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:
Differentiate $y_1(t)=e^t, y_2(t)=te^t$ respect to t,
$$y_1'=e^t, y_2'=e^t+te^t$$
$$y_1''=e^t,y_2=2e^t+te^t$$
Substitute back to the differential equation,
$$e^t-2e^t+e^t=0$$
$$2e^t+te^t-2(e^t+te^t)+te^t=0$$
So both $y_1(t)$ and $y_2(t)$ are two valid solutions to the equation.
To check whether it is a fundamental set of solution:
We want $W(y_1,y_2)(t) \neq 0$
$$W=\begin{bmatrix}e^t&te^t\\e^t&e^t+te^t\end{bmatrix}$$
$$e^t*(e^t+te^t)-te^t*e^t=e^{2t}$$
The determinant is not zero, so it is a fundamental set of solutions.
