What is the Correct Domain for this ODE Solution?

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SUMMARY

The discussion centers on the correct domain for the solution of the ordinary differential equation (ODE) y' = 1 + y^2. The solution is derived as y = tan(x + C), with an initial proposed domain of x ∈ (-π/2 - C, π/2 - C). However, it is clarified that for a general solution, the domain should be expressed as x ∈ (-π/2 - C + kπ, π/2 - C + kπ), where k ∈ ℤ, to account for all possible solutions. This distinction is crucial for accurately representing the solution's behavior across its periodic nature.

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twoflower
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Hi all,

here is one ODE I solved now

[tex] y' = 1 + y^2[/tex]

So

[tex] \frac{y'}{1+y^2} = 1[/tex]

[tex] \int \frac{dy}{1+y^2} = \int 1 dx[/tex]

[tex] \arctan y = x + C \leftrightarrow y = \tan (x + C)[/tex]

[tex] x \in (-\frac{\pi}{2} - C, \frac{\pi}{2} - C)[/tex]

The last line is what I'm unsure about.

Shouldn't it rather be

[tex] x \in (-\frac{\pi}{2} - C + k\pi, \frac{\pi}{2} - C + k\pi), k \in \mathbb{Z}[/tex]

or is it ok as I wrote it originally?

Thank you.
 
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It depends. Usually you would determine the constant by an initial value, ie, y(x0)=y0, in which case the solution would be valid within whichever region (ie, of width [itex]\pi[/tex]) contains x<sub>0</sub>. If you're just looking for a completely general form, one that represents every possible solution, then your second form is correct.[/itex]
 

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