**Proof plan.** We compute the derivative of the Wronskian $W = y_1 y_2' - y_2 y_1'$ directly, substitute the ODE to eliminate the second [derivatives](/page/Derivative), and recognise the resulting expression as $W' = -p(x)W$. This [separable](/page/Separable) ODE is solved to give Abel's identity, from which the all-or-nothing vanishing property follows because the exponential factor is never zero. **Step 1: Differentiate the Wronskian.** For two solutions $y_1, y_2$ of the ODE, the Wronskian is $W(x) = y_1 y_2' - y_2 y_1'$. Differentiating: \begin{align*} W' = y_1 y_2'' - y_2 y_1''. \end{align*} (The terms $y_1' y_2'$ cancel.) **Step 2: Substitute the ODE.** Since $y_1$ and $y_2$ both satisfy $y'' + p(x)y' + q(x)y = 0$, we have $y_i'' = -p(x)y_i' - q(x)y_i$. Substituting: \begin{align*} W' &= y_1(-p y_2' - q y_2) - y_2(-p y_1' - q y_1) \\ &= -p(y_1 y_2' - y_2 y_1') - q(y_1 y_2 - y_2 y_1) \\ &= -p(x) W. \end{align*} **Step 3: Solve the ODE for $W$.** The equation $W' = -p(x)W$ is separable. Separating and integrating: \begin{align*} \frac{dW}{W} = -p(x) \, dx \quad \implies \quad W(x) = W(x_0) \exp\left(-\int_{x_0}^{x} p(u) \, du\right). \end{align*} **Step 4: Deduce the all-or-nothing property.** The exponential factor $\exp\bigl(-\int_{x_0}^x p(u) \, du\bigr)$ is strictly positive for all $x \in I$. Therefore $W(x) = 0$ for some $x \in I$ if and only if $W(x_0) = 0$, which holds if and only if $W(x) = 0$ for all $x \in I$.