I am unsure what I am doing wrong with difference equations
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The first one is:
$x(n+2) - 4x(n) = 27n^2$, $x(0) = 1$, $x(1) = 3$.
So I work out the complimentary and particular solutions and add them, and I get:
$$x(n) = 5(2)^n + 2(-2)^n - 9n^2 + 12n - 6.$$
However, because I was unsure, I used this to check https://www.wolframalpha.com/input/?i=g(0)%3D1,+g(n%2B1)%3Dn%5E2%2Bg(n)&lk=3
and it says the answer is:
$$x(n) = frac14 (e^{2 i pi n} (-9 n^2 + 6 n + 1) + 43 cdot 2^{n + 1} - 2^{n + 1} e^{i pi n} - 27 (n (n + 2) + 3))$$
I also have a problem with:
$$x(n+1)-4x(n)+3x(n-1)=36n^2$$
When solving for the constants in the particular, I am getting $0=36n^2$,
and because of this, I feel that I am missing something.
Thanks for any help.
differential-equations
edited Nov 22 at 4:23
Rócherz
2,7412721
asked Nov 22 at 4:11
Plz help me
766
add a comment |
up vote
0
down vote
favorite
The first one is:
$x(n+2) - 4x(n) = 27n^2$, $x(0) = 1$, $x(1) = 3$.
So I work out the complimentary and particular solutions and add them, and I get:
$$x(n) = 5(2)^n + 2(-2)^n - 9n^2 + 12n - 6.$$
However, because I was unsure, I used this to check https://www.wolframalpha.com/input/?i=g(0)%3D1,+g(n%2B1)%3Dn%5E2%2Bg(n)&lk=3
and it says the answer is:
$$x(n) = frac14 (e^{2 i pi n} (-9 n^2 + 6 n + 1) + 43 cdot 2^{n + 1} - 2^{n + 1} e^{i pi n} - 27 (n (n + 2) + 3))$$
I also have a problem with:
$$x(n+1)-4x(n)+3x(n-1)=36n^2$$
When solving for the constants in the particular, I am getting $0=36n^2$,
and because of this, I feel that I am missing something.
Thanks for any help.
differential-equations
edited Nov 22 at 4:23
Rócherz
2,7412721
asked Nov 22 at 4:11
Plz help me
766
As a start, you know that $e^{2pi i n} = 1$ and $e^{pi i n} = -1$. Maybe you can simplify the answer from Wolfram a bit now.
– D.B.
Nov 22 at 4:41
1
1. $mathrm e^{2mathrm i pi n} = 1, mathrm e^{mathrm i pi n} = (-1)^n$. 2. If you cannot get a feasible solution for the particular ones, raise the greatest power [2 to 3] and do it again.
– xbh
Nov 22 at 4:43
So I have tried simplifying using e2πin=1 and eπin=−1, I end up getting -9n^2 + 12n + (-1)^n * 2^(n-1) -43 * 2^(n-1) +20
– Plz help me
Nov 22 at 11:41
I've managed to do it now, I realised I made a very stupid mistake, thank you for the help
– Plz help me
Nov 22 at 12:39
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The first one is:
$x(n+2) - 4x(n) = 27n^2$, $x(0) = 1$, $x(1) = 3$.
So I work out the complimentary and particular solutions and add them, and I get:
$$x(n) = 5(2)^n + 2(-2)^n - 9n^2 + 12n - 6.$$
However, because I was unsure, I used this to check https://www.wolframalpha.com/input/?i=g(0)%3D1,+g(n%2B1)%3Dn%5E2%2Bg(n)&lk=3
and it says the answer is:
$$x(n) = frac14 (e^{2 i pi n} (-9 n^2 + 6 n + 1) + 43 cdot 2^{n + 1} - 2^{n + 1} e^{i pi n} - 27 (n (n + 2) + 3))$$
I also have a problem with:
$$x(n+1)-4x(n)+3x(n-1)=36n^2$$
When solving for the constants in the particular, I am getting $0=36n^2$,
and because of this, I feel that I am missing something.
Thanks for any help.
differential-equations
edited Nov 22 at 4:23
Rócherz
2,7412721
asked Nov 22 at 4:11
Plz help me
766
The first one is:
$x(n+2) - 4x(n) = 27n^2$, $x(0) = 1$, $x(1) = 3$.
So I work out the complimentary and particular solutions and add them, and I get:
$$x(n) = 5(2)^n + 2(-2)^n - 9n^2 + 12n - 6.$$
However, because I was unsure, I used this to check https://www.wolframalpha.com/input/?i=g(0)%3D1,+g(n%2B1)%3Dn%5E2%2Bg(n)&lk=3
and it says the answer is:
$$x(n) = frac14 (e^{2 i pi n} (-9 n^2 + 6 n + 1) + 43 cdot 2^{n + 1} - 2^{n + 1} e^{i pi n} - 27 (n (n + 2) + 3))$$
I also have a problem with:
$$x(n+1)-4x(n)+3x(n-1)=36n^2$$
When solving for the constants in the particular, I am getting $0=36n^2$,
and because of this, I feel that I am missing something.
Thanks for any help.
differential-equations
differential-equations
edited Nov 22 at 4:23
Rócherz
2,7412721
asked Nov 22 at 4:11
Plz help me
766
edited Nov 22 at 4:23
Rócherz
2,7412721
asked Nov 22 at 4:11
Plz help me
766
edited Nov 22 at 4:23
Rócherz
2,7412721
edited Nov 22 at 4:23
Rócherz
2,7412721
edited Nov 22 at 4:23
Rócherz
2,7412721
2,7412721
asked Nov 22 at 4:11
Plz help me
766
asked Nov 22 at 4:11
Plz help me
766
asked Nov 22 at 4:11
Plz help me
766
766
As a start, you know that $e^{2pi i n} = 1$ and $e^{pi i n} = -1$. Maybe you can simplify the answer from Wolfram a bit now.
– D.B.
Nov 22 at 4:41
1
1. $mathrm e^{2mathrm i pi n} = 1, mathrm e^{mathrm i pi n} = (-1)^n$. 2. If you cannot get a feasible solution for the particular ones, raise the greatest power [2 to 3] and do it again.
– xbh
Nov 22 at 4:43
So I have tried simplifying using e2πin=1 and eπin=−1, I end up getting -9n^2 + 12n + (-1)^n * 2^(n-1) -43 * 2^(n-1) +20
– Plz help me
Nov 22 at 11:41
I've managed to do it now, I realised I made a very stupid mistake, thank you for the help
– Plz help me
Nov 22 at 12:39
add a comment |
As a start, you know that $e^{2pi i n} = 1$ and $e^{pi i n} = -1$. Maybe you can simplify the answer from Wolfram a bit now.
– D.B.
Nov 22 at 4:41
1
1. $mathrm e^{2mathrm i pi n} = 1, mathrm e^{mathrm i pi n} = (-1)^n$. 2. If you cannot get a feasible solution for the particular ones, raise the greatest power [2 to 3] and do it again.
– xbh
Nov 22 at 4:43
So I have tried simplifying using e2πin=1 and eπin=−1, I end up getting -9n^2 + 12n + (-1)^n * 2^(n-1) -43 * 2^(n-1) +20
– Plz help me
Nov 22 at 11:41
I've managed to do it now, I realised I made a very stupid mistake, thank you for the help
– Plz help me
Nov 22 at 12:39
As a start, you know that $e^{2pi i n} = 1$ and $e^{pi i n} = -1$. Maybe you can simplify the answer from Wolfram a bit now.
– D.B.
Nov 22 at 4:41
As a start, you know that $e^{2pi i n} = 1$ and $e^{pi i n} = -1$. Maybe you can simplify the answer from Wolfram a bit now.
– D.B.
Nov 22 at 4:41
1
1
1. $mathrm e^{2mathrm i pi n} = 1, mathrm e^{mathrm i pi n} = (-1)^n$. 2. If you cannot get a feasible solution for the particular ones, raise the greatest power [2 to 3] and do it again.
– xbh
Nov 22 at 4:43
1. $mathrm e^{2mathrm i pi n} = 1, mathrm e^{mathrm i pi n} = (-1)^n$. 2. If you cannot get a feasible solution for the particular ones, raise the greatest power [2 to 3] and do it again.
– xbh
Nov 22 at 4:43
So I have tried simplifying using e2πin=1 and eπin=−1, I end up getting -9n^2 + 12n + (-1)^n * 2^(n-1) -43 * 2^(n-1) +20
– Plz help me
Nov 22 at 11:41
So I have tried simplifying using e2πin=1 and eπin=−1, I end up getting -9n^2 + 12n + (-1)^n * 2^(n-1) -43 * 2^(n-1) +20
– Plz help me
Nov 22 at 11:41
I've managed to do it now, I realised I made a very stupid mistake, thank you for the help
– Plz help me
Nov 22 at 12:39
I've managed to do it now, I realised I made a very stupid mistake, thank you for the help
– Plz help me
Nov 22 at 12:39
add a comment |
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As a start, you know that $e^{2pi i n} = 1$ and $e^{pi i n} = -1$. Maybe you can simplify the answer from Wolfram a bit now.
– D.B.
Nov 22 at 4:41
1
1. $mathrm e^{2mathrm i pi n} = 1, mathrm e^{mathrm i pi n} = (-1)^n$. 2. If you cannot get a feasible solution for the particular ones, raise the greatest power [2 to 3] and do it again.
– xbh
Nov 22 at 4:43
So I have tried simplifying using e2πin=1 and eπin=−1, I end up getting -9n^2 + 12n + (-1)^n * 2^(n-1) -43 * 2^(n-1) +20
– Plz help me
Nov 22 at 11:41
I've managed to do it now, I realised I made a very stupid mistake, thank you for the help
– Plz help me
Nov 22 at 12:39