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I'm supposed to derive a formula for mortgages, $f(t)$, which represents the current debt at time $t$. It's given that ($m$ and $r$ are constants):
$\displaystyle \frac{df(t)}{dt}= r\cdot f(t) - 12m$, I've found the solutions $\displaystyle f(t) = K\cdot e^{rt} +\frac{12m}{r}$ (With $K, m$ and $r$ as constants).

But all of these solutions clash with the fact that $f(N)=0$ and $f(0)=B, \ B \ne 0$.

Solving for $K$ in the term $\displaystyle f(N)=0 =K\cdot e^{rt}+\frac{12m}{r}$ yields zero for every possible input to the function $f$, while it's only supposed to yield $0$ for $t=N$ and it's supposed to be at a maximum for $t=0$, what am I doing wrong here?

Edwin
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  • The solution is $\displaystyle f(t)=K\cdot{e^{rt}}+\frac{12m}{r^2}$. Makes sense? – Galc127 Oct 18 '14 at 20:32
  • The solution is almost exactly equal to the solution Wolfram Alpha came up with, according to the paper it's supposed to yield a positive value for B.

    I'm supposed to find a value for K, so that filling in m and r will yield values, but all solutions I find give 0, which is wrong.

     – Edwin Oct 18 '14 at 20:35

1 Answers1

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By solving the ODE is simple to find that $\displaystyle f(t)=\frac{1}{r}e^{rt}+\frac{12m}{r}$.

We want $f(0)=B \rightarrow \displaystyle \frac{1}{r}e^{0}+\frac{12m}{r}=B \Rightarrow \displaystyle m=\frac{Br-1}{12}$.

We got $\displaystyle f(t)=\frac{1}{r}e^{rt}+\frac{Br-1}{r}$.

Now we want $\displaystyle f(N)=0$, so $$\displaystyle \frac{1}{r}e^{rN}+\frac{Br-1}{r}=0 \rightarrow e^{rN}+Br-1=0 \rightarrow rN=\ln(1-Br) \Rightarrow N=\frac{\ln(1-Br)}{r}$$

You asked for B in terms of N, so we can write $\displaystyle Br=1-e^{rN} \Rightarrow B=\frac{1-e^{rN}}{r}$.

Hope this answer helps.

If the problem is solving the ODE, I would love to help.

Galc127
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  • I'm so sorry, I noticed that I made a typo in typing the ODE here, It's 12m, not 12m/r, so that'd make the solution I proposed in my post correct ,right? I'm supposed to express the B in terms of, so solving f(N) = 0 and f(0)=B should lead to a formule: B(N)=..... – Edwin Oct 18 '14 at 21:13
  • Please see the edited solution. – Galc127 Oct 18 '14 at 21:26