formulation of first order linear and nonlinear 2nd order differential equation

APPLIED MATHEMATICS IN CHEMICAL ENGG
SEMINAR ON :
SUBMITTED BY
MAHASWARI JOGIA
11/9/2017DSCE,CHEM ENGG DEPT
1

11/9/2017 DSCE,CHEMICAL DEPT 2
CONTENTS:
•INTRODUCTION
•FIRST ORDER DIFFERENTIALS
•TYPES OF 1ST ORDER DIFFERENTIALS
•SECOND ORDER DIFFERENTIALS
•TYPES OF 2ND ORDER DIFFERENTIALS
•REFERENCES

INTRODUCTION:
Equations which are composed of an unknown function and its derivatives are called
differential equations.
Differential equations play a fundamental role in engineering because many physical
phenomena are best formulated mathematically in terms of their rate of change.
When a function involves one dependent variable, the equation is called an ordinary
differential equation (ODE).
A partial differential equation (PDE) involves two or more independent variables.
v
m
c
g
dt
dv

v - dependent variable
t - independent variable
11/9/2017 3DSCE,CHEMICAL DEPT

( , )y f x y 

FIRST ORDER DIFFERENTIAL EQUATION:
FIRST ORDER LINEAR AND NON LINEAR EQUATION:
•A first order equation includes a first derivative as its highest derivative.
- Linear 1st order ODE:
Where P and Q are functions of x.

TYPES OF LINEAR DIFFERENTIAL
EQUATION:
1. Separable Variable
2. Homogeneous Equation
3. Exact Equation
4. Linear Equation

•SEPARABLE VARIABLE:
The first-order differential equation:
Is called separable provided that f(x,y) can be written as the product of a function of x and a
function of y.
Suppose we can write this equation as:
 ,
dy
f x y
dx


)()( yhxg
dx
dy

We then say we have “separated” the variables. By taking h(y) to the LHS, the equation becomes:
Integrating, we get the solution as:
1
( )
( )
dy g x dx
h y

1
( )
( )
dy g x dx c
h y
  

OR
EXAMPLE 1.
Consider the DE :
Separating the variables, we get
Integrating we get the solution as:
Where C is an arbitrary constant.
y
dx
dy

dxdy
y

1
kxy ||ln
x
cey or

HOMOGENEOUS EQUATIONS:
Definition: A function f(x, y) is said to be homogeneous of degree n in x, y if
for all t, x, y
Examples
Is a homogenous degree of 2
A first order Differential equation is called homogeneous if are homogeneous functions
of x and y of the same degree.
),(),( yxfttytxf n

22
2),( yxyxyxf 
0),(),(  dyyxNdxyxM

WORKING RULE TO SOLVE A HDE:
•Put the given equation in the form
•Check M and N are Homogeneous function of the same degree.
•Let y=zx
•Differentiate y = z x to get
•Put this value of dy/dx into (1) and solve the equation for z by separating the variables.
•Replace z by y/x and simplify.
)1(0),(),(  dyyxNdxyxM
dx
dz
xz
dx
dy


EXACT DIFFERENTIAL EQUATIONS:
A first order differential equation
is called an exact DE if
The solution is given by:
EXAMPLE:
The DE is exact as it is d (x2+ y2) = 0
Hence the solution is: x2+ y2 = c
0),(),(  dyyxNdxyxM
M N
y x
 

 
   cNdyMdx
0xdx ydy 

•LINEAR EQUATIONS:
A linear first order equation is an equation that can be expressed in the form:
where a1(x), a0(x), and b(x) depend only on the independent variable x, not on y.
We assume that the function a1(x), a0(x), and b(x) are continuous on an interval and that a1(x)  0
on that interval. Then, on dividing by a1(x), we can rewrite equation (1) in the standard form
where P(x), Q(x) are continuous functions on the interval.
1 0( ) ( ) ( ), (1)
dy
a x a x y b x
dx
 
( ) ( ) (2)
dy
P x y Q x
dx
 

RULES TO SOLVE A LINEAR DE:
1. Write the equation in the standard form
2. Calculate the IF (x) by the formula
3. Multiply the equation by (x).
4. Integrate the last equation.
( ) ( )
dy
P x y Q x
dx
 
 ( ) exp ( )x P x dx  

SECOND ORDER LINEAR AND NON LINEAR
DIFFERENTIAL EQUATIONS
A second order differential equation is an equation involving the unknown function y, its
derivatives y' and y” and the variable x:
We will only consider explicit differential equations of the form,
Homogeneous Equations: If g(t) = 0, then the equation above becomes
y″ + p(t) y′ + q(t) y = 0.
It is called a homogeneous equation. Otherwise, the equation is nonhomogeneous (or
inhomogeneous).
0),,,(  yyyxF
),,( yyxfy 

SECOND ORDER LINEAR HOMOGENEOUS
DIFFERENTIAL EQUATIONS WITH
CONSTANT COEFFICIENTS
For the most part, we will only learn how to solve second order linear equation with constant
coefficients (that is, when p(t) and q(t) are constants). Since a homogeneous equation is easier to
solve compares to its non homogeneous counterpart, we start with second order linear
homogeneous equations that contain constant coefficients only:
a y″ + b y′ + c y = 0.
Where a, b, and c are constants, a ≠ 0.

THE CHARACTERISTIC POLYNOMIAL
a y″ + b y′ + c y = 0, a ≠ 0. (1)
This polynomial, a r2 + b r + c, is called the characteristic polynomial of the differential equation
(1). The equation a r2 + b r + c = 0 is called the characteristic equation of (1). Each and every
root, sometimes called a characteristic root, r, of the characteristic polynomial gives rise to a
solution y = e rt of (1).
We will take a more detailed look of the 3 possible cases of the solutions thusly found:
1. (When b2 − 4 ac > 0) There are two distinct real roots r1, r2.
2. (When b2 − 4 ac < 0) There are two complex conjugate roots r = λ ± μi.
3. (When b2 − 4 ac = 0) There is one repeated real root r.

CASE 1: TWO DISTINCT REAL ROOTS
When b2 − 4 ac > 0, the characteristic polynomial have two distinct real roots r1, r2. They give
two distinct solutions and OR
Therefore, a general solution of (*) is

Example: y″ + 5 y′ + 4 y = 0
The characteristic equation is r2 + 5 r + 4 = (r + 1)(r + 4) = 0, the roots of the polynomial are r =
−1 and −4. The general solution is then
y = C1 e−t + C2 e−4t.
Suppose there are initial conditions y(0) = 1, y′(0) = −7. A unique particular solution can be found
by solving for C1 and C2 using the initial conditions.
First we need to calculate y′ = −C1 e −t − 4C2 e −4t, then apply the initial values:
1 = y(0) = C1 e 0 + C2 e 0 = C1 + C2
−7 = y′(0) = −C1 e 0 − 4C2 e 0 = −C1 − 4C2
The solution is C1 = −1, and C2 = 2 → y = −e −t + 2 e −4t.

CASE 2 :TWO COMPLEX CONJUGATE ROOTS
When b2 − 4 ac < 0, the characteristic polynomial has two complex roots, which are conjugates,
r1 = λ + μi and r2 = λ − μi (λ, μ are real numbers,μ > 0). As before they give two linearly
independent solutions
Consequently the linear combination will be a general solution. At this juncture
you might have this question: “but aren’t r1 and r2 complex numbers; what would become of the
exponential function with a complex number exponent?” The answer to that question is given by
the Euler’s formula.

Hence, when r is a complex number λ + μi, the exponential function e rt becomes
CASE 3 ONE REPEATED REAL ROOT
When b2 − 4 ac = 0, the characteristic polynomial has a single repeated real
root, This causes a problem, because unlike the previous two cases the roots of characteristic
polynomial presently only give us one distinct solution y1 = e rt. It is not enough to give us a
general solution. We would need to come up with a second solution, linearly independent with y1,
on our own. How do we find a second solution?

By the Abel’s Theorem, the fact C ≠ 0 guarantees that y1 and y2 are going to be linearly
independent. Now, we have two expressions for the Wronskian of the same pair of solutions. The
two expressions must be equal:

This is a first order linear differential equation with y2 as the unknown. Put it into its standard
form and solve by the integrating factor method.

Any such a function would be a second, linearly independent solution of the differential equation.
We just need one instance of such a function. The only condition for the coefficients in the above
expression is C ≠ 0. Pick, say, C = 1, and C1 = 0 would work nicely.
Therefore, the general solution in the case of a repeated real root r is

REFERENCES
•advanced engineering mathematics by erwin kreyszig
•engineering mathematics iv by dr. k.s.c
•lecture notes on mathematical methods bymihir sen and joseph m. powers,department of
aerospace and mechanical engineering, university of notre dame
notre dame, indiana 46556-5637,usa
•first and second order linear differential equations,dr. radhakant padhi, asst.
professor,dept. of aerospace engineering,indian institute of science – bangalore
•linear differential equation by nofal umair

formulation of first order linear and nonlinear 2nd order differential equation

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formulation of first order linear and nonlinear 2nd order differential equation