I have the ODE $$ x'(t) = C\cdot t + D\cdot \sqrt{x(t)}$$
with $x(0) = 0$.
This doesn't have a closed form solution for generic constants $C,D\in \mathbb R$, but I am more interested in the growth behavior of $x$ for $t\to\infty$.
It makes sense that something quadratically growing will work, as $x'$ will be linearly growing and all terms on the right hand side do, too. But how can I do this properly? A reference to a book or paper would be highly appreciated.
So far, I have done the following: Make the ansatz $x'(t) = Et + F$ or $x(t) = \frac{1}{2}Et^2 + Ft$ (because $x(0) = 0$, we don't get another constant).
Then we get
$$Et + F = Ct + D\cdot \sqrt{E/2\cdot t^2 + Ft}.$$
Setting $t=0$ results in $F=0$ and we get only factors of $t$, hence $$E = C + D\cdot \sqrt{E/2}$$ or $$E = \frac{D^2 + 4C + \sqrt{D^4+8CD}}{4}.$$ Of course, this is all very rough and totally wrong but seems to recover some kind of asymptotic behaviour of $x$. I am interested in the correct way of doing that.
To do this formally, expand $x(t)$ in a "power series at infinity", multiplied by an unknown power of $t$: $$ x(t) = t^\alpha \sum_{n=0}^\infty \frac{a_n}{t^n}. $$ We can assume WLOG that $a_0 \neq 0$, since if $a_0 = 0$ we could just rewrite this series with a lower power of $t$ in front and the new leading-order coefficient $a_0 \neq 0$. We then have $$ x'(t) = t^{\alpha} \sum_{n=0}^{\infty} \frac{(\alpha - n) a_n}{t^{n+1}} $$ and $$ \sqrt{x(t)} = t^{\alpha/2} \sqrt{a_0} \left[ 1 + \frac{a_1}{2a_0} t^{-1} + \left(\frac{a_2}{a_0} - \frac{a_1^2}{4a_0^2}\right) t^{-2} + \dots \right] $$ This latter expression can be derived by expanding the power series $\sqrt{1 + x}$, where $x = \frac{1}{a_0} \sum_{n=1}^\infty a_n t^{-n}$. It is straightforward, if tedious, to calculate it to any order.
Once we have these series in place, we can examine the possible asymptotic behavior. We have \begin{multline} \alpha a_0 t^{\alpha-1} + (\alpha - 1) a_1 t^{\alpha - 2} + (\alpha - 2) a_2 t^{\alpha - 3} + \dots\\ - C t - D \left[ \sqrt{a_0} t^{\alpha/2} + \frac{a_1}{2 \sqrt{a_0} }t^{\alpha/2 - 1} + \left(\frac{a_2}{a_0} - \frac{a_1^2}{4a_0^2}\right) t^{\alpha/2-2} + \dots \right] = 0. \end{multline}
The idea now is to ensure that these expressions vanish to all orders present in this big, messy series. If $\alpha \neq 0$, the leading-order power present is either $t^{\alpha - 1}$, $t^1$, or $t^{\alpha/2}$. If any one of these is greater than the other two, then there will be a large power in this series as $t \to \infty$ that dominates all of the other terms, and the series will not vanish. Thus, at least two of $\alpha - 1$, $\alpha/2$, and $1$ must be equal—but as it happens, if two of them are equal, so is the third, with $\alpha = 2$. Thus, we can rewrite this whole mess (in this case) as \begin{multline} (2 a_0 - C - D \sqrt{a_0} ) t^{1} + \left(a_1 - D \frac{a_1}{2 \sqrt{a_0} }\right)t^{0} + \left[ - D \left(\frac{a_2}{a_0} - \frac{a_1^2}{4a_0^2}\right) \right] t^{-1} + \dots = 0 \end{multline} This implies that $$ 2 a_0 - D \sqrt{a_0} - C = 0 \\ a_1 \left( 1 - \frac{D}{2 \sqrt{a_0}} \right) = 0 \\ D \left(\frac{a_2}{a_0} - \frac{a_1^2}{4a_0^2}\right) = 0 \\ \vdots $$ where we could carry this to higher order if desired. These equations can be solved to determine the unknown coefficients $a_0$, $a_1$, and $a_2$.
Note finally that the case $\alpha = 0$ doesn't actually work: in the original expanded power series, if $\alpha = 0$ then the leading-order power will be $t^1$ with coefficient $C$. So unless $C = 0$, there will be a non-vanishing term in the asymptotic expansion.
EDIT: This whole procedure turns out to be overkill in this particular case, since as was pointed out by Kiryl Pesotski in the comments we have $x(t) = A^2 t^2$ is a solution, where $A$ is a positive root (if any) of the quadratic $2 A^2 - DA - C = 0$. This is exactly the quadratic satisfied by $\sqrt{a_0}$ in my solution above, which implies that all of the higher-order coefficients $\{ a_1, a_2, \dots \}$ will vanish. Still it's a useful technique for more general ODEs where there really isn't a closed-form solution.