MHB Please give me an idea for Reduction of order

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What have you tried? If y= uv then what is y'? What is y''? What do you get when you put those into the differential equation? And then use the fact that u itself satisfies the equation, that u''+ V(x)u= 0.
 
Country Boy said:
What have you tried? If y= uv then what is y'? What is y''? What do you get when you put those into the differential equation? And then use the fact that u itself satisfies the equation, that u''+ V(x)u= 0.
y'' = uv'' +2u'v'+ u''v

so

y''+ Vy = uv'' +2u'v'+ u''v + Vuv = 0
 
Another said:
y'' = uv'' +2u'v'+ u''v

so

y''+ Vy = uv'' +2u'v'+ u''v + Vuv = 0
Okay, and since u satisfies u''+ Vu= 0, that is
uv''+ 2u'v'+ v(u''+ Vu)= uv''+ 2u'v'= 0.

uv''= -2u'v'.

Let p= v'. Then up'= -2u'p so that p'/p= -2u'/u.
The order of the equation has been "reduced" from 2 to 1.

Integrating both sides ln(p)= -2ln(u)+ c= ln(u^{-2})+ c.

Taking the exponential of both sides, p= v'= Cu^{-2},
$v(x)=\int^x \frac{d\chi}{\chi^2}$
 
Last edited:
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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