Solve the recurrence relation $a_n-3a_{n-1}=5(7^n)$ with $a_0=2$
I solved this recurrence relation using the undetermined coefficiente method:
- First I found the solution for $a_n^{(h)}$ (homogeneous equation):
$a_n-3a_{n-1}=0$ so $a_n^{(h)}=c3^n$
- And then I found the solution for $a_n^{(p)}$ (particular equation):
$a_n^{(p)}=A(7^n)$, and then I substitute in the original equation:
$A(7^n)-3A(7^{n-1})=5(7^n)$, and I end up with $A=\frac{35}{4}$.
So $a_n^{(p)}=\frac{35}{4}(7^n)$.
However, my book's solution says $a_n^{(p)}=(\frac{5}{4})7^{n+1}$.
I know this may seem really silly, but how did they get that?
Best Answer
$$\frac{35}{4}(7^n)=\frac{5\cdot7}{4}(7^n)=\frac{5}{4}(7^{n+1})$$