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Difference Calculus: Obtaining 2nd solution to Homogeneous Equation

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0

$begingroup$

Consider the 2nd order difference equation:

$u(n + 2) - frac(n + 3)(n + 2) u(n + 1) + frac2(n + 2) u(n) = 0$

I verified that the given solution $u_1(n) = frac2^nn!$ is a solution to the above since:

$frac2^n + 2(n + 2)! - frac(n + 3)(n + 2) * frac2^n + 1(n + 1)! + frac2(n + 2) * frac2^nn! =$

$frac4*2^n(n + 2)! - frac(n + 3)2^n + 1(n+2)*(n + 1)! + frac(n + 1)(n + 1) * frac2*2^n(n + 2)n! =$

$frac4*2^n(n + 2)! - frac2*2^n(n + 3)(n + 2)! + frac2*2^n(n + 1)(n + 2)! = $

$frac2^n(4 - 2n - 6 + 2n + 2)(n + 2)! = frac2^n*0(n + 2)! = 0.$

The difficulty I am having is in the follow up part of the problem of obtaining another solution $u_2(n)$ to the 2nd order homogeneous equation. The instructor assigning this problem wanted us to use the following formula to solve for the other solution $u_2(n)$:

$u_2(n) = u_1(n)[sumlimits_r = 0^n - 1 fracw(r)u_1(r)u_1(r + 1)]$

where $w(r)$ is the casaratian defined at $r$, for this problem it's just

$beginvmatrix
u_1(r) & u_2(r) \
u_1(r+1) & u_2(r+1) \
endvmatrix = beginvmatrix
frac2^rr! & u_2(r) \
frac2*2^r(r+1)! & u_2(r+1) \
endvmatrix$

Here are the steps I have taken so far in applying the given formula on $u_1(n) = frac2^nn!$

$u_2(n) = u_1(n)[sumlimits_r = 0^n - 1 fracw(r)u_1(r)u_1(r + 1)] =$

$u_2(n) = frac2^nn! left[ sumlimits_r = 0^n - 1 w(r)*(frac1(2^n/n!)*(2*2^n/(n + 1)!)) right] =$

$frac2^nn! left[ sumlimits_r = 0^n - 1 w(r)*fracn!(n + 1)!2*2^2n right] = frac2^nn!*frac(n + 1)(n!)^22*2^2n left[ sumlimits_r = 0^n - 1 w(r) right] =$

$frac2^nn!frac(n + 1)(n!)^22*2^2nleft[ sumlimits_r = 0^n - 1 w(r) right] = frac(n + 1)(n!)2^n + 1left[ sumlimits_r = 0^n - 1 beginvmatrix
frac2^rr! & u_2(r) \
frac2*2^r(r+1)! & u_2(r+1) \
endvmatrix right] =$

$frac(n + 1)n!2*2^nleft[ sumlimits_r = 0^n - 1 frac2^rr! u_2(r+1) - frac2^r+1(r+1)!u_2(r) right] = u_2(n)$

I am having trouble proceeding on from here to simplifying and solving for $u_2(n)$, any sort of help would be much appreciated. The 2nd solution $u_2(n)$ must also satisfy the given 2nd order homogeneous equation.

proof-verification homogeneous-equation summation-method

share|cite|improve this question

asked Mar 12 at 20:02

Jmath99Jmath99

877

$endgroup$

add a comment | 

0

$begingroup$

Consider the 2nd order difference equation:

$u(n + 2) - frac(n + 3)(n + 2) u(n + 1) + frac2(n + 2) u(n) = 0$

I verified that the given solution $u_1(n) = frac2^nn!$ is a solution to the above since:

$frac2^n + 2(n + 2)! - frac(n + 3)(n + 2) * frac2^n + 1(n + 1)! + frac2(n + 2) * frac2^nn! =$

$frac4*2^n(n + 2)! - frac(n + 3)2^n + 1(n+2)*(n + 1)! + frac(n + 1)(n + 1) * frac2*2^n(n + 2)n! =$

$frac4*2^n(n + 2)! - frac2*2^n(n + 3)(n + 2)! + frac2*2^n(n + 1)(n + 2)! = $

$frac2^n(4 - 2n - 6 + 2n + 2)(n + 2)! = frac2^n*0(n + 2)! = 0.$

The difficulty I am having is in the follow up part of the problem of obtaining another solution $u_2(n)$ to the 2nd order homogeneous equation. The instructor assigning this problem wanted us to use the following formula to solve for the other solution $u_2(n)$:

$u_2(n) = u_1(n)[sumlimits_r = 0^n - 1 fracw(r)u_1(r)u_1(r + 1)]$

where $w(r)$ is the casaratian defined at $r$, for this problem it's just

$beginvmatrix
u_1(r) & u_2(r) \
u_1(r+1) & u_2(r+1) \
endvmatrix = beginvmatrix
frac2^rr! & u_2(r) \
frac2*2^r(r+1)! & u_2(r+1) \
endvmatrix$

Here are the steps I have taken so far in applying the given formula on $u_1(n) = frac2^nn!$

$u_2(n) = u_1(n)[sumlimits_r = 0^n - 1 fracw(r)u_1(r)u_1(r + 1)] =$

$u_2(n) = frac2^nn! left[ sumlimits_r = 0^n - 1 w(r)*(frac1(2^n/n!)*(2*2^n/(n + 1)!)) right] =$

$frac2^nn! left[ sumlimits_r = 0^n - 1 w(r)*fracn!(n + 1)!2*2^2n right] = frac2^nn!*frac(n + 1)(n!)^22*2^2n left[ sumlimits_r = 0^n - 1 w(r) right] =$

$frac2^nn!frac(n + 1)(n!)^22*2^2nleft[ sumlimits_r = 0^n - 1 w(r) right] = frac(n + 1)(n!)2^n + 1left[ sumlimits_r = 0^n - 1 beginvmatrix
frac2^rr! & u_2(r) \
frac2*2^r(r+1)! & u_2(r+1) \
endvmatrix right] =$

$frac(n + 1)n!2*2^nleft[ sumlimits_r = 0^n - 1 frac2^rr! u_2(r+1) - frac2^r+1(r+1)!u_2(r) right] = u_2(n)$

I am having trouble proceeding on from here to simplifying and solving for $u_2(n)$, any sort of help would be much appreciated. The 2nd solution $u_2(n)$ must also satisfy the given 2nd order homogeneous equation.

proof-verification homogeneous-equation summation-method

share|cite|improve this question

asked Mar 12 at 20:02

Jmath99Jmath99

877

$endgroup$

add a comment | 

$begingroup$

Consider the 2nd order difference equation:

$u(n + 2) - frac(n + 3)(n + 2) u(n + 1) + frac2(n + 2) u(n) = 0$

I verified that the given solution $u_1(n) = frac2^nn!$ is a solution to the above since:

$frac2^n + 2(n + 2)! - frac(n + 3)(n + 2) * frac2^n + 1(n + 1)! + frac2(n + 2) * frac2^nn! =$

$frac4*2^n(n + 2)! - frac(n + 3)2^n + 1(n+2)*(n + 1)! + frac(n + 1)(n + 1) * frac2*2^n(n + 2)n! =$

$frac4*2^n(n + 2)! - frac2*2^n(n + 3)(n + 2)! + frac2*2^n(n + 1)(n + 2)! = $

$frac2^n(4 - 2n - 6 + 2n + 2)(n + 2)! = frac2^n*0(n + 2)! = 0.$

The difficulty I am having is in the follow up part of the problem of obtaining another solution $u_2(n)$ to the 2nd order homogeneous equation. The instructor assigning this problem wanted us to use the following formula to solve for the other solution $u_2(n)$:

$u_2(n) = u_1(n)[sumlimits_r = 0^n - 1 fracw(r)u_1(r)u_1(r + 1)]$

where $w(r)$ is the casaratian defined at $r$, for this problem it's just

$beginvmatrix
u_1(r) & u_2(r) \
u_1(r+1) & u_2(r+1) \
endvmatrix = beginvmatrix
frac2^rr! & u_2(r) \
frac2*2^r(r+1)! & u_2(r+1) \
endvmatrix$

Here are the steps I have taken so far in applying the given formula on $u_1(n) = frac2^nn!$

$u_2(n) = u_1(n)[sumlimits_r = 0^n - 1 fracw(r)u_1(r)u_1(r + 1)] =$

$u_2(n) = frac2^nn! left[ sumlimits_r = 0^n - 1 w(r)*(frac1(2^n/n!)*(2*2^n/(n + 1)!)) right] =$

$frac2^nn! left[ sumlimits_r = 0^n - 1 w(r)*fracn!(n + 1)!2*2^2n right] = frac2^nn!*frac(n + 1)(n!)^22*2^2n left[ sumlimits_r = 0^n - 1 w(r) right] =$

$frac2^nn!frac(n + 1)(n!)^22*2^2nleft[ sumlimits_r = 0^n - 1 w(r) right] = frac(n + 1)(n!)2^n + 1left[ sumlimits_r = 0^n - 1 beginvmatrix
frac2^rr! & u_2(r) \
frac2*2^r(r+1)! & u_2(r+1) \
endvmatrix right] =$

$frac(n + 1)n!2*2^nleft[ sumlimits_r = 0^n - 1 frac2^rr! u_2(r+1) - frac2^r+1(r+1)!u_2(r) right] = u_2(n)$

I am having trouble proceeding on from here to simplifying and solving for $u_2(n)$, any sort of help would be much appreciated. The 2nd solution $u_2(n)$ must also satisfy the given 2nd order homogeneous equation.

proof-verification homogeneous-equation summation-method

share|cite|improve this question

asked Mar 12 at 20:02

Jmath99Jmath99

877

$endgroup$

share|cite|improve this question

asked Mar 12 at 20:02

Jmath99Jmath99

877

share|cite|improve this question

asked Mar 12 at 20:02

Jmath99Jmath99

877

asked Mar 12 at 20:02

Jmath99Jmath99

877

asked Mar 12 at 20:02

Jmath99Jmath99

877