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ajaday · Oct 28, 2008, 08:41 PM

ln(y-2x)=y
I am trying to figure out how to differentiate this equation twice and solve for d^2/dx^2 but I am stuck on solving for dy/dx . What I got so far is 1/(y-2x) *(dy/dx-2)=dy/dx

galactus · Oct 29, 2008, 11:02 AM

$y=ln(y-2x)$

$y'=\frac{1}{y-2x}\cdot (y'-2)$

$y'=\frac{y'-2}{y-2x}$

Now, factor out y' and solve for y' and get $y'=\frac{2}{2x-y+1}$

Now, the thing most people fail to realize is to differentiate again. When you do, you will have

y' in the second derivative. Then, you must sub in the y' you found and simplify.

Let us know how you progress.

The algebra is the trickiest thing.