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

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hba
 4 years ago
Best ResponseYou've already chosen the best response.0Is it \[y^{n+1} \ or \ y_{n+1}\]

ajprincess
 4 years ago
Best ResponseYou've already chosen the best response.0I 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, \(y^(p)=c_1(4)^n\)

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

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.0@experimentX Can you give a hand here?

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1any info given on y_0 ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0do you have initial cond.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it is not hom. diff. eq, but we make hom. diff. eq..

ajprincess
 4 years ago
Best ResponseYou've already chosen the best response.0we 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
 4 years ago
Best ResponseYou've already chosen the best response.0this is the solution ..\[\Large y_{n}=a4^{n}+b1^{n}\]

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1that looks pretty nice.

ajprincess
 4 years ago
Best ResponseYou've already chosen the best response.0i dnt see that screenshot properly

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

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

ajprincess
 4 years ago
Best ResponseYou've already chosen the best response.0I dnt get the second step?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0multiply top eq. by 1 and add them

ajprincess
 4 years ago
Best ResponseYou've already chosen the best response.0I mean y do u make it y_n+24y_n+1=8

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0in order to get rid of 8

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0to make it hom. diff. eq..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but I dont know how to find a and b if initial condition is not given..

ajprincess
 4 years ago
Best ResponseYou've already chosen the best response.0Really sorry. I still dnt get the conversion part

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1yep ... 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
 4 years ago
Best ResponseYou've already chosen the best response.0How do u get y_n+25y_n+1+4y_n=0

experimentX
 4 years ago
Best 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.

experimentX
 4 years ago
Best 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.

ajprincess
 4 years ago
Best ResponseYou've already chosen the best response.0so there will not be any particular solutions right?

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1no ... unless you have initial value i.e y_0

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

ajprincess
 4 years ago
Best ResponseYou've already chosen the best response.0Ya. I checked it out. Bt got confused seeing 32 and all that:(

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

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.1no probs at all. using this method, you can solve for nth Fibonacci number.
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