In Verrill's paper preprint here, she has the following theorem which is from a paper of Stiller. It states that
Let $\Gamma$ be a discrete subgroup of $SL_{2}(\mathbb{R})$ commensurable with $SL_{2}(\mathbb{Z})$. For $f \in M_{k}(\Gamma)$ (the space of weight $k$ modular forms) and $t \in M_{0}(\Gamma)$ (the space of meromorphic weight 0 modular forms), if $f = \sum_{n \geq 0}b_{n}t^{n}$ near $t = 0$, then there is a linear order $k + 1$ differential equation satisfied by $g(x) = \sum_{n \geq 0} b_{n}x^{n}$, of the form
$$P_{k + 1}(x)\frac{d^{k + 1}g}{dx^{k + 1}} + P_{k}(x)\frac{d^{k}g}{dx^{k}} + \cdots + P_{0}(x)g = 0
\tag{1}$$
where $P_{i}(x)$ are algebraic functions in $x$.
If we take $t$ to be a Hauptmodul for $\Gamma$, then $P_{i}(x)$ are rational functions. Hence by multiplying by a suitable polynomial, we can in fact assume that the $P_{i}(x)$'s are polynomials. My question is that how does one get explicit bounds on the degrees of these $P_{i}(x)$'s (specifically in the case when $k = 1$)?
added
(It seems Mr. 120487 has not been back here since the day he posted this.
So I will take over the question… — G. Edgar)
Mr. User's comment seems to say this is impossible. It's not.
The prototype for this may be
$$
f = \eta(\tau)^2 j(\tau)^{1/12} = 1+60q-4860q^2+660480q^3+\dots
$$
with weight $1$, so we expect an order 2 differential equation. The
relation we want is
$$
f = \;{}_2F_1\left(\frac{1}{12},\frac{5}{12};1;\frac{1728}{j}\right)
\tag{2}$$
and the hypergeometric differential equation indeed has order 2.
I do not know how to predict the degree of the coefficients in
the differential equation. It would be useful to know for computer
calculations, though. So we can tell when it will do no good to
run it again with higher degree…
added 2
Dror asks about writing $\Delta$as a function of $1/j$. This is practiacally already done in (2). Raise (2) to the power $12$, and use $\Delta = (2\pi)^{12} \eta^{24}$ to conclude
$$
\Delta = \frac{(2\pi)^{12}}{j}\;\;{}_2F_1\left(\frac{1}{12},\frac{5}{12};1;\frac{1728}{j}\right)^{12}
$$
This may illustrate the question for us. Suppose we know $\Delta$ and $j$ to many terms of the Fourier series in $q$. (Integer coefficients!) Then can we find the differential equation? If we know a bound on the degree of the coefficients in the differential equation, then we can. Plug in the two $q$-expansions and solve a system of linear equations for the coefficients in the DE. Knowing the degree of the coefficients tells us how many coefficients of the $q$-series will suffice. Since $\Delta$ has weight $12$, we expect a differential equation of order $13$. If the coefficients have degree at most $5$, then we have to solve for around $(13+1)\cdot(5+1) = 84$ unknowns. Should be possible by computer. But if the coefficients have degree $25$? Maybe not.
Best Answer
If we write the differential equation in terms of $D_t = t\frac{\partial}{\partial_t}$ then the coefficients in the equation are polynomials in certain functions $p_1, p_2$, their derivatives and $t$. Calculation of these polynomials can be done by a polynomial time algorithm. For $k=1$ we get $D_t^2 f + p_1(t)D_tf + p_2f = 0$.
The functions $p_i$ are given explicitly as $$ p_1(t) = \frac{D_q G_1 - 2G_1G_2/k}{G_1^2}, \quad p_2(t) = -\frac{D_q G_2 - G_2^2/k}{G_1^2}, $$ where $D_t = q\frac{\partial}{\partial_q}$ and $$ G_1 = \frac{D_q t}{t}, \quad G_2 = \frac{D_q f}{f}. $$
So the question boils down to bounding degrees in $p_1$ and $p_2$.
See Yifan Yang, On differential equations satisfied by modular forms, Math. Z. 246, 1–19 (2004) for details and examples. The calculations seem to be tricky.