Question regarding usage of absolute value within natural log in solution of differential equation

You should have, as your general solution, $$ -\ln|6-y|=x+C\ \quad\iff\quad |6-y|=e^C e^{-x} . $$

If $y-6>0$, you have the solution $$y-6= e^Ce^{-x}\ \quad\iff\quad y=6+ e^Ce^{-x} . $$

If $y-6<0$, you have the solution $$6-y= e^Ce^{-x}\ \quad\iff\quad y=6- e^Ce^{-x} . $$

In either case, the solution can be written as $y=6- Ce^{-x} $, for some constant $C$ (different from the $C$ above).

Tags:

Absolute Value

Ordinary Differential Equations

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