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Please help:)
Solve the difference equation
\(\Large y_{n+1}4y_{n}=8\)
 one year ago
 one year ago
Please help:) Solve the difference equation \(\Large y_{n+1}4y_{n}=8\)
 one year ago
 one year ago

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hbaBest ResponseYou've already chosen the best response.0
Is it \[y^{n+1} \ or \ y_{n+1}\]
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
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 \(A\zeta^{n+1}4A\zeta^n=0\) \(\zeta4=0\) whose root is \(\zeta=4\) Therefore, \(y^(p)=c_1(4)^n\)
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
*therefore, \(y^{(p)}=c_1(4)^n\)
 one year ago

CallistoBest ResponseYou've already chosen the best response.0
@experimentX Can you give a hand here?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
any info given on y_0 ?
 one year ago

cinarBest ResponseYou've already chosen the best response.1
do you have initial cond.
 one year ago

cinarBest ResponseYou've already chosen the best response.1
it is not hom. diff. eq, but we make hom. diff. eq..
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
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
 one year ago

cinarBest ResponseYou've already chosen the best response.1
this is the solution ..\[\Large y_{n}=a4^{n}+b1^{n}\]
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
that looks pretty nice.
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
i dnt see that screenshot properly
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
http://www.wolframalpha.com/input/?i=y_%28n%2B1%29++4+y_%28n%29+%3D+8 Wolf says only one variable.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
\[ y_{n+2} = 5 y_{n+1} 4y_n \\ y_{n+1} = y_{n+1} \\ \] \[u_{n+1} = \begin{bmatrix} 5 & 4\\ 1 & 0 \end{bmatrix} u_n\]
 one year ago

cinarBest ResponseYou've already chosen the best response.1
http://imageshack.us/photo/myimages/836/fsedfe.png/
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
I dnt get the second step?
 one year ago

cinarBest ResponseYou've already chosen the best response.1
multiply top eq. by 1 and add them
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
I mean y do u make it y_n+24y_n+1=8
 one year ago

cinarBest ResponseYou've already chosen the best response.1
in order to get rid of 8
 one year ago

cinarBest ResponseYou've already chosen the best response.1
to make it hom. diff. eq..
 one year ago

cinarBest ResponseYou've already chosen the best response.1
but I dont know how to find a and b if initial condition is not given..
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
Really sorry. I still dnt get the conversion part
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
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.
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
How do u get y_n+25y_n+1+4y_n=0
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
\[ y_0 = c_1 + c_2 \\ y_1 = 8 + 4y_0\\ y_1 = 4c_1 +c_2 \\ \]  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.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
\[ u_{n+1} = \begin{bmatrix} 5 & 4\\ 1 & 0 \end{bmatrix} u_n \] 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.
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
so there will not be any particular solutions right?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
no ... unless you have initial value i.e y_0
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
did you check the final answer? http://www.wolframalpha.com/input/?i=y_%28n%2B1%29++4+y_%28n%29+%3D+8
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
Ya. I checked it out. Bt got confused seeing 32 and all that:(
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
just put c_1 = k + 8/3 and c_2 = 8/3 .. rest is just manipulation of numbers by WA
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
oh k. Thanxx a lot.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
no probs at all. using this method, you can solve for nth Fibonacci number.
 one year ago
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