Author Topic: Bonus problem for week 5b  (Read 2996 times)

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 1332
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Bonus problem for week 5b
« on: February 07, 2013, 11:54:56 PM »
Write down an $m$-th order homogeneous linear equation with constant coefficients (with the smallest possible $m$) such that it has solutions
\begin{equation*}
y_1= e^t, \qquad y_2= te^{-t}.
\end{equation*}

Brian Bi

  • Moderator
  • Full Member
  • *****
  • Posts: 31
  • Karma: 13
    • View Profile
Re: Bonus problem for week 5b
« Reply #1 on: February 08, 2013, 01:31:32 PM »
The general solution to a homogeneous linear ODE with constant coefficients is given by
\begin{equation}
y = \sum_{i=1}^n \sum_{j=0}^{p_i - 1} A_{ij} t^j e^{r_i t} \label{eqn:general}
\end{equation}
where $r_1, r_2, ..., r_n$ are the distinct roots of the characteristic polynomial and $p_i$ is the multiplicity of the $i$th distinct root.

We see that a term of the form $Ae^t$ is given by $j = 0$ and $r_i = 1$ in ($\ref{eqn:general}$), and that a term of the form $Bte^{-t}$ is likewise given by $j = 1$ and $r_i = -1$. Since $j = 1$, the multiplicity of the root $r_i = -1$ must be at least 2. So the minimal characteristic polynomial ought to be $$(x - 1)(x + 1)^2 = x^3 + x^2 - x - 1$$Conclude that the desired ODE is $$y''' + y'' - y' - y = 0$$

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 1332
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: Bonus problem for week 5b
« Reply #2 on: February 08, 2013, 02:46:58 PM »
Yes, comparing with the "sister problem 5a" we see that constant coefficients condition changes the game.