Find equivalent interest rate for cash advance fee + promo rate

Question

I take a cash advance of 'amount' from my credit card, paying an up-front 'fee' (given as a percentage), with a promotional rate 'int' for time 'len'. I must pay at least 'min'% of the owed amount monthly.

I put 'amount' into an investment account earning 'p'% interest, and also make the monthly payments from this account.

Question: for what value of 'p' will I break even after time 'len'?

Here's how I set it up in Mathematica:

DSolve[{ 

(* I start off owing amount plus the fee *) 
owed[0] == amount*(1+fee), 

(* The amount I owe increases due to credit card interest, 
   but decreases due to monthly payments *) 
owed'[t] == int*owed[t]-min*12*owed[t], 

(* I start off having amount *) 
have[0] == amount, 

(* The amount I have increases due to investment interest, 
   but decreases due to monthly payments *) 
have'[t] == p*have[t]-min*12*owed[t], 

(* After len, I want to break even *) 
owed[len] == have[len] 
}, 
{owed[t], have[t]}, {t}] 

Mathematica returns "DSolve::bvnul: For some branches of the general solution, the given boundary conditions lead to an empty solution", which is actually reasonable: there's only one value of 'p' that will yield a solution for the differential equations above.

How do I coerce Mathematica into finding this value?

I tried solving for owed[t], then substituting owed[t] into have[t], and then solving owed[len] == have[len], but this yield a similar error. Running Reduce on "owed[len] == have[len]" yielded something complex and ugly.

Answer 1

The equation:

owed'[t] == int owed[t]-min 12 owed[t] 

if both int and min are constants, is just a exponential function. With the initial condition

owed[0] == amount*(1 + fee)  

gives

owed[t_] := amount E^((int - 12 min) t) (1 + fee)  

And that's the solution for owed[t]

Now for have[t] you may use:

DSolve[{
  have'[t] == p*have[t] - min*12*owed[t],
  have[len] == owed[len]},
 {have[t]}, {t}]  

That gives you the expression for have[t] that meets your break even condition.

For obtaining the value of p, you must use the last equation:

 have[0] == amount  

or, after replacing have[0] for it's value:

(amount E^(-len p) (1 + fee) (12 E^(len p) min + 
   E^(len (int - 12 min)) (-int + p)))/(-int + 12 min + p) == amount 

This last equation seems not easily solved for p. I tried a few things (not too much, certainly) and it resists strong.

But ... given numerical values for the rest of the parameters is trivially solved by any numerical method (I guess)