# Differential Equation

• November 17th 2010, 07:31 AM
BabyMilo
Differential Equation
Given $dy/dx = -\frac{8y^2}{(x-1)^3}$ at point (x,y)
and the curve goes through (2, -1/3)

Prove the equation of the curve,

$y=\frac{x^2-2x+1}{x^2-2x-3}$
• November 17th 2010, 07:34 AM
Unknown008
Separate the variables and integrate...

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

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

Or, you can find directly:

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

EDIT: You should know by now that you should post what you have tried.
• November 17th 2010, 07:36 AM
HallsofIvy
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 $y^{-2}dy= -\frac{8 dx}{(x- 1)^3}$ and those are easy to integrate.
• November 17th 2010, 07:40 AM
BabyMilo
i have

$\displaystyle \dfrac{-1}{y} =\dfrac{-4}{(x-1)^2}-1$
• November 17th 2010, 07:46 AM
Unknown008
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.
• November 17th 2010, 07:54 AM
BabyMilo
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.