Let $\alpha < \beta$ be consecutive zeros of $y_2$, and suppose $y_2 > 0$ on $(\alpha, \beta)$. Assume for contradiction that $y_1 > 0$ on $(\alpha, \beta)$ (no zeros).

Consider the Wronskian-type expression: $W(x) = p(x)[y_1(x)y_2'(x) - y_1'(x)y_2(x)]$.

Differentiate: $W'(x) = p[y_1 y_2'' - y_1'' y_2] + p'[y_1 y_2' - y_1' y_2]$.

From the differential equations: $(py_1')' = q_1 y_1$ and $(py_2')' = q_2 y_2$, so $py_i'' + p'y_i' = q_i y_i$.

Therefore: $W'(x) = y_1[(py_2')' - p'y_2'] - y_2[(py_1')' - p'y_1'] + p'(y_1y_2' - y_1'y_2)$

$= y_1(q_2 y_2) - y_2(q_1 y_1) = (q_2 - q_1)y_1 y_2 \geq 0$ on $(\alpha, \beta)$.

Integrating: $W(\beta) - W(\alpha) = \int_\alpha^\beta (q_2 - q_1)y_1 y_2\, dx \geq 0$.

At $x = \alpha$: $y_2(\alpha) = 0$, so $W(\alpha) = p(\alpha)y_1(\alpha)y_2'(\alpha)$. Since $y_2 > 0$ on $(\alpha, \beta)$ and $y_2(\alpha) = 0$: $y_2'(\alpha) \geq 0$. With $y_1(\alpha) > 0$: $W(\alpha) \geq 0$.

At $x = \beta$: $y_2(\beta) = 0$, so $W(\beta) = p(\beta)y_1(\beta)y_2'(\beta)$. Since $y_2 > 0$ on $(\alpha,\beta)$ and $y_2(\beta) = 0$: $y_2'(\beta) \leq 0$. With $y_1(\beta) > 0$: $W(\beta) \leq 0$.

Thus $W(\beta) - W(\alpha) \leq 0$, but we showed $\geq 0$. Equality holds only if $q_1 = q_2$ a.e. and $W(\alpha) = W(\beta) = 0$. If $q_1 < q_2$ somewhere, strict inequality gives a contradiction. The conclusion: $y_1$ must have a zero in $(\alpha, \beta)$.

Oscillation theorem (sketch). Apply the comparison theorem with $q_1 = q - \lambda_n w$ and $q_2 = q - \lambda_{n-1}w$. Since $\lambda_n > \lambda_{n-1}$: $q_1 < q_2$. The eigenfunction $\phi_{n-1}$ has $n-2$ zeros; the comparison theorem forces $\phi_n$ to have at least $n-1$ zeros. An upper bound argument (using $\phi_n$ vs. $\phi_{n+1}$) shows exactly $n-1$ zeros. $\square$