Explain a step for solving differential equation

Question

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.

Answer 1

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 $y$.

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

Answer 2

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.

Answer 3

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