Here's my attempt at a solution to the differential equation $y' - y\tan(x) = 0$:

$y' - y\tan(x) = 0 \Rightarrow \frac{y'}{y} = \tan(x) \\ \Rightarrow \int \frac{y'}{y} = \int \tan(x) + C \\ \Rightarrow \log y = -\log|\cos x| + C \\ \Rightarrow y = \frac{e^C}{|cosx|} = e^C|\sec x|$

My book lists the answer as $y = C \sec x$. I have two slightly embarrassing misunderstandings about why this is true:

  1. First, since the logarithm is defined only for positive real numbers, why is it that we can ignore the absolute value? For a fixed $C$, the two solutions clearly differ on points like $x = 3\pi/2$, where the secant is negative. Which one is correct?
  2. Why is it that we can reduce $e^C$ to just the constant $C$? $e^C$ doesn't take on negative values, so it seems like, for example, $y = - \sec x$ is a particular solution of the second general solution, but not the first.


  • $\begingroup$ Because $-log(cosx)$ is the same thing as $log(secx)$, that's the first part of your question $\endgroup$ – imranfat Aug 9 '19 at 2:00

First, and very importantly $$\int\frac{y'}{y}=\log |y|$$If you fix this, your solution is $$|y|=e^C |\sec(x)|$$

(2) Since $C$ is constant, so is $C'=e^{C}$. With this, your solution becomes $$|y|=C' |\sec(x)|$$ ($C'$ is positive)

(1) Remember that you solve the DE on an interval where everything makes sense.

Since $\tan(x)$ is not defined at $\frac{\pi}{2}+k \pi$, you are looking for the solution between two consecutive such numbers.

On any such interval, $\sec(x)$ does not change signs.

Therefore, for every $x$ your solution is $$y(x)= \pm C' \sec(x)$$ where the choice of $\pm$ could depend on $x$. But since the LHS is differentiable, hence continuous, the RHS is also continuous. Since $\sec(x)$ does not change sigh, this is only possible if the choice of $\pm$ is the same at all $x$ in the interval [this follows immediately from Intermediate Value Theorem]

  • $\begingroup$ Thanks! Your answer to (1) makes sense to me. I still don't understand your answer to (2); an arbitrary real choice of $C$ doesn't allow for $e^C < 0$, so you'd still have $C' \geq 0$, no? $\endgroup$ – TheProofisTrivium Aug 9 '19 at 3:20
  • 1
    $\begingroup$ @TheProofisTrivium That would be correct if your computation didn't contain a small mistake: What is $\int \frac{y'}{y}$? ;) $\endgroup$ – N. S. Aug 9 '19 at 3:22
  • $\begingroup$ @TheProofisTrivium See the edit $\endgroup$ – N. S. Aug 9 '19 at 3:28
  • $\begingroup$ Oops, my mistake. Thank you very much! $\endgroup$ – TheProofisTrivium Aug 9 '19 at 3:30

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