Solving inhomogeneous second-order linear difference equation
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In general, how does one solve an inhomogeneous second-order linear difference equation of the form
$$ a f(n+1) + b f(n) + c f(n-1) = d(n) $$
where $a, b, c$ are constants but $d(n)$ may also depend on $n$?
For instance, consider as in here:
$ 2 f(n+1) - 7 f(n) + 3 f(n-1) = 2 + 2^n $
with boundary conditions $q_0 = 1$ and $q_1 = 2$.
I understand that the general solution can be written as the sum of a particular solution to the inhomogeneous equation plus
the general solution of the corresponding homogeneous equation, and I am able to find the latter, but how does one come up with a particular solution to the inhomogeneous equation in this case?
recurrence-relations
asked Dec 25 '18 at 3:40
p-valuep-value
234111
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add a comment |
$begingroup$
In general, how does one solve an inhomogeneous second-order linear difference equation of the form
$$ a f(n+1) + b f(n) + c f(n-1) = d(n) $$
where $a, b, c$ are constants but $d(n)$ may also depend on $n$?
For instance, consider as in here:
$ 2 f(n+1) - 7 f(n) + 3 f(n-1) = 2 + 2^n $
with boundary conditions $q_0 = 1$ and $q_1 = 2$.
I understand that the general solution can be written as the sum of a particular solution to the inhomogeneous equation plus
the general solution of the corresponding homogeneous equation, and I am able to find the latter, but how does one come up with a particular solution to the inhomogeneous equation in this case?
recurrence-relations
asked Dec 25 '18 at 3:40
p-valuep-value
234111
$endgroup$
$begingroup$
obtaining a particular solution can be really complicated. It depends a lot on the function type of $d (n)$. In the polynomial case it is usually direct. Also for some exponential cases.
$endgroup$
– Cesareo
Dec 25 '18 at 9:02
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@Cesareo Thanks; could you please elaborate how to do so for the polynomial and exponential cases such as $2 + 2^n$ in an answer?
$endgroup$
– p-value
Dec 25 '18 at 21:10
add a comment |
$begingroup$
In general, how does one solve an inhomogeneous second-order linear difference equation of the form
$$ a f(n+1) + b f(n) + c f(n-1) = d(n) $$
where $a, b, c$ are constants but $d(n)$ may also depend on $n$?
For instance, consider as in here:
$ 2 f(n+1) - 7 f(n) + 3 f(n-1) = 2 + 2^n $
with boundary conditions $q_0 = 1$ and $q_1 = 2$.
I understand that the general solution can be written as the sum of a particular solution to the inhomogeneous equation plus
the general solution of the corresponding homogeneous equation, and I am able to find the latter, but how does one come up with a particular solution to the inhomogeneous equation in this case?
recurrence-relations
asked Dec 25 '18 at 3:40
p-valuep-value
234111
$endgroup$
In general, how does one solve an inhomogeneous second-order linear difference equation of the form
$$ a f(n+1) + b f(n) + c f(n-1) = d(n) $$
where $a, b, c$ are constants but $d(n)$ may also depend on $n$?
For instance, consider as in here:
$ 2 f(n+1) - 7 f(n) + 3 f(n-1) = 2 + 2^n $
with boundary conditions $q_0 = 1$ and $q_1 = 2$.
I understand that the general solution can be written as the sum of a particular solution to the inhomogeneous equation plus
the general solution of the corresponding homogeneous equation, and I am able to find the latter, but how does one come up with a particular solution to the inhomogeneous equation in this case?
recurrence-relations
recurrence-relations
asked Dec 25 '18 at 3:40
p-valuep-value
234111
asked Dec 25 '18 at 3:40
p-valuep-value
234111
asked Dec 25 '18 at 3:40
p-valuep-value
234111
asked Dec 25 '18 at 3:40
p-valuep-value
234111
asked Dec 25 '18 at 3:40
p-valuep-value
234111
234111
$begingroup$
obtaining a particular solution can be really complicated. It depends a lot on the function type of $d (n)$. In the polynomial case it is usually direct. Also for some exponential cases.
$endgroup$
– Cesareo
Dec 25 '18 at 9:02
$begingroup$
@Cesareo Thanks; could you please elaborate how to do so for the polynomial and exponential cases such as $2 + 2^n$ in an answer?
$endgroup$
– p-value
Dec 25 '18 at 21:10
add a comment |
$begingroup$
obtaining a particular solution can be really complicated. It depends a lot on the function type of $d (n)$. In the polynomial case it is usually direct. Also for some exponential cases.
$endgroup$
– Cesareo
Dec 25 '18 at 9:02
$begingroup$
@Cesareo Thanks; could you please elaborate how to do so for the polynomial and exponential cases such as $2 + 2^n$ in an answer?
$endgroup$
– p-value
Dec 25 '18 at 21:10
$begingroup$
obtaining a particular solution can be really complicated. It depends a lot on the function type of $d (n)$. In the polynomial case it is usually direct. Also for some exponential cases.
$endgroup$
– Cesareo
Dec 25 '18 at 9:02
$begingroup$
obtaining a particular solution can be really complicated. It depends a lot on the function type of $d (n)$. In the polynomial case it is usually direct. Also for some exponential cases.
$endgroup$
– Cesareo
Dec 25 '18 at 9:02
$begingroup$
@Cesareo Thanks; could you please elaborate how to do so for the polynomial and exponential cases such as $2 + 2^n$ in an answer?
$endgroup$
– p-value
Dec 25 '18 at 21:10
$begingroup$
@Cesareo Thanks; could you please elaborate how to do so for the polynomial and exponential cases such as $2 + 2^n$ in an answer?
$endgroup$
– p-value
Dec 25 '18 at 21:10
add a comment |
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$begingroup$
obtaining a particular solution can be really complicated. It depends a lot on the function type of $d (n)$. In the polynomial case it is usually direct. Also for some exponential cases.
$endgroup$
– Cesareo
Dec 25 '18 at 9:02
$begingroup$
@Cesareo Thanks; could you please elaborate how to do so for the polynomial and exponential cases such as $2 + 2^n$ in an answer?
$endgroup$
– p-value
Dec 25 '18 at 21:10