# Thread: using a transformation to simplify a differential equation

1. ## using a transformation to simplify a differential equation

Hi all,
Is it just me or is there something strange going on here.

Consider the following differential equation:

$\displaystyle 0=\frac{1}{2}-f(x)+f'(x)[3(\frac{1}{2}-x)-2(\frac{1}{2}-f(x))]$

Consider the transformation (x<1/2)

$\displaystyle h(x)=\frac{1/2-f(x)}{1/2-x}$.

Then the differential equation is supposed to become:

$\displaystyle h'(2h-x)(\frac{1}{2}-x)=2h(h-2)$

However, when I did it I got:

$\displaystyle h'(2h-3)(\frac{1}{2}-x)=2h(h-2)$.

I did the following:

Calculate f'(x):

$\displaystyle f(x)=\frac{1}{2}-h(x)(\frac{1}{2}-x)$
$\displaystyle => f'(x)=h(x)-h'(x)(\frac{1}{2}-x)$

Substitute f'(x) into the DE, also replacing (1/2-f(x)), to get

$\displaystyle 0=h(\frac{1}{2}-x)+(h-h'(\frac{1}{2}-x))[3(\frac{1}{2}-x)-2h(\frac{1}{2}-x)]$
$\displaystyle 0=(\frac{1}{2}-x)[h+(h-h'(\frac{1}{2}-x))(3-2h)]$
$\displaystyle 0=(\frac{1}{2}-x)[4h-2h^{2}-h'(\frac{1}{2}-x))(3-2h)]$
$\displaystyle h'(\frac{1}{2}-x)(3-2h)=2h(2-h)$
$\displaystyle h'(\frac{1}{2}-x)(2h-3)=2h(h-2)$

So what am I missing here (I'm 99% sure it's me who's wrong)?

Thanks very much!

2. Originally Posted by honda
Hi all,
Is it just me or is there something strange going on here.

Consider the following differential equation:

$\displaystyle 0=\frac{1}{2}-f(x)+f'(x)[3(\frac{1}{2}-x)-2(\frac{1}{2}-f(x))]$

Consider the transformation (x<1/2)

$\displaystyle h(x)=\frac{1/2-f(x)}{1/2-x}$.

Then the differential equation is supposed to become:

$\displaystyle h'(2h-x)(\frac{1}{2}-x)=2h(h-2)$

However, when I did it I got:

$\displaystyle h'(2h-3)(\frac{1}{2}-x)=2h(h-2)$.

I did the following:

Calculate f'(x):

$\displaystyle f(x)=\frac{1}{2}-h(x)(\frac{1}{2}-x)$
$\displaystyle => f'(x)=h(x)-h'(x)(\frac{1}{2}-x)$

Substitute f'(x) into the DE, also replacing (1/2-f(x)), to get

$\displaystyle 0=h(\frac{1}{2}-x)+(h-h'(\frac{1}{2}-x))[3(\frac{1}{2}-x)-2h(\frac{1}{2}-x)]$
$\displaystyle 0=(\frac{1}{2}-x)[h+(h-h'(\frac{1}{2}-x))(3-2h)]$
$\displaystyle 0=(\frac{1}{2}-x)[4h-2h^{2}-h'(\frac{1}{2}-x))(3-2h)]$
$\displaystyle h'(\frac{1}{2}-x)(3-2h)=2h(2-h)$
$\displaystyle h'(\frac{1}{2}-x)(2h-3)=2h(h-2)$

So what am I missing here (I'm 99% sure it's me who's wrong)?

Thanks very much!
Dear honda,