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- Jan 30, 2018

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y'' = uv'' +2u'v'+ u''vyoutried? 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.

so

y''+ Vy = uv'' +2u'v'+ u''v + Vuv = 0

- Jan 30, 2018

- 731

Okay, and since u satisfies u''+ Vu= 0, that isy'' = uv'' +2u'v'+ u''v

so

y''+ Vy = uv'' +2u'v'+ u''v + Vuv = 0

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},

$v(x)=\int^x \frac{d\chi}{\chi^2}$

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