Given $\displaystyle dy/dx = -\frac{8y^2}{(x-1)^3}$ at point (x,y)

and the curve goes through (2, -1/3)

Prove the equation of the curve,

$\displaystyle y=\frac{x^2-2x+1}{x^2-2x-3}$

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- Nov 17th 2010, 07:31 AMBabyMiloDifferential Equation
Given $\displaystyle dy/dx = -\frac{8y^2}{(x-1)^3}$ at point (x,y)

and the curve goes through (2, -1/3)

Prove the equation of the curve,

$\displaystyle y=\frac{x^2-2x+1}{x^2-2x-3}$ - Nov 17th 2010, 07:34 AMUnknown008
Separate the variables and integrate...

$\displaystyle \displaystyle \int \dfrac{1}{y^2}\ dy =-8 \int \dfrac{1}{(x-1)^3}\ dx$

Then, use the values of (2, -1/3) to find the value of the constant of integration.

Or, you can find directly:

$\displaystyle \displaystyle \int^y_{-\frac13} \dfrac{1}{y^2}\ dy =-8 \int^x_2 \dfrac{1}{(x-1)^3}\ dx$

EDIT: You should know by now that you should post what you have tried. - Nov 17th 2010, 07:36 AMHallsofIvy
Have you

**tried**anything your self? To show that a given function satisifies a differential equation is typically much simpler than actually solving the equation- just differentiate the given y and show that it satisfies the equation.

However, in this case the equation "separates" as $\displaystyle y^{-2}dy= -\frac{8 dx}{(x- 1)^3}$ and those are easy to integrate. - Nov 17th 2010, 07:40 AMBabyMilo
i have

$\displaystyle \displaystyle \dfrac{-1}{y} =\dfrac{-4}{(x-1)^2}-1$ - Nov 17th 2010, 07:46 AMUnknown008
EDIT: Hm... I'm not getting this. Either I made a mistake or you made one.

Make y the subject now, then expand the square that you have. - Nov 17th 2010, 07:54 AMBabyMilo
Got it. Ha you guys didnt really help.

It was just the rearrange i got a bit confused with tbh.

but thank you as always.