engineering mathematics differential equations

8/12/2019 Engineering Mathematics Differential Equations

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Dr. Ir. Harinaldi, M.EngMechanical Engineering Department

Faculty of Engineering University of Indonesia

DIFFERENTIALEQUATIONS


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Differential Equations

An equation which involves unknown function and itsderivatives

ordinary differential equation (ode) : not involve partialderivativespartial differential equation (pde) : involves partial

derivativesorder of the differential equation is the order of thehighest derivatives

Examples:

second order ordinarydifferential equation

first order partial differentialequation

2

23 sin

d y dy x y

dx dx

y y x t x

t x x t


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Modeling via Differential Equations

Note that the set of equations is called aModel for the system.How do we build a Model?The basic steps in building a model are:Step 1: Clearly state the assumptions on which the model will be

based. These assumptions should describe therelationships among the quantities to be studied.

Step 2: Completely describe the parameters and variables to be

used in the model. Step 3: Use the assumptions (from Step 1) to derivemathematical equations relating the parameters andvariables (from Step 2).


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Laju alir : e liter/min

Tercampur sempurna

laju alir : f liter/min

Awal:Vol. air : Vo literMassa garam :

a gram

Konsentrasi garam : b gram/liter

Massa garam setiap saat (menit) ?

?

Larutan

Air-Garam


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A linear differential equation of order n is a differential equatiowritten in the following form:

1

1

1 01 ( )n n

n nn n

d y d y dy a x a x a x a x y f x

dx dx dx

where an(x) is not the zero functionGeneral solution : looking for the unknown function of adifferential equation

Particular solution (Initial Value Problem) : looking forthe unknown function of a differential equation where thevalues of the unknown function and its derivatives at somepoint are known

Issues in finding solution :existence and uniqueness


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1 st Order DE Separable Equations

The differential equationM(x,y)dx + N(x,y)dy= 0 is separable ifthe equation can be written in the form:

02211 dy y g x f dx y g x f

Solution :

1. Multiply the equation by integrating factor: y g x f 121

2. The variable are separated :

01

2

2

1 dy y g y g

dx x f x f

3. Integrating to find the solution:

C dy y g y g

dx x f x f

1

2

2

1


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Examples:

1. Solve : dy x dx y dy x 2

4 Answer:


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Examples:

1. Solve : y x

dxdy 22

Answer:


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Examples:

2. Find the particular solution of : 21;

12 y

x

y

dx

dy

Answer:


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1 st Order DE Hom ogeneous Equations

Homogeneous Function

f (x,y) is called homogenous of degreen if : y , x f y , x f n

Examples:

y x x y , x f 34 homogeneous of degree 4

y x f y x x

y x x y x f ,

,4344

34

y x x y x f cossin, 2 non-homogeneous

y x f

y x x

y x x y x f

n ,

cossin

cossin,22

2


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Examples:

1. Solve : 03 233 dy xydx y x

Answer:


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Examples:

2. Solve : 2

52

y x

y x

dx

dy

Answer:


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1 st Order DE Exact Equation

The differential equationM(x,y)dx + N(x,y)dy= 0 is an exactequation if :

Solution :

The solutions are given by the implicit equation x N

y M

C y x F ,

1. Integrate eitherM(x,y) with respect to x or N(x,y)to y . Assume integratingM(x,y), then :

where : F/ x = M(x,y)and F/ y = N(x,y)

y dx y x M y x F ,,

2. Now : y x N y dx y x M y y

F ,',

or : dx y x M y

y x N y ,,'


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1 st Order DE Exact Equation

3. Integrate (y)to get (y) and write down the resultF(x,y) = C

Examples:1. Solve : 01332 3 dy y x dx y x

Answer:


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1 st Order DE Non Exact Equation

The differential equationM(x,y)dx + N(x,y)dy= 0 is a non exactequation if :

Solution :

The solutions are given by using integrating factor to change theequation into exact equation

x N

y M

1. Check if : only x of function x f N x N

y M

then integrating factor is dx x f

e

only yof function y g M

y M

x N

or if :

then integrating factor is dy y g

e


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1 st Order DE Non Exact Equation

Examples:

1. Solve : 022

dy xy dx x y x Answer:


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1 st Order DE Linear Equation

Afirst order linear differential equation has the followinggeneral form:

x qy x pdx dy Solution :

1. Find the integrating factor: dx x pe x u2. Evaluate :

3. Find the solution:

x u

C dx x q x uy

dx x q x u


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Examples:

1. Solve : x xy

dx

dy 42

Answer:


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2tan cos , 0 2y x y x y

Examples:

2. Find the particular solution of :

Answer:

d


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2 nd Order DE Linear Equation

Asecond order differential equation is an equation involving the

unknown functiony , its derivativesy ' and y '', and the variable x .We will consider explicit differential equations of the form :

x y y f dx

y d ,',2

2

Alinear second order differential equations is written as:

x d y x c y x by x a

Whend ( x ) = 0, the equation is calledhomogeneous , otherwise itis callednonhomogeneous

d


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To a nonhomogeneous equation

x d y x c y x by x a we associate the so called associated homogeneous equation

0 y x c y x by x a

(NH)

(H)

Main result: The general solution to the equation (NH) is given by:

where:(i) y h is the general solution to the associated homogeneous

equation (H );

(ii) y p is a particular solution to the equation (NH ).

ph y y y

d


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Basic property Consider the homogeneous second order linear equation

0 y x c y x by x aor the explicit one

0 y x qy x py

Property:Ify 1 and y 2 are two solutions, then:

x y c x y c x y 2211

is also a solution for any arbitrary constantsc 1 ,c 2

nd d d f d


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2 nd Order DE Reduction of Order

This technique is very important since it helps one to finda secondsolution independent froma known one .

Reduction of Order Technique

Lety 1 be a non-zero solution of: 0 y x qy x py Then, a second solutiony 2 independent ofy 1 can be found as:

x v x y x y 12 Where:

dx e x y

x v dx x p

21

1

The general solution is then given by

x y c x y c x y 2211

2 nd O d DE R d i f O d


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2 nd Order DE Reduction of Order

Examples:1. Find the general solution to the Legendre equation

0221 2 y y x y x

Using the fact that :y 1 = x is a solution.

2 nd O d DE H LE i h


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2 nd Order DE Hom ogeneous LE withConstant Coefficients

Homogeneous Linear Equations with Constant Coefficients

A second order homogeneous equation with constant coefficientsis written as: 0 0 acy y by awhere a, b and c are constant

The steps to follow in order to find the general solution is as follow(1) Write down thecharacteristic equation

0 02 ac ba

This is a quadratic. Let 1 and 2 be its roots we have

aac bb

242

2,1



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(2) If 1 and 2 are distinct real numbers (ifb2 - 4ac > 0 ), then thegeneral solution is: x x ec ec y 21 21

(3) If 1 = 2 (ifb2 - 4ac = 0 ), then the general solution is: x x

xec ec y 11 21

(4) If 1 and 2 are complex numbers (ifb2 - 4ac < 0 ), then thegeneral solution is:

x ec x ec y x x

sincos 21 Where:

abac

ab

24

and 2

2



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1. Find the solution to the Initial Value Problem 24 ; 24 ; 022 y y y y y


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2 nd O d DE N H LE


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2 nd Order DE Non Homogeneous LEMethod Undeterm ined CoefficientsWe will guess the form ofy p and then plug it in the equation to find itHowever, it works only under the following two conditions:

the associated homogeneous equations has constant coefficientsthe nonhomogeneous termd ( x ) is a special form

x e x L x d x e x P x d x

n

x

n sin or cos

Note: we may assume thatd ( x ) is a sum of such functionswhere P n( x ) and Ln( x ) are polynomial functions of degreen

Then a particular solutionyp is given by:

x e x R x e x T x x y x n x ns p sincos

nnn x A x A x A A x T ...2210

n

nn x B x B x BB x R ...2

210

Where:

2 nd O d DE N H g LE


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2 nd Order DE Non Homogeneous LEMethod Undeterm ined CoefficientsThe steps to follow in applying this method:

1. Check that the two conditions are satisfied2. Write down the characteristic equation and find its root

3. Write down the number4. Compare this number to the roots of the characteristic equation

5. Write down the form of particular solution

6. Find constant A and B by pluggingy p solution to original equation

02 c ba i

If : 0 rootstheof onenotis si 1 rootsdistinctheof oneis si 2 rootsbothtoequalis si

x e x R x e x T x x y x n x ns p sincos nnn x A x A x A A x T ...2210

nnn x B x B x BB x R ...

2210Where:

2 nd O d DE N H g LE


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2 nd Order DE Non Homogeneous LEMethod Undeterm ined Coefficients

1. Find a particular solution to the equation x y y y sin243

2 nd Order DE Non Homogeneous LE


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If the nonhomogeneous termd ( x ) consist of several terms:

N i

i i N x d x d x d x d x d

121 ...

We split the original equation intoN equations


x d y x c y x by x a x d y x c y x by x a

N

2

1

Then find a particular solutiony pi A particular solution to the original equation is given by:

N i

i pi pN p p p x y x y x y x y x y

121 ...



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If the nonhomogeneous termd ( x ) consist of several terms:

N i

i i N x d x d x d x d x d

121 ...

We split the original equation intoN equations


x d y x c y x by x a x d y x c y x by x a

N

2

1

Then find a particular solutiony pi A particular solution to the original equation is given by:

N i

i pi pN p p p x y x y x y x y x y

121 ...



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1. Find a particular solution to the equation x x eey y y 8343 2

engineering mathematics differential equations

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