From part (a), we have
![{\displaystyle {\frac {dy}{dt}}=0.05y-A,}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/d61f11d21b3d18b7e71897377def00016cf8e469)
which we will solve by separation of variables. Knowing the solution will allow us to give a condition on
so that after 25 years, the balance owed is zero.
We notice that the equation is separable and rearrange to find
.
Now we integrate to get
.
Therefore,
![{\displaystyle {\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}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/442a3e2fe84350e487737409493f9fa6614ee209)
where
is an arbitrary constant that comes out of the integration,
![{\displaystyle \displaystyle \pm e^{C/20}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/413c1e303ef5658a8918fa9a76f0b6105d3452a4)
in this case.
After one last rearrangement, we see that the amount of money owed after
years is
![{\displaystyle \displaystyle {}y(t)=20A+20Fe^{t/20}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/9240fa0e99b1d94234bc91a9bf8a20f61f45075f)
or
![{\displaystyle \displaystyle {}y(t)=20A+Ge^{20t}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/dcd5846efe19a7da3bc2b4a853b6a882fd70be92)
for another arbitary constant
.
From the initial condition,
so
or
. Finally then our full equation for the amount of money owing,
is,
.
Setting
(which corresponds to having no money owing after 25 years) forces
![{\displaystyle \displaystyle {}(240000-20A)e^{25/20}+20A=0}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/dccd66629e1e9e6f3fbc2f1447e07c1985ec7436)
so
.
Therefore, the annual payments are
= $16818.61.