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How do I differentiate $y(t)=-[t]+\frac{1}{2}\cdot(1-3^{t-[t]})$, $t \ge 0$, ($[x]$ is the integer part of $x$) in order to verify that it is the solution of the ODE $y' = \log(3) \cdot (y-[y]-\frac{3}{2})$, $y(0)=0$?

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up vote 5 down vote accepted

The same way you differentiate any other expression.

$$\begin{align} \frac{dy(t)}{dt} &= -\frac{d[t]}{dt} + \frac12\frac{d}{dt}\left(1 - 3^{t-[t]}\right) \\ &= -\frac{d[t]}{dt} + \frac12 \left(0 - 3^{t-[t]} \log 3 \cdot \frac{d}{dt}(t-[t])\right) \\ &= -\frac{d[t]}{dt} - \frac12 3^{t-[t]}\log 3\cdot\left(1 - \frac{d[t]}{dt}\right). \end{align}$$

When $t$ is an integer, $[t]$ has a discontinuity and its derivative is undefined. So, for the differentiation to make sense, we can only consider the case when $t$ is not an integer. Then, $[t]$ is locally constant, so $d[t]/dt = 0$, and we have $$\begin{align} \frac{dy(t)}{dt} &= -0 - \frac12 3^{t-[t]}\log 3\cdot\left(1 - 0\right) \\ &= -\frac123^{t-[t]}\log 3. \end{align}$$

Edit: Okay, let's see if we can plug this into the ODE and verify the solution. We have $$y(t) = -[t] + \frac12\left(1 - 3^{t-[t]}\right),$$ and we want to know what $\log3\cdot\left(y - [y] - \frac32\right)$ is, so we should figure out something about $y - [y]$. Knowing that $t - [t]$ lies between $0$ and $1$, it is straightforward to find that $\frac12\left(1 - 3^{t-[t]}\right)$ lies between $-1$ and $0$. So $y(t)$ is a little lower than $-[t]$, but not so low that it passes $-[t]-1$. Since $-[t]$ is an integer, that tells you exactly what $[y(t)]$ is. Plug that in and I expect you should arrive at the solution pretty quickly.

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The function $y(t)$ is actually continuous everywhere. It's precisely this that makes it a valid solution to the ODE. (The derivative is not continuous at integer arguments, but we already knew this from the RHS of the ODE.) – TonyK Jan 26 '12 at 13:21
Thank you Rahul and TonyK. But how would I simplify y-[y] ? I still can't see y, as given, is the solution of the ODE. This: [x+y]=[x]+[y],however, is wrong. – yuanwei Jan 26 '12 at 13:51
@yuanwei, please see my edit. – Rahul Jan 26 '12 at 14:40

Let's denote $[t]$ as : $[t]=trunc(t)$

According to Maple solution exists only for real numbers that are not non-zero integers and it is given by following expression :

$y'(t)=-trunc(1,t)-\frac{1}{2}\cdot 3^{t-trunc(t)}\cdot (1-trunc(1,t))\cdot \ln 3$

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