I have a question on my homework in Differential Equations and Linear Algebra that I'm not quite sure on. Here is the background given to describe the question and the one following it:
"[Here] we explore a relationship between determinants and solutions to a differential equation. The $3\times 3$ matrix consisting of solutions to a differential equation and their derivatives is called the Wronskian and, as we will see in later chapters, plays a pivotal role in the theory of differential equations."
This is the question following the description:
"Verify that $y_1(x)=\cos(2x), y_2(x)=\sin(2x), y_3(x)=e^x$ are solutions to the differential equation: $y'''-y''+4y'-4y=0$, and show that $\{\{y_1, y_2, y_3\},\{y_1', y_2', y_3'\},\{y_1'', y_2'', y_3''\}\}$ is nonzero on any interval."
Now I'm really just looking for an explanation on how I would show that it is nonzero on any interval. I completed the first part, I believe, by just taking up to the third derivative for each and plugging each set in to verify an identity of 0=0 at the end. I would then put the terms into the matrix which would give me:
$$\{\{\cos(2x), \sin(2x), e^x\}, \{-2\sin(2x), 2\cos(2x), e^x\}, \{-4\cos(2x), -4\sin(2x), e^x\}\}$$
So I've gotten this far and if anyone could point me in the right direction on where to go from here it would be greatly appreciated. Thanks guys.
Calculating the Wronskian involves taking the determinant of the matrix you've calculated. The typical formula given looks like $$ \left| \begin{array}{ccc} a&b&c\\d&e&f\\g&h&i \end{array}\right| = a\left| \begin{array}{cc} e&f\\h&i \end{array}\right| - b\left| \begin{array}{cc} d&f\\g&i \end{array}\right| + c\left| \begin{array}{cc} d&e\\g&h \end{array}\right| $$ $$ = aei - afh - bdi + bfg + cdh - ceg.$$ When you calculate this for your matrix you'll get some function in $x$ (since each entry is a function in $x$). This function is the Wronskian, and you simply examine it to make sure it is never zero.