Science:Math Exam Resources/Courses/MATH105/April 2010/Question 05 (b)/Solution 2

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!