Absolute values resulting in diff-eqs

[process91]

Homework Statement

Consider a simple first-order linear differential equation, such as

$$y' + \tan x y = 0$$

With initial condition y(0)=C for some constant C. Find all solutions y which satisfy this differential equation on the entire real line.

Homework Equations

General method for solving such an equation:
Given
$$y' + P(x)y = 0$$
and an initial condition y(a)=b for some constant b, the solution will be given by
$$y=b \exp{\left(-\int_a^x P(t) dt\right)}$$

The Attempt at a Solution

$$\int_0^x \tan x dx = -ln|\cos x|$$
so
$$y=C \exp(ln|\cos x|) = C | \cos x |$$

However, the absolute value actually presents a problem. Leaving it in means that this is, in fact, not a solution for the differential equation, however nothing in the original question indicates that it should be removed.


To quote from http://tutorial.math.lamar.edu/Classes/DE/Linear.aspx" ,

Often [absolute value bars] can’t be dropped so be careful with them and don’t drop them unless you know that you can.

A similar problem is present with something like this:

$$y' - \frac {1}{x} y = 0$$

Wolfram Alpha has removed the absolute value signs on both of these, but they often remove absolute value signs when it is convenient in the answer without justification.

 

[LCKurtz]

In your first example, you can't get a solution beyond (-π/2,π/2) because tan(x) isn't defined at the end points. And on that interval you can indeed drop the || signs because cos(x) is positive there.

Similarly with 1/x in the equation, you aren't going to get a solution crossing the y axis. I would say you have to look at your particular equation and be careful about it.

 

[process91]

OK, sounds good. Thanks for the clarification.