ENGINEERING MATHEMATICS II

MA-2001

Dr. Umber Sheikh

Course Outlines
Week Activity
1 Basic Concepts and Ideas, Differential Equations and their
Classifications
2 Formation of Differential Equations, Initial and Boundary
Conditions, Geometrical Meaning of y’=f(x,y), Separable
Differential Equations
3 Homogeneous Equations, Differential Equations reducible
to Homogeneous Form, Exact Equations, Integrating
Factors
4 Linear Equations, Bernoulli Equations, Orthogonal
Trajectories of Curves, Homogeneous Linear Equations of
Second Order, Differential Operators
5 Euler-Cauchy Equation, Nonhomogeneous Linear
Equations, Reduction of Order, Existance and Uniqueness
Theory, Wronskian
6 Variation of Parameters
7
Higher Order Differential Equations, Higher Order
Nonhomogeneous Equations, Introduction to Vectors,
Matrices, Eigenvalues
8 Homogeneous Systems with Constant Coefficients, Phase
Plane, Critical Points, Criteria for Critical Points
9 Stability, Nonhomogeneous Linear Systems, Power Series
Method, Theory of Power Series Method, Legendre
Equations
10 Legendre Polynomials, Frobenius Method,
11 Laplace Transforms, Properties of Laplace Transforms,
Table of Some Laplace Transforms, Inverse Transform,
Linearity, Shifting
12 Transforms of Derivative and Integral, Differential
Equations Solution of Initial Value Problems
Basic Concepts
 
Equation: Equations describe the relations between the
dependent and independent variables. An equal sign "=" is
required in every equation.

Differential Equations: Equations that involve dependent

variables and their derivatives with respect to the independent
variables are called differential equations.

Example: The following are differential equations involving the

unknown function y.
Some Other Examples:
Newton’s Second Law of Motion
Maxwell Equations
Kirchhoff’s Laws
Heat Equation
Wave Equation

Ordinary Differential Equation: Differential equations that

involve only one independent variable are called ordinary
differential equations.

Partial Differential Equation: Differential equations that

involve two or more independent variables are called partial
differential equations.

Examples: Equations 1 and 2 are examples of ordinary

differential equations, since the unknown function y depends
solely on the variable
x. Equation 3 is a partial differential equation, since y depends
on both the independent variables t and x.
Order: The order of a differential equation is the highest
derivative that appears in the differential equation.

Degree: The degree of a differential equation is the power of

the highest derivative term.

Linear Equations: A differential equation is called linear if

there are no multiplications among dependent variables and
their derivatives. In other words, all coefficients are functions
of independent variables.

Non-linear: Differential equations that do not satisfy the

definition of linear are non-linear.
Equation Indepen-dent Depen-dent Order Degree
Variable Variable
1 2

12 1
2

4 1
2
1 11

2 1
4 1

1 1

2 1
 
Linear Homogeneous: A differential equation is homogeneous if
every single term contains the dependent variables or their
derivatives.

Linear Non-homogeneous: Differential equations which do not

satisfy the definition of homogeneous are considered to be non-
homogeneous.
Solutions
General Solution: Solutions obtained from integrating the
differential equations are called general solutions. The general
solution of a order ordinary differential equation contains
arbitrary constants resulting from integrating times.

Particular Solution: Particular solutions are the solutions

obtained by assigning specific values to the arbitrary constants
in the general solutions.

Singular Solutions: Solutions that can not be expressed by the

general solutions are called singular solutions.
 Explicit Solution: Any solution that is given in the form y = y
(independent variable).

Examples: etc.

Implicit Solution: Any solution that isn’t in explicit form.

Examples: etc.
Conditions
Initial Condition: Constrains that are specified at the initial
point, generally time point, are called initial conditions.
Problems with specified initial conditions are called initial value
problems.

Examples: y’+y=0 y(0)=1,

z’=1, z(1)=-1.

Boundary Condition: Constrains that are specified at the

boundary points, generally space points, are called boundary
conditions. Problems with specified boundary conditions are
called boundary value problems.

Examples: z”+z’+z=1, z(0)=2, z(2)=-1,

f”+f=0, f(0)=0, f(/2)=1.
 Worksheet 1
Equation Independent Dependent Order Degree Homo-genity
Variable Variable
 Worksheet 2
Check whether the given expression a solution of the corresponding equation?

Equation Solution to Check

First Order Differential Equations
Standard Form:
Standard form for a first-order differential equation in the
unknown function y(x) is
y’=f(x,y).

Interpretation:
The form shows the slope of the tangent at each point of the
xy-plane. Thus the standard form of the first order differential
equation give the slope field. Solving the equation gives the
original curve family having the slope field expressed by the
differential equation.
Separable Equations
 Form of Equation:

Method of Solution: Integrate both sides will give the solution

of such type of equations.
Examples
Example 1: Solve Example 2: Solve

Solution: Solution:

 Exercise: Solve
Homogeneous Equations
(of degree zero)
 Form of Equation:
is a function homogeneous of degree zero, i.e.,

A function homogeneous of degree n can be defined as

Method of Solution: Substituting reduces the equation in form

separable in variables v and x. Integrate both sides and then
substituting the value of v will give the solution.
Examples
Example 1: Solve Example 2: Solve

Solution: Solution:

Put
Put

Integrating we get

or

Exercise: Solve
First Order Linear Equations

 Form of Equation:

Method of Solution: The integrating factor is I.F.= Multiplying

the equation with this factor gives

Solving this gives the solution y.

Examples
Example 1: Solve Example 2: Solve

Solution: Here Solution:

Substituting values in

Substituting values in

 Exercise: Solve
Bernoulli Equations
 Form of Equation:

Method of Solution: The substitution transform the Bernoulli’s

equation into a linear equation in z. This linear equation, after
solving and back substituting the value of z gives the solution
of the Bernoulli’s equation.
Examples
Example 1: Solve Exercise: Solve

Solution: Put Solution: Put

Substituting values in Substituting values in

+c
Exact Equations
 A differential equation

is exact if there exists a function such that

Test for exactness: If and are continuous functions and have continuous
first partial derivatives on some rectangle of the -plane, then

is exact if and only if

How to obtain an exact equation from a function?

 
Method of Solution
To solve an exact equation, first solve the equations

for . The solution to the exact equation is then given implicitly by

where represents an arbitrary constant.

Examples

 Example 1: Make an exact ODE using

Sol: If , then
 Example 2:
Solve

Sol: Here
Since therefore equation is exact.

+c
+c
Integrating Factors
  an equation is inexact (not exact), it is possible to transform
If
Such equation into an exact differential equation by a judicious
multiplication.

A function is an integrating factor for an inexact equation if the equation

is an exact equation.

The famous formulas to obtain an integrating factor are as follows:

If then
If then
If then
Examples and Exercises
 Make an exact differential equation from the functions

Check for the exactness and solve

Orthogonal Trajectories
 
Given a one-parameter family of curves .
A curve that intersects each member of the family at right angles
(orthogonally) is called an orthogonal trajectory of the family.

The one-parameter families and are orthogonal trajectories if each member

of one family is an orthogonal trajectory of the other family.

A procedure for finding a family of orthogonal trajectories for a given

family of curves is as follows:
Step 1. Determine the differential equation for the given family .
Step 2. Replace in that equation by ; the resulting equation is the
differential equation for the family of orthogonal trajectories.
Step 3. Find the general solution of the new differential equation. This is the
family of orthogonal trajectories.
Example
 
Find family of curves orthogonal to one parameter family of
quadratic parabolas .

Solution:

Now the slope of the family of new curves is

is required family of curves.

Exercises
 1. Find the family of orthogonal trajectories of:

Answer:

2. Find the orthogonal trajectories of the family of parabolas with

vertical axis and vertex at the point .

Answer:
Linear Second Order DEs
 The most general linear second order differential equation
is in the form.

 The constant coefficient linear second order differential

equation is

where a, b, c are all constants.

 
Initially we will make our life easier by looking at differential
equations with   When  we call the differential
equation homogeneous and when  we call the differential
equation nonhomogeneous. 
 Solving Second Order, Linear, Homogeneous
ODE with Constant Coefficients
 Consider the solution of the type Substituting in
We have

As
This equation is typically called the characteristic equation 
 
This will be a quadratic equation and so we should expect two
roots, r1 and r2.  Once we have these two roots we have two
solutions to the differential equation.
and
 The roots will have three possible forms.  These are
 
Real, distinct roots,  .  
Complex root,  .
Double roots, 
 
Real Roots
Example: Find two solutions to
 
Solution: The characteristic equation is
 
The two roots are 3 and -3. Therefore, two solutions are and
The general solution is  
 Example: Solve the following IVP
 
Solution: The characteristic equation is
The general solution and its derivative is

Putting the initial conditions, we have the following system

of equations

Solving we get and Thus the solution is

 Complex Roots
Example: Solve the following IVP
 
Solution: The characteristic equation is
The general solution and its derivative is

 
Putting the initial conditions, we have the following system of
equations

so

 Thus the solution is

 Repeated Roots
Example: Solve the following IVP
 
Solution: The characteristic equation is
The general solution and its derivative is

Putting the initial conditions, we have the following system of

equations

Solving we get and Thus the solution is

 Exercises: Solve the following IVPs

                                               
 Reduction of Order
 Let be the known solution of second order DE

Assume is the other solution. Then

Since
therefore

or
Examples and Exercises

 Solve
 Solutions of Linear Homogeneous
Equations; the Wronskian
 Theorem . (Existence and Uniqueness Theorem)
Consider the initial value problem

If , and are continuous and bounded on an open interval

containing , then there exists exactly one solution of this
equation, valid on .
 Wronskian: Given two functions , , the Wronskian is defined as

Remark: One way to remember this definition could be using the

determinant,

Main property of the Wronskian:

• If , then anf are linearly dependent.
• Otherwise, they are linearly independent.
 Example:Check if the given pair of functions are linearly
dependent or not.
(a)
(b)
(c)
(d)

Suppose𝑦_1 (t),𝑦_2 (t) are two solutions of

𝑦′′ + 𝑝(𝑡)𝑦′ + 𝑞(𝑡)𝑦 = 0
Then
(I) We have either W(𝑦_1,𝑦_2)≡0 or W(𝑦_1,𝑦_2) never zero;
(II) If W(𝑦_1,𝑦_2)≠0, then y =𝑐_1 𝑦_1+𝑐_2 𝑦_2is the general
solution. They are also called to form a fundamental set of
solutions. As a consequence, for any Ics 𝑦(𝑡_0) = 𝑦_0, 𝑦′
(𝑡_0)= 〖𝑦′〗 _0, there is a unique set of (𝑐_1,𝑐_2) that give a
unique solution.
 
Theorem (Abel’s Theorem):Let be two (linearly independent) solutions
to on an open interval I. Then, the Wronskian on I is given by

for some constant depending on , but independent on in .

 Example: Given

Find without solving the equation.

Answer. We first find the
p(t) = −(t + 2)t
which is
which is valid for . By Abel’s Theorem, we have
=

Example: If y1, y2 are two solutions of

ty′′ + 2y′ + tety = 0,
and W(y1, y2)(1) = 2, find W(y1, y2)(5).

Example 5. If W(f, g) = 3e4t, and f = e2t, find g.

 Example: For the equation

given one solution find a second linearly independent solution.

Solution: Use Abel’s Theorem and Wronskian. By Abel’s
Theorem, and choose C = 1, we have
=.
By definition of the Wronskian,

Solve this for (taking c=0):

Example: Consider the equation

Given find the general solution.

Example: Given the equation

and , find .
Linear Differential Equations
An nth-order linear differential equation has the form
(4.1)
where g(x) and the coefficients bj(x) ( j = 0,1,2,..., n) depend
solely on the variable x. In other words, they do
not depend on y or any derivative of y.
If g(x) = 0, then above Equation is homogeneous; if not, 4.1 is
nonhomogeneous.
A linear differential equation has constant coefficients if all
the coefficients bj(x) in above equation are constants; if one or more
of these coefficients
is not constant, the above equation has variable coefficients.
Theorem 4.1. Consider the initial-value problem given by the
linear differential
equation 4.1 and the n initial conditions
y(x0)=c0, y’(x0)=c1, y’’(x0)=c2,…, y(n-1)(x0)=cn-1
Second Order Non-homgeneous Linear
Differential Equations

 
The general solution to the linear differential equation where
denotes one solution to the differential equation and is the
general solution to the associated homogeneous equation, .
Methods for obtaining when the differential equation has
constant coefficients are given in previous lectures. In this lecture,
we give methods for obtaining a particular solution once is
known.
Method of Undetermined Coefficients

 
This method is applicable only if and all of its derivatives can be written
in terms of the same finite set of linearly independent functions, which
we denote by . The method is initiated by assuming a particular solution
of the form

where denote arbitrary multiplicative constants. These arbitrary

constants are then evaluated by substituting the proposed solution into
the given differential equation and equating the coefficients of like
terms.
 an nth-degree polynomial in .

Assume a solution of the form

where is a constant to be determined.

where and are known constants.

Assume a solution of the form

where is a constant to be determined.

where and are known constants.

Assume a solution of the form

,
where and are constants to be determined.
 Generalizations

If is the product of terms considered in all the cases given above, take
to be the product of the corresponding assumed solutions and
algebraically
combine arbitrary constants where possible. In particular, if is the
product of a polynomial with an exponential, assume

where Aj is as in 1st case. If, instead, is the product of a polynomial,

exponential, and sine term, or if is the product of a polynomial,
exponential, and cosine term, then assume

where and are constants which still must be determined.

If is the sum (or difference) of terms already considered, then we take
to be the sum (or difference) of the corresponding assumed solutions
and algebraically combine arbitrary constants where possible.
 Modifications

If any term of the assumed solution, disregarding multiplicative

constants, is also a term of
(the homogeneous solution), then the assumed solution must be
modified by multiplying it by , where is the smallest positive integer
such that the product of with the assumed solution has no terms in
common with .

Limitations of the Method

In general, if is not one of the types of functions considered above, or

if the differential equation does not have constant coefficients, then
the method of variations of parameters is preferred.
Variation of Parameters
 
Variation of parameters is another method for finding a particular solution of
the nth-order linear differential equation , once the solution of the associated
homogeneous equation is known. If are linearly independent solutions of then
the general solution of is

Methodology:
A particular solution of has the form

where is given in above Equation and is an unknown function of which still

must be determined.
 To find , first solve the following linear equations simultaneously for :

Then integrate each to obtain , disregarding all constants of integration.

This is permissible because we are seeking only one particular solution.
Example 6.1: For the special case , Equations reduce to

For the case , Equations become

and for the case , we obtain the single equation

 
Since are linearly independent solutions of the same equation L(y)=0, their
Wronskian is not zero. This means that the system 6.9 has a nonzero
determinant and can be solved uniquely for .

Scope of the Method

The method of variation of parameters can be applied to all linear differential
equations. It is therefore more powerful than the method of undetermined
coefficients, which is restricted to linear differential equations with constant
coefficients and particular forms of
Nonetheless, in those cases where both methods are applicable, the method
of undetermined coefficients is usually the more efficient and, hence,
preferable.
As a practical matter, the integration of may be impossible to perform. In
such an event other methods (in particular, numerical techniques) must be
employed.
 
Initial-Value Problems

Initial-value problems are solved by applying the initial conditions

to the general solution of the differential equation. It must be
emphasized that the initial conditions are applied only to the
general solution and not to the homogeneous solution that
possesses all the arbitrary constants that must be evaluated. The
one exception is when the general solution is the homogeneous
solution; that is, when the differential equation under consideration
is itself homogeneous.
 
Example: Solve
Solution: Here a second degree polynomial. We assume that

Thus and Substituting these results into the differential equation, we have

Equating the coefficients of like powers of , we obtain

Solving this system, we find that

Hence and the general solution is
 
Example: Solve
Solution: Again
Assume that
Thus, and Substituting these results into the differential equation,
we have
Equating coefficients of like terms, we obtain

Solving this system, we find that and . Then

and the general solution is

 Example: Solve
This is a third-order equation with
It follows that

So the set of Equations becomes

Solving this set of equations simultaneously, we obtain

Thus
Gives
The general solution is therefore
Laplace Transform
 Let be defined for and let denote an arbitrary real variable. The Laplace transform of ,
designated by either or ,is

for all values of for which the improper integral converges. Convergence occurs when
the limit

exists. If this limit does not exist, the improper integral diverges and has no Laplace
transform. When evaluating the integral, the variable is treated as a constant because
the integration is with respect to .
 
The Laplace transforms for a number of elementary functions can be found in Annexure
of the Book.
 Properties of Laplace Transforms
 ☺ (Linearity). If and then
for any two constants and

☺ If , then for any constant

☺ If , then for any positive integer
☺ If and if exists, then
☺ If , then
☺ If is periodic with period , that is, , then
 Inverse Laplace Transform
 An inverse Laplace transform of designated by , is another function having
the property that .

Methodology

The simplest technique for identifying inverse Laplace transforms is to

recognize them, either from memory or from a table.
If is not in a recognizable form, then occasionally it can be transformed into
such a form by algebraic manipulation.
 Manipulating Denominators
 
-The method of completing the square deals with quadratic polynomials
-The method of partial fractions

Manipulating Numerators
 
A factor in the numerator may be written in terms of the factor , where both
and are constants, through the identity

♥ (Linearity). If the inverse Laplace transforms of two functions and exist,

then for any constants and,
 Convolution
 The convolution of two functions and is

 Theorem
Theorem. (Convolution Theorem). If and , then

The inverse Laplace transform of a product is computed using a

convolution.

 
If one of the two convolutions in above Equation is simpler to calculate,
then that convolution is chosen when determining the inverse Laplace
transform of a product.