Homogeneous second-order differential equation with constant Wronskian 
Problem Prove that if the Wronskian of any two solutions of differential equation $y''+p(x)y'+q(x)y=0$ is constant, then $p(x)$ is zero.

My attempt. : Let $y_1$ and $y_2$ be two solutions of given differential equation. Note that the Wronskian $W=W[y_1,y_2]$ satisfies $W'+p(x)W=0$. Since $W$ is constant, we get $p(x)W=0$.
Question. How do I show that $W$ cannot be equal to zero? If there are two solutions $y_1$, $y_2$ that are linearly independent, then $W[y_1,y_2]\neq 0$. But I am not certain of the existence of such solutions.

My question is : If a second-order ODE has a solution, do there exist two solutions that are linearly independent?
 A: You can directly use Abel's identity to show that if the Wronskian of any two solutions of the differential equation $y''+p(x)y'+q(x)y = 0$ (on an interval $I$) is constant, then $p(x) = 0$. (If $\int_{x_0}^{x} p(t) dt$ is constant $\forall$ $x \in I$, then $p(t) = 0 $)
The answer to your last question can be found here.
A: We know that if $W(x)$ be the Wronskian of the differential equation $y''+p(x)y'+q(x)y=0$, then $$W(x)=Ae^{\int p(x) dx}\qquad \text{where $A$ is constant.}$$
Now if $W(x)=constant = c$(say), then $$Ae^{\int p(x) dx}=c$$
$$\implies e^{\int p(x) dx}=d$$
$$\implies {\int p(x) dx}= \log d=k \text{(say)$\qquad$ where $k$ is constant}$$
Differentiating both side with respect to $x$ $$p(x)=0$$

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If a second-order ODE has a solution, then there exist two solutions that are linearly independent. In fact, in general, an $n^{th}$-order ODE has $n$ linearly independent solutions.
For more details about this you may find the following link
An nth-order ODE has n linearly independent solutions
