Parametric ODEs - when do there exist solutions independent of the parameter? I have a complicated 3rd-order ODE of the form
$P(y, y', y'', y''') = 0$, where $P$ is a complicated polynomial (5th-order with 24 terms) and coefficients that are (unknown) functions of a parameter $\lambda$, say
$$ P(z_0, z_1, z_2, z_3) = \sum c_{j_0 j_1 j_2 j_3}(\lambda) z_0^{j_0} z_1^{j_1} z_2^{j_2} z_3^{j_3}. $$
What I want to know is: Suppose that there exists a non-constant solution $y(x)$ that is independent of $\lambda$.  What conditions does this force the coefficient functions $c_{j_0 j_1 j_2 j_3}(\lambda)$ to satisfy?
(An easy analog would be something like: a 1st-order ODE of the form
$$y' + c_0(\lambda) y = 0 $$
has nonconstant parameter-independent solutions if and only if $c_0(\lambda)$ is a constant function.)
I'm sure that the specific conditions I'm looking for depend on the precise form of the ODE, which is pretty daunting in this case.  I'm just wondering if there's a reasonable algorithm I could apply to find them.
 A: Depending on what you know about the coefficients $c_{j_0j_1j_2j_3}(\lambda)$, I think that it's not as hopeless as all that.
First of all, such a curve would have to lie in the common zero locus $Z\subset\mathbb{C}^4$ of the polynomials $P_\lambda(z)$.  This locus is the same as the space of common zeroes of all the polynomials in the linear span $L$ of the $P_\lambda(z)$ in the space of quintic polynomials.  Thus, consider the ideal $I\subset\mathbb{C}[z_0,z_1,z_2,z_3]$ generated by this linear span of quintics.  Assuming that you can compute this (i.e., find a basis for it), there are fast algorithms (using Gröbner bases and Macaulay for example) for determining the dimension $Z$.
Case 0: If $Z$ is empty or its dimension is zero, then there is no nonconstant solution $y(x)$ whose graph $\bigl(y(x),y'(x),y''(x),y'''(x)\bigr)$ lies in $Z$.
Otherwise, decompose $Z$ into its irreducible components (again, Gröbner and Macaulay can be very helpful here) and treat each component of $Z$ separately.  From now on, I'll assume that $Z$ is irreducible and is defined by a reduced ideal (i.e., you have found a basis for the ideal of polynomials that vanish on $Z$).  
Suppose that the dimension of $Z$ is at least $1$, and consider the Pfaffian system $I$ generated by the $1$-forms
$$
\zeta_0 = z_2\,\mathrm{d}z_0 - z_1\,\mathrm{d}z_1\,,
\quad\text{and}\quad
\zeta_1 = z_3\,\mathrm{d}z_0 - z_1\,\mathrm{d}z_2\,,
\quad\text{and}\quad
\zeta_2 = z_3\,\mathrm{d}z_1 - z_2\,\mathrm{d}z_2\,.
$$
This system will have rank $2$ everywhere except along the locus $z_1=z_2=z_3=0$,
which you don't care about anyway, since this would correspond only to $y(x)$ being constant.  In fact, you only care about the part of $Z$ that is not contained in the hyperplane $z_1=0$, so I'll assume from now on that we have removed this hyperplane.  Nearly all of the integral curves of $I$ that have $z_1$ not identically vanishing are locally graphs of the form $\bigl(y(x),y'(x),y''(x),y'''(x)\bigr)$, and you can easily characterize the exceptions (such as, for example, $z_0$ and $z_1$ and $z_2$ are constant), so you can incorporate the test to throw those out into your algorithm, so that you get only the so-called 'admissable curves'.
Case 1: If the dimension of $Z$ is $1$ and it is not an integral curve of $I$ (i.e., the $1$-forms $\zeta_j$ are not in the differential ideal generated by the polynomials that vanish on $Z$), then there is no admissable curve in $Z$.
Case 2: If the dimension of $Z$ is $2$, then you need to look at the locus $Z'\subset Z$ on which either $Z$ is singular or on which $I$ pulls back to have rank at most $1$.  If $Z'$ has dimension $2$, then $Z'$ is foliated by integral curves of $I$, and the admissable ones will be the $3$-jet graphs of the curves you seek.  If $Z'$ has dimension $1$, then repeat Case 1 with $Z'$ in the place of $Z$.
Case 3: If the dimension of $Z$ is $3$, then the smooth part of $Z$ is foliated by integral curves of $I$, and each admissable integral will correspond to a $y(x)$ that satisfies all the $P_\lambda(z)$.  The singular locus of $Z$ will have dimension at most $2$, so, for that, you are reduced to Case 2 or Case 1 (or Case 0).
