(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Given the IVP problem

[tex]\rm{(1+x^2)^{-1} \frac{dx}{dt} = \frac{\alpha \cdot 2 \pi}{4} \cdot cos(t)} \iff [/tex]

[tex]\rm{(1+x^2)^{-1} dx = \frac{\alpha \cdot 2 \pi}{4} \cdot cos(t)} dt[/tex]

Find the maximal solution with the initial conditio condition x(0) = 0.

Then alpha is either [tex]\alpha_{1} = \frac{1}{4}[/tex] or [tex]\alpha_{2} = \frac{1}{2}[/tex]

3. The attempt at a solution

I am fairly new to differential equation theory like this, but couldn't my professor mean "Maximum interval of existence"? By that meaning "maximal solution" is equivalent to the statement that finding the maximal interval on which the solution curve is defined.

Thusly if I try to solve the above equation I get

[tex]\rm{tan(x)^{-1} = \frac{\alpha \cdot 2 \pi}{4} \cdot sin(t) + C}[/tex]

[tex]x = \rm{tan(\frac{\alpha \cdot \pi}{2} \cdot sin(t) + C)}[/tex]

What I don't get here is that if I insert the values of first or second alpha then I get an interval which is centered around zero.

But that can be right can it?? Here I am stuck and need help.....

Best Regards

Mathboy

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Maximum interval of existence

**Physics Forums | Science Articles, Homework Help, Discussion**