I need to do this exercise but i can't find the solution, i have tryed to substitute $\phi_1(t)$, $\phi_2(t)$ with their solution of the differential equation when the delta is positive but i was wondering if there is a method for solve this without sostitution and for general case.
we have $y''+by'+cy=0$ a differential linear equation with $b, c \in C(I)$, $ I\subset R $, $I$ interval
$W(t)=det\begin{bmatrix}\phi_1(t) & \phi_2(t)\\\phi_1'(t) & \phi_2'(t)\end{bmatrix}$ where $\phi_1(t)$, $\phi_2(t)$ $\in V_2$ ($\phi_1(t)$, $\phi_2(t)$) are solutions for $y''+by'+cy=0$)
Proof that W satisfy this differential equation:
$W'=-bW$
Proof that {$\phi_1(t)$, $\phi_2(t)$} is a base for $V_2$ if and only if exist $t\in I$ such that $W(t)\ne 0$
Starting from: $$W(t)=\begin{vmatrix}\phi_1(t) & \phi_2(t)\\\phi_1'(t) & \phi_2'(t)\end{vmatrix}$$ Differentiate the determinant: $$W'(t)=\begin{vmatrix}\phi_1'(t) & \phi_2'(t)\\\phi_1'(t) & \phi_2'(t)\end{vmatrix}+\begin{vmatrix}\phi_1(t) & \phi_2(t)\\\phi_1''(t) & \phi_2''(t)\end{vmatrix}$$ Since the first determinant is zero, we have: $$ \implies W'(t)=\begin{vmatrix}\phi_1(t) & \phi_2(t)\\\phi_1''(t) & \phi_2''(t)\end{vmatrix}$$ Now check that you have: $$\begin{vmatrix}\phi_1(t) & \phi_2(t)\\\phi_1''(t) & \phi_2''(t)\end{vmatrix}=-b\begin{vmatrix}\phi_1(t) & \phi_2(t)\\\phi_1'(t) & \phi_2'(t)\end{vmatrix}$$ $$\phi_2''(t)\phi_1(t)+b\phi_2'(t)\phi_1(t)=\phi_1''(t)\phi_2(t)+b\phi_1'(t)\phi_2(t)$$ $$\phi_1(t)(\phi_2''(t)+b\phi_2'(t))=\phi_2(t)(\phi_1''(t)+b\phi_1'(t))$$ Using the DE: $$ \begin{cases} \phi_2''(t)+b\phi_2'(t)+c\phi_2(t)=0 \\ \phi_1''(t)+b\phi_1'(t)+c\phi_1(t)=0 \end{cases} $$ It gives us: $$\phi_1(t)(-c\phi_2(t))=\phi_2(t)(-c\phi_1)$$ Which is true.