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

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}((1t)p + tq + (t2)
\end{align}
and dividing through by $e^{t}$ we obtain $(1t)p + tq + (t2) = 0$.
Solving this linear system for $p$ and $q$ we arrive at
\begin{align}
p &= \frac{2}{2t1} \\
q &= \frac{32t}{2t1}
\end{align}
so we conclude that the desired ODE is $$(2t1)y''  2y'  (2t3)y = 0$$

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 &(1t)e^{t}\\
y'' & e^t& (t2)e^{t}
\end{matrix}\right
\end{equation*}
which results in the same equation (up to some factor)