Differential equation hamzah asyrani sulaiman at .

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description

what is differential equation?

Transcript of Differential equation hamzah asyrani sulaiman at .

Page 1: Differential equation hamzah asyrani sulaiman at  .

differential equation

hamzah asyrani sulaiman at www.asyrani.com

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topic

β€’ basic conceptβ€’ first order differential equationβ€’ second order linear differential equation with

constant coefficientsβ€’ the Laplace transformβ€’ Fourier series

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what is differential equation?

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equation involving an unknown function and its derivatives

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example𝑑𝑦𝑑π‘₯ =3 π‘₯βˆ’1

π‘₯2 𝑑2 𝑦

𝑑 π‘₯2+( 𝑑𝑦𝑑π‘₯ )

2

+2 𝑦= lnπ‘₯

𝑑3 𝑦𝑑 π‘₯3

βˆ’4 π‘₯ 𝑑2 𝑦

𝑑 π‘₯2=cos2 π‘₯

πœ• π‘¦πœ•π‘‘ +

πœ• π‘¦πœ• π‘₯=4 𝑑2+2π‘₯

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there are two types of differential equations (DE)

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ordinary differential equation

β€œif the unknown function depends on only one independent variable”

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partial differential equation

β€œif the unknown function depends on two or more independent variables”

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𝑑𝑦𝑑π‘₯ =3 π‘₯βˆ’1

π‘₯2 𝑑2 𝑦

𝑑π‘₯2+( 𝑑𝑦𝑑π‘₯ )

2

+2 𝑦= lnπ‘₯

𝑑3 𝑦𝑑 π‘₯3

βˆ’4 π‘₯ 𝑑2 𝑦

𝑑π‘₯2=cos2 π‘₯

πœ• π‘¦πœ•π‘‘ +

πœ• π‘¦πœ• π‘₯=4 𝑑2+2π‘₯

ordinary differential equation

partial differential equation

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order

β€œthe order of a DE is the order of the highest derivative in the equation”

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order?𝑑𝑦𝑑π‘₯=3 π‘₯βˆ’1

π‘₯2 𝑑2 𝑦

𝑑 π‘₯2+( 𝑑𝑦𝑑π‘₯ )

2

+2 𝑦= lnπ‘₯

𝑑3 𝑦𝑑 π‘₯3

βˆ’4 π‘₯ 𝑑2 𝑦

𝑑 π‘₯2=cos2 π‘₯

πœ• π‘¦πœ•π‘‘ +

πœ• π‘¦πœ• π‘₯=4 𝑑2+2π‘₯

1st order

2nd order

1st order

3rd order

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notation

β€œyou can represented the order by using a simple notation”

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notation

𝑦 β€²=𝑑𝑦𝑑π‘₯=𝑦 β€² (π‘₯ )

𝑦 β€² β€²= 𝑑2 𝑦𝑑 π‘₯2

=𝑦 β€² β€² (π‘₯)

𝑦 𝑛=𝑑𝑛 𝑦𝑑π‘₯𝑛=𝑦𝑛 (π‘₯)

If the independent variable is

𝑦 β€²=𝑑𝑦𝑑𝑑 =𝑦 β€² (𝑑)

𝑦 β€² β€²= 𝑑2 𝑦𝑑𝑑 2

=𝑦 β€² β€² (𝑑)

𝑦 𝑛=𝑑𝑛 𝑦𝑑𝑑𝑛

=𝑦𝑛 (𝑑)

If the independent variable is

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Linear Differential Equations

β€’ A linear differential equation is any differential equation that can be written in the following form

π‘Žπ‘› (𝑑 ) 𝑦 (𝑛 ) (𝑑 )+π‘Žπ‘›βˆ’1 (𝑑 ) 𝑦 (π‘›βˆ’1 ) (𝑑 )+…+π‘Ž1 (𝑑 ) 𝑦 β€² (𝑑 )+π‘Ž0 (𝑑 ) 𝑦 (𝑑 )=𝑔 (𝑑)

*Important:-no products of the function,

y(t), and its derivatives and neither the function or its derivatives occur to any power other than the first power

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linearity

β€œa crucial classification of differential equations is whether they are linear or nonlinear”

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linear𝑑𝑦𝑑π‘₯ + 𝑦=sin2π‘₯

𝑑2 𝑦𝑑π‘₯2

+3 π‘₯2 𝑦=0

π‘₯3 𝑑3 𝑦

𝑑 π‘₯3+3 π‘₯ 𝑑𝑦𝑑π‘₯ βˆ’5 𝑦=𝑒π‘₯

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non linear

( 𝑑𝑦𝑑π‘₯ )2

+ 𝑦=𝑠𝑖𝑛π‘₯

𝑑2 𝑦𝑑π‘₯2

+3 π‘₯ 𝑦2=0

(1+ 𝑦)𝑑3 𝑦𝑑π‘₯3

=sin 𝑦

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Got the idea? No?

β€’ We try to look again

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why linear?

𝑑𝑦𝑑π‘₯ + 𝑦=sin2π‘₯

𝑑2 𝑦𝑑π‘₯2

+3 π‘₯2 𝑦=0

π‘₯3 𝑑3 𝑦

𝑑 π‘₯3+3 π‘₯ 𝑑𝑦𝑑π‘₯ βˆ’5 𝑦=𝑒π‘₯

It has a form just like the previous one, we have:-y’(t), y(t)

It has y’’(t), and y(t)

It has y’’’(t), y’(t), and y(t)

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why non linear?

( 𝑑𝑦𝑑π‘₯ )2

+ 𝑦=𝑠𝑖𝑛π‘₯

𝑑2 𝑦𝑑π‘₯2

+3 π‘₯ 𝑦2=0

(1+ 𝑦)𝑑3 𝑦𝑑π‘₯3

=sin 𝑦

( 𝑦 ’ (𝑑 ) )2

( 𝑦 (𝑑 ) )2

𝑦 (𝑑 ) 𝑦 β€² β€² β€² (𝑑)

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how to detect?

β€œthe dependent variable y and all its derivatives are of the first degree”

β€œthe coefficient depends on constants or independent variables or both of them”

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Solution to a differential equation

β€’ A solution to a differential equation on an interval is any function which satisfies the differential equation in question on the interval .

β€’ It is important to note that solutions are often accompanied by intervals and these intervals can impart some important information about solution.

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example

β€’ Show that

is a solution to

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Solution

β€’ We’ll need the first and second derivatives to do this

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Solution

β€’ Plug these as well as the function into the differential equation.

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Solution

β€’ So, does satisfy the differential equation and hence is a solution.

β€’ Why then did I include the condition that x > 0?

RECALL BACK

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Solution

β€’ To see why recall that

β€’ In this form, it is clear that we’ll need to avoid x=0 at the least as this would give division by zero.

β€’ And then, x > 0 (forever).

𝑦 (π‘₯ )=π‘₯βˆ’ 23= 1

√π‘₯3

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Initial Value Problem

β€’ Initial condition(s) is a condition or set of conditions that help you to solve the differential equation problem

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Initial Condition(s) example

β€’ Show that

is a solution to

where

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Interval of Validity

β€’ The interval of validity for an Initial Value Problem with initial condition(s)

And/or

Is the largest possible interval on which the solution is valid and contains

We will see it when we actually solve the differential equation (trust me I will tell you)

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General Solution of DE

β€’ The general solution to a differential equation is the most general form that the solution can take and doesn’t take any initial conditions into account.

β€’ Just a regular DE

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Actual Solution of DE

β€’ What is the actual solution to the following IVP?

Is the general solution to

where

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Actual Solution of DE

Put in

So, the actual solution to the IVP is

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Implicit/Explicit Solution

Explicit solutionβ€œis any solution that is given in the form ”

Implicit solutionβ€œany solution that is not in explicit form”

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Explicit function

β€’ When an equation can be written in the form , It is said to be an explicit function of x.

β€’ Examples

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Implicit function

β€’ Sometimes with equations involving, say , y and, it is impossible to make the subject of the formula. This is called implicit function

β€’ Examples

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Homogeneous Function

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How to detect homogeneous or not

Given that Then, we put for any power of x and y

Separate it

Arrange it back so is the only remainder

Then, it can be called as homogeneous function as we still got the same equation as

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Homogenity

β€’ Homogenity

β€’ HOMOGENEOUS, 2ND ORDER, AND LINEAR

β€’ NON HOMOGENEOUS, 3RD ORDER, AND NON

LINEAR