots, then the corresponding terms of the complementary
function are
Equations with undetermined coefficients
In the previous section we learnt that solution to
the complete linear differential equation is composed of the sum of
the complementary function and the particular integral. Techniques
for obtaining the complementary function
were
developed in sections 2.1.4-2.1.6 with many working examples. What
remains is only to provide techniques for finding a particular
integral in order to obtain a complete solution. In this section we
shall discuss the technique called the method of undetermined
coefficients.
Although the method of undetermined coefficients
is not applicable in all cases, it may be used if the right- hand
side
,contains
only terms which have a finite number of linearly independent
derivatives such as
,
,
,
or products of these.
Procedure for the techniques of undetermined coefficients
The general procedure in this technique is to
assume the particular integral
of
a form similar to the right member
in equation (1). The necessary derivatives of
are
then obtained and substituted into the given differential equation.
This results in an identity for the independent variable, and
consequently the coefficients of the like terms are equated. The
values of the undetermined coefficients are found from the resulting
system of linear equations. The procedure is best illustrated with an
example
Table 1 below summarizes a general rule for the formulation of the particular integral
-
If is of the form
Choose
to be
Table 1
Example 13: Find the particular integral of
Solution: You
actually worked out the complementary function of this equation to
be
.That
is, the auxiliary equation is
,
giving
or
.Hence
the complementary function is
as
before. Looking at the right hand side of the above differential
equation example 13.3.1 and the general rule in table 1, the
particular integral is
,
,
Substituting in the original equation
Comparing the coefficients on both sides
The general solution is then the complementary function + particular integral, which is
.
LECTURE 12.
Numerical series. The convergence of the sum of the series. Alternating series. Absolute and conditional convergence of the series. Functional series. The area of series convergence. Power series. The radius of convergence
