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Suppose Ms. Lee is buying a new house and must borrow 150,000. She wants a 30-year mortgage and she has two choices. She can either borrow money at 7% per year with no points, or she can borrow the money at 6.5% per year with a charge of 3 points. (A "point" is a fee of 1% of the loan amount that the borrower pays the lender at the beginning of the loan. For example, a mortgage with 3 points requires Ms. Lee to pay 4,500 extra to get the loan.) As an approximation, we assume that interest is compounded and payments are made continuously. Let

$$M(t) = \text{amount owed at time } t\ \left(\text{measured in years}\right)$$ $$r= \text{annual interest rate, and}$$ $$p= \text{annual payment}$$

Then the model for the amount owed is

$$ \frac{dM}{dt}=rM-p$$

Q.How much does Ms Lee has to pay in each case?

I have tried solving the DE, and i get
$$ M(t)=C_1e^{rt} + \frac{p}{r}$$

Now what to do?

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What is the s in $\frac{s}{r}$? Now you know what M($0$) should be, right? It is the amount owed at the beginning, which should give you a value for $C_1$, so that you have a formula for $M(t)$. If I understood the problem well, the payments will be made over 30 years. If this is the case, the formula M(t) for t=30 and r=$7%$% and for M(t) for t=30 and $6.5%$%+ $4,500 should give you different values, which you can compare to each other. – gary Jul 1 '11 at 11:07
thanks, Gary. $\frac{s}{r}$ should actually be $$\frac{p}{r}$$. As for $M(0)$ is the amount she owed at the beginning of the time, it's an unknown. we only know $M(30)$ which is equal to $0$. – nihilisticgeek Jul 1 '11 at 11:55
Why do you assume continuous compounding and payments? I've always seen this kind of problem handled by difference equations (recurrence relations), rather than differential equations. – Gerry Myerson Jul 1 '11 at 12:55
@Gerry: Maybe you don't want to use continuous compounding, but the bank does. Or maybe a writer wants to use it in his differential equations textbook. – GEdgar Jul 1 '11 at 16:06
With the exception of consumer and small business loans, continuous compounding is standard. – André Nicolas Jul 1 '11 at 16:51

We can work with concrete numbers, or develop a general formula. Ideally, you should do both, as an exercise and a partial check. Let's develop a general formula. I will use your notation, but introduce two new symbols. Let $N$ be the amortization period, that is, the number of years until the mortgage is paid off. In our case, $N=30$. Let $A$ be the initial amount owed. With no "points", $A=150000$. With $3$ points, she needs to borrow $150000/(1.03)$ in order to have $150000$ left after paying the points. The general equation for the amount owed is, as you wrote, $$M(t)=C_1e^{rt}+\frac{p}{r}$$ This is the general solution of the equation, but it is incomplete until we evaluate the constant $C_1$. (Technical note: It will turn out, of course, that $C_1$ is negative, else what we owe would increase rapidly forever. I would have preferred to arrange things so that any constant is positive.)

Note that $M(0)=A$ and $M(N)=0$. We obtain the two equations $$A=C_1+\frac{p}{r}$$ $$0=C_1e^{rN} +\frac{p}{r}$$ Subtract, to get rid of the $p/r$ term. We get $$A=C_1(1-e^{rN})$$ So $C_1=-\frac{A}{e^{rN}-1}$ and we obtain the equation $$M(t)=\frac{p}{r} -\frac{A}{e^{rN}-1}e^{rt}$$

Now that we have full information about $M(t)$, we should be able to answer any question. In particular, by taking $t=N$, we have $$0=\frac{p}{r} -\frac{A}{e^{rN}-1}e^{rN}$$ Now we can solve for the payment $p$: $$p=rA\frac{e^{rN}}{e^{rN}-1}=\frac{rA}{1-e^{-rN}}$$ and easily find $p$ given any $r$, $N$, and $A$.

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To calculate how much she borrows with $3$ points, she gets $0.97$ of the amount borrowed, so $M(0)=150,000/0.97$, which is slightly higher than $154,500$. Now if you incorporate $M(0)$ and $M(30)$ you can solve for $C_1$ and $p$ and pick the choice with lower $p$

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