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

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