# How to derive and solve a difference equation.

I am having terrible trouble with the following question:

Suppose that $x_n$ is the amount owed on a mortgage after n years, £$m$ is the monthly repayment and $r$ is the annual percentage interest rate charged on the amount of the mortgage outstanding.

i) Derive a difference equation satisfied by $x_n$.

ii) Solve the difference equation derived in i) for a loan of $£m$ to be repaid over $N$ years and hence determine what the monthly payment should be.

iii) If the interest rate is 5% show that the monthly repayment on a loan of £50,000 to be repaid over $25$ years is £295.64.

iv) What is the total amount paid back on the loan?

Any help would be great thanks :)

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What do you know about these sort of problems? What have you tried so far? We can be a lot more helpful if you let us know more specifically where you're stuck. –  KReiser Aug 9 '12 at 22:34
I am having problems deriving a difference equation from the information. Once I have an equation I should be able to solve it myself. –  Apeman Aug 9 '12 at 22:43
Also please add the homework tag, if appropriate. –  Alexander Thumm Aug 9 '12 at 23:14
Welcome to Math.StackExchange! I've posted an answer for (i), so I hope that will get you started. Let me know if there's anything else that you could use help with. –  Matt Groff Aug 9 '12 at 23:29

As stated in the problem, $£m$, or for simplicity's sake, $x_0$, is the loan amount. We want to derive an equation for $x_n$, which is the amount owed after $n$ years. That's why we make the loan amount $x_0$, because it is the amount owed after 0 years. We do this by writing it in terms of $x_{n-1}$, the amount owed from the previous year. Doing this is the same as writing a difference equation; they are the same thing.

We start with $x_n$: $$x_n = ?$$

So we know that $x_{n-1}$ is the amount owed from the previous year. So we will say that $x_n$, the amount owed for the $n$th year, is related to $x_{n-1}$, the amount owed for the previous year: $$x_n = f(x_{n-1}) = x_{n-1} + \text{adjustments}$$

We make 12 monthly payments of $m$, so we will subtract $12m$ from this: $$x_n = x_{n-1} - 12m + \text{adjustments}$$

Then we are charged $r$ times the amount owed the previous year, which is $x_{n-1}$: $$x_n = x_{n-1} - 12m + r\cdot x_{n-1}$$

That gives (i).

An alternate form of (i) can be obtained by just using arithmetic: $$x_n = (1 + r)x_{n-1} - 12m$$

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Probably the interest charge changes by the month. Then it is not natural to use a recurrence that goes by the year. Can be done, but messier. –  André Nicolas Aug 9 '12 at 23:46
Thank you for your help. I can't seem to get the correct value for the monthly repayment. I keep getting an answer of £292.30 not £295.64. –  Apeman Aug 11 '12 at 22:15
@Apeman: I'm going to try to go through the rest of the problem to see what answer I get. I'm wondering if we may need to go through André Nicolas's idea to get the correct value for the repayment... –  Matt Groff Aug 12 '12 at 21:49
@Apeman: I got the correct answer for (iii), and we don't need André Nicolas's method. Let's see if our answers for (ii) match... What did you get for (ii)? –  Matt Groff Aug 13 '12 at 1:18

i appreciate this might be a bit late to reply, but it seems that you should only be using the calculated value of a to 2 decimal places, i.e. a =1.05, rather than a=1.0511...

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Welcome to MSE! While it might be a good idea to round numbers this is not really an answer to the question. –  Jay Feb 26 '14 at 0:51
This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post - you can always comment on your own posts, and once you have sufficient reputation you will be able to comment on any post. –  PVAL Feb 26 '14 at 0:55