For student 4 (differential equations - reduced version)



Linearity



First order equation and its application



Second order equation

Introduction

Definition: An equation involving derivatives of one or more dependent variables with respect to one or more independent variables is called a differential equation.

For examples of differential equations we list the following:

Classification by type: A differential equation involving ordinary derivatives of one or more dependent variables with respect to a single independent variable is called an ordinary differential equation (ODE).

Classification by Order: The order of the highest ordered derivative involved in a differential equation is called the order of the differential equation.

In symbols we can express an n th order ordinary differential equation in one dependent variable by the general form

F

(

x

,

y

,

y

',...

y

)

=

0

Classification by Linearity: An n-th order ordinary differential equation is linear if it can be written in the form

an(x) y(n) + an-1(x) y(n-1)+--- --- ---+a1(x) y′+a0(x) y = g(x) ---(5)

The differential equation is said to be homogeneous if g (x) = 0 and non-homogeneous

if g(x) is not identically zero.

Two important special cases of (5) are linear first-order (n=1) and linear second- order(n=2)

Two properties of linearity:

→ The dependent variable y and all its derivatives y′, y′′,…,y(n) are of the first degree. → The coefficients a0, a1,…, an of y, y′, y′′,…,y(n) respectively depend at most on the

independent variable x.

Examples

(1-y) y′ + 2y = ex y′′ + y2 _{= 0}

↑ ↑

Non-linear term non-linear term coefficient depends on y power not 1

Origin of Differential Equations:

Having classified differential equations in various ways, let us now consider where and how such equations actually originate.

Differential equations occur in connection with numerous problems that are encountered in various branches of science and engineering. We indicate a few such problems in the following list:

→ The problem of determining the motion of a projectile, rocket, satellite or planet → The problem of determining the charge or current in an electric circuit

→ The problem of the conduction of heat in a rod → The study of the rate of growth of a population → The study of the reactions of chemicals

Solutions: A solution of an n-th order differential equation is an n times differentiable function y = f(x).

Example: Verify that the function y (x) = 3 e2x is a solution of the differential equation

dy/dx – 2y = 0 for all x. Solution:

Tasks.Verify that the functions

(i) y(x) =sinx–cosx+1 is a solution of the equation y” + y=1

(ii) y1(x)=e5x and y2(x) = e-3x are solutions of y′′- 2 y′-15y

= 0

General solution: The general solution of an n-th order differential equation is a family of solutions consisting of n essential parameters.

Existence and Uniqueness

Theorem 1.1 Existence and Uniqueness for a Linear Differential Equation

If the functions a0(x), a1(x),…, an(x) and g(x) are continuous on the interval I and

an(x) ≠ 0 ∀x ∈ I , then the initial value problem

an(x) y(n) + an-1(x) y(n-1)+--- --- ---+a1(x) y′+a0(x) y = g(x)

y(x₀ ) = y_0, y′(x₀ ) = y_1, ….., y (n-1)_(x

0 ) = yn −1, x0 ∈ I has a unique solution on the interval I.

First order equation and its application

First Order Equations:

The first order differential equation may be expressed in either the derivative form

dy/dx = f(x,y)---(i) or the differential form

M(x,y)dx + N(x,y)dy = 0---(ii)

For example, the equation dy/dx = (x2 _{+ y}2_{)/ (x-y) is of the form (i). It may be written}

 Separable Equation:

A first order differential equation of the form

dy/dx = g(x) h(y) is said to be separable or to have separable variables.

Examples

Verify if the following equations are separable and non separable:

1- dy/dx = y2 _{x e}3x+4y

2- dy/dx = y + sinx

Solving separable equations:

Step 1: Rewrite the separable equation dy/dx = g(x) .h(y) in separated (or differential)

form g x dx

y h

dy

) ( )

( 

Step 2: Integrate each side of this equation with respect to its respective variable.

Solve: dy/dx = 1+y2

Solve the initial value problem: dy/dx = - x/y, y (0) = 1

Solution

Solve :

2 ₁

2

dy x

dx y

 



Solution

Tasks. Solve: (i) dy/dx = xy, y(0) = 1 (ii) (1 + x ) dy – y dx = 0

(iii) dy/dx = - x/y, y(4) = -3.

Homogeneous Equation:

Definition

If a function f possesses the property f( x, y)kf x y( , ) for some real k, then f is said to be a homogeneous function of degree k.

Example

2 2

( , )

f x y x y is a homogeneous function of degree 2 since

Example

f (x,y) = x2 _{+ y}2_{+ 1 is not a homogeneous.}

Theorem

Consider a first order DE in the differential form M(x,y)dx + N(x,y)dy = 0.

If both M(x,y) and N(x,y) are both homogeneous of the same degree, then, the function ( , ) / ( , )

M x y N x y is homogeneous of degree zero.

Remark:

If M(x,y)dx + N(x,y)dy = 0 is a homogeneous equation, then the change of variables y = vx

Solve the equation:

4 4 '

3

2

y

x

y

xy





Solution

Tasks. Solve (i)

(

y



x



y

)

dx



xdy



0, (1)

y



0

Exact Equation:

A differential expression M(x,y)dx + N(x,y)dy is an exact differential in a region R of the xy-plane if it corresponds to the differential of some function f (x,y) defined in R. A first order differential equation of the form M(x,y)dx + N(x,y)dy = 0 is said to be exact equation if the expression on the left hand side is an exact differential.

For example, x2_y3 _{dx + x}3_y2 _{dy = 0 is an exact equation, because its left hand side is an}

exact differential:

1

3

d



_

x y

 

_

x y dx



x y dy





Notice that if we make the identities M(x,y) = x2_y3_{and N (x,y) = x}3_y2_{, then}

2 2

3

M

N

x y

y

x



_

_







Solution of exact equation:

There exists a function f(x,y) such that

 

f

y

x M x

 _

 and

 

f

y

y N x

  

Integrate the first equation with respect to x then differentiate with respect to y to find the value h(y) (the integration constant c).

Example: Solve:

'

4

0

Example

2 2 2

(y x sinxy)dy xysinxy cosxy e x

dx

   

An alternative method of Solution: First integrate the terms in M dx as if y were constant, then integrate the terms in N dy considering x as constant and rejecting the terms already obtained, equate the sum of these integrals to a constant. This will be the solution of the required equation

Example: The given equation can be written in the form as (y + 4)dx + x dy = 0 Here M (x,y) = y + 4, N(x,y) = x

1

M

N

y

x







 





hence it is an exact equation.

(

4)

4

( )

( )

Mdx

y

dx

xy

x

i

Ndy

xdy

xy

ii









   





         









Rejecting the term xy in (ii) which already occurs in (i) and then adding (i) and (ii) and equating the sum to a constant, we get the general solution to be

xy + 4x = c.

Tasks. Solve : (i) (3x2 + 4xy)dx + ( 2x2 + 2y) dy = 0

A System of Linear Algebraic Equations

Homogeneous Linear System with Constant Coefficients (we use three equations for simplicity):

3 13 2 12 1 11

1 a x a x a x

x  

, (S1a)

3 33 2 32 1 31

3 a x a x a x

x   

is a homogeneous system of equations with constant coefficients

Solution Procedure

This system of equations can be written in matrix form as

(S1b)

In short this system may be written as (S1c)

where                3 2 1 x x x x            33 32 31 23 22 21 13 12 11 a a a a a a a a a A

Solution method: We seek the nontrivial solution in the following form:

(S2a)

Writing this in vector form, the trial solution is of the form

x  et (S2b)

Important Comments: Thus we see that, if the vector differential equation

x Ax (S1c) has a solution of the form

(S2b)

then the number  must be a characteristic value_i of the coefficient matrix A and the vector  must be characteristic vector (i) corresponding to the characteristic value_i.

3 23 2 22 1 21

2 a x a x a x

x   

                                  3 2 1 33 32 31 23 22 21 13 12 11 3 2 1 x x x a a a a a a a a a x x x x A x 

Case 1. Distinct characteristic values:

Suppose that are distinct eigenvalues of the coefficient matrix A and the

corresponding characteristic vectors are linearly independent. Then the distinct

vectors defined by

 t t e t

x x x e x x x e x x x 3 2 1 33 23 13 ) 3 ( 3 ) 3 ( 2 ) 3 ( 1 32 22 12 ) 2 ( 3 ) 2 ( 2 ) 2 ( 1 31 21 11 ) 1 ( 3 ) 1 ( 2 ) 1 ( 1 , ,                                                                                 

are the solutions of the differential equations (S1).

and the general solution is

x c₁x(1)  c₂x(2)  c₃x(3)

where are arbitrary constants.

Example: Solve

3 2 1, ,

 ) 3 ( ) 2 ( ) 1 ( _{, ,}  t t

t _{x t e x t e}

e t

x(1)₍₎(1) 1 _, (2)₍₎(2) 2 _, (3)₍₎(3) 3

Case 2. Repeated characteristic values:

Two equal eigenvalues: Let, out of the characteristic values ₁,₂, ₃ of the coefficient matrix

A, is different. There may be two cases now: for , we may have (a)

two linearly independent eigenvectors or (b) only one eigenvector .

Case (a): In this case corresponding to , the two solutions

x(1) (1)et and are linearly independent.

Case (b): Let we have only one characteristic vector, say , and hence we have

only one solution given by

where  is determined from

or

So we must look of another linearly independent solution corresponding to . The second linearly independent solution is given by

x(2) (t  )et

where  is determined from

Example 5. Solve x _x

        2 1 4 6 Ans: 3 2

1  and 

  ₁ ₂  

) 2 ( ) 1 ( and     ₁ ₂ 

t

e x(2) (2) 

 

₁ ₂  

t

e x(1)  

   

A (A I)  0

  ₁ ₂ 

 

 

 )

Case (c): In this case we have only one characteristic vector, say , and hence we have only one solution given by

where  is determined from

or

So we must look of two another linearly independent solution corresponding to 

 

₁ ₂  ₃  . The second linearly independent solution is given by

where  is determined from

The third linearly independent solution is given by

where  is determined from

(A I)  

t

e x(1)  

   

A (A I)  0

t

e t

x(2)  (  )   

 

 )

(A I

t e t t

x   ) 

Introduction to Second Order Linear Equations Second order Linear Differential Equation:

2

0

( )

( )

( )

( )

d y

dy

a x

a x

a x y

g x

dx

dx







where a0(x)0, a1(x), a2(x) and g(x) are continuous functions on some interval of interest

I. When the functions a0(x), a1(x), a2(x) and g(x) are constants, the differential equation is

said to have constant coefficients; otherwise it is said to have variable coefficients. Standard form:

2

2

( )

( )

( )

d y

dy

p x

q x y

f x

dx

dx







or in the compact form

'' '

( )

( )

( )

y



p x y



q x y



f x

Homogeneous form:

2 2

'' '

( )

( )

0

( )

( )

0

d y

dy

p x

q x y

dx

dx

y

p x y

q x y













Theorem: Principle of Superposition for Homogeneous Equation If y1(x) and y2(x) are any two solutions of the linear homogeneous equation

'' '

( )

( )

0

y



p x y



q x y



y(x) = c1 y1(x) + c2 y2(x) where c1 and c2 are arbitrary constants, is also a solution on the

interval I.

Definition: Linear Independence and Dependence

Two functions f and g are said to be linearly dependent on some interval I if there exist two constants c1 and c2 , not both zero, that satisfy

c1f + c2g =0--- (i), for all x in the interval I. If the set of functions is not linearly

dependent on some interval, it is said to be linearly independent. That is, eq.(i) holds for all x in I only for c1 =c2 =0.

Note: c1f + c2g =0 f = -(c2/c1)g, that is if two functions are linearly dependent, then one

function is simply a constant multiple of the other.

Useful tool for determining linear independence or dependence Definition: The Wronskian

Given two functions y1and y2, the function

1 2 _{' '}

1 2 _{' '} 1 2 1 2

1 2

( ,

)

y

y

W y y

y y

y y

y

y







is called the Wronskian of y1 and y2.

Example: Show that the two functions y1(x) = sinx and y2(x) = cosx are linear

independent solutions of y + y = 0 on (-,).

Soln: By direct verification, y1 and y2both satisfy given equation.

Now

(sin ,cos )

sin

cos

sin

cos

1 0

cos

sin

x

x

W

x

x

x

x

x

x



 



  



Since the Wronskian is never zero, therefore y1 and y2 are linearly independent.

General solution of Homogeneous Equations

Let y1 and y2are linearly independent solutions of

y



p x y

( )



q x y

( )



0

interval I. Then the general solution of the equation on the interval is

y = c1y1 (x) + c2y2(x), where c1 and c2 are arbitrary constants

Homogeneous equations with constant coefficients

ay + by + cy = 0

Nature of the unknown solution y : emx

Auxiliary Equation: am2 + bm + c = 0

Case (i) Real and unequal roots:

1 2

1

,

m x m x

y



e

y



e

are linearly independent solutions of (i), and the general solution is

1 2

1 2

( )

Example: Find the general solution of y + 5y -6y=0 Soln:

Solve: 3y- y - 2y = 0 Solution:

Initial value problem: Solve y -y =0, y(0)=1, y (0)=0

Case (ii): Repeated roots: when m1=m2

General solution 1 1

1 2

( )

y x



c e



c xe

Example: Find the general solution of y + 4y +4y=0 Soln:

Solve the initial value problem:4y -20y +25y, y (0) = 2 and y (0) = 5 Solution:

Case (iii) Conjugate complex roots: If m1= +i and m2= -i, then the general

solution is y = c1e(+i) x + c2 e(-i) x = ex[c1eix+c2e-ix]

Nonhomogeneous Equations

The general solution of the second order linear nonhomogeneous equation

'' '

( )

( )

( )

y



p x y



q x y



f x

has the form

y(x) = c

y

(x) + c

y

(x) + y

(x),

Where:

- y1(x) and y2(x) are linear independent solutions of the corresponding

homogeneous equation

y



p x y

( )



q x y

( )



0

- yp (x) is any single solution of the nonhomogeneous equation.

Example: From inspection we see that yp (x) =1 is a particular solution of y + y = 1

Find the general solution. Solution:

Method of Undetermined Coefficients

Simple procedure for finding a particular solution yp (x) of the linear nonhomogeneous

equation

2

2

( )

( )

( )

d y

dy

p x

q x y

f x

dx

dx







either a, xn, eax, sinx, cosx, or products of these functions. Example: Find a particular solution of y + 3y = 2

Example: An exponential nonhomogeneous term: Find a particular solution of y + 3y + 2y = 2e3x

Soln:

General solution using undetermined coefficients: Solve:y - y - 2y = e3x

Example: A polynomial nonhomogeneous term: Find a particular solution of y - 4y + 2y = 2x2

Soln:

Example: Sine or Cosine nonhomogeneous term: Find a particular solution of y - 3y + 2y = 10sin2x

Soln:

Table:

UC function UC set

xn eax

sin(ax+b) or cos(ax+b) xn _eax

{xn,xn-1,….x,1} {eax}