Fraction Problem

Jason76

Senior Member
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Oct 19, 2012
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This is an excerpt from a diff equation, but one problem with algebra.

Goal is to get x's on one side and v's on the other, but I understand that part.

xdvdx=2vv21v\displaystyle x\dfrac{dv}{dx} = \dfrac{2v}{v^{2} - 1} - v

xdvdx=2vv21v1\displaystyle x\dfrac{dv}{dx} = \dfrac{2v}{v^{2} - 1} - \dfrac{v}{1}

xdvdx=2vv21v1v21v21\displaystyle x\dfrac{dv}{dx} = \dfrac{2v}{v^{2} - 1} - \dfrac{v}{1}\dfrac{v^{2} - 1}{v^{2} - 1} Ok, solving the top here makes sense, but not the bottom.

xdvdx=3vv3v21\displaystyle x\dfrac{dv}{dx} = \dfrac{3v - v^{3}}{v^{2} - 1} As I said, what's on top makes sense, but not the bottom.

UPDATED: Oh wait, I now see that subtraction of fractions was going on, so that explains the v21\displaystyle v^{2} - 1 on the bottom, as that is LCD.
 
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I don't understand whatever do you mean by "top here makes sense, but not the bottom". Continuing...

v213vv3dv=dxx\displaystyle \dfrac{v^{2} - 1}{3v - v^{3}}{dv} = \dfrac{dx}{x}

Now use partial fractions and finish it....
 

It doesn't make sense to me that 1) apparently you don't understand that multiplying by (v^2 - 1)/(v^2 - 1) is multiplying by a form of 1, and that 2) apparently, you don't understand when subtracting (or adding) fractions with like denominators that one copy of the denominator remains as the sole denominator. Edit: Subhotosh Khan, I don't see evidence here that Jason76 knows how to combine algebraic fractions, much less continue with the integration portion.
 
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I don't understand whatever do you mean by "top here makes sense, but not the bottom". Continuing...

v213vv3dv=dxx\displaystyle \dfrac{v^{2} - 1}{3v - v^{3}}{dv} = \dfrac{dx}{x}

Now use partial fractions and finish it....

I didn't see that the v21\displaystyle v^{2} - 1 on the bottom came from "subtraction of fractions", it being the LCD.
 
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