Graphical examination for existence of 2nd order ODE given a solution curve Given any solution curve $y(t)$, what are some graphical criterion to determine whether there exists some 2nd order linear, homogeneous, with continuous but possibly non-constant coefficients, to which the curve is a solution? 
Each of the plots below show graphs of three functions $y(t)$. For which options do there exist a second order linear homogeneous ode with continuous but possibly non-constant coefficients to which all three functions are solutions of?

 A: Not a complete answer, but too long for a comment.
Upon closer inspection, I'll concede that (b) is in fact not a solution of the original question.
Note that the solution curves in (b) all differ by a constant. If there exists a second-order ODE that admits all of these curves as a solution, then it must also admit a constant as a solution.
In general, an IVP that admits a constant solution looks something like this

$$ y''(t) + p(t)y'(t) = 0, \quad y(t_0) = y_0, \ y'(t_0) = 0 \tag{1} $$

where $p(t)\ne 0$. 
If the ODE in $(1)$ has a non-constant solution, said solution must not have a zero derivative at any point (since if a solution has a zero derivative, it must be constant due to the Uniqueness Theorem). We see that the non-constant curves in (b) do not satisfy this, therefore (b) is not correct.
A: The answers are (a), (c), (f). 
Notice that the red graph $y_3$ of option (f) is exactly in the middle, so it is a linear combination of $y_1$ and $y_2$.In comparison of option (e), the red graph is not a linear combination of the other two, so we get three linearly independent solutions. But the solution space of a second order ODE has dimension two, so option (e) is not possible.
In option (d), the green curve and the red curve have common zeros, which is not allowed unless they are constant multiples of each other. If they were constants multiples of each other, they would have the same zeros.
*Please view my other answer for why option (b) is incorrect. 
Therefore options (b), (d) and (e) are incorrect.
A: The Existence and Uniqueness Theorem for second order ODE states that 

For the initial value problem 
  
$$
y''+p(t)y'+q(t)y = g(t),\qquad y(t_0)=y_0, y'(t_0)=y_0'
$$

  where $p,q$ and $g$ are continuous on an open interval $I$ that contains the point $t_0$, there is exactly one solution $y=\phi(t)$ of this problem and the solution exists throughout the interval $I$.

For option (b) at $t=0$, we see that the solutions each take on different values, hence they cannot be solutions to the same initial value problem.
