Inhomogeneous Difference Equation $p_{n}+\frac{1}{4}p_{n-1}=\frac{3}{4}$ with $p_0=0, p_1=3/4$

Question

I'm trying to solve the above equation, and this is where I've got to so far:

Consider homogeneous problem $p_{n}+\frac{1}{4}p_{n-1}=0$. We then have characteristic equation $\omega^{n-1}(\omega + 1/4)=0$ and hence $\omega=-1/4$. So we have $p_{n}=A(-\frac{1}{4})^n$ for some constant $A$.

At this point I'm completely stuck on where to go, I don't have much experience with difference equations. I believe we find a particular solution but not sure what form this will take?

Answer 1

The general solution to the inhomogeneous equation is the general solution to the homogenous equation, $A(-1/4)^n$, plus any solution solution to the inhomogeneous equation. So, we have to guess a solution to the inhomogeneous equation. Since the right-hand side is a constant, we test whether there is a constant solution.$$C+\frac14C=\frac34\implies C=\frac35$$ so the general solution is $$p_n=A\left(-\frac14\right)^n+\frac35$$ Now we just have to find $A$ such that $p_0=0$ which is clearly $A=-\frac35$. So, $$p_n=-\frac35\left(-\frac14\right)^n+\frac35$$

Answer 2

Note that if another sequence $q_n$ is equal to $p_n$ shifted by a constant $k$, then $$q_n+k+\frac{1}{4}\left(q_{n-1}+k\right)=\frac34$$ Equating the constant terms: $$k+\frac14k=\frac34$$ from which you find that $k=\frac35$. And now you have the homogeneous relation $$q_n+\frac14q_{n-1}=0$$ Solve this, then remember that $p_n=q_n+k$.