ERTH/OCN312: ODE's: Linear 1st order, homogeneous cases

ERTH/OCN312: Advanced Mathematics for Engineers and Scientists
Prof. Garrett Apuzen-Ito
University of Hawaii, Dept. of Earth Sciences
School of Ocean and Earth Science and Technology (SOEST)

[garrettapuzen-ito]

9:30 How do we prove the integral of 1/y is ln(|y|)+C?
If y>0, then the integral(1/y*dy) = ln(y)+C.
However, if y<0, then we cannot say that the integral (1/y*dy)= ln(y)+C because the log of a negative number is not defined.
Instead we can do a change of variables with u=-y and (remember du=-dx).
Integral(1/u*du)=ln(u)+C. Re-substituting u=-y, we get integral(1/y*dy)=ln(-y)+C, and since y<0, then ln(-y) is the log of a positive number, which is defined.
Thus integral(1/y*dy)=ln(y)+C if y>0 and =ln(-y)+C if y<0. In short integral(1/y*dy)=ln(|y|)+C.