Specifically, I want to show that $$y_2(t) = [y_2 (t^*)/y_1'(t^*)]y_1(t)$$
Here's how far I got.
$W[y_1,y_2](t^*)=y_1'(t^*)y_2(t^*)= 0$. Now, we have two choices.
If $y_1'(t^*)=0$ then $y_1(t)=0$, because of a theorem.
If, on the other hand, $y_2(t^*)=0$, we can rewrite $y_2$ using the Wronskian. Since the Wronskian is either always zero or never, it is always zero since $W(t^*)=0$;
$$y_1(t)y'_2(t)-y_1'(t)y_2(t)=0$$ and $$y_2(t)=[y_2'(t)/y_1'(t)]\ y(t).$$
Now, my question. Why is the answer $[y_2 (t^*)/y_1'(t^*)]y_1(t)$? That is, with $t^*$ filled in.
Context: this is question 2.1.19 from Braun's Differential Equations and their Applications.
If $y'_1(t^*)=0$ as well, then $y_1\equiv 0$, because the IVP $$ y''+p(x)y'+q(x)y=0, \quad y(t^*)=y'(t^*)=0 $$ possesses a unique solution, the zero one, and the conclusion holds.
If $y'_1(t^*)=a\ne0$ then $$ 0=w(y_1,y_2)(t^*)=y_1(t^*)y_2'(t^*)-y_1'(t^*)y_2(t^*)=-y_1'(t^*)y_2(t^*) $$ and hence $y_2(t^*)=0$. But $y_2$ is the unique solution of the IVP $$ y''+p(x)y'+q(x)y=0, \quad y(t^*)=0,\,\,y'(t^*)=y'_1(t^*)\ne 0, \tag{1} $$ Whereas $y_2$ is the unique solution of the IVP $$ y''+p(x)y'+q(x)y=0, \quad y(t^*)=0,\,\,y'(t^*)=y_2'(t^*). \tag{2} $$ Observe now the if $y$ is a solution of $(1)$, then $\big( y'_2(t^*)/y_1'(t^*)\big)y(t)$ is a solution of $(2)$. Hence, uniqueness of solutions of both IVPs implies that $$ y_2(t)=\big( y'_2(t^*)/y_1'(t^*)\big)y_1(t) $$