I Reduction of order in solving second order differential equations

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chwala
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TL;DR
Why is the constant dropped when determining solutions to second order differential equations. (See highlight in red -attached). Otherwise, the reduction of order approach is pretty straightforward.
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chwala said:
TL;DR Summary: Why is the constant dropped when determining solutions to second order differential equations. (See highlight in red -attached). Otherwise, the reduction of order approach is pretty straightforward.

View attachment 338047
Because keeping ##k## is the same thing as adjusting the value of the arbitrary constant ##c_1## in the general solution?
 
renormalize said:
Because keeping ##k## is the same thing as adjusting the value of the arbitrary constant ##c_1## in the general solution?
Thanks noted...was wondering why they substituted for the constant ##c=-3## and dropped the other constant ##k##. The constant ##c## was not dropped as indicated rather a value was assigned to it. .

Is this not for convenience? to perhaps have " nice solutions'.

Cheers man.
 
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chwala said:
Thanks noted...was wondering why they substituted for the constants ##c=-3## and dropped ##k##. The constant ##c## was not dropped as indicated rather a value was assigned to it. .
Just like keeping ##k## amounts to redefining ##c_1##, wouldn't keeping ##-c/3## simply adjust the value of the arbitrary constant ##c_2##? These are the types of simplifications that come naturally once you're practiced enough solving differential questions. When solving a 2nd-order ODE, as long as you're left in the end with two arbitrary constants, you know you've found the general solution.
 
I have the equation ##F^x=m\frac {d}{dt}(\gamma v^x)##, where ##\gamma## is the Lorentz factor, and ##x## is a superscript, not an exponent. In my textbook the solution is given as ##\frac {F^x}{m}t=\frac {v^x}{\sqrt {1-v^{x^2}/c^2}}##. What bothers me is, when I separate the variables I get ##\frac {F^x}{m}dt=d(\gamma v^x)##. Can I simply consider ##d(\gamma v^x)## the variable of integration without any further considerations? Can I simply make the substitution ##\gamma v^x = u## and then...

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