# How do I solve this second-order homogeneous differential equation with piecewise constant coefficients?

I am trying to solve the following differential equation (for fun, mind you):

$$\lfloor x \rfloor y'' + 2\lfloor x \rfloor y' + \lfloor x \rfloor y = 0$$

Because of the piece wise coefficients, my gut instinct is to first solve the associated implied differential equation.

This means that I can now (legally) perform the substitution:

$$\lfloor x \rfloor = a$$

$$ay'' + 2ay' + ay = 0$$

The associated auxiliary equation is:

$$ar^2 + 2ar + a = 0$$

$$r_1 = \frac {-2a + \sqrt{4a^2 - 4a^2}}{2a}$$

$$r_2 = \frac {-2a - \sqrt{4a^2 - 4a^2}}{2a}$$

It would appear as if I have a double root r = -1. This serves as a problem though. If I write the solution to the implied differential equation as follows:

$$y(x) = C_1(x)e^{-x} + C_2(x)xe^{-x}$$

At this point, I would usually seek out the two solutions $C_1$ and $C_2$ to the system of functional equations where y(x) and y'(x) are continuous (referring to the implied derivative). However, in this situation, either floor(x) simply drops off the map, so to speak, and has no role in the solution or I am misreading the final solution because it would appear as if y(x) is continuous with C1 and C2 regular constants. Did something I do in treating floor like a constant prevent me from solving the equation or does this equation just have a "nice solution"?

I have no idea what this implied differential equation is which you talk about, but it seems you are making things way too difficult for yourself.

Let me first of all note that $\lfloor x \rfloor$ appears in every term in your differential equation, so when $x \notin [0,1)$ you may simply divide by $\lfloor x \rfloor$ and obtain the differential equation $$y'' + 2y' + y = 0$$ which has solution $$y(x) = C_1e^{-x}+C_2xe^{-x}$$ where $C_1,C_2$ are just constants.

Next, we have to solve the differential equation when $x \in [0,1)$. In this case $\lfloor x \rfloor = 0$, so your differential equation reduces to $0 = 0$. In other words, every function $y(x)$ is a solution here - with the restriction that $y'$ and $y''$ exist, of course.

So, we would like to say that $$y(x) = \left\{\begin{array}{ll} f(x) = C_1e^{-x} + C_2xe^{-x} & \text{ if } x \notin [0,1)\\ g(x) & \text{ if } x \in [0,1) \end{array}\right.$$ where $g(x)$ is any twice differentiable function (else $y''$ does not exist and the equation makes no sense), is the general solution to your differential equation.

However, you'll need to make sure $y'(0)$, $y'(1)$, $y''(0)$ and $y''(1)$ exist, which will again place some conditions on $g(x)$.

First, you'll need continuity of $y$, meaning that $$\lim_{x \to 1\\x < 1}g(x) = f(1) = \frac{C_1+C_2}{e},$$ and $$g(0) = \lim_{x \to 0\\x < 0}f(x) = C_1 .$$ And similar limits for the first and second derivatives of $f$ and $g$.

• An implied differential equation is a differential equation consisting of implied derivatives and implied integrals (which are described in that paper) rather than derivatives and integrals. (A pretty strong opinion I have is that the continuous solutions to those are the same as a regular differential equations solution of the exact same construction) I hope that makes sense! :-) – The Great Duck Jan 12 '17 at 22:35