If $y''+p(x)y'+q(x)y=0$ and two solutions have null second-derivative, then the coefficients are both zero I have the following differential equation:
$$ y'' + p(x) y' + q(x) y = 0, \quad x\in(a,b) $$
such that $y_1$ and $y_2$ are both linearly independent solutions and $p$ and $q$ are continuous. I would like to prove that if $y_i''(x_0) = 0$, $i=1,2$ for some $x_0=0$, then necessarily, $p(x_0)=q(x_0) = 0$. However, I'm dealing with some issues. First, the fact that $y_i(x_0)''= 0$ means that
$$ p(x_0)[y_2'(x_0)-y_1'(x_0)] + q(x_0)[y_2(x_0)-y_1(x_0)] = 0 $$
and $y_3 = y_2-y_1 \neq 0$ also satisfies, for $x\in(a,b)$:
$$ y_3'' + p(x) y_3' + q(x) y_3 = 0 $$
If I suppose that $p(x_0)=0\neq q(x_0)$, $y_2(x_0) = y_1(x_0)$, which is possible, and also if $p(x_0)\neq 0 = q(x_0)$ or if $p(x_0)\neq 0\neq q(x_0)$, so I'm a little bit lost since I can't find out the behaviour of the solution around $x_0$. I would appreciate some hint. Thanks
 A: You can use the Wronskian. Let's consider the following matrix :
$$S(x)=\begin{pmatrix} y_1(x) & y_2(x) \\ y'_1(x) & y'_2(x) \end{pmatrix}$$
Define the Wronskian $W(x)=\det S(x)=y_1(x)y'_2(x)-y_2(x)y'_1(x)$. Then, you have :
$$\begin{array}{rcl}
W'(x) & = & y_1(x)y''_2(x)+y'_1(x)y'_2(x)-y_2(x)y''_1(x)-y'_2(x)y'_1(x) \\
      & = & y_1(x)y''_2(x) - y_2(x)y''_1(x) \\
      & = & -y_1(x)\left(p(x)y'_2(x)+q(x)y_2(x)\right)+y_2(x)\left(p(x)y'_1(x)+q(x)y_1(x)\right) \\
      & = & p(x) \left(y_2(x)y'_1(x)-y_1(x)y'_2(x)\right)
\end{array}$$
So, in the end, you have :
$$W'(x)=-p(x)W(x)$$
This differential equation is solved by $W(x)=W(x_1) \exp \left( \int_{x_1}^x p(y) \text{d}y \right)$, for any $x_1$.
Now the linear independance of $y_1$ and $y_2$ means that there exists $x_1$ such that $W(x_1)\neq 0$. Thus, considering the expression of $W$, we have $W(x) \neq 0$, for all $x$.
Now to conclude, define $V(x)=\left( q(x) \; \; p(x) \right)$ and observe that the differential equation can be written :
$$V(x)S(x)=-\left( y''_1(x) \; \; y''_2(x) \right)$$
Thus, you have $V(x_0)S(x_0)=0$. Yet, $W(x_0)=\det S(x_0)\neq 0$ so $S(x_0)$ is invertible, which means that $V(x_0)=0$. QED
