# Difference Equation Initial Value Problem

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?

• Set $x_k=y_k^2.$ Then it becomes a second-order linear inhomogeneous difference equation with constant coefficients. Look here – saulspatz Feb 3 at 12:34
• In your equation for $k=0$, you need to set $k=0$ on both sides. Then $m=3$ for the given initial values, and per the theory of linear recursion, the basis solutions are $1$ and $3^k$, so that by undetermined coefficients the general solution has the form $y^2_k=A+B3^k+(Ck+Dk^2)$ with $A,B,C,D$ completely determined by equation and initial values. – Dr. Lutz Lehmann Feb 3 at 17:59
• Hahaha what a silly mistake! Thanks – VakiPitsi Feb 3 at 19:47
• Are you sure that $k$ in the RHS is also the subscript $k$ used in the LHS of the recursion? – Did Feb 5 at 8:08

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.

• When $m>4$, the roots of the characteristic polynomial are complex and the solution of the homogeneous equation takes a different form. More precisely, if $r = \rho e^{i \theta}$ is one of the roots, you get $x_k = \rho^k (c_1 \cos (\theta k) + c_2 \sin (\theta k))$. Although this can be recovered from the mentioned formula for $k<4$, it is more direct to have this expression. – PierreCarre Feb 3 at 14:10
• With the given 3 initial values for an order-2 recursion equation, the value of $m$ is fixed as $m=3$. – Dr. Lutz Lehmann Feb 3 at 17:07
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 $$m$$.