Linear Differential Equations in Kinematics

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SUMMARY

The discussion focuses on solving linear differential equations in kinematics, specifically the equations \(\ddot{x}(t)+\alpha \dot{x}(t)^2=0\) and \(\ddot{y}(t)+\alpha\dot{y}(t)^2=-g\). Participants highlight the challenges in integrating the equation \(\int \frac{1}{-g-\alpha \dot{y}^2}\:d\dot{y}\) and suggest using techniques such as arctan and arctanh for resolution. The conversation emphasizes the importance of correctly applying integration techniques and understanding the signs of forces involved in projectile motion.

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  • Understanding of linear differential equations
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  • Knowledge of integration techniques, including arctan and arctanh
  • Basic principles of air resistance and gravitational forces
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  • #31
P3X-018 said:
Havn't the gravity and aerodynamic forces same signs, when the projectil is on it's way to the top, and on the way down, they have opposite signs?

That's something totally different.I was considering free fall.Yes,the aerodynamics force changes the sign...

Daniel.
 
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  • #32
P3X-018 said:
Havn't the gravity and aerodynamic forces same signs, when the projectil is on it's way to the top, and on the way down, they have opposite signs?
Precisely!
However, you didn't really make it clear from the start that you were considering BOTH cases; I well understand Daniel's objection (he has worked under the most natural assumption given your info).
 
  • #33
Yea I know it was my fault. I didn't consider the 2 cases, but in the second equation then you got to use arctanh.
 
  • #34
P3X-018 said:
I didn't consider the 2 cases, but in the second equation then you got to use arctanh.

Since partial fraction decomposition disgusts me, go for artanh.
(I guess Daniel has a different view on this..:wink:)
 
  • #35
Take both methods... :-p You'll get the expression for "arctanh" in terms of "ln"...A useful result...A nice proof to it,also...

Daniel.

P.S.Arildno,i LOVE HYPERBOLIC FUNCTIONS... :!) :-p
 
  • #36
Well, but I don't have so much experience in seprating fractions :'(. How do you seprate the fraction
\frac{1}{1\pm u}
 
  • #37
P3X-018 said:
Well, but I don't have so much experience in seprating fractions :'(. How do you seprate the fraction
\frac{1}{1\pm u}
Those can't be decomposed further, but:
\frac{1}{1-u^{2}}=\frac{1}{2}(\frac{1}{1+u}+\frac{1}{1-u})
\frac{1}{1+u^{2}} cannot be partially fractiondecomposed in real partial fractions.
 

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