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