How Do You Solve an Annuity Differential Equation?

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SUMMARY

The discussion focuses on solving the annuity differential equation represented by \(\frac{dS}{dt} = rS + d\) with the initial condition \(S(0) = S_0\). The correct solution is derived using the integrating factor method, leading to the future value of an annuity equation: \(S(t) = S_0e^{rt} + \frac{d}{r}(e^{rt} - 1)\). A common mistake identified was neglecting the initial value, which is crucial for determining the constant \(C\). The alternative approach using the substitution \(V = S + \frac{d}{r}\) simplifies the problem significantly.

PREREQUISITES
  • Understanding of differential equations, specifically first-order linear equations.
  • Familiarity with the integrating factor method for solving differential equations.
  • Knowledge of initial value problems and their significance in determining constants in solutions.
  • Basic grasp of annuities and their mathematical representation.
NEXT STEPS
  • Study the integrating factor method in detail to solve similar differential equations.
  • Learn about initial value problems and their role in differential equations.
  • Explore the mathematical foundations of annuities and their applications in finance.
  • Investigate alternative methods for solving differential equations, such as substitution techniques.
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Students studying differential equations, finance professionals dealing with annuities, and educators teaching mathematical methods in applied contexts.

_N3WTON_
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Homework Statement


Solve the annuity problem:
\frac{dS}{dt} = rS + d
S(0) = S_0

Homework Equations


Integrating factor method equation
Future value of an annuity equation (this should be the final answer):
S(t) = S_0e^{rt} + \frac{d}{r}(e^{rt} - 1)

The Attempt at a Solution


Ok, I am getting quite close to doing this derivation correctly. However, I keep ending up with a negative that should not be there.
First, I set:
p(x) = -r
Then:
u(x) = e^{-rt}
This means that I need to take the integral of:
\frac{d}{dt} (e^{-rt}S(t)) = de^{-rt}
After taking the integral of both sides I end up with:
(e^{-rt} * S(t)) = - \frac{d}{r} e^{-rt} + C
Therefore:
S(t) = - \frac{d}{r} + Ce^{rt}
At this point I am not sure what to do because I believe that the negative symbol should not be there. If somebody could point out where my mistake is I would greatly appreciate it.
 
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_N3WTON_ said:
At this point I am not sure what to do because I believe that the negative symbol should not be there...

Why? Your answer seems to agree with ##
S(t) = S_0e^{rt} + \frac{d}{r}(e^{rt} - 1)
## if ##C=S_0+\frac{d}{r}##.
 
ZetaOfThree said:
Why? Your answer seems to agree with ##
S(t) = S_0e^{rt} + \frac{d}{r}(e^{rt} - 1)
## if ##C=S_0+\frac{d}{r}##.
ok I see it now, I guess I am just an idiot XD ...sorry for the waste of time, I forgot to do the initial value portion of the problem
 
_N3WTON_,
It's a good habit to get into to check a solution you get. If the solution you get 1) satisfies the initial condition, and 2) satisfies the differential equation, you're golden. You don't need us to verify that your solution is correct.
 
_N3WTON_ said:
ok I see it now, I guess I am just an idiot XD ...sorry for the waste of time, I forgot to do the initial value portion of the problem

An easier way is to note that if ##V = S + (d/r)## then ##dV/dt = dS/dt = r V##, so ##V(t) = V_0 e^{rt}##, where ##V_0 = S_0 + d/r##.
 

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