Understanding Zero Value Differential in Euler Number Equations

[Kampret]

However, $\ \ln (0) \$ is undefined as is $\ \displaystyle \frac{|v-3|}{|v+3|} \$ when the denominator is zero, i.e. when $\ v=-3 \,.$

okay, I am grateful for your consideration , please answer my last question, I am know it somewhat annoying to deal with me, but at least if i get answer of this question i feel i could understand matter regarding absolute value in relation between integrating 1/x dx so please accompany me to the end
- as delta² already explained the maximum value of v is 3 and once v became 3 this object would move at constant speed since a=0 and it should not greater than 3 because a will turn negative (all from a=9-v²). so v is must smaller than 3.then v-3 turning into 3-v because 3 is must bigger than v
- on what condition we need review our work so we can check whether the result need be altered or not (i mean altered is like v-3 to become 3-v and not altered is like the answer i gave above)

 

[SammyS]

okay, I am grateful for your consideration , please answer my last question, I am know it somewhat annoying to deal with me, but at least if i get answer of this question i feel i could understand matter regarding absolute value in relation between integrating 1/x dx so please accompany me to the end
- as delta² already explained the maximum value of v is 3 and once v became 3 this object would move at constant speed since a=0 and it should not greater than 3 because a will turn negative (all from a=9-v²). so v is must smaller than 3.then v-3 turning into 3-v because 3 is must bigger than v
- on what condition we need review our work so we can check whether the result need be altered or not (i mean altered is like v-3 to become 3-v and not altered is like the answer i gave above)

Well, I was attempting to make a point by my question to you. (Remains un answered.) Let's leave that for now.

please answer my last question

What question is that? In which post ?

Edit:

Maybe this will help with the answer.

You get equivalent results, whether working with $\ \displaystyle \frac{1}{v^2-9} \$ or working with $\ \displaystyle \frac{1}{9-v^2} \$. Each is just the negative of the other.

(Easier if using 6 for numerators.)

Decomposition of $\ \displaystyle \frac{6}{v^2-9} \$ is $\displaystyle \ \frac{1}{(v-3)} - \frac{1}{(v+3)} \,.$

Decomposition of $\ \displaystyle \frac{6}{9-v^2} \$ is $\displaystyle \ \frac{1}{(3-v)} + \frac{1}{(3+v)} \,.$

Can you see that they are opposites?

 

[Delta2]

Its because we have the absolute value . We need to check what happens to the expression inside the absolute value, that's why we check for v<3 or v>3.
The other way to remove the absolute values is to raise both sides of the equation to the power of 2, but this will create more problems than it will solve.