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Consider the equation $\int \frac{dy}{y}=k\int dt$. According to my book, integrating both sides gives $\ln|y|=kt+\ln(C)$.

Could someone explain me why does one get $\ln{C}$ after integrating right side of the equation? I'm getting $+C$ instead.

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    Possible duplicate of [Differential equation / Separable](http://math.stackexchange.com/questions/2112148/differential-equation-separable) – 2017-01-24
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    Both are constant, so your solution is same as textbook. Taking on ln c as a constant gives a form as explained by fly by night. Intact you can even take it as e^c if you like. – 2017-01-24

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If $\ln|y|=kt+\ln C$ then $|y| = \mathrm e^{kt + \ln C}$.

Using basic laws of exponentials: $|y|=\mathrm e^{kt + \ln C} = \mathrm e^{kt}\times \mathrm e^{\ln C} = \mathrm e^{kt} \times C$, where $C>0$.

We can drop the condition that $C>0$ as long as we drop the modulus from $y$.

$$y=C\mathrm e^{kt}$$

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Using $\ln{C}$ as the constant of integration is fine since the constant is arbitrary.

Here, I can obtain the answer without using $\ln{C}$ as the constant if integration:

$$\int \frac{dy}{y}=k\int dt$$ $$\ln{|y|}=kt+C$$ Exponentiating both sides: $$e^{\ln{|y|}}=e^{kt+C}$$ $$|y|=e^{kt}\cdot e^C$$ Since $e^C$ is arbitrary, we can substitute it for some new arbitrary constant $A$. $$|y|=Ae^{kt}$$ Since the right hand side will always be greater than $0$, we may remove the modulus sign ($e^C$ and $e^{kt}$ both positive) to give the general solution: $$y=Ae^{kt}$$ To conclude: The $\ln{C}$ was used to avoid the substitution at the end with $A$. This allows for a simpler form for the solution to our differential equation.

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Actually, the use of the logarithm has one other crucial effect: it enforces that $C$ cannot be zero, since $\log (0)$ is undefined.

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    I don't think this addresses the question in the OP directly; she's asking why we can equivalently write $C$ and $\ln(C)$. – 2017-01-24