ajprincess
  • ajprincess
Please help:) Solve the difference equation \(\Large y_{n+1}-4y_{n}=8\)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
hba
  • hba
Is it \[y^{n+1} \ or \ y_{n+1}\]
ajprincess
  • ajprincess
The second one
ajprincess
  • ajprincess
I found homogeneous solution in this way \(y_{n+1}-4y_n=0\) Substituting \(y_n=A\zeta^n\) where \(A\ne0\), a constant we get the characteristic equation as \(A\zeta^{n+1}-4A\zeta^n=0\) \(\zeta-4=0\) whose root is \(\zeta=4\) Therefore, \(y^(p)=c_1(4)^n\)

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ajprincess
  • ajprincess
@Callisto
ajprincess
  • ajprincess
*therefore, \(y^{(p)}=c_1(4)^n\)
Callisto
  • Callisto
@experimentX Can you give a hand here?
experimentX
  • experimentX
any info given on y_0 ?
anonymous
  • anonymous
yn=a4^n+b1^n
anonymous
  • anonymous
do you have initial cond.
ajprincess
  • ajprincess
no it is not given
anonymous
  • anonymous
it is done then..
anonymous
  • anonymous
it is not hom. diff. eq, but we make hom. diff. eq..
ajprincess
  • ajprincess
we need to find the particular solution too. right? I assumed it to be of the form an+b. then found a=0 and b=-8/3
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
is it clear..
anonymous
  • anonymous
this is the solution ..\[\Large y_{n}=a4^{n}+b1^{n}\]
experimentX
  • experimentX
that looks pretty nice.
ajprincess
  • ajprincess
i dnt see that screenshot properly
experimentX
  • experimentX
http://www.wolframalpha.com/input/?i=y_%28n%2B1%29+-+4+y_%28n%29+%3D+8 Wolf says only one variable.
experimentX
  • experimentX
\[ y_{n+2} = 5 y_{n+1} -4y_n \\ y_{n+1} = y_{n+1} \\ \] \[u_{n+1} = \begin{bmatrix} 5 & -4\\ 1 & 0 \end{bmatrix} u_n\]
anonymous
  • anonymous
http://imageshack.us/photo/my-images/836/fsedfe.png/
ajprincess
  • ajprincess
I dnt get the second step?
anonymous
  • anonymous
multiply top eq. by -1 and add them
ajprincess
  • ajprincess
I mean y do u make it y_n+2-4y_n+1=8
anonymous
  • anonymous
in order to get rid of 8
anonymous
  • anonymous
to make it hom. diff. eq..
anonymous
  • anonymous
but I dont know how to find a and b if initial condition is not given..
ajprincess
  • ajprincess
Really sorry. I still dnt get the conversion part
experimentX
  • experimentX
yep ... the answer is \[ y_n = c_1 4^n + c_2 1^n\] I think we should be able to eliminate one constant. since the initial condition is not given, there will be one constant.
ajprincess
  • ajprincess
How do u get y_n+2-5y_n+1+4y_n=0
experimentX
  • experimentX
\[ y_0 = c_1 + c_2 \\ y_1 = 8 + 4y_0\\ y_1 = 4c_1 +c_2 \\ \] --------------------- solving this you get, http://www.wolframalpha.com/input/?i=a+%3D+x%2By+%2C+b+%3D+8+%2B+4a%2C+b+%3D+4x+%2By+ where x=c1 and y=c2 and a =y_0 and b=y1 you get the same answer as Wolf. there is one constant since you do not have initial condition.
anonymous
  • anonymous
1 Attachment
experimentX
  • experimentX
\[ u_{n+1} = \begin{bmatrix} 5 & -4\\ 1 & 0 \end{bmatrix} u_n \] where \[ u_n = \begin{bmatrix}{ y_{n+1}\\ y_n }\end{bmatrix}\] the solution will be \( u_n = c_1 \lambda _1^n x_1 + c_2 \lambda_2^2 x_2 \) where \( \lambda \) and x are the eigen value and eigen vector of given matrix.
ajprincess
  • ajprincess
so there will not be any particular solutions right?
experimentX
  • experimentX
no ... unless you have initial value i.e y_0
experimentX
  • experimentX
did you check the final answer? http://www.wolframalpha.com/input/?i=y_%28n%2B1%29+-+4+y_%28n%29+%3D+8
ajprincess
  • ajprincess
Ya. I checked it out. Bt got confused seeing 32 and all that:(
experimentX
  • experimentX
just put c_1 = k + 8/3 and c_2 = -8/3 .. rest is just manipulation of numbers by WA
ajprincess
  • ajprincess
oh k. Thanxx a lot.
experimentX
  • experimentX
no probs at all. using this method, you can solve for n-th Fibonacci number.

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