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Find the solution of the difference equation∆Yt=2 assuming an initial value yo=15 (b). Yt+1=0.9yt assuming an initial value yo=yo?
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Find the solution of the difference equation∆Yt=2 assuming an initial ...
Difference Equation:
A difference equation is a mathematical equation that defines a sequence recursively in terms of its previous terms. It describes how the values of a sequence change from one term to the next.
The given difference equation is:
∆Yt = 2
This equation represents a constant change of 2 between consecutive terms.
Initial Value:
The initial value is given as yo = 15. It represents the value of the sequence at the starting point.
Solution of the Difference Equation:
To find the solution of the difference equation, we need to apply the difference operator (∆) repeatedly, starting from the initial value yo.
Step 1: Applying the Difference Operator
∆Yt = 2
Applying the difference operator (∆) to both sides of the equation, we get:
∆(∆Yt) = ∆(2)
Simplifying the equation:
∆(∆Yt) = 0
Step 2: Applying the Difference Operator Again
∆(∆Yt) = 0
Applying the difference operator (∆) to both sides of the equation, we get:
∆(∆(∆Yt)) = ∆(0)
Simplifying the equation:
∆(∆(∆Yt)) = 0
Step 3: Continuing the Process
We need to continue applying the difference operator (∆) until we reach the desired term.
∆(∆(∆Yt)) = 0
Continuing the process, we apply the difference operator (∆) again:
∆(∆(∆(∆Yt))) = ∆(0)
Simplifying the equation:
∆(∆(∆(∆Yt))) = 0
Step 4: Solving for Yt
To solve for Yt, we need to integrate the equation obtained from Step 3.
∆(∆(∆(∆Yt))) = 0
Integrating the equation:
∆(∆(∆Yt)) = C
Where C is the constant of integration.
Step 5: Applying Initial Value
To determine the value of the constant C, we substitute the initial value yo = 15 into the equation obtained from Step 4.
∆(∆(∆Yt)) = C
∆(∆(∆(15))) = C
Simplifying the equation:
∆(∆(2)) = C
∆(0) = C
C = 0
Step 6: Final Solution
Now that we have determined the value of the constant C, we can write the final solution to the difference equation.
∆(∆(∆Yt)) = 0
Integrating the equation:
∆(∆Yt) = 0
Integrating the equation again:
∆Yt = C
Integrating the equation one last time:
Yt = Ct + D
Where C and D are constants of integration.
Substituting the initial value yo = 15, we get:
15 = C(0) + D
D =
A difference equation is a mathematical equation that defines a sequence recursively in terms of its previous terms. It describes how the values of a sequence change from one term to the next.
The given difference equation is:
∆Yt = 2
This equation represents a constant change of 2 between consecutive terms.
Initial Value:
The initial value is given as yo = 15. It represents the value of the sequence at the starting point.
Solution of the Difference Equation:
To find the solution of the difference equation, we need to apply the difference operator (∆) repeatedly, starting from the initial value yo.
Step 1: Applying the Difference Operator
∆Yt = 2
Applying the difference operator (∆) to both sides of the equation, we get:
∆(∆Yt) = ∆(2)
Simplifying the equation:
∆(∆Yt) = 0
Step 2: Applying the Difference Operator Again
∆(∆Yt) = 0
Applying the difference operator (∆) to both sides of the equation, we get:
∆(∆(∆Yt)) = ∆(0)
Simplifying the equation:
∆(∆(∆Yt)) = 0
Step 3: Continuing the Process
We need to continue applying the difference operator (∆) until we reach the desired term.
∆(∆(∆Yt)) = 0
Continuing the process, we apply the difference operator (∆) again:
∆(∆(∆(∆Yt))) = ∆(0)
Simplifying the equation:
∆(∆(∆(∆Yt))) = 0
Step 4: Solving for Yt
To solve for Yt, we need to integrate the equation obtained from Step 3.
∆(∆(∆(∆Yt))) = 0
Integrating the equation:
∆(∆(∆Yt)) = C
Where C is the constant of integration.
Step 5: Applying Initial Value
To determine the value of the constant C, we substitute the initial value yo = 15 into the equation obtained from Step 4.
∆(∆(∆Yt)) = C
∆(∆(∆(15))) = C
Simplifying the equation:
∆(∆(2)) = C
∆(0) = C
C = 0
Step 6: Final Solution
Now that we have determined the value of the constant C, we can write the final solution to the difference equation.
∆(∆(∆Yt)) = 0
Integrating the equation:
∆(∆Yt) = 0
Integrating the equation again:
∆Yt = C
Integrating the equation one last time:
Yt = Ct + D
Where C and D are constants of integration.
Substituting the initial value yo = 15, we get:
15 = C(0) + D
D =
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Find the solution of the difference equation∆Yt=2 assuming an initial value yo=15 (b). Yt+1=0.9yt assuming an initial value yo=yo? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about Find the solution of the difference equation∆Yt=2 assuming an initial value yo=15 (b). Yt+1=0.9yt assuming an initial value yo=yo? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Find the solution of the difference equation∆Yt=2 assuming an initial value yo=15 (b). Yt+1=0.9yt assuming an initial value yo=yo?.
Find the solution of the difference equation∆Yt=2 assuming an initial value yo=15 (b). Yt+1=0.9yt assuming an initial value yo=yo? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about Find the solution of the difference equation∆Yt=2 assuming an initial value yo=15 (b). Yt+1=0.9yt assuming an initial value yo=yo? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Find the solution of the difference equation∆Yt=2 assuming an initial value yo=15 (b). Yt+1=0.9yt assuming an initial value yo=yo?.
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