The role of the Wronskian I am a physics student taking ODE this semester.
Let's consider an initial value problem. Consider a homogenous second-order linear differential equation. Suppose we have two solutions $\varphi_1(x)$ and $\varphi_2(x)$. To my understanding, we check the Wronskian to confirm whether we can span the entire solution space of the differential equation with just those two solutions. This is because if the Wronskian is zero, we know that the Wronskian matrix is non-invertible and so is inconsistent or has multiple solutions.
Intuitively, I thought we'd just need to check that $\varphi_1$ and $\varphi_2$ are linearly independent. However, checking the Wronskian seems to impose stricter requirements, i.e., that the vector $(\varphi_1 \ \varphi_2)$ is linearly independent from $(\partial_x\varphi_1 \ \partial_x\varphi_2)$ and equivalently (it seems) that $\partial_x\varphi_1$ is not a multiple of $\partial_x\varphi_2$ and that $\varphi_1$ is not a multiple of $\varphi_2$ (other than trivially when $x_1 = x_2 = 0$). Where the latter statement is what I intuitively think is all that need be checked.
Why do we need to check this stricter Wronskian requirement as opposed to the linear independence of our solutions? Does it have to do with the fact that we need to allow freedom in what the initial value of the first derivative of a solution is? And thus is not about the nature of linear independence, but more about the fact that we're dealing with an initial value problem?
 A: To show that the two functions $\varphi_1$ and $\varphi_2$ are linearly independent you make
$$
A\varphi_1(x)+B\varphi_2(x)=0
$$
and you want to prove that $A$ an $B$ are null.
But you need another condition to force $A$ and $B$ to be zero. And you get it by deriving
$$
A\varphi_1'(x)+B\varphi_2'(x)=0
$$
This is a linear homogeneous system
$$
\left\{
\begin{array}{l}
A\varphi_1(x)+B\varphi_2(x)=0\\
A\varphi_1'(x)+B\varphi_2'(x)=0\\
\end{array}
\right.
$$
which have only the solution $A=B=0$ if the determinant is non-null. But this determinant is precisely the Wroskian!!
Added
I agree with you that, for second order equations, $W\neq 0$ seems to be too strong a requirement, since I think you can prove it "by hand". But for higher order ODEs the Wroskian is more important.
A: The Wronskian requirement is in fact not any stricter than linear independence. We can show that linear dependence implies vanishing of the Wronskian. If $\varphi_1$ and $\varphi_2$ are two linearly dependent solutions, then there exist nonzero scalars $a_1$, $a_2$ such that
\begin{align}
a_1 \varphi_1 + a_2 \varphi_2 = 0
\end{align}
Differentiating, we get
\begin{align}
a_1 \partial_x\varphi_1 + a_2 \partial_x\varphi_2 = 0
\end{align}
so that
\begin{align}
a_2 \partial_x \varphi_2 = -a_1 \partial_x\varphi_1
\end{align}
Multiplying the first equation by $a_2 \partial_x \varphi_2$ gives
\begin{align}
0&= a_1 a_2 \varphi_1\partial_x \varphi_2 + a_2^2 \varphi_2 \partial_x \varphi_2\\
&= a_1 a_2 \left(\varphi_1\partial_x \varphi_2 - \varphi_2 \partial_x \varphi_1\right)\\
\rightarrow &\varphi_1\partial_x \varphi_2 - \varphi_2 \partial_x \varphi_1 = 0
\end{align}
A: Let's first define the Wronskian of $n$ functions.
Let $\phi_1,\ldots,\phi_n\in C^n(I)$, and set
$$W(\phi_1,\ldots,\phi_n)(t)=\big(\phi_i^{(j-1)}(t)\big).$$
We define the Wronskian of $\phi_1,\ldots,\phi_n$ as $w(t)=\det W(t)$.
Fact I. If $w(t)$ is not identically zero in $I$, then
$\phi_1,\ldots,\phi_n$ are linearly independent in $I$.
Fact II. If $w(t)\ne 0$, for all $t\in I$, then the functions $\phi_1,\ldots,\phi_n$ constitute a basis of the space of the solutions of the $n-$order linear ODE:
$$
\det W(\phi_1,\ldots,\phi_n,x)=0. \qquad\qquad (\star)
$$
However, if $\det W(\phi_1,\ldots,\phi_n)(t_0)=0$, is zero, for some $t_0\in I$, then $\phi_1,\ldots,\phi_n$ do not constitute a basis of $(\star)$ in $I$.
For example $\{t,t^2\}$ are linearly independent over $\mathbb R$. Their Wronskian is
$$
\left|\begin{array}{rr} t & t^2 \\ 1 & 2t\end{array}\right|=t^2 
$$
which vanishes at $t=0$. Meanwhile $\{t,t^2\}$ constitute a basis of the solution space of
$$
\left|\begin{array}{rrr} t & t^2 & x\\ 1 & 2t & x' \\ 0&2&x'' \end{array}\right|=t^2x''-2tx'+2x=0. \qquad\qquad (\star\star)
$$
However, strictly speaking, the domain of any solution of $t^2x''-2tx'+2x=0$,  CANNOT include $t=0$, since for $t=0$, the higher order coefficient vanishes.
In fact, the domain solutions of $(\star\star)$ could be either $(-\infty,0)$ or $(0,\infty)$. Not their union.
A: Suppose that $ φ_1 $ and $ φ_2 $ are two solutions of the differential equation. If
$  c_1 φ_1(t) + c_2 φ_2(t) = 0 $ then we can differentiate both sides to get $  c_1 φ'_1(t) + c_2 φ'_2(t) = 0 \ $.
$ \left\{\begin{matrix}
c_1 φ_1(t) + c_2 φ_2(t) = 0
 &  \\
c_1φ'_1(t) + c_2 φ'_2(t) = 0 
 &  \\
\end{matrix}\right.
\
$
This is a system of two equations with two unknowns. The determinant of the corresponding matrix is the Wronskian. So, if the Wronskian is nonzero at some $t_0$ only the trivial solution exists. Hence  $ φ_1 $ and $ φ_2 $ are linearly independent. In other words, it is always really handy proving linear independence (or dependence) using some kind of determinant because it provides a systematic way of checking. It wouldn't be wrong if you proved linear independence of $ φ_1 $ and $ φ_2 $ exclusively (edit: and in fact this implies that their derivatives are also independent because differentiation is a linear operator), but generally it will not always be easy.
