# Solve the recurrence relation $a_n-3a_{n-1}=5(7^n)$ with $a_0=2$

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?

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

$$\frac{35}{4}(7^n)=\frac{5\cdot7}{4}(7^n)=\frac{5}{4}(7^{n+1})$$