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zizi's question at Yahoo! Answers regarding an inhomogeneous linear difference equation
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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:
\(\displaystyle u_n=5^n\left(4\cos(nA)+7\sin(nA) \right)+2\cdot3^n\)
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:
\(\displaystyle u_n=5^n\left(4\cos(nA)+7\sin(nA) \right)+2\cdot3^n\)