Science:Math Exam Resources/Courses/MATH105/April 2010/Question 05 (b)/Solution 1

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.