Is it possible to do separation of variables in this manner? dx = ?(? + 1)(? − ?)

[akn2244]

Is it possible to do separation of variables in this manner? dx = ?(? + 1)(? − ?)

This the equation I have and I am attempting to use separation of variables in order to integrate it
dx = ?(? + 1)(? − ?)

So, the integral of 1/(? + 1)(? − ?) dx = the integral of k dt
Then,
A/?+1 + B/?−? will therefore = 1/ (?+1) (?−?)
and (after multiplying the LHS and RHS by (x+1)(n-x), I achieved
?(?−?) +?(?+1) =1 (Eq. 1)

I have been told that the following is a method used I should use to solve for A and B

(?−?)?+?+??=1
⇒?=? and ?+??=1
⇒?+??=1⇒?=1/(1+n)⇒?= 1/(1+n)

However, I am confused about how it is known that B=A.

Instead, I thought it was possible to substitute 'useful' values for x in order to find a and then b. For example, sub in x=-1 to find A, and substitute x=n to find B.

So my question is, how does A=B and also, it is possible to let x=n in order to find B (I.e. use 'useful' values to cancel out B when finding A and A when finding B?)

Thank you.

 

[nasi112]

Is it possible to do separation of variables in this manner? dx = ?(? + 1)(? − ?)

This the equation I have and I am attempting to use separation of variables in order to integrate it
dx = ?(? + 1)(? − ?)

So, the integral of 1/(? + 1)(? − ?) dx = the integral of k dt
Then,
A/?+1 + B/?−? will therefore = 1/ (?+1) (?−?)
and (after multiplying the LHS and RHS by (x+1)(n-x), I achieved
?(?−?) +?(?+1) =1 (Eq. 1)

I have been told that the following is a method used I should use to solve for A and B

(?−?)?+?+??=1
⇒?=? and ?+??=1
⇒?+??=1⇒?=1/(1+n)⇒?= 1/(1+n)

However, I am confused about how it is known that B=A.

Instead, I thought it was possible to substitute 'useful' values for x in order to find a and then b. For example, sub in x=-1 to find A, and substitute x=n to find B.

So my question is, how does A=B and also, it is possible to let x=n in order to find B (I.e. use 'useful' values to cancel out B when finding A and A when finding B?)

Thank you.

No. $\displaystyle A \neq B$

$\displaystyle A = \frac{-1}{n + 1}$

$\displaystyle B = \frac{1}{n + 1}$

Re-visit your partial fraction calculations again.

 

[topsquark]

Is it possible to do separation of variables in this manner? dx = ?(? + 1)(? − ?)

This the equation I have and I am attempting to use separation of variables in order to integrate it
dx = ?(? + 1)(? − ?)

So, the integral of 1/(? + 1)(? − ?) dx = the integral of k dt
Then,
A/?+1 + B/?−? will therefore = 1/ (?+1) (?−?)
and (after multiplying the LHS and RHS by (x+1)(n-x), I achieved
?(?−?) +?(?+1) =1 (Eq. 1)

I have been told that the following is a method used I should use to solve for A and B

(?−?)?+?+??=1
⇒?=? and ?+??=1
⇒?+??=1⇒?=1/(1+n)⇒?= 1/(1+n)

However, I am confused about how it is known that B=A.

Instead, I thought it was possible to substitute 'useful' values for x in order to find a and then b. For example, sub in x=-1 to find A, and substitute x=n to find B.

So my question is, how does A=B and also, it is possible to let x=n in order to find B (I.e. use 'useful' values to cancel out B when finding A and A when finding B?)

Thank you.

A useful LaTeX note: The integral is \int: $\int$.

Also, please use parenthesis. A/x + 1 is $\dfrac{A}{x} + 1$, not $\dfrac{A}{x + 1}$. You need to write this as A/(x + 1).

Now,
A(n - x) + B(x + 1) = 1

(-A + B)x + (An + B) = 1

This has to hold for any value of x, so if you just plug, say, x = 1 into the equation you will get a solution for A and B that depends on x = 1. So, we have to require that the coefficient -A + B = 0. Thus B = A.

-Dan

 

[Dr.Peterson]

No. $\displaystyle A \neq B$

$\displaystyle A = \frac{-1}{n + 1}$

$\displaystyle B = \frac{1}{n + 1}$

Re-visit your partial fraction calculations again.

You did notice that the question is 7 years old, right? And do you see that your answer is wrong? For the sake of completeness, let's finish it.

To integrate 1/(x+1)(n-x), we want to set it equal to A/(x+1) + B/(n-x). Multiplying both sides by (x+1)(n-x), we get $$\frac{1}{(x+1)(n-x)}=\frac{A}{x+1}+\frac{B}{n-x}\\\\1=A(n-x)+B(x+1)\\\\1=-Ax+nA+Bx+B\\\\(A-B)x+(1-nA-B)=0$$
Since we need this polynomial to be zero for all x, both coefficients must be 0: $$A-B=0\\1-nA-B=0$$
The first equation implies that A=B; the second then becomes $1-nA-A=0$, so that $1=A(n+1)$ and $A=\frac{1}{n+1}$.

So our partial fractions are $$\frac{1}{(x+1)(n-x)}=\frac{\frac{1}{n+1}}{x+1}+\frac{\frac{1}{n+1}}{n-x}$$
To verify this, we can factor out $\frac{1}{n+1}$ and have $$\frac{\frac{1}{n+1}}{x+1}+\frac{\frac{1}{n+1}}{n-x}=\frac{1}{n+1}\left(\frac{1}{x+1}+\frac{1}{n-x}\right)\\=\frac{1}{n+1}\left(\frac{n-x}{(x+1)(n-x)}+\frac{x+1}{(x+1)(n-x)}\right)\\=\frac{1}{n+1}\left(\frac{n+1}{(x+1)(n-x)}\right)=\frac{1}{(x+1)(n-x)}$$ which is what we wanted.

Instead, I thought it was possible to substitute 'useful' values for x in order to find a and then b. For example, sub in x=-1 to find A, and substitute x=n to find B.

So my question is, ... it is possible to let x=n in order to find B (I.e. use 'useful' values to cancel out B when finding A and A when finding B?)

Yes, this is a valid method for a simple problem like this; and since n is just a stand-in for a number, there's no reason not to let x = n to eliminate A, and let x = -1 to eliminate B.. The equations you'll get are $1=A(n+1)$ and $1=B(n+1)$, which gives the same results more quickly.

I hope this is worth the 7-year wait .

 

[Steven G]

The 4th line in the original post has a dt in it. Where did that come from?

 

[khansaheb]

You did notice that the question is 7 years old, right? And do you see that your answer is wrong? For the sake of completeness, let's finish it.

To integrate 1/(x+1)(n-x), we want to set it equal to A/(x+1) + B/(n-x). Multiplying both sides by (x+1)(n-x), we get $$\frac{1}{(x+1)(n-x)}=\frac{A}{x+1}+\frac{B}{n-x}\\\\1=A(n-x)+B(x+1)\\\\1=-Ax+nA+Bx+B\\\\(A-B)x+(1-nA-B)=0$$
Since we need this polynomial to be zero for all x, both coefficients must be 0: $$A-B=0\\1-nA-B=0$$
The first equation implies that A=B; the second then becomes $1-nA-A=0$, so that $1=A(n+1)$ and $A=\frac{1}{n+1}$.

So our partial fractions are $$\frac{1}{(x+1)(n-x)}=\frac{\frac{1}{n+1}}{x+1}+\frac{\frac{1}{n+1}}{n-x}$$
To verify this, we can factor out $\frac{1}{n+1}$ and have $$\frac{\frac{1}{n+1}}{x+1}+\frac{\frac{1}{n+1}}{n-x}=\frac{1}{n+1}\left(\frac{1}{x+1}+\frac{1}{n-x}\right)\\=\frac{1}{n+1}\left(\frac{n-x}{(x+1)(n-x)}+\frac{x+1}{(x+1)(n-x)}\right)\\=\frac{1}{n+1}\left(\frac{n+1}{(x+1)(n-x)}\right)=\frac{1}{(x+1)(n-x)}$$ which is what we wanted.

Yes, this is a valid method for a simple problem like this; and since n is just a stand-in for a number, there's no reason not to let x = n to eliminate A, and let x = -1 to eliminate B.. The equations you'll get are $1=A(n+1)$ and $1=B(n+1)$, which gives the same results more quickly.

I hope this is worth the 7-year wait .

Only caveat → n <>(-1)

 

[nasi112]

You did notice that the question is 7 years old, right? And do you see that your answer is wrong? For the sake of completeness, let's finish it.

To integrate 1/(x+1)(n-x), we want to set it equal to A/(x+1) + B/(n-x). Multiplying both sides by (x+1)(n-x), we get $$\frac{1}{(x+1)(n-x)}=\frac{A}{x+1}+\frac{B}{n-x}\\\\1=A(n-x)+B(x+1)\\\\1=-Ax+nA+Bx+B\\\\(A-B)x+(1-nA-B)=0$$
Since we need this polynomial to be zero for all x, both coefficients must be 0: $$A-B=0\\1-nA-B=0$$
The first equation implies that A=B; the second then becomes $1-nA-A=0$, so that $1=A(n+1)$ and $A=\frac{1}{n+1}$.

So our partial fractions are $$\frac{1}{(x+1)(n-x)}=\frac{\frac{1}{n+1}}{x+1}+\frac{\frac{1}{n+1}}{n-x}$$
To verify this, we can factor out $\frac{1}{n+1}$ and have $$\frac{\frac{1}{n+1}}{x+1}+\frac{\frac{1}{n+1}}{n-x}=\frac{1}{n+1}\left(\frac{1}{x+1}+\frac{1}{n-x}\right)\\=\frac{1}{n+1}\left(\frac{n-x}{(x+1)(n-x)}+\frac{x+1}{(x+1)(n-x)}\right)\\=\frac{1}{n+1}\left(\frac{n+1}{(x+1)(n-x)}\right)=\frac{1}{(x+1)(n-x)}$$ which is what we wanted.

Yes, this is a valid method for a simple problem like this; and since n is just a stand-in for a number, there's no reason not to let x = n to eliminate A, and let x = -1 to eliminate B.. The equations you'll get are $1=A(n+1)$ and $1=B(n+1)$, which gives the same results more quickly.

I hope this is worth the 7-year wait .

My answer is wrong because I was lazy to do the partial fraction work. I let my calculator to do it.

It gave me

$\displaystyle \frac{1}{(n + 1)(x - n)} - \frac{1}{(n + 1)(x + 1)}$

I did not notice that it changed $\displaystyle (n - x)$ to $\displaystyle (x - n)$

If I saw that, I would say $\displaystyle A = B = \frac{-1}{(n + 1)}$.

The interesting thing is your answer is the same but with a positive sign. I am sure that your calculations were so precise, so the calculator was wrong to give the negative sign.

 

[Dr.Peterson]

It gave me $\displaystyle \frac{1}{(n + 1)(x - n)} - \frac{1}{(n + 1)(x + 1)}$. I did not notice that it changed $\displaystyle (n - x)$ to $\displaystyle (x - n)$.

If I saw that, I would say $\displaystyle A = B = \frac{-1}{(n + 1)}$.

The interesting thing is your answer is the same but with a positive sign. I am sure that your calculations were so precise, so the calculator was wrong to give the negative sign.

I wonder if you accidentally entered x-n in place of n-x? That would explain it.

 

[nasi112]

I wonder if you accidentally entered x-n in place of n-x? That would explain it.

Ohhh I cannot believe it. I thought the calculator change it to $\displaystyle (x - n)$.

It seems that you are right. I don't remember if I entered $\displaystyle (x - n)$ or $\displaystyle (n - x)$ yesterday, but now when I do it with $\displaystyle (n - x)$, it gives me a postive answer.

I have to change my glasses as soon as possible!?