Finally, we find that it takes *t*=21.638 days from 3 days ago for the exchange rate to drop to 6.1:

$\begin{array}{c}{r}_{t}=6.1=6+0.5\times {\mathrm{e}}^{-0.07438t}\\ t=\frac{1}{-0.07438}\mathrm{ln}\frac{0.1}{0.5}=21.638\end{array}$

4.13.

a. The equation is separable. We will divide both sides by *r*(*μ*−*r*) to separate and integrate:

$\int \frac{\mathrm{d}r}{r(\mu -r)}=\int \lambda \mathrm{d}t$

b. Using partial fractions, this is written:

$\int \left(\frac{1}{\mu r}+\frac{1}{\mu (\mu -r)}\right)\mathrm{d}r=\int \lambda \mathrm{d}t$

It is easy ...

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