# Zero value differential help

• Kampret
In summary, the post is trying to say that if you want to solve for v in the equation given for ##a##, you need to understand how to solve ordinary differential equations, and they provide some symbols you can use in that process.

## Homework Statement

If a = 9-v² then prove that v = 3 (e^6t - 1)/(e^6t + 1) the condition when t=0 also v has zero value

## Homework Equations

I don't quite understand in this but general equation should be dv/dt = a

## The Attempt at a Solution

Actually i don't don't have any idea in this problem since i only encountered something similar problem like determine v when a=v+1 or a=-bv² where b is coefficient but well i have gist since the final form is including euler number so natural logaritm should have something to do in this for me i write like this
dv/dt = a
dv/dt = 9-v²
dv/(9-v²) = dt since the only way I am know to attain answer that contain e is do something abou ln and go to form
dy/y = ln y
v² need become positive so
both sides multipled by -1
dv/(v²-9)=-dt (im I am sorry idont have integrer symbol so i will write it as intg)
dv/(v+3)(v-3) = -dt just from here I am not sure when forced its become like
then both side integrated so it become like ln (v+3) (v-3) = -t + C then i don't know anymore (frankly even this part I am not sure whether its i correct or just my foolishness) so please help I am really restless until i can solve this Last edited:
You're given the result, so you don't need to derive it. Start from the equation you are given for ##v## and show that it satisfies dv/dt = 9-v².

Well personally I am say that's isn't matter from where i start cause in the end all is is related and linked from v to a , could you give further explanation for this problem well if if not bother i really hope given same same model of problem and the answer so i could analyze it by myself and it should not violate the purpose from this forum ,frankly i don't have dependable teacher to relied upon so I am trying sought guide from here

I don't understand what you mean. I told how to solve the problem at hand, where you have to prove that the equation given for ##v## solves ##a = 9 -v^2##.

If you want to know how to arrive at that equation for ##v##, then you need to pick up a textbook that covers solving ordinary differential equations.

Basically what you did was not wrong, only in the final step the integration gives ##\frac{1}{6}\ln\frac{3-v}{3+v}=-t+C## (1) (can give you details if you want about the integration steps)

Using the initial condition that for t=0 , v=0, you can conclude that C=0 (by plugging t=0 and v=0 in (1)).
hence from (1) for C=0 and by taking the powers with respect to ##e## in both sides (first we transfer the 6 to the other side, making that term -6t), we get
$$e^{\ln\frac{3-v}{3+v}}=e^{-6t}\Rightarrow \frac{3-v}{3+v}=e^{-6t} (2)$$,
where we used the basic property that ##e^{\ln y}=y##.

From the equation (2) we can get after some algebraic manipulation that v is indeed equal as given by the exercise.

Last edited:
Hello @Kampret . Welcome to PF.

I will make some corrections to typos in the quoted text of Post #1 below, as well as inserting a few comments.
Kampret said:

## Homework Statement

If a = 9-v² then prove that v = 3 (e^(6t) - 1)/(e^(6t) + 1) the condition when t=0 also v has zero value

## Homework Equations

I don't quite understand in this but general equation should be dv/dt = a

## The Attempt at a Solution

Actually i don't have any idea in this problem since i only encountered something similar problem like determine v when a=v+1 or a=-bv² where b is coefficient but well i have gist since the final form is including Euler number so natural logarithm should have something to do in this. for me i write like this
dv/dt = a
dv/dt = 9-v²
dv/(9-v²) = dt since the only way I am know to attain answer that contain e is do something about ln and go to form
dy/y = ln y
Actually that should be: d/dy (ln y) = 1/y .
v² need become positive so
There is no need to multiply by −1, although it is allowed.
both sides multiplied by -1
dv/(v²-9)=-dt (I'm sorry i don't have integrer integration symbol so i will write it as intg)
dv/((v+3)(v-3)) = -dt just from here I'm not sure when forced its become like
then both side integrated so it become like ln ((v+3) (v-3)) = -t + C
That is not correct. Check that by differentiation..
then i don't know anymore (frankly even this part I am not sure whether its i correct or just my foolishness) so please help I'm really restless until i can solve this You can find some useful symbols by clicking on the big Sigma ( ∑ ) in the blue banner at the top of the Input window used to enter your Post. Included there is the integration symbol, ∫ .

You can compose higher quality mathematical expressions by use of Latex.
LaTeX FAQ: https://www.physicsforums.com/help/latexhelp/
e.g. ## \displaystyle \ \int \frac{dv}{(v-3)(v+3)} = -\int dt ##

Now to the problem at hand:​
If you are to solve the differential equation by the above integration, you need to review some integration techniques. In this case use "partial fraction decomposition" to break the fraction, ## \displaystyle \ \frac{1}{(v-3)(v+3)} \ ## into the sum of two fractions, one with denominator, ##\ v-3\ ##, the other with denominator, ##\ v+3\, . ## Continue from there following Δ2's post(s).

On the other hand, if you only need to verify that the given solution is correct, DrClaud has given help along those lines. For this: Basically differentiate ##\ v\,,\ ## which is ## \displaystyle \ 3 \frac{e^{6t} - 1} {e^{6t} + 1} \,, \ ## and compare that to ##\ 9-v^2\,.\ ## Also, verify that the boundary condition is met.

• Delta2 and DrClaude
Delta² said:
Basically what you did was not wrong, only in the final step the integration gives ##\frac{1}{6}\ln\frac{3-v}{3+v}=-t+C## (1) (can give you details if you want about the integration steps)

Using the initial condition that for t=0 , v=0, you can conclude that C=0 (by plugging t=0 and v=0 in (1)).
hence from (1) for C=0 and by taking the powers with respect to ##e## in both sides (first we transfer the 6 to the other side, making that term -6t), we get
$$e^{\ln\frac{3-v}{3+v}}=e^{-6t}\Rightarrow \frac{3-v}{3+v}=e^{-6t} (2)$$,
where we used the basic property that ##e^{\ln y}=y##.

From the equation (2) we can get after some algebraic manipulation that v is indeed equal as given by the exercise.
thanks that was very helpful after reading sammys post about "partial fraction decomposition) this is my conclusion
dv(1/2(v-3) - (1/2(v+3)) = -dt
both sides multipled by 6 then integrated
3 ln(v-3)/ln(v+3) = -6t please correct me if there some mistake in it or if this turned out completely wrong that it even seems so weird. I humbly ask you please explain the integrating process that you offered above.
im just fool who want understand the beauty of physics

Last edited:
SammyS said:
Hello @Kampret . Welcome to PF.

I will make some corrections to typos in the quoted text of Post #1 below, as well as inserting a few comments.

Actually that should be: d/dy (ln y) = 1/y .
There is no need to multiply by −1, although it is allowed.
That is not correct. Check that by differentiation..
You can find some useful symbols by clicking on the big Sigma ( ∑ ) in the blue banner at the top of the Input window used to enter your Post. Included there is the integration symbol, ∫ .

You can compose higher quality mathematical expressions by use of Latex.
LaTeX FAQ: https://www.physicsforums.com/help/latexhelp/
e.g. ## \displaystyle \ \int \frac{dv}{(v-3)(v+3)} = -\int dt ##

Now to the problem at hand:​
If you are to solve the differential equation by the above integration, you need to review some integration techniques. In this case use "partial fraction decomposition" to break the fraction, ## \displaystyle \ \frac{1}{(v-3)(v+3)} \ ## into the sum of two fractions, one with denominator, ##\ v-3\ ##, the other with denominator, ##\ v+3\, . ## Continue from there following Δ2's post(s).

On the other hand, if you only need to verify that the given solution is correct, DrClaud has given help along those lines. For this: Basically differentiate ##\ v\,,\ ## which is ## \displaystyle \ 3 \frac{e^{6t} - 1} {e^{6t} + 1} \,, \ ## and compare that to ##\ 9-v^2\,.\ ## Also, verify that the boundary condition is met.
thanks a lot i can't believe someone want to waste their precious time to correct my petty mistakes your help about "partial fraction decomposition" was really helpful to turn v²-9 into 2 familiar form that comprehensible for me so i can do something about integration process

Kampret said:
thanks a lot i can't believe someone want to waste their precious time to correct my petty mistakes your help about "partial fraction decomposition" was really helpful to turn v²-9 into 2 familiar form that comprehensible for me so i can do something about integration process
I hoped I was not wasting my time. I didn't mention nor indicate all of the "editorial" changes I made. I did insert some important sets of parentheses in RED but they don't show up all that well.

Also, I like to use the 'Reply' feature to quote the post I am referring to, especially I like to see the entire Original Post (OP) quoted. So I thought, why not edit it a little.

Kampret said:
thanks that was very helpful after reading sammys post about "partial fraction decomposition) this is my conclusion
dv(1/2(v-3) - (1/2(v+3)) = -dt
both sides multipled by 6 then integrated
3 ln(v-3)/ln(v+3) = -6t please correct me if there some mistake in it or if this turned out completely wrong that it even seems so weird. I humbly ask you please explain the integrating process that you offered above.
im just fool who want understand the beauty of physics
Probably @Delta² is sleeping at this time. (He lives in Athens, Greece.)

You have an error in your partial fraction decomposition of ##\displaystyle \ \frac{1}{(v-3)(v+3)} \,. ##

If you add ##\displaystyle \ \frac{1}{2(v-3)} - \frac{1}{2(v+3)} \,, \ ## the result is ##\displaystyle \ \frac{6}{2(v-3)(v+3)} \,.\ ## So rather than having 2 in the
numerators, each needs a 6 instead.

A pointer on algebra notation:
You have the differential equation: dv(1/2(v−3) −(1/2(v+3)) = −dt, which has unbalanced parentheses, but more important than that, if you want to write the fractional expression, ##\displaystyle \ \frac{1}{2(v-3)} - \frac{1}{2(v+3)} \,, \ ## using "linear" form for the fractions, you should enclose each denominator in parentheses.
1/(2(v−3)) − 1/(2(v+3))​
.

• Kampret
SammyS said:
Probably @Delta² is sleeping at this time. (He lives in Athens, Greece.)

You have an error in your partial fraction decomposition of ##\displaystyle \ \frac{1}{(v-3)(v+3)} \,. ##

If you add ##\displaystyle \ \frac{1}{2(v-3)} - \frac{1}{2(v+3)} \,, \ ## the result is ##\displaystyle \ \frac{6}{2(v-3)(v+3)} \,.\ ## So rather than having 2 in the
numerators, each needs a 6 instead.

A pointer on algebra notation:
You have the differential equation: dv(1/2(v−3) −(1/2(v+3)) = −dt, which has unbalanced parentheses, but more important than that, if you want to write the fractional expression, ##\displaystyle \ \frac{1}{2(v-3)} - \frac{1}{2(v+3)} \,, \ ## using "linear" form for the fractions, you should enclose each denominator in parentheses.
1/(2(v−3)) − 1/(2(v+3))​
.
i can't believe i make an embarassing mistake the truth is when i look for more zuitable example for my question I've found x²-1 model since it has value 1/2 and -1/2 for a and b just and i write in unconsciously, no wonder it feels weird x_x

Yes I went to bed early (8pm local time=6pm GMT), cause I had slept only 4 hours monday's night.

The partial fraction decomposition as suggested and corrected by @SammyS needs a 6 in the denominator of both fractions.

After that , we have to integrate . So ##\int \frac{1}{6}\frac{dv}{v-3}-\int \frac{1}{6}\frac{dv}{v+3}##

The first integral is equal to ##\frac{1}{6}\ln|v-3|##. From the problem data given we can conclude that ##0<v<3## (1), so ##|v-3|=3-v##.
Similarly for the second integral we ll have it equal to ##\frac{1}{6}\ln|v+3|## with ##|v+3|=v+3## because of (1).
After that you can combine the two results using the property of the logarithms that ##\ln A-\ln B=\ln\frac{A}{B}##.

please take a rest , I am not rushed as long i can conpletely understand it, so here my next attempt please take a look when you at leasure time and please correct it if I am done something wrong
ln(v-3)/(v+3)=-6t
v/v= e^-6t+3/e^-6t-3 then swapped between
Numerator and denominator because it has negative exponent but really I am not very sure about this part,but at least
ln(v-3)/(v+3)=-6t part should be correct since it follow principle of logaritm like you wrote above so how should i do since on left side between numerator and denominator both has v variable ? could you show me the final step of algebra that you mentioned on your first post?

Last edited:
Ok we have ##\ln\frac{3-v}{3+v}=-6t##. From this we can infer that ##e^{\ln\frac{3-v}{3+v}}=e^{-6t}## . Is this step clear? it is like saying ##A=B\Rightarrow e^{A}=e^B##

Then we use the property that ##e^{\ln C}=C## so we ll get (for ##C=\frac{3-v}{3+v}##)
$$e^{\ln\frac{3-v}{3+v}}=\frac{3-v}{3+v}=e^{-6t}$$

From this last equation it is typical algebraic manipulation to solve for v. First we take the inverse in both parts of the equation so we ll have
$$\frac{1}{\frac{3-v}{3+v}}=\frac{1}{e^{-6t}}\Rightarrow \frac{3+v}{3-v}=e^{6t}$$

And then $$3+v=(3-v)e^{6t}$$ and I believe you can see the rest of algebra.

Last edited:
Delta² said:
Ok we have ##\ln\frac{3-v}{3+v}=-6t##. From this we can infer that ##e^{\ln\frac{3-v}{3+v}}=e^{-6t}## . Is this step clear? it is like saying ##A=B\Rightarrow e^{A}=e^B##

Then we use the property that ##e^{\ln C}=C## so we ll get (for ##C=\frac{3-v}{3+v}##)
$$e^{\ln\frac{3-v}{3+v}}=\frac{3-v}{3+v}=e^{-6t}$$

From this last equation it is typical algebraic manipulation to solve for v. First we take the inverse in both parts of the equation so we ll have
$$\frac{1}{\frac{3-v}{3+v}}=\frac{1}{e^{-6t}}\Rightarrow \frac{3+v}{3-v}=e^{6t}$$

And then $$3+v=(3-v)e^{6t}$$ and I believe you can see the rest of algebra.
thanks , your explanation is really easy to understand just still have something stuck my mind
after ##e^{\ln\frac{3-v}{3+v}}=e^{-6t}##
im sure C is number that determined by the given condition before hand, on my case when t=0 v also 0 so if i try to subtitute these numbers left side should became 1 (from3/3) and left side also became one since has zero degree, then C value should be zero. so could you please tell me why on your equation ##e^{\ln C}=C## it has value
##C=\frac{3-v}{3+v}##)
$$e^{\ln\frac{3-v}{3+v}}=\frac{3-v}{3+v}=e^{-6t}$$ instead of zero for the rest , i cannot say anything except from gratitude that was outstanding explanation

Last edited by a moderator:
Oh I used the same symbol for C the constant of integration and for that C to demonstrate the property . Just call it D then so that ##e^{\ln D}=D## where D can be anything, in our case we apply the property for ##D=\frac{3-v}{3+v}##

• Kampret
Delta² said:
Ok we have ##\ln\frac{3-v}{3+v}=-6t##. From this we can infer that ##e^{\ln\frac{3-v}{3+v}}=e^{-6t}## . Is this step clear? it is like saying ##A=B\Rightarrow e^{A}=e^B##

Then we use the property that ##e^{\ln C}=C## so we ll get (for ##C=\frac{3-v}{3+v}##)
$$e^{\ln\frac{3-v}{3+v}}=\frac{3-v}{3+v}=e^{-6t}$$

From this last equation it is typical algebraic manipulation to solve for v. First we take the inverse in both parts of the equation so we ll have
$$\frac{1}{\frac{3-v}{3+v}}=\frac{1}{e^{-6t}}\Rightarrow \frac{3+v}{3-v}=e^{6t}$$

And then $$3+v=(3-v)e^{6t}$$ and I believe you can see the rest of algebra.
Delta² said:
Oh I used the same symbol for C the constant of integration and for that C to demonstrate the property . Just call it D then so that ##e^{\ln D}=D## where D can be anything, in our case we apply the property for ##D=\frac{3-v}{3+v}##
uhm I am very sorry if this rather trivial and out of date but after i examine this carefully I am sure in order to split 9-v² we need both positive value for v so it became v²-9 with multipling both right and left side by -1, well even though sammys say its unnecessary to do even it allowed, so my point is why on your post it became ln((3-v)/(3+v)) instead of ln((v-3)/(v+3))

You can do anything you want with ##9-v^2## or ##v^2-9## till the moment we have to take the integrals

##\int \frac{1}{v-3}dv, \int\frac{1}{v+3}dv##.

Basic integral calculus theory tell us that these integrals are equal to,##\ln|v-3|,\ln|v+3|## respectively.

Now because we know that ##0<v<3## (I can explain why this inequality is true), we have that ##0<3-v<3+v##.

hence ##|v-3|=3-v## and ##|v+3|=v+3##

Delta² said:
You can do anything you want with ##9-v^2## or ##v^2-9## till the moment we have to take the integrals

##\int \frac{1}{v-3}dv, \int\frac{1}{v+3}dv##.

Basic integral calculus theory tell us that these integrals are equal to,##\ln|v-3|,\ln|v+3|## respectively.

Now because we know that ##0<v<3## (I can explain why this inequality is true), we have that ##0<3-v<3+v##.

hence ##|v-3|=3-v## and ##|v+3|=v+3##
could you kindly tell me how do you know that ##0<v<3## cause if only i know that v<3, 『3-v 』form would be understandable for me

Kampret said:
could you kindly tell me how do you know that ##0<v<3## cause if only i know that v<3, 『3-v 』form would be understandable for me
yes , but i first have to correct myself abit it is actually ##0\leq v\leq 3##

We know at ##t=0## , that ##v=0##.
We also know that the acceleration is ##a=9-v^2## so at ##t=0## ,##a=9-0^2=9##
So ##v## will start increasing from 0 (cause ##a=9>0 ##at ##t=0## so we have positive acceleration), but can never become greater than 3:in order to become greater than 3, ##v## has to reach 3 first (and then increase) (##v## is a continuous variable) but if ##v## becomes 3 then the acceleration will become ##a=9-3^2=0##, and with zero acceleration ##v## cannot increase neither decrease, so if ##v## reaches 3 it will stay constant at 3 and the acceleration will stay constant at 0. So we proved what we wanted that is inequality (1).

EDIT: The above is mostly intuitive proof. For a formal proof use the theorems about monotonicity and extrema involving the first derivative ##dv/dt##. You know that for ##v<3## , the first derivative ##dv/dt=9-v^2>0## and for ##v>3## the first derivative ##dv/dt=9-v^2<0## and for ##v=3## it is ##dv/dt=0##

Last edited:
Delta² said:
yes , but i first have to correct myself abit it is actually ##0\leq v\leq 3##

We know at ##t=0## , that ##v=0##.
We also know that the acceleration is ##a=9-v^2## so at ##t=0## ,##a=9-0^2=9##
So ##v## will start increasing from 0 (cause ##a=9>0 ##at ##t=0## so we have positive acceleration), but can never become greater than 3:in order to become greater than 3, ##v## has to reach 3 first (and then increase) (##v## is a continuous variable) but if ##v## becomes 3 then the acceleration will become ##a=9-3^2=0##, and with zero acceleration ##v## cannot increase neither decrease, so if ##v## reaches 3 it will stay constant at 3 and the acceleration will stay constant at 0. So we proved what we wanted that is inequality (1).

EDIT: The above is mostly intuitive proof. For a formal proof use the theorems about monotonicity and extrema involving the first derivative ##dv/dt##. You know that for ##v<3## , the first derivative ##dv/dt=9-v^2>0## and for ##v>3## the first derivative ##dv/dt=9-v^2<0## and for ##v=3## it is ##dv/dt=0##
ohh that's really make sense when you explain it like that, so if by any chance this problem does not reveal the orginal condition (i mean there is no word about v=0 when t=0 and we add C in the end of result)
would it be still ##ln(3-v)## or it somewhat became ##ln(v-3)## because from what i see, the result original result from partial fraction decomposition are both ##v-3## and ##v+3## and not ##3-v## and ##v+3## but of course we know that a still 9-v² so naturally if v is 3 a=0 and if v more than 3 a became negative. and the second, after we find the result of partial fraction decomposition, it is still needed to make inequality from original equation? (9-v² on my case) in order to determine the range of related variable (0≤v≤3 on my case) so we can find part that to be altered (v-3 became 3-v on my case because if v less than 3 and we still use v-3 the result would became (-)) so i can use this knowledge when encountered altered result in the future.
this is the example that i found
https://socratic.org/questions/how-do-you-integrate-1-x-2-4-using-partial-fractions
this question has exact pattern like mine except given condition and the final result is like what you can see there it remain ##v-2## and not##2-v##

Kampret said:
ohh that's really make sense when you explain it like that, so if by any chance this problem does not reveal the orginal condition (i mean there is no word about v=0 when t=0 and we add C in the end of result)
would it be still ##ln(3-v)## or it somewhat became ##ln(v-3)## because from what i see, the result original result from partial fraction decomposition are both ##v-3## and ##v+3## and not ##3-v## and ##v+3## but of course we know that a still 9-v² so naturally if v is 3 a=0 and if v more than 3 a became negative. and the second, after we find the result of partial fraction decomposition, it is still needed to make inequality from original equation? (9-v² on my case) in order to determine the range of related variable (0≤v≤3 on my case) so we can find part that to be altered (v-3 became 3-v on my case because if v less than 3 and we still use v-3 the result would became (-)) so i can use this knowledge when encountered altered result in the future.
this is the example that i found
https://socratic.org/questions/how-do-you-integrate-1-x-2-4-using-partial-fractions
this question has exact pattern like mine except given condition and the final result is like what you can see there it remain ##v-2## and not##2-v##
Look again at Post #18 by @Delta² . Here is the relevant part of that:

Delta² said:
You can do anything you want with ##9-v^2## or ##v^2-9## till the moment we have to take the integrals

##\int \frac{1}{v-3}dv, \int\frac{1}{v+3}dv##.

Basic integral calculus theory tell us that these integrals are equal to,##\ln|v-3|,\ln|v+3|## respectively.
...
@Delta² also mentioned the use of absolute value, briefly, in Post #12 .

Upon doing the integration

##\displaystyle \int \frac{1}{v-3}dv - \int\frac{1}{v+3}dv = -\int 6\ dt \,,##

we get the following.

##\displaystyle \ln|v-3|-\ln|v+3| = -6 t + \text{Constant of Integration} \,.##

Using properties of logarithms on the left, we get:

##\displaystyle \ln \frac{|v-3|}{|v+3|} = -6 t + \text{Constant of Integration} \,.##

I now ask: What is the domain of ##\displaystyle \ln \frac{|v-3|}{|v+3|} \,?##

• Delta2
SammyS said:
Look again at Post #18 by @Delta² . Here is the relevant part of that:

@Delta² also mentioned the use of absolute value, briefly, in Post #12 .

Upon doing the integration

##\displaystyle \int \frac{1}{v-3}dv - \int\frac{1}{v+3}dv = -\int 6\ dt \,,##

we get the following.

##\displaystyle \ln|v-3|-\ln|v+3| = -6 t + \text{Constant of Integration} \,.##

Using properties of logarithms on the left, we get:

##\displaystyle \ln \frac{|v-3|}{|v+3|} = -6 t + \text{Constant of Integration} \,.##

I now ask: What is the domain of ##\displaystyle \ln \frac{|v-3|}{|v+3|} \,?##
well, since it can't be negative nor zero it should be (3,∞) and(-3,-∞)

Kampret said:
well, since it can't be negative nor zero it should be (3,∞) and(-3,-∞)
I suppose you mean ##\ (-\infty, \,-3 )\ ## rather than ##\ (-3,\, -\infty )\ ## for the second interval.

So, you're saying that ##\ v\ ## cannot be between −3 and 3 ? That would be incorrect.

Does ##\ \displaystyle \frac{|v-3|}{|v+3|} \ ## ever give a negative value ?

Last edited:
SammyS said:
I suppose you mean ##\ (-\infty, \,-3 )\ ## rather than ##\ (-3,\, -\infty )\ ## for the second interval.

So, you're saying that ##\ v\ ## cannot be between −3 and 3 ? That would be incorrect.

Does ##\ \displaystyle \frac{|v-3|}{|v+3|} \ ## ever give a negative value ?
ill try insert 1 on v which one is number between -3 and 3, the numerator would became -2 and denominator became 4 and the result is -0.5

Kampret said:
ill try insert 1 on v which one is number between -3 and 3, the numerator would became -2 and denominator became 4 and the result is -0.5
Those are absolute values.

##|1-3| = |-2| = 2\,.\ ## That's positive.

Does ##\ \displaystyle \frac{|v-3|}{|v+3|} \ ## ever give a negative value ?

Last edited:
SammyS said:
Those are absolute values.

##|1-3| = |-2| = 2\,.\ ## That's positive.

Does ##\ \displaystyle \frac{|v-3|}{|v+3|} \ ## ever give a negative value ?
[/QUOTE]
no, absolute value only gave positive result

no, absolute value only gave positive result[/QUOTE]
Actually it gives a non-negative result. (There is the possibility that it gives 0 .)

Now back to the previous question:

What is the domain of ## \displaystyle \ln \frac{|v-3|}{|v+3|} \,? ##

SammyS said:
no, absolute value only gave positive result
Actually it gives a non-negative result. (There is the possibility that it gives 0 .)

Now back to the previous question:

What is the domain of ## \displaystyle \ln \frac{|v-3|}{|v+3|} \,? ##[/QUOTE]
(-∞,∞) ? because in absolute value only gave non negative result...

Kampret said:
SammyS said:
Actually it gives a non-negative result. (There is the possibility that it gives 0 .)

Now back to the previous question:

What is the domain of ## \displaystyle \ln \frac{|v-3|}{|v+3|} \,? ##
(-∞,∞) ? because in absolute value only gave non negative result...
However, ##\ \ln (0) \ ## is undefined as is ##\ \displaystyle \frac{|v-3|}{|v+3|} \ ## when the denominator is zero, i.e. when ##\ v=-3 \,.##

Therefore, ##\ -3 \ ## and ##\ 3\ ## are not in the domain of ## \displaystyle \ln \frac{|v-3|}{|v+3|} \, . ##

Last edited:
SammyS said:
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)

Kampret said:
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.

Kampret said:
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?

Last edited:
• Kampret
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.

• Kampret