# 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?

• The wronskian can be $0$, just set $y_1=y_2$ and $p$ not constant to find a counterexample to the statement. You need to require that the solutions are linearly independent. Apr 8 '14 at 11:15
• @GitGud I know it. My question is 'it is possible that W is not zero?' Apr 8 '14 at 11:16
• The wronskian isn't zero if, and only if, the two two solutions are linearly independent. Does this answer your question? Apr 8 '14 at 11:18
• @gitgud No. There is no second-order ODE that every two solutions of given ODE is linearly dependent? Apr 8 '14 at 12:21
• There is no second-order ODE that every two solutions of given ODE is linearly dependent This statement is false. Just take any non trivial solution and the trivial solution which are linearly dependent. I really don't see what you want to ask, as far as I can tell I've answered your question. Maybe you're not asking what you want to ask. Apr 8 '14 at 12:24

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 $$p(t) = 0$$)

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 $$x$$ $$p(x)=0$$
$${}$$
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.