Science:Math Exam Resources/Courses/MATH105/April 2010/Question 05 (b)

MATH105 April 2010

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Question 05 (b)

A person purchased a home at the price of $ 300,000, paid a down payment equal to 20% of the purchase price, and financed the remaining balance with a 25 year term mortgage. Assume that the person makes payments continuously at a constant annual rate $A$ and that the interest is compounded continuously at the rate of 5%. Determine A, the rate of annual payments that is required to pay off the loan in 25 years.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
We have a differential equation to model the rate of the amount of money. What techniques do we have to turn that rate, ${\frac {{\textrm {d}}y}{{\textrm {d}}t}}$ into the actual amount of money left owed, $y(t)$?

Hint 2
How can you use separation of variables to write your differential equation in terms of some integrals that you know how to evaluate?

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. From part (a), we have ${\frac {dy}{dt}}=0.05y-A,$ which we will solve by separation of variables. Knowing the solution will allow us to give a condition on $A$ so that after 25 years, the balance owed is zero. We notice that the equation is separable and rearrange to find ${\frac {dy}{0.05y-A}}=dt$. Now we integrate to get $\int {\frac {dy}{0.05y-A}}=20\ln |0.05y-A|=t+C$. Therefore, ${\begin{aligned}\ln |0.05y-A|&=t/20+C/20\\|0.05y-A|&=e^{t/20+C/20}=e^{C/20}e^{t/20}\\0.05y-A&=Fe^{t/20}\end{aligned}}$ where $F$ is an arbitrary constant that comes out of the integration, $\displaystyle \pm e^{C/20}$ in this case. After one last rearrangement, we see that the amount of money owed after $t$ years is $\displaystyle {}y(t)=20A+20Fe^{t/20}$ or $\displaystyle {}y(t)=20A+Ge^{20t}$ for another arbitary constant $G$. From the initial condition, $y(0)=240000$ so $20A+G=240000$ or $G=240,000-20A$. Finally then our full equation for the amount of money owing, $y(t)$ is, $\displaystyle {}y(t)=(240000-20A)e^{t/20}+20A$. Setting $y(25)=0$ (which corresponds to having no money owing after 25 years) forces $\displaystyle {}(240000-20A)e^{25/20}+20A=0$ so $A={\frac {240000e^{25/20}}{20(e^{25/20}-1)}}={\frac {12000e^{5/4}}{e^{5/4}-1}}=16818.61$. Therefore, the annual payments are $A$ = $16818.61.

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. From part (a) we have that ${\frac {{\textrm {d}}y}{{\textrm {d}}t}}=0.05y(t)-A$ with the initial amount of money owed being $y(0)=240000$. The goal is to solve this equation for the rate of $y$ so that we can actually obtain the quantity we want, $y(t)$ which is the money owed after $t$ years. Consider the substitution, $\displaystyle {}u(t)=0.05y-A$ which implies that ${\frac {{\textrm {d}}u}{{\textrm {d}}t}}=0.05{\frac {{\textrm {d}}y}{{\textrm {d}}t}}.$ Therefore we can rewrite (and simplify) our differential equation as ${\frac {{\textrm {d}}u}{{\textrm {d}}t}}=ru$ which we recognize as the exponential model and so we know the solution is $\displaystyle {}u(t)=B\exp(0.05t)$ with $B$ an arbitrary constant. This solution can also be determined from separation of variables and integration. Putting back into our original variable, $y(t)$, we get $y(t)={\frac {B\exp(0.05t)+A}{0.05}}.$ which is our desired equation for the amount owed on the mortgage after time $t$. Using the initial condition $y(0)={\frac {B+A}{0.05}}=240000$ we get that $B=12000-A$. Therefore we are able to write the amount of money owed solely in terms of the annual payments, $A$, $y(t)={\frac {(12000-A)\exp(0.05t)+A}{0.05}}.$ Now when $t=25$, we want that the mortgage is paid off, i.e., we want the amount of money we owe to be zero. Therefore we seek that $y(25)=0$. Therefore, ${\begin{aligned}0&={\frac {(12000-A)\exp(0.05\cdot 25)}{0.05}}\\A&={\frac {12000\exp(0.05\cdot 25)}{\exp(0.05\cdot 25)-1}}=16818.61.\end{aligned}}$ Therefore, in order to pay off the mortgage in 25 years, we require that $A$ = 16818.61. Recall that when we started paying the mortgage at year zero we owed 80% of the value or 240 000 dollars. If we want to find out how much we really paid, we just need to multiply our annual rate, $A$ by 25 years to get $420 465.25. Add this to the $60 000 we already paid up front then our $300 000 dollar house has actually cost us $480 465.25!

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