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I f a transformation $y = u v$ transforms the given differential equation $f(x) y'' - 4 f'(x) y'+ g(x) y = 0 $ into the equation of the form $v'' + h(x) v = 0 $ then $u$ must be

  1. $ \frac{1}{f^2} $
  2. $xf $
  3. $ \frac{1}{2f} $
  4. $f^2$

I am stuck on this problem. Can anyone help me please...............

1 Answers 1

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take $y=uv$ we have

$y'=u'v+v'u$

and

$y''=u''v+2u'v'+uv''$

replace in equation we have

$(f(x)u''-4f'(x)u')v+f(x)uv''+(f(x)2u'-4f'(x)u)v'=0$

Then if we want get the equation

$v''+h(x)v=0$

we can assume that

$f(x)2u'-4f'(x)u=0$ solving this simple equaton to u

we have

$u=e^{2lnf}=f^2$,

and we get the equation desired,

$v''+h(x)v=0$,

where $h(x)=\frac{f(x)u''-4f'(x)u'}{f(x)u}$