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Has somebody please explain for me how to derive 2.6 equation?

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    What do you know about integration? Have you tried anything? – 2017-01-07
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    I tried to separate into $\frac{dS}{S} = R_o dR$, integrate both sides we have $lnS = \int R_odR$. What to next since to convert $R_o$ to R? and t parameter? – 2017-01-07
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    It depends, is $R_0$ a function of $R$ or is it a constant? Well, I know what it is according to the solution given but do you? – 2017-01-07

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You might be confused because $S$ and $R$ are functions of $t$, but the derivative is with respect to $R$. You're correct (except for dropping the minus sign) at $\ln S = \int -R_0 \; dR.$ But $R_0$ is a constant, so the integral is easy: $\ln S = -R R_0 +C$. Exponentiate to get

$$S = e^{-RR_0+C} = Ae^{-RR_0}.$$

We've done this without thinking about $t$, but we can put it in now:

$$S(t) = Ae^{R(t)R_0}.$$

Since I don't have the text, I have to assume that $R(0)=0$. That is probably evident from the type of problem. So plug in $t=0$ and we get

$$S(0) = A e^0 = A$$. Plug that in for $A$ above and you get your equation.

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You're already in the right direction. Using separation of variables: $$ \frac{dS}{dR} = -\frac{\beta S}{\gamma} \implies \frac{1}{S}dS = -\frac{\beta}{\gamma}dR$$ Integrate both sides:

$$ \ln S(t) = -\frac{\beta R(t)}{\gamma} + C $$ where $C$ is a constant $$ S(t) = e^c \cdot e^{-\frac{\beta}{\gamma}R(t)}$$

Note that $R_0 = \frac{\beta}{\gamma}$ and $e^c$ will be your $S(0)$