Different results to a basic question ( Newton's law and perservation of momentum) 
Trolley with mass of $m_0=1 \ kg$ is moving without friction on the railway track. It is raining so there is a constant mass flow of water $\Phi_m=0.1\ kg/s$. Constant force $F=0.1 \ N$ is accelerating the trolly horizontally.
What is the velocity at time $t$ if the trolly is stationary initially ?

I tried two different aproaches and got different results. I graphed the both functions  and noticed that both were similar at $t=0$.
1. Newton's  law
$$F=m(t)a$$
$$F=(m+\Phi t) \frac {dv}{dt}$$
$$\int dv=F \int\frac{dt}{m+\Phi t}$$
..integrated 0 to v; and 0 to t
$$v=\frac F\Phi \ln(m+\Phi t)$$
2. Momentum
$$(m+\Phi t)v - 0 = \int Fdt$$ as $F=const.$
$$v=\frac{Ft}{m+\Phi t}$$
Am I missing some concept behind differential equations?
 A: Newton's 2nd law in differential form (ignoring vectors) is
$$F_\text{net}=\frac{dp}{dt}=\frac{d}{dt}\left(p_\text{train} + p_\text{water in trolley} + p_\text{rain just hitting trolley}\right). \tag{1}$$
You must take into account the change in momentum of the rain that occurs when it falls into the trolley and accelerates up to the speed of the train. This is in addition to knowing that the force exerted on the train acts on a body whose mass is increasing. (There are subtleties associated with this wording; see comments below.) It seems in method 1 you're not accounting for everything.
Now, Eqn. 1 can be re-written as
$$dp = F_\text{net}\,dt$$
then you can integrate both sides to get the general form of Newton's second law in integral form. After that, take into account the momentum of the rain + train at both the initial and final times. Luckily $p_\text{i,rain}=0$, so this simplifies some things.
A: You should be careful, since you need to take account of the force that the rain exerts on the trolley, or the momentum of the rain. Your second approach does this rather nicely, (with the assumption that the rain falls vertically, and hence doesn't contribute to the initial momentum).
In the first approach, you could redo it to add the force that the system must exert on the rain that is just landing in the trolley, and hence (by NIIL) the extra resistance that provides.
A: It looks like in this problem the rain moves at whatever $v$ the train is currently at, giving an infinite amount of energy as $lim_{t\rightarrow \infty}$. However it doesn't look like your using energy anyway, just be careful if you do.
As BMS said, use the product rule. This gives you $F= {\Phi}{v(t)}+\frac{dv}{dt}{(m+\Phi)}$
then subtract $\Phi v(t)$ to the other side and then move things around and integrate over $t$ to get $$\frac{1}{m+\Phi}\int{dt}=\int\frac{dv}{F-\Phi v(t)}$$ I ended up with $$v=\frac{F}{\Phi}\left(1-exp\left[\frac{-\Phi t}{(m+\Phi)}\right]\right)$$ Not too pretty with the $\Phi$ in 3 places but it satisfies $v(0)=0$ and has a convergent final velocity which seems to appeal to my intuition.It could be a useful exercise to solve this problem but assuming instead that the rain always falls vertically.
