Please give me an idea for Reduction of order

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Discussion Overview

The discussion revolves around the method of reduction of order in solving differential equations, specifically focusing on the transformation of a second-order differential equation into a first-order equation. Participants explore the implications of substituting a function into the equation and the resulting mathematical manipulations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • Some participants suggest starting with the substitution y = uv and inquire about the derivatives y' and y''.
  • There is a reiteration of the substitution leading to the expression y'' + Vy = uv'' + 2u'v' + u''v + Vuv = 0.
  • One participant notes that since u satisfies the equation u'' + Vu = 0, this leads to a simplification of the terms.
  • It is proposed that the order of the equation has been reduced from 2 to 1 through the manipulation of the resulting expressions.
  • Integration is discussed, leading to the expression for p = v' as Cu^{-2} and the subsequent integral for v(x).

Areas of Agreement / Disagreement

The discussion appears to be exploratory with no clear consensus reached. Participants build upon each other's contributions but do not explicitly agree on a single approach or solution.

Contextual Notes

Participants do not clarify all assumptions made during the derivations, and the discussion includes various steps that may depend on specific definitions or interpretations of the functions involved.

Another1
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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:

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