Question:
Verify that the functions y1 and y2 are solutions of the given differential equation. Do they constitute a fundamental set of solutions?
$$ y" -2y'+y=0, y_1(t) = e^t, y_2(t) = te^t$$
Solution: $$
y_1'(t) = e^t$$
$$y_1"(t) = e^t$$
Substitude:\begin{equation}\begin{split} y" -2y'+y &= e^t-2(e^t)+e^t\\
&=2e^t-2e^t\\
&= 0
\end{split}\end{equation}
Thus y1 is a solution.
$$
y_2'(t) = e^t+te^t$$
$$y_1"(t) = 2e^t+te^t$$
Substitude:\begin{split} y" -2y'+y &= 2e^t+te^t-2(e^t+te^t)+te^t\\
&=2e^t+te^t-2e^t-2te^t+te^t\\
&= 0
\end{split}
Thus y2is a solution.
\begin{equation}
\begin{split}
W &= y_1y_2'-y_2y_1'\\
&=e^t(e^t+te^t)-e^t(te^t)\\
&=e^{2t}+te^{2t}-te^{2t}\\
&=e^{2t}
\end{split}\end{equation}
Since W is not 0 for every value of t, y1 and y2 form a fundamental set of solution.
