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This is a rather trivial question but differential equations isn't my field and my professor advised me to do something I don't completely buy. We are given

$$x'=\frac{x^2-1}{2}$$

whose general solution is

$$\left| \frac{x-1}{x+1} \right| = e^{t+c}$$

Now, the goal is to find the maximal interval of existence for the initial conditions $(0,0)$ and $(\ln(2),-3)$. However, when the absolute value bars these give the same constant and, by uniqueness, hence the two initial conditions must be describing the same IVP. However, my professor just dropped the absolute value and continued on. Why is this legal?

Thanks!

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    $\begingroup$ Small note: this equation isn't linear $\endgroup$ – themathandlanguagetutor Sep 24 '17 at 16:26
  • $\begingroup$ Your solution should be the natural log of the LHS $\endgroup$ – themathandlanguagetutor Sep 24 '17 at 16:31
  • $\begingroup$ Thanks for the notes. Was not being particularly attentive. Fixed both issues. $\endgroup$ – Aaron Zolotor Sep 24 '17 at 16:36
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These do NOT describe the same IVP; observe that the solution $$\left\lvert\frac{x+1}{x-1}\right\rvert=Ce^t$$ is undefined for $x=1$ and $x=-1$, thus any particular solution cannot cross those lines. Since your first initial conditions lies between these lines observe that $\left\lvert\frac{x+1}{x-1}\right\rvert<0$ so the absolute value can be replaced with a negative (which can then be attached to the constant when it's pulled out of the exponential), whereas, since your second initial condition is below the lower of the lines, the expression is positive and the absolute value can be ignored straight out.

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