Author Topic: Bonus problem for week 5a  (Read 2913 times)

Victor Ivrii

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Bonus problem for week 5a
« on: February 07, 2013, 11:52:36 PM »
Write down a 2nd order homogeneous linear equation such that it has solutions
\begin{equation*}
y_1= e^t, \qquad y_2= te^{-t}.
\end{equation*}
« Last Edit: February 07, 2013, 11:55:14 PM by Victor Ivrii »

Brian Bi

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Re: Bonus problem for week 5a
« Reply #1 on: February 08, 2013, 01:23:31 PM »
We can tackle this by "brute force":
\begin{align}
0 &= y_1'' + py_1' + qy_1 \\
&= (e^t)'' + p(e^t)' + q(e^t) \\
&= e^t + pe^t + qe^t \\
&= e^t(1 + p + q)
\end{align}
and dividing through by $e^t$ we obtain $1 + p + q = 0$.
Also
\begin{align}
0 &= y_2'' + py_2' + qy_2 \\
&= (te^{-t})'' + p(te^{-t})' + q(te^{-t}) \\
&= (e^{-t} - te^{-t})' + p(e^{-t} - te^{-t}) + qte^{-t} \\
&= (-e^{-t} - e^{-t} + te^{-t}) + pe^{-t} - pte^{-t} + qte^{-t} \\
&= e^{-t}((1-t)p + tq + (t-2)
\end{align}
and dividing through by $e^{-t}$ we obtain $(1-t)p + tq + (t-2) = 0$.

Solving this linear system for $p$ and $q$ we arrive at
\begin{align}
p &= -\frac{2}{2t-1} \\
q &= \frac{3-2t}{2t-1}
\end{align}

so we conclude that the desired ODE is $$(2t-1)y'' - 2y' - (2t-3)y = 0$$

Victor Ivrii

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Re: Bonus problem for week 5a
« Reply #2 on: February 08, 2013, 02:45:03 PM »
Another way is to use Wronskian
\begin{equation*}
0=\left|\begin{matrix}
y  & y_1 &y_2\\
y' & y_1' &y_2'\\
y'' & y_1''& y_2''
\end{matrix}\right|=
\left|\begin{matrix}
y  & e^t & te^{-t}\\
y' &e^t &(1-t)e^{-t}\\
y'' &  e^t& (t-2)e^{-t}
\end{matrix}\right|
\end{equation*}
which results in the same equation (up to some factor)