First-Order Difference Equations in Mathematics
Εξισώσεις Διαφορών
2
•
General Form:
•
Step 1:
Solve the Homogenous Part
3
We solve with Forward Substitution.
4
Step 2:
Solve for the non-Homogenous Part (Particular Solution)
The candidate particular solution has the same algebraic form as g(t).
(Where
α
is a constant)
Α)
First Case
5
Guess:
In order for
μ
to be a solution we need
If then this is not a solution and we have to
multiply by t our guess.
New Guess:
6
B
)
Second Case
Guess:
7
Guess:
Γ)
Third Case
New Guess:
8
So:
9
If then this is not a solution and we have to
multiply by t our guess.
New Guess:
•
Step 3:
Sum
the solution of Homogenous part and the
particular solution
10
In order to find A we need an initial condition. If we know that
for, we substitute the initial condition into
the general solution and we take A.
•
Step 4:
Find the value of A
11
•
Step 1:
Homogenous Solution
•
Step 2:
Particular Solution
Guess:
12
So:
13
•
Step 3:
General Solution
•
Step
4
:
Find value of A
Οπότε
:
14
•
Step 1:
Homogenous Solution
•
Step 2:
Particular Solution
Guess:
15
So:
16
•
Step 3:
General Solution
•
Step
4
:
Find value of A
Οπότε
:
17
•
General Form:
•
Step 1:
Solve the Homogenous Part
18
Guess:
Λύνουμε λοιπόν το πολυώνυμο δευτέρου βαθμού
:
The solution depends on the roots
1) If
Δ>0
So, the solution is:
19
2) If
Δ
=
0
Προφανώς η λύση είναι το
Όμως έχουμε
second-order
εξίσωση οπότε χρειαζόμαστε δύο
ρίζες. Άρα η δεύτερη ρίζα θα είναι
So, the solution is:
20
3) If
Δ
<
0
So, the solution is:
21
•
General Form:
•
Step 2:
Particular Part
A) If g(t)=G
Guess:
22
23
•
Step 1:
Homogenous Solution
άρα
24
•
Step 2:
Find value of A
Οπότε
:
25
•
Step 1:
Homogenous Solution
•
Step 2:
Particular Solution
Guess:
26
So:
27
•
Step 3:
General Solution
•
Step
4
:
Find value of A
Οπότε
:
Σας Ευχαριστώ πολύ
28
First-order difference equations involve solving both the homogenous and non-homogenous parts to find the general and particular solutions. By using forward substitution and candidate particular solutions, techniques like guessing and summing up the solutions help in solving these equations. This comprehensive guide will walk you through the steps with detailed explanations and examples.
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Presentation Transcript
First-Order Difference Equation General Form: c Y + = ( ) c Y g t 1 0 1 t t Step 1: Solve the Homogenous Part c c c c + = + = 0 0 0 Y Y Y Y 1 1 1 t t t t 1 1 2
First-Order Difference Equation We solve with Forward Substitution. = = = = = = 1: 2: 3: t t t Y Y Y Y 1 0 Y Y 2 1 = = t h t ( ) ( ) Y Y Y A 3 2 0 t t t T = = : Y Y 1 T t 3
First-Order Difference Equation Step 2: Solve for the non-Homogenous Part (Particular Solution) c Y + = ( ) c Y g t 1 0 1 t t The candidate particular solution has the same algebraic form as g(t). ) First Case = ( ) g t (Where is a constant) c Y + = c Y 1 0 1 t t 4
First-Order Difference Equation = ( ) y t Guess: + + = + = = ( ) c c c c 1 0 0 1 ( ) c c 0 1 + ( ) 0 c c In order for to be a solution we need 0 1 + = ( ) 0 c c If multiply by t our guess. then this is not a solution and we have to 0 1 = ( ) y t t New Guess: 5
First-Order Difference Equation B) Second Case = + ( ) g t t 0 1 = + ( ) y t t Guess: 0 1 + + + + + + = + = = [ c c ] [ ( 1)] + c t c t c t 1 0 1 0 + = 0 1 0 c 1 + + c c + t c t t t 1 0 1 1 0 0 0 1 t 0 1 t 0 1 + = c c c 1 0 0 0 0 1 1 1 0 1 0 1 ( ) c c c 0 1 + 0 0 1 0 c c 1 1 0 1 1 6
First-Order Difference Equation = + 2 ( ) y t t t New Guess: 0 1 ) Third Case = B d t ( ) g t = C d t ( ) y t Guess: 7
First-Order Difference Equation + = 1 t t t c C d c C d d c C c C d c + + B d d 1 0 + = 1 t t t C d C d B d B d 1 0 c C d c C + = 1 t t ( ) B 1 0 = 1 ( ) c B d 1 0 = C ( ) c d c 1 0 B d = C d = p t p t ( ) t ( ) t y y d So: + ( ) c d c 1 0 8
First-Order Difference Equation + = ( ) 0 c d c If multiply by t our guess. then this is not a solution and we have to 1 0 t C d = t ( ) y t New Guess: Step 3: Sum the solution of Homogenous part and the particular solution = + GS t h t p y y y t 9
First-Order Difference Equation Step 4: Find the value of A = + = + GS t h t p t p t ( ) y y y A y t In order to find A we need an initial condition. If we know that for, we substitute the initial condition into the general solution and we take A. = = * * * ( ) y t t t y 10
1 = t 3 4 t=0, y(0)=0 Y Y 1 t t Step 1: Homogenous Solution = h t t h t t 3 0 =(- ) =3 Y Y y A y A 1 t t Step 2: Particular Solution = 4t ( ) y t C Guess: 11
1 = 1 t t t 4 4 3 3 4 4 = 4 C C C C C C C = 1 t t t 4 4 1 3 4 4 4 = 1 C = 3 4 C ( ) t = p t 4 4 y So: 12
1 Step 3: General Solution = + = + 4 4 GS t h t p t t 3 y y y A t Step 4: Find value of A = 4 4 + = + = = 0 0 (0) 0 3 0 4 0 4 y A A A = 3 ( 4) + 4 4 GS t t t y : 13
2 = 2 5 t Y Y 1 t t Step 1: Homogenous Solution = h t t h t t 2 0 =(- ) =2 Y Y y A y A 1 t t Step 2: Particular Solution = + ( ) y t a t b Guess: 14
2 + + + = 2 [ 2 5 0 = ( 1) ] = 5 t a t a t = b b a + t b b t 2 2 5 a t t a a b = = 5 10 a t a 2 b = p ( ) t 5 10 y t So: 15
2 Step 3: General Solution = + = GS t h t p t 2 5 10 y y y A t t Step 4: Find value of A = 5 0 10 1 = = = 0 (0) 1 2 10 1 11 y A A A = GS t 11 2 5 10 t y t : 16
Second-Order Difference Equation General Form: + c Y + = ( ) c Y c Y g t 2 1 1 0 2 t t t Step 1: Solve the Homogenous Part c c c c c c + + = + + = 0 0 0 Y Y Y Y a Y a Y 2 1 1 2 1 1 2 2 t t t t t t 2 2 2 17
Second-Order Difference Equation = t ty Guess: : + + = 2 0 a a 1 2 The solution depends on the roots = = 2 1 4 1 1) If >0 2 1,2 2 = + h t t ty A A So, the solution is: 1 1 2 2 18
Second-Order Difference Equation = = = 2 1 * 4 1 2) If =0 2 1,2 2 t second-order t t . = + h t t t y A A t So, the solution is: 1 2 19
Second-Order Difference Equation = = 2 1 4 a i 3) If <0 2 1,2 1 2 = 1 1[4 2 = 2 1 ] 2 = + h t [ cos( ) sin( )] ty r B tw B tw So, the solution is: 1 2 20
Second-Order Difference Equation General Form: + c Y + = ( ) c Y c Y g t 2 1 1 0 2 t t t Step 2: Particular Part A) If g(t)=G t y = Guess: 21
Second-Order Difference Equation + + = c c c G 2 1 0 G c = + + c c 0 1 2 = t y t ?? ?0+ ?1+ ?2=0 then new guess: 22
1 + = = = 5 6 0 y 1 y Y Y Y 1 1 2 0 1 t t t Step 1: Homogenous Solution + = + = = = 2 5 6 0 5 6 0 2 Y Y Y 3 1 2 1 2 t t t = + h t t 2 3 t y A A 1 2 23
1 Step 2: Find value of A = + = = + = = = 0 0 (0) 1 = 2 3 1 1 4 y A A + A A A A 1 A 2 1 2 1 3 1 1 ( 1) 1 2 3 1 y A 2 1 2 = 4 2 3 3 GS t t t y : 24
2 + = = = t 4 3 5 y 1 y Y Y Y 2 + + 2 1 0 1 t t t Step 1: Homogenous Solution + = + = = = 2 4 3 0 4 3 0 3 Y Y Y 1 + + 2 1 1 2 t t t = + = + h t t t 3 1 3 t y A A A A 1 2 1 2 Step 2: Particular Solution = 5t ( ) y t C Guess: 25
2 + + + = = 2 1 t t t t 5 4 5 C C 3 5 5 5 C C C C + 2 1 t t t t 25 5 5 C 4 + 5 5 1 = 3 5 C 20 3 C 1 8 = C 1 8 ( ) t = p t 5 y So: 26
2 Step 3: General Solution 1 8 = + = + + GS t h t p t t 3 5 y y y A A 1 2 t Step 4: Find value of A 1 8 1 8 7 8 1 4 5 8 = + + = + = = 0 0 (0) 1 3 5 1 y A A A A A 1 2 1 2 1 11 8 = = + + = + = 1 1 (1) 2 3 5 2 3 A y A A A A 2 1 2 1 2 1 4 5 8 1 8 = + + GS t t 3 5 t y : 27
