Solution for differential equation describing linear input and exponential decay I try to model the concentration over time when there's an linear input and an exponential output (i.e. exponential decay) with known half-life.  
Input(t) = a+b*t

, where t is the time (in years), a the input at year 0 and b the yearly growth to that input.   
Output(t) = lambda*N(t)

, where t is the time (in years), lambda the exponential decay constant and N(t) the concentration at year t. 
lambda is defined as 
lambda = ln(2)/th

, where th is the half-life in years.
The differential equation is therefore:
dN/dt = Input - Output = a+b*t-lambda*N(t)

Using wolframAlpha I get this solution:
N(t) = C*exp(-lambda*t) + (b/lambda)*t + (a/lambda) - (b/lambda^2)

, where C is the initial concentration at t=0.
For small values of th this makes sense, but if th assumes higher values, the concentration is negative. How can this be?  
As example:   
a = 300
b = 20
th = 2 
C = 300

For t = 10 I get a concentration of 1285.561.
But if I set th to 200, I get -1520536 at t = 10
 A: You assumed that $C$ was the initial amount ("concentration" is a term indicating the ratio of "active" amount to total amount, which is not something so easily controlled). But this is not the case. The initial amount is $N(0)$, and that value is $$N(0) = C + \frac a\lambda - \frac b{\lambda^2}$$
When $a = 300, b = 20, C = 300$ and the half-life is $200$ years, that value is about
$-1578233.5$. The reason that after 10 years you have a negative amount is that you started with an even more negative amount and have not yet managed to overcome it.
If you want an initial amount $N(0)$ to be $300$, then you need to set $C = 300 - \frac a\lambda + \frac b{\lambda^2} \approx 1578833.5$

However, as I indicated in a comment, I am doubtful this equation is modelling what you think it is. 
By your description, the repository starts out with an amount of $300$. The first year, an additional $300$ are added to it, for a total of $600$, not including decay. The next year, an additional $320$ are added, the year after $340$ are added, then $360$ the year after that, etc. Since you described it as "linear", I don't think this is what you are after.
Instead, I am guessing that $a$ is supposed to be the initial amount, and each year an additional amount $b$ is added. This gives the differential equation
$$\frac {dN}{dt} = b -\lambda N$$ whose solution is
$$N(t) = \frac b{\lambda} + Ce^{-\lambda t}$$
Since $a = N(0) = b/\lambda + C$, we have $C = a - b/\lambda$. Therefore
$$N(t) = \frac b{\lambda} + \left(a - \frac b\lambda\right)e^{-\lambda t}$$
Is the equation you are after.
