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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?

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  • $\begingroup$ Set $x_k=y_k^2.$ Then it becomes a second-order linear inhomogeneous difference equation with constant coefficients. Look here $\endgroup$ – saulspatz Feb 3 at 12:34
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    $\begingroup$ 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. $\endgroup$ – Dr. Lutz Lehmann Feb 3 at 17:59
  • $\begingroup$ Hahaha what a silly mistake! Thanks $\endgroup$ – VakiPitsi Feb 3 at 19:47
  • $\begingroup$ Are you sure that $k$ in the RHS is also the subscript $k$ used in the LHS of the recursion? $\endgroup$ – Did Feb 5 at 8:08
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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.

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  • $\begingroup$ Thank you for your answer! Very helpfull $\endgroup$ – VakiPitsi Feb 3 at 13:31
  • $\begingroup$ 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. $\endgroup$ – PierreCarre Feb 3 at 14:10
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    $\begingroup$ With the given 3 initial values for an order-2 recursion equation, the value of $m$ is fixed as $m=3$. $\endgroup$ – Dr. Lutz Lehmann Feb 3 at 17:07
  • $\begingroup$ @LutzL Good point. I didn't even think about that, though I admit I wondered why they gave three initial values. I feel so stupid. $\endgroup$ – saulspatz Feb 3 at 17:14
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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 $m$.

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