Differential equation hamzah asyrani sulaiman at .
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Transcript of Differential equation hamzah asyrani sulaiman at .
topic
β’ basic conceptβ’ first order differential equationβ’ second order linear differential equation with
constant coefficientsβ’ the Laplace transformβ’ Fourier series
what is differential equation?
equation involving an unknown function and its derivatives
exampleππ¦ππ₯ =3 π₯β1
π₯2 π2 π¦
π π₯2+( ππ¦ππ₯ )
2
+2 π¦= lnπ₯
π3 π¦π π₯3
β4 π₯ π2 π¦
π π₯2=cos2 π₯
π π¦ππ‘ +
π π¦π π₯=4 π‘2+2π₯
there are two types of differential equations (DE)
ordinary differential equation
βif the unknown function depends on only one independent variableβ
partial differential equation
βif the unknown function depends on two or more independent variablesβ
ππ¦ππ₯ =3 π₯β1
π₯2 π2 π¦
ππ₯2+( ππ¦ππ₯ )
2
+2 π¦= lnπ₯
π3 π¦π π₯3
β4 π₯ π2 π¦
ππ₯2=cos2 π₯
π π¦ππ‘ +
π π¦π π₯=4 π‘2+2π₯
ordinary differential equation
partial differential equation
order
βthe order of a DE is the order of the highest derivative in the equationβ
order?ππ¦ππ₯=3 π₯β1
π₯2 π2 π¦
π π₯2+( ππ¦ππ₯ )
2
+2 π¦= lnπ₯
π3 π¦π π₯3
β4 π₯ π2 π¦
π π₯2=cos2 π₯
π π¦ππ‘ +
π π¦π π₯=4 π‘2+2π₯
1st order
2nd order
1st order
3rd order
notation
βyou can represented the order by using a simple notationβ
notation
π¦ β²=ππ¦ππ₯=π¦ β² (π₯ )
π¦ β² β²= π2 π¦π π₯2
=π¦ β² β² (π₯)
π¦ π=ππ π¦ππ₯π=π¦π (π₯)
If the independent variable is
π¦ β²=ππ¦ππ‘ =π¦ β² (π‘)
π¦ β² β²= π2 π¦ππ‘ 2
=π¦ β² β² (π‘)
π¦ π=ππ π¦ππ‘π
=π¦π (π‘)
If the independent variable is
Linear Differential Equations
β’ A linear differential equation is any differential equation that can be written in the following form
ππ (π‘ ) π¦ (π ) (π‘ )+ππβ1 (π‘ ) π¦ (πβ1 ) (π‘ )+β¦+π1 (π‘ ) π¦ β² (π‘ )+π0 (π‘ ) π¦ (π‘ )=π (π‘)
*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
linearity
βa crucial classification of differential equations is whether they are linear or nonlinearβ
linearππ¦ππ₯ + π¦=sin2π₯
π2 π¦ππ₯2
+3 π₯2 π¦=0
π₯3 π3 π¦
π π₯3+3 π₯ ππ¦ππ₯ β5 π¦=ππ₯
non linear
( ππ¦ππ₯ )2
+ π¦=π πππ₯
π2 π¦ππ₯2
+3 π₯ π¦2=0
(1+ π¦)π3 π¦ππ₯3
=sin π¦
Got the idea? No?
β’ We try to look again
why linear?
ππ¦ππ₯ + π¦=sin2π₯
π2 π¦ππ₯2
+3 π₯2 π¦=0
π₯3 π3 π¦
π π₯3+3 π₯ ππ¦ππ₯ β5 π¦=ππ₯
It has a form just like the previous one, we have:-yβ(t), y(t)
It has yββ(t), and y(t)
It has yβββ(t), yβ(t), and y(t)
why non linear?
( ππ¦ππ₯ )2
+ π¦=π πππ₯
π2 π¦ππ₯2
+3 π₯ π¦2=0
(1+ π¦)π3 π¦ππ₯3
=sin π¦
( π¦ β (π‘ ) )2
( π¦ (π‘ ) )2
π¦ (π‘ ) π¦ β² β² β² (π‘)
how to detect?
βthe dependent variable y and all its derivatives are of the first degreeβ
βthe coefficient depends on constants or independent variables or both of themβ
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.
example
β’ Show that
is a solution to
Solution
β’ Weβll need the first and second derivatives to do this
Solution
β’ Plug these as well as the function into the differential equation.
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
Solution
β’ To see why recall that
β’ In this form, it is clear that weβll need to avoid x=0 at the least as this would give division by zero.
β’ And then, x > 0 (forever).
π¦ (π₯ )=π₯β 23= 1
βπ₯3
Initial Value Problem
β’ Initial condition(s) is a condition or set of conditions that help you to solve the differential equation problem
Initial Condition(s) example
β’ Show that
is a solution to
where
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)
General Solution of DE
β’ The general solution to a differential equation is the most general form that the solution can take and doesnβt take any initial conditions into account.
β’ Just a regular DE
Actual Solution of DE
β’ What is the actual solution to the following IVP?
Is the general solution to
where
Actual Solution of DE
Put in
So, the actual solution to the IVP is
Implicit/Explicit Solution
Explicit solutionβis any solution that is given in the form β
Implicit solutionβany solution that is not in explicit formβ
Explicit function
β’ When an equation can be written in the form , It is said to be an explicit function of x.
β’ Examples
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
Homogeneous Function
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
Homogenity
β’ Homogenity
β’ HOMOGENEOUS, 2ND ORDER, AND LINEAR
β’ NON HOMOGENEOUS, 3RD ORDER, AND NON
LINEAR