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We can use the integral of $\frac{1}{x}$ in order to solve a separable first-order linear equation like this:

$\frac{dy}{dt} + f(t) y = 0$

$ ln |y| = \left(-\int f(t)\,dt\right) + C $

and then:

$y = \pm e^{\left(-\int f(t)\,dt\right) + C} = \pm e^{C} e^{-\int f(t)\,dt}$

As far as I know, we are able to remove the absolute value because $y=0$ it's a solution, and, thanks to Cauchy theorem, no solution of the ode may change its sign.

I'm okay with this, the problem is that I found some exercise on my textbook where the absolute value is removed, even if $y=0$ is not a constant solution of the ODE:

E.G. $$ x' = \frac{3x-2}{t^2+1} $$

$ \frac{1}{3} log|3x-2| = arctan(t) + c$

$ 3x - 2 = ce^{3arctan (t)}$ where $ c \neq 0$

Why it was possible to remove the absolute value here? Why $3x-2$ doesn't change its sign?

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  • $\begingroup$ In your example, $x=2/3$ is a solution. Since solutions lines do not cross, $x-3/2$ does not change its sign. $\endgroup$ Commented Jan 21, 2013 at 18:04
  • $\begingroup$ can you make clear to me "Cauchy theorem, no solution of the ode may change its sign" ? I am not getting this one $\endgroup$ Commented Jan 22, 2013 at 15:41
  • $\begingroup$ @RealHilbert: I think the OP wanted to know why the sign of absolut value is ommited while we are used to consider it routinely? For example, for an OE we got $|y+1|=\exp(x)+c$. This is what we always write after solving an first order OE. Now, consider a writer noted $y+1=\exp(x)+c$. It is acceptable that someone asks why didn't he put the ||. This is what, I was trying to say the OP, but it seems I have been fail in it. :( $\endgroup$ Commented Jan 22, 2013 at 19:52

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Indeed, if you want write it as $\ln(3x-2)$, you must consider the region $I: x\neq 2/3$ instead. Unless, the term $\ln(3x-2)$ doesn't make sense at all. I think, if the writer didn't noted the right region as $I$, or take a positive sign as you noted, probably he assumed the right region before or he wanted to take one 1-parameter of solutions. For example, in OE $$yy'=(y+1)^2$$ by solving we get $$\frac{1}{y+1}+\ln|y+1|=x+c$$ we just need to indicate $y\neq-1$.

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  • $\begingroup$ I hope I didn't misunderstood your answer. Your are saying that this solution if only valid for $x > 2/3$, aren't you? If it is true, does it mean that we are ignoring the solutions defined for $ x \in (-oo, 2/3) $ $\endgroup$ Commented Jan 21, 2013 at 17:24
  • $\begingroup$ I fixed the answer. $\endgroup$ Commented Jan 21, 2013 at 17:43
  • $\begingroup$ I'm sorry, but I still didn't understand if solution defined for $ x \in (−oo,2/3) $ are possible, and if the author chose just to ignore them. The exercise that I wrote here doesn't add any costraint, it just ask for a solution of the ODE. $\endgroup$ Commented Jan 21, 2013 at 17:55
  • $\begingroup$ @user47581: Leave the absolute value and just consider $x=2/3$. Always, you can take a positive or the negative one by ignoring $|...|$. We must just care about excluding 0. $\endgroup$ Commented Jan 21, 2013 at 18:03
  • $\begingroup$ @user47581: I don't know your source, but probably the writer chose one of 1-parameter family of solutions. Another one may arise by $$-(3x-2)=..., x\neq 2/3,c\neq 0$$ $\endgroup$ Commented Jan 21, 2013 at 18:05

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