## Related questions with answers

Demetrios opens an account with an initial investment of $2000. The annual interest rate is 5%. (a) If the interest is compounded continuously and Demetrios makes an additional$1000 deposit every year, what will be the balance at the end of 10 years? (b) If the interest is compounded quarterly (four times per year) and $250 is deposited at the end of each compounding period, what will be the balance after 10 years? (c) What happens if the interest is compounded daily?

Solution

Verified$\textbf{a)}$

$\textbf{Our model}$

$\frac{dP(t)}{dt}=rP(t)+D$

$P(t)-\text{ balance on the bank account in moment $t$}$

$r=0.05-\text{ annual interest rate}$

$D=1000\$-\text{ annual deposit}$

Our model equation is first order linear differential equation. To $\text{solve it, we must find integration factor. $\mu (t).$ First define function }$

$\text{ $p(t)$ from our model equation}$

$\frac{dP(t)}{dt}-rP(t)=D$

$\Rightarrow p(t)=-r$

$\Rightarrow$

$\mu (t)=\mathrm{e}^{\int p(t)dt}=\mathrm{e}^{\int (-r)dt}=\mathrm{e}^{-rt}$

$\text{Now, we multiply all equation with integration factor }\mu (t)$

$\frac{dP(t)}{dt}-rP(t)=D\quad\bigg/ \cdot \mathrm{e}^{-rt}$

$\frac{dP(t)}{dt}\mathrm{e}^{-rt}-rP(t)\mathrm{e}^{-rt}=D\mathrm{e}^{-rt}$

$\frac{d}{dt}\bigg(P(t)\mathrm{e}^{-rt} \bigg)=D\mathrm{e}^{-rt}$

$\text{After integration both sides of our equation for $t$, we get:}$

$\int\frac{d}{dt}\bigg(P(t)\mathrm{e}^{-rt} \bigg)dt=\int D\mathrm{e}^{-rt}dt$

$P(t)\mathrm{e}^{-rt}=-\frac{D}{r}\mathrm{e}^{-rt}+C,\quad C=const$

$\text{After we multiply last expression with }\mathrm{e}^{rt}\text{ we get}$

$\boxed{P(t)=-\frac{D}{r}+C\mathrm{e}^{rt}}\dots (**)$

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