(instead of the dot above, i used ' and ", am I correct in thinking that these are equivalent?)
Consider the system of equations,
$$x'_1=x_2$$ $$x'_2=-q(t)x_1-p(t)x_2$$
where $q(t)$ and $p(t)$ are continuous functions on all of the real numbers. Find an expression for the Wronskian of a fundamental set of solutions.
I know what a wronskian is, $W(t)=detM(t)$ but I guess I am confused about how to find the fundamental set of solutions.
I was looking at a similar example in the book but I couldn't quite follow it either:
The Euler differential equation with nonconstant coefficients $t^2y"-2ty'+2y=0$ has 2 solutions $y^1(t)=t$ and $y^2(t)=t^2$. (now i realize this might be obvious, but is the superscript number just symbolizing "first" solution and "second" solution, not actually squaring?) This second order differential equation corresponds to the first-order system of differential equations $x'_1=x_2$ and $x'_2=\frac{2}{t^2}x_1+\frac{2}{t}x_2$ (so I don't understand how they got these equations?)
The preceding scalar solutions correspond to the vector solutions $x^1(t)=(t,1)^T$ and $x^2(t)=(t^2,2t)^T$ (what is this T?? and how did they get these answers?)
Yes, there are two solutions, $y_1(t)$ and $y_2(t)$.
$T$ = Transpose
To write the DEQ as a system of first order equations, we let:
$x_1 =y $, so:
$$x_1' = y' = x_2$$
$$x_2' = y'' = \dfrac{2 y'}{t} - \dfrac{2 y }{t^2} = - \dfrac{2 x_1}{t^2} + \dfrac{2 x_2}{t} $$
To solve the Euler-Cauchy equation, we let:
$$y = t^m \implies y' = m t^{m-1} \implies y'' = m(m-1)t^{m-2}$$
Substituting this into the DEQ, yields:
$$(m^2-3m+2)t^m = 0 \implies m^2 - 3m + 2 = 0 \implies m_1 = 1, m_2 = 2$$
Thus, we have two solutions:
$$y = y_1(t) + y_2(t) = c_1 t + c_2 t^2$$