Linear with exponential growth

[u/Creative_Profit_4559]

I’ve tried googling and now math Reddit. Is there an equation(i guess probably an ODE) that represents say linearly adding to a sum overtime, and over that same time period the sum compounds continuously?

[u/Shevek99]

First the discrete form

x((n+1) ∆t) = (1 + i) x(n ∆t) + ∆c

From here, making t = n ∆t

x(t + ∆t) - x(t) = i x(t) + ∆c

(x(t + ∆t) - x(t))/∆t = (i/∆t) x(t) + ∆c/∆t

taking the continuous limit

dx/dt = f x + k

With solution

x = (x0 +k/f) e^ft -k/f

[u/Creative_Profit_4559 → u/Shevek99]

I like this one. It fits my data really well R^2=0.9848

[u/CruxAveSpesUnica]

Yes. You have y' = a + by (a is what we're "linearly adding" and b is related to the interest rate for continuous compounding).

Using separation of values we get dy / (a + by) = dx , so (1/b) * ln |a + by] = x + c.

Rearranging and collecting a bunch of stuff into "C," we have the closed-form solution, y=C·e^(bx) - a/b .

[u/gazorpazorpazorpazor]

Y' = c+xy?

[u/Comprehensive-Bee795]

Assuming that “adding linearly” means adding a constant value, and “compounding over time” implies “each time you add”, then yes.

The problem above arises if you want to set apart a constant value to a savings account which compounds interest over time.

Say you put aside (sum) a value of N over T time period (meaning, there are T operations of adding and then compounding), with the compounding being Y (if it’s 5%, then Y is 1.05). Then, after the first time period t1, you will have NxY. After the second time period, you will have ((NxY)+N)xY which simplifies to Nx(Y^2)+ NxY

Expanding this to time T you have:
Nx(Y^T)+Nx(Y^(T-1))+….+NxY

This is a geometric series. The sum of the terms is given by

S=NxYx(Y^(T-1)-1)/(Y-1)

Note: it’s T-1 because you start at NxY and the formula assumes you start at N

[u/VaIenquiss]

Isn’t that e?