Are $e^t$ and $\sin(t)$ dependent since the Wronskiian vanishes for certain t-inputs? I calculated the Wronskiian of $e^t$ and $\sin(t)$, and got $e^t\cos(t)-e^t\sin(t)$. This would be zero at $t = \sqrt{2}/2$. 
I know the Wronskiian has to be non-zero for the solutions to be independent; however, I still think the two solutions are independent in this case. 
Can someone please explain why or why I'm wrong?
The differential equation is $y'''' - y = 0$.
Thank you in advance! 
 A: If a set of functions are linearly dependent, the Wronskian will be zero everywhere. That is, for all $t$, we will have
$e^t\cos(t)-e^t\sin(t)=0.$
That this is not the case at $t=0$ tells us the functions are independent. Moreover, the Wronskian being zero everywhere is not a sufficient condition for functions to be linearly dependent - Wikipedia gives $x^2$ and $x|x|$ as an example where the Wronskian is identically zero, but which are linearly independent.
We can also observe more directly that $e^t$ and $\sin(t)$ are linearly independent: If that were so, we could find some $c$ such that for all $t$ we had:
$$e^t=c\sin(t)$$
which is absurd since the right hand is bounded by $c$ and the left hand is unbounded.
A: A set of functions is linearly dependent on an interval $I$ if the Wronskian is $\textit {identically}$ zero there.  
In your case, if you pass to the characteristic equation $r^4-1=(r^2+1)(r-1)(r+1)$, you can read off the fundamental set: $\left \{ \cos t,\sin t, e^{t},e^{-t} \right \}$. Now a calculation shows that the Wronskian is not identically zero on any interval.  
