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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?

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  • $\begingroup$ They are same!. $\endgroup$ – Mostafa Ayaz Mar 30 at 15:37
  • $\begingroup$ $\frac 54 7^{n+1}=\frac{5\times 7}4\,7^n$ $\endgroup$ – zwim Mar 30 at 15:38
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$$\frac{35}{4}(7^n)=\frac{5\cdot7}{4}(7^n)=\frac{5}{4}(7^{n+1})$$

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