How Can Diffe Equations Determine Loan Payments for a House Purchase?

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Homework Statement



# 39 on page 36, specifically part a and case 1

http://books.google.com/books?id=mw...e&q=suppose ms lee is buying a house&f=false"

M(0) = $150000 ------>initial amount due
M(30) = $0 -------> final amount due
r = .07
p=?


Homework Equations



anything to do with diffe equations


The Attempt at a Solution



Well I don't want to show all the work for it but one can find a general solution to the equation listed in the problem by dividing both sides by rM-p and by multilpying both sides by dt. So

dM/(rM-p)=dt

From here you can integrate and eventually get the general solution of

M = c*e^(r*t)+p/r

With M(0) = 150000 and M(30) = 0 I can solve for both c and p but I seem to be missing something. How do I find out how much she pays? I'm lost. Even by after checking with the answers in the back of the book I can't get the answer that they got. :cry:
 
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That's the easy part! You are told that "p is the annual payment". Since she holds this loan for 30 years, she makes constant payment, p, 30 times: her total cost is 30p. Don't forget to add the three "points", .03(150,000)= $4,500 she has to make up front for the 6.5% loan.
 
Wow I really overlooked this... I feel silly now. I think I was treating this problem harder than it really was because it has to do with diffe equations.

Thanks
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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