Differential Equation - Water dripping from roof Just need some help with a DE. Here goes:

The roof is leaking water with a constant speed of 1 litre per hour. A bucket is placed under the hole to catch the water. The water is evaporating with a speed proportional to the amount of water in the bucket.
When the bucket has 1 litre of water, the speed of evaporation is 0.2 litres per hour.

So, assuming that the bucket is empty at t = 0, we have two values we can use:
y(0) = 0,
y'(1) = 4/5

I've tried various things but I'm getting too tired to keep this up so I'm going to sleep a bit. But this is something I had in mind, that I would try if I wasn't too tired:
y' = (t - 0.2y)t

I read this as the speed of one litre per hour minus a fifth per litre of water in the bucket, per hour. Is this a correct approach? I started writing this post before I came up with this (probably came up with it since I read it and had to write it here, making me think about it again), but since I'm going to bed I would want some feedback on this and maybe some help if this isn't correct. I'm doing my exam tomorrow quite early so I'm getting up in just a few hours again.
Anyway, I hope somebody can help me! Thanks
 A: Let $y(t)$ be the amount of water in the bucket in liters, and let $t$ be the time in hours. The rate (liters/hour) at which water in the bucket is increasing due to the leak is a constant $1$ liters/hour. The rate at which the water volume is decreasing is proportional to the liters $y(t)$, with a proportionality constant of 0.2. So
$$
                \frac{dy}{dt}=1-0.2y,\;\;\;y(0)=0.
$$
The differential equation can be written as
$$
              y'+0.2y = 1,\;\;\; y(0)=0.
$$
Multiplying by the integrating factor $e^{0.2t}$ gives
$$
               e^{0.2t}y'+0.2e^{0.2t}y = e^{0.2t} \\
              \frac{d}{dt}(e^{0.2t}y)=e^{0.2t}
$$
Now replace $t$ by $s$ and integrate over $[0,t]$ to obtain
$$
                    \left.e^{0.2s}y(s)\right|_{s=0}^{s=t}= \int_{0}^{t}e^{0.2s}\,ds.
$$
Use the fact that $y(0)=0$:
$$
                    e^{0.2t}y(t) = 5e^{0.2t}-5 \\
                     y(t) = 5-5e^{-0.2t}.
$$
A: 
The roof is leaking water with a constant speed of 1 litre per hour. 

The leak is: $1$ a constant rate of water added to the level

A bucket is placed under the hole to catch the water. The water is evaporating with a speed proportional to the amount of water in the bucket.

The evaporation is: $-c\; y(t)$ for some constant of proportionality.
So: $y'(t) = 1 - c y(t)$

When the bucket has 1 litre of water, the speed of evaporation is 0.2 litres per hour.

That is: $c=1/5$ 
Hence: $y'(t) = 1 -y(t)/5$
Our DE is then:
$$\int_{y(0)}^{y(t)} \frac{1}{1-\gamma/5}  \operatorname{d}\gamma = \int_0^t \operatorname{d}\tau$$
Assuming the bucket is empty at time zero: $y(0)=0$ then
$$\begin{align}
-5 \ln(1-y(t)/5)&= t
\\[1ex]
- t/ 5 &= \ln(1-y(t)/5)
\\[1ex]
e^{-t/5} &= 1 - y(t)/5
\\[2ex]
\therefore y(t) &= 5-5e^{-t/5}
\end{align}$$
