Solve Inhomogeneous Linear Diff Eq: Zizi’s Question @ Yahoo Answers

[MarkFL]

Here is the question:

Recurrence relation math/question?

Hi

Please check my attempt at solving the problem:

http://oi39.tinypic.com/2cfwfm1.jpg

Would appreciate any help with the questions I have written in red, thanks

I have posted a link there to this thread so the OP can view my work.

 

[MarkFL]

Hello zizi,

We are given the linear difference equation:

\(\displaystyle u_{n}=6u_{n-1}-25u_{n-2}+\frac{32}{9}3^n\) where \(\displaystyle u_0=6,\,u_1=46\)

and we are told to define \(\displaystyle A=\sin^{-1}\left(\frac{4}{5} \right)\) and we are allowed to express the solution in terms of $A$.

We first want to solve the associated homogeneous equation:

\(\displaystyle u_{n}-6u_{n-1}+25u_{n-2}=0\)

The characteristic equation is:

\(\displaystyle r^2-6r+25=0\)

Application of the quadratic formula yields the characteristic roots:

\(\displaystyle r=3\pm4i\)

Expressing the roots in polar form, we find:

\(\displaystyle r=5e^{\pm Ai}\)

And so the homogenous solution is given by:

\(\displaystyle h_n=c_15^n\cos(nA)+c_25^n\sin(nA)\)

Now, we may assume there is a particular solution of the form:

\(\displaystyle p_n=c_33^n\)

We may substitute this solution into the difference equation to determine the value of the parameter $c_3$:

\(\displaystyle c_33^n-6c_33^{n-1}+25c_33^{n-2}=\frac{32}{9}3^n\)

Multiply through by 9 and factor the left side:

\(\displaystyle c_3\left(9\cdot3^n-18\cdot3^{n}+25\cdot3^{n} \right)=32\cdot3^n\)

Divide through by $3^n$:

\(\displaystyle c_3\left(9-18+25 \right)=32\)

\(\displaystyle 16c_3=32\)

\(\displaystyle c_3=2\)

And so we find the particular solution is:

\(\displaystyle p_n=2\cdot3^n\)

Thus, by superposition, we find the general solution to the difference equation is given by:

\(\displaystyle u_n=h_n+p_n=c_15^n\cos(nA)+c_25^n\sin(nA)+2\cdot3^n=5^n\left(c_1\cos(nA)+c_2\sin(nA) \right)+2\cdot3^n\)

Now we may use the initial values to determine the values of the two parameters.

\(\displaystyle u_0=5^0\left(c_1\cos(0\cdot A)+c_2\sin(0\cdot A) \right)+2\cdot3^0=c_1+2=6\implies c_1=4\)

\(\displaystyle u_1=5^1\left(c_1\cos(A)+c_2\sin(A) \right)+2\cdot3^1=3c_1+4c_2+6=46\implies c_2=7\)

Hence, the solution satisfying all given conditions is:

\(\displaystyle u_n=5^n\left(4\cos(nA)+7\sin(nA) \right)+2\cdot3^n\)