Fraction Problem

[Jason76]

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.

\(\displaystyle x\dfrac{dv}{dx} = \dfrac{2v}{v^{2} - 1} - v\)

\(\displaystyle x\dfrac{dv}{dx} = \dfrac{2v}{v^{2} - 1} - \dfrac{v}{1}\)

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

\(\displaystyle x\dfrac{dv}{dx} = \dfrac{3v - v^{3}}{v^{2} - 1}\) 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 \(\displaystyle v^{2} - 1\) on the bottom, as that is LCD.

 

[Deleted member 4993]

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.

\(\displaystyle x\dfrac{dv}{dx} = \dfrac{2v}{v^{2} - 1} - v\)

\(\displaystyle x\dfrac{dv}{dx} = \dfrac{2v}{v^{2} - 1} - \dfrac{v}{1}\)

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

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

I don't understand whatever do you mean by "top here makes sense, but not the bottom". Continuing...

\(\displaystyle \dfrac{v^{2} - 1}{3v - v^{3}}{dv} = \dfrac{dx}{x}\)

Now use partial fractions and finish it....

 

[lookagain]

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.

\(\displaystyle x\dfrac{dv}{dx} = \dfrac{2v}{v^{2} - 1} - v\)

\(\displaystyle x\dfrac{dv}{dx} = \dfrac{2v}{v^{2} - 1} - \dfrac{v}{1}\)

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

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

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.

 

[Jason76]

I don't understand whatever do you mean by "top here makes sense, but not the bottom". Continuing...

\(\displaystyle \dfrac{v^{2} - 1}{3v - v^{3}}{dv} = \dfrac{dx}{x}\)

Now use partial fractions and finish it....

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