Difference Equation Initial Value Problem

Question

Solve the IVP:

$$y^2_{k+2}-4y^2_{k+1}+m\cdot y^2_k=k, m \in \mathbb{R},$$ $y_0=1, y_1=2, y_2=\sqrt{13}$

I started by taking $k=0$

$y^2_2-4y^2_1+my^2_0=k$ $\Rightarrow13-16+m=k$ $\Rightarrow m=k+3$

But I don't know how to proceed any help?

Answer 1

I realized that the source I cited in my comment doesn't have an example quite like this one, so I'll sketch the procedure, leaving the details for you.

First, as I said, set $x_k=y_k^2$ giving $$x_{k+2}-4x_{k+1}+mx_k=k\tag{1}$$ with initial data $$\begin{align} x_0&=1\\ x_1&=4\\ x_2&=13\end{align}$$ The general solution of $(1)$ is the general solution of the associated homogeneous equation $$x_{k+2}-4x_{k+1}+mx_k=0\tag{2}$$ plus any particular solution of $(1).$

The general solution of $(2)$ is $$x_k=c_1r_1^k+c_2r_2^k$$ where $r_1$ and $r_2$ are the roots of the characteristic polynomial $$x^2-4x+m$$ If there is a double root, which will happen if $m=4,$ then the general solution is a bit different. When $m=4,\ 2$ is a double root of the characteristic polynomial, and the general solution of $(2)$ becomes $$x_k=c_12^k+c_2k2^k$$

Now we have to guess a particular solution to $(1)$. Since the right-hand side is a linear polynomial in $k$ we guess that a solution in the form of a linear polynomial exists $$x_k=ak+b$$ and we substitute this guess into $(1)$ to solve for $a$ and $b$. When I did did, I found that it worked except in the case $m=3.$ If $m=3$ you should guess that there is a quadratic solution. I didn't actually do this, but it should work.

Now you have three cases, $m=3,$ $m=4,$ and $m\neq 3,4,$ and in each case you have a slightly different formula for the general solution of $(1).$ In each case, you can substitute the initial data to find the undetermined coefficients.

EDIT

As LutzL has pointed out, setting $k=0$ and substituting the initial values in $(1)$ gives $m=3,$ so there is really only one case to consider.

Answer 2

Substituting $w_k = y_k^2$ you get the linear difference equation $$ w_{k+2}-4 w_{k+1}+m w_k = m. $$

The general solution of this linear equation can be written as $w_K = x_k + x_k^*$, where $x_k$ is the general solution of the homogeneous equation (set rhs to zero) and $x_k^*$ is a particular solution of the difference equation.

The characteristic polynomial is $p(\lambda) = \lambda^2-4 \lambda + m$, whose roots are $2 \pm \sqrt{4-m}$, and so depending on whether $m<4$, $m>4$ or $m=4$ you have the expression for $x_k$.

The particular solution $x_k^*$ can be obtained assuming that it is similar to the rhs and plugging into the equation. The choice of particular solution may also depend on $m$.