Toronto Math Forum
MAT2442013S => MAT244 MathLectures => Ch 4 => Topic started by: Victor Ivrii 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*}

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 $$y''' + y''  y'  y = 0$$

Yes, comparing with the "sister problem 5a" we see that constant coefficients condition changes the game.