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anonymous
 5 years ago
x{n+1} = (1/2)x{n} + n when n = 06 the values of x{n} are 0, 1, 2.5, 4.25, 6.125, 8.0625, and 10.03125. find a particular solution {pn} of the difference equation
anonymous
 5 years ago
x{n+1} = (1/2)x{n} + n when n = 06 the values of x{n} are 0, 1, 2.5, 4.25, 6.125, 8.0625, and 10.03125. find a particular solution {pn} of the difference equation

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0particularvalues of coefficients in trial solution

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0btm...i'm sorry, but i have no idea what the question is even asking... i thought you had meant differential equations.... i'm sorry, i can't help you ><

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Step 1: What you want to do is drop the constant, the same as simply making the constant zero to get the homogeneous difference equation. Then you get a general solution to that homogenous difference equation. Step 2: Then go back and come up with a trial solution to the nonhomogenous difference equation, the style of the trial solution being based on the kind of terms that are creating the nonhomogeneous part (this will probably be the hardest part, so read that section of material carefully). Then put in particular values that you know solve the difference equation, specific values you got from working with the difference equation such as what xo is, what x1 is, allowing you to determine what the constants in your trial solution need to be. This gives you your particular solution. Step 3: As the last step you combine your general solution and your particular solution to get your complete solution for this specific problems, and of course verify that when applied it creates the correct values for x0, x1, .... xn. Overall you will have started with a "difference equation" expressing how to get x n+1 if you know x n, and you are ending with a formula that will calculate any x n by just knowing n. The formula allows you to calculate any x n term without working out all of the terms ahead of it.
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