Seperable DE with Initial Condition

[PsychonautQQ]

Homework Statement

dy/dx =4yx^3-y y(1)=-3
dy/y = (4x^3-1)dx
ln(y) = x^4-x+C
y = e^(x^4-x+C)

But an answer source says that after the integration I get
ln(y) = x^4 - x + ln(C)
so then..
ln(y/c) = x^4 - x
y = Ce^(x^4 - x)

which makes it much easier to solve for the constant given the initial condition... My question is why when you take the integral is the constant ln(C) instead of just C...?

 

[Chestermiller]

Homework Statement

dy/dx =4yx^3-y y(1)=-3
dy/y = (4x^3-1)dx
ln(y) = x^4-x+C
y = e^(x^4-x+C)

But an answer source says that after the integration I get
ln(y) = x^4 - x + ln(C)
so then..
ln(y/c) = x^4 - x
y = Ce^(x^4 - x)

which makes it much easier to solve for the constant given the initial condition... My question is why when you take the integral is the constant ln(C) instead of just C...?

It's really the same thing, but it makes the final result look simpler. The two things are equivalent.