I am working on:
Let $\phi_1,\phi_2$, so that $\phi_1(x)\phi_2'(x)-\phi_1'(x)\phi_2(x)\neq 0.$ for all $x\in\mathbb{R}$. Then there exists exactly on homogeneous differential equation of second order $$y''(x)=f(x)y'(x)+g(x)y(x)$$ so that the functions $\phi_1,\phi_2$ are fundamental solutions.
My idea is to suppose that there are two homogeneous ODE of second order. Is there a possibility to combine the terms, so I can use $W(x)\neq 0$?
Thanks for your answers!
I think your idea works to prove uniqueness. Suppose there exist functions $f_1, f_2, g_1, g_2$ such that $\phi_1$ and $\phi_2$ are fundamental solutions to the second order, homogeneous ordinary differential equations $y'' + f_1(x)y' + g_1(x)y = 0$ and $y'' + f_2(x)y' + g_2(x)y = 0$. This means that the general solution of these differential equations are of the form $y(x) = c_1\phi_1(x) + c_2\phi_2(x)$ for constants $c_1, c_2$. In particular, $y(x) = \phi_1(x)$ and $y(x) = \phi_2(x)$ are solutions to these differential equations. So $$\phi_1'' + f_1(x)\phi_1' + g_1(x)\phi_1 = 0 \quad \text{and} \quad \phi_2'' + f_1(x)\phi_2' + g_1(x)\phi_2 = 0,$$ and $$\phi_1'' + f_2(x)\phi_1' + g_2(x)\phi_1 = 0 \quad \text{and} \quad \phi_2'' + f_2(x)\phi_2' + g_2(x)\phi_2 = 0.$$ Subtracting, you get $$(f_1(x) - f_2(x))\phi_1' + (g_1(x) - g_2(x))\phi_1 = 0 \quad \text{and} \quad (f_1(x) - f_2(x))\phi_2' + (g_1(x) - g_2(x))\phi_2 = 0.$$ You can the rewrite this as a matrix equation: $$\begin{bmatrix} \phi_1'(x) & \phi_1(x) \\ \phi_2'(x) & \phi_2(x) \end{bmatrix}\begin{bmatrix} f_1(x) - f_2(x) \\ g_1(x) - g_2(x) \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}.$$ For any fixed $x \in \mathbb{R}$, the $2 \times 2$ matrix is invertible as your Wronskian is non-zero. This means that for any $x \in \mathbb{R}$, $f_1(x) = f_2(x)$ and $g_1(x) = g_2(x)$, which shows that if there exists a second order homogeneous ordinary differential equation for which $\phi_1$ and $\phi_2$ are fundamental solutions, then it must be unique.
