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).

somesolution y(x), but I don't have any restrictions on the form of the solution. $\endgroup$