Mixing
Finding homogennous and particular solution of an equation
$begingroup$
Let's say I have an equation like this:
$Sn = 2 cdot S(n-1) + 2^{(n+1)} - 2$
Now I know that I need to transform it to look like this:
$S(n+1) - 2 cdot Sn = 2^{(n+2)} - 2$
From this I can get the characteristic equation: $lambda - 2$ and from that the characteristic number 2. But how do I find the particular solution? And how do I then add it all together? Any explanations or links to materials are appreciated. I did find some but it did not make sense to me.
discrete-mathematics homogeneous-equation
edited Jan 21 at 18:34
EdOverflow
25119
asked Jan 21 at 10:09
ponikoliponikoli
416
$endgroup$
add a comment |
$begingroup$
Let's say I have an equation like this:
$Sn = 2 cdot S(n-1) + 2^{(n+1)} - 2$
Now I know that I need to transform it to look like this:
$S(n+1) - 2 cdot Sn = 2^{(n+2)} - 2$
From this I can get the characteristic equation: $lambda - 2$ and from that the characteristic number 2. But how do I find the particular solution? And how do I then add it all together? Any explanations or links to materials are appreciated. I did find some but it did not make sense to me.
discrete-mathematics homogeneous-equation
edited Jan 21 at 18:34
EdOverflow
25119
asked Jan 21 at 10:09
ponikoliponikoli
416
$endgroup$
$begingroup$
Please typeset the equations using MathJax.
$endgroup$
– Yves Daoust
Jan 21 at 10:12
add a comment |
$begingroup$
Let's say I have an equation like this:
$Sn = 2 cdot S(n-1) + 2^{(n+1)} - 2$
Now I know that I need to transform it to look like this:
$S(n+1) - 2 cdot Sn = 2^{(n+2)} - 2$
From this I can get the characteristic equation: $lambda - 2$ and from that the characteristic number 2. But how do I find the particular solution? And how do I then add it all together? Any explanations or links to materials are appreciated. I did find some but it did not make sense to me.
discrete-mathematics homogeneous-equation
edited Jan 21 at 18:34
EdOverflow
25119
asked Jan 21 at 10:09
ponikoliponikoli
416
$endgroup$
Let's say I have an equation like this:
$Sn = 2 cdot S(n-1) + 2^{(n+1)} - 2$
Now I know that I need to transform it to look like this:
$S(n+1) - 2 cdot Sn = 2^{(n+2)} - 2$
From this I can get the characteristic equation: $lambda - 2$ and from that the characteristic number 2. But how do I find the particular solution? And how do I then add it all together? Any explanations or links to materials are appreciated. I did find some but it did not make sense to me.
discrete-mathematics homogeneous-equation
discrete-mathematics homogeneous-equation
edited Jan 21 at 18:34
EdOverflow
25119
asked Jan 21 at 10:09
ponikoliponikoli
416
edited Jan 21 at 18:34
EdOverflow
25119
asked Jan 21 at 10:09
ponikoliponikoli
416
edited Jan 21 at 18:34
EdOverflow
25119
edited Jan 21 at 18:34
EdOverflow
25119
edited Jan 21 at 18:34
EdOverflow
25119
25119
asked Jan 21 at 10:09
ponikoliponikoli
416
asked Jan 21 at 10:09
ponikoliponikoli
416
asked Jan 21 at 10:09
ponikoliponikoli
416
416
$begingroup$
Please typeset the equations using MathJax.
$endgroup$
– Yves Daoust
Jan 21 at 10:12
add a comment |
$begingroup$
Please typeset the equations using MathJax.
$endgroup$
– Yves Daoust
Jan 21 at 10:12
$begingroup$
Please typeset the equations using MathJax.
$endgroup$
– Yves Daoust
Jan 21 at 10:12
$begingroup$
Please typeset the equations using MathJax.
$endgroup$
– Yves Daoust
Jan 21 at 10:12
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
The homogeneous equation reads
$$S_n=2S_{n-1}$$ and by induction
$$S_n=2^nS_0.$$
Now let us find a particular solution for the term $-2$, and let us try a constant $a$:
$$a=2a-2$$ yields the solution $a=2$.
Now finding a particular solution for the term $2^{n+1}$ is more tricky. You will be tempted to use the Ansatz $b,2^n$, but as this is of the same form as the homogeneous solution, it won't work:
$$b,2^n=2b,2^{n-1}+2^{n+1}$$ has no solution.
Inspired by the theory of linear ODEs, we will try $bn,2^n$, and
$$bn,2^n=2b(n-1)2^{n-1}+2^{n+1},$$ giving $b=2$.
The global solution is just the sum of these contributions,
$$S_n=c,2^n+2n,2^n+2.$$
Indeed,
$$S_n-2S_{n-1}=c,2^n-c,2^n+2n,2^n-2(n-1),2^n+2-4=2^{n+1}-2.$$
You determine $c$ from the initial condition,
$$S_0=c+2.$$
answered Jan 21 at 10:28
Yves DaoustYves Daoust
129k676227
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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oldest
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active
oldest
votes
$begingroup$
The homogeneous equation reads
$$S_n=2S_{n-1}$$ and by induction
$$S_n=2^nS_0.$$
Now let us find a particular solution for the term $-2$, and let us try a constant $a$:
$$a=2a-2$$ yields the solution $a=2$.
Now finding a particular solution for the term $2^{n+1}$ is more tricky. You will be tempted to use the Ansatz $b,2^n$, but as this is of the same form as the homogeneous solution, it won't work:
$$b,2^n=2b,2^{n-1}+2^{n+1}$$ has no solution.
Inspired by the theory of linear ODEs, we will try $bn,2^n$, and
$$bn,2^n=2b(n-1)2^{n-1}+2^{n+1},$$ giving $b=2$.
The global solution is just the sum of these contributions,
$$S_n=c,2^n+2n,2^n+2.$$
Indeed,
$$S_n-2S_{n-1}=c,2^n-c,2^n+2n,2^n-2(n-1),2^n+2-4=2^{n+1}-2.$$
You determine $c$ from the initial condition,
$$S_0=c+2.$$
answered Jan 21 at 10:28
Yves DaoustYves Daoust
129k676227
$endgroup$
add a comment |
$begingroup$
The homogeneous equation reads
$$S_n=2S_{n-1}$$ and by induction
$$S_n=2^nS_0.$$
Now let us find a particular solution for the term $-2$, and let us try a constant $a$:
$$a=2a-2$$ yields the solution $a=2$.
Now finding a particular solution for the term $2^{n+1}$ is more tricky. You will be tempted to use the Ansatz $b,2^n$, but as this is of the same form as the homogeneous solution, it won't work:
$$b,2^n=2b,2^{n-1}+2^{n+1}$$ has no solution.
Inspired by the theory of linear ODEs, we will try $bn,2^n$, and
$$bn,2^n=2b(n-1)2^{n-1}+2^{n+1},$$ giving $b=2$.
The global solution is just the sum of these contributions,
$$S_n=c,2^n+2n,2^n+2.$$
Indeed,
$$S_n-2S_{n-1}=c,2^n-c,2^n+2n,2^n-2(n-1),2^n+2-4=2^{n+1}-2.$$
You determine $c$ from the initial condition,
$$S_0=c+2.$$
answered Jan 21 at 10:28
Yves DaoustYves Daoust
129k676227
$endgroup$
add a comment |
$begingroup$
The homogeneous equation reads
$$S_n=2S_{n-1}$$ and by induction
$$S_n=2^nS_0.$$
Now let us find a particular solution for the term $-2$, and let us try a constant $a$:
$$a=2a-2$$ yields the solution $a=2$.
Now finding a particular solution for the term $2^{n+1}$ is more tricky. You will be tempted to use the Ansatz $b,2^n$, but as this is of the same form as the homogeneous solution, it won't work:
$$b,2^n=2b,2^{n-1}+2^{n+1}$$ has no solution.
Inspired by the theory of linear ODEs, we will try $bn,2^n$, and
$$bn,2^n=2b(n-1)2^{n-1}+2^{n+1},$$ giving $b=2$.
The global solution is just the sum of these contributions,
$$S_n=c,2^n+2n,2^n+2.$$
Indeed,
$$S_n-2S_{n-1}=c,2^n-c,2^n+2n,2^n-2(n-1),2^n+2-4=2^{n+1}-2.$$
You determine $c$ from the initial condition,
$$S_0=c+2.$$
answered Jan 21 at 10:28
Yves DaoustYves Daoust
129k676227
$endgroup$
The homogeneous equation reads
$$S_n=2S_{n-1}$$ and by induction
$$S_n=2^nS_0.$$
Now let us find a particular solution for the term $-2$, and let us try a constant $a$:
$$a=2a-2$$ yields the solution $a=2$.
Now finding a particular solution for the term $2^{n+1}$ is more tricky. You will be tempted to use the Ansatz $b,2^n$, but as this is of the same form as the homogeneous solution, it won't work:
$$b,2^n=2b,2^{n-1}+2^{n+1}$$ has no solution.
Inspired by the theory of linear ODEs, we will try $bn,2^n$, and
$$bn,2^n=2b(n-1)2^{n-1}+2^{n+1},$$ giving $b=2$.
The global solution is just the sum of these contributions,
$$S_n=c,2^n+2n,2^n+2.$$
Indeed,
$$S_n-2S_{n-1}=c,2^n-c,2^n+2n,2^n-2(n-1),2^n+2-4=2^{n+1}-2.$$
You determine $c$ from the initial condition,
$$S_0=c+2.$$
answered Jan 21 at 10:28
Yves DaoustYves Daoust
129k676227
answered Jan 21 at 10:28
Yves DaoustYves Daoust
129k676227
answered Jan 21 at 10:28
Yves DaoustYves Daoust
129k676227
answered Jan 21 at 10:28
Yves DaoustYves Daoust
129k676227
129k676227
add a comment |
add a comment |
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$begingroup$
Please typeset the equations using MathJax.
$endgroup$
– Yves Daoust
Jan 21 at 10:12