- #1
Lord Anoobis
- 131
- 22
Homework Statement
##y' = \frac{cos x}{sin y}##
##y' = \frac{6x^2}{y(1 + x^3)}##
Homework Equations
The Attempt at a Solution
So I was working through some textbook problems and there's something about the solutions of the above equations I don't quite understand.
The first one:
##\int sin y\, dy = \int cos x\, dx##
##-cos y = sin x + C##
##y = cos^{-1}(-sin x - C)##
Here the book shows the answer as ##cos y = -sin x + C##
The second equation:
With ##u = 1 + x^3## as a substitution we get
##\int y\, dy = 2\int \frac{du}{u}##
##\frac{1}{2}y^2 = 2(\ln |u| + C)##
##y = \pm\sqrt{4(\ln |1 + x^3| + C)}##
Book answer: ##y = \pm \sqrt{4\ln|1 + x^3| + C}##
So in the first one the sign of the constant remains positive, in the second it remains as ##C## instead of being multiplied. My question then is: does the ##C## in the book answer simply represent a constant inclusive of its sign and multiple(s) or have I overlooked something?