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What is integration of y/(x^2-y^2) dx

rsoy

New member
Mar 3, 2013
14
Hi all

can please explaine to me what is integration of y/(x^2-y^2) dx

step by step ...
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: what is integration of y/(x^2-y^2) dx

Hey Ahmed :

I think you have already studied partial fraction decomposition when you learned about integration .

Here we can treat y as constant hence, we don't need to worry about it , because we are integrating w.r.t x ...

we know by the difference of two squares that :

\(\displaystyle x^2-y^2=(x-y)(x+y)\)

\(\displaystyle \frac{y}{(x-y)(x+y)}=\frac{A}{x-y}+\frac{B}{x+y}\)

Hence we have the following :

\(\displaystyle y = A(x+y) + B(x-y) \)

Now we need to find both A and B so do the following :

1- put x= y so the equation becomes :

\(\displaystyle y = 2y\, A\) , \(\displaystyle A =\frac{1}{2}\)

2-To find B we put x=-y

\(\displaystyle y = -2y \,B \) , \(\displaystyle B =\frac{-1}{2}\)

\(\displaystyle \frac{y}{(x-y)(x+y)}=\frac{\frac{1}{2}}{x-y}+\frac{\frac{-1}{2}}{x+y}\)

Now can you integrate the right hand side ?
 

rsoy

New member
Mar 3, 2013
14
Re: what is integration of y/(x^2-y^2) dx

1/2ln(x-y) + -1/2(x+y)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: what is integration of y/(x^2-y^2) dx

Hi all

can please explaine to me what is integration of y/(x^2-y^2) dx

step by step ...
Did you check out the Heaviside cover-up method I pointed you to the other day for partial fraction decomposition?
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: what is integration of y/(x^2-y^2) dx

1/2ln(x-y) + -1/2(x+y)
No, this isn't correct , you are missing an absolute value and a ln in the second part !

Also , don't forget that there should be a constant which is a function of y ...
 

rsoy

New member
Mar 3, 2013
14
Re: what is integration of y/(x^2-y^2) dx

1/2ln(x-y) + -1/2ln(x+y) + c

- - - Updated - - -

Did you check out the Heaviside cover-up method I pointed you to the other day for partial fraction decomposition?
Yes
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
Re: what is integration of y/(x^2-y^2) dx

1/2ln(x-y) + -1/2ln(x+y) + c
Still , you are missing the absolute value , also C here is a function of y it is is usually wirtten \(\displaystyle \phi(y)\)
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Re: what is integration of y/(x^2-y^2) dx

Hi all

can please explaine to me what is integration of y/(x^2-y^2) dx

step by step ...
Has this question come from trying to solve a differential equation, which would make y a function of x? Or are you doing a "partial integral", in other words, holding y constant while trying to integrate with respect to x?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,779
Re: what is integration of y/(x^2-y^2) dx

Hi all

can please explaine to me what is integration of y/(x^2-y^2) dx

step by step ...
Another approach.
If you have a list of derivatives of trigonometric hyperbolic functions, you should have:
$$\frac{d}{dx} \text{ artanh } x = \frac 1 {1-x^2}$$
 
Last edited:

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
Re: what is integration of y/(x^2-y^2) dx

Another approach.
If you have a list of derivatives of trigonometric functions, you should have:
$$\frac{d}{dx} \text{ artanh } x = \frac 1 {1-x^2}$$
Hyperbolic functions...
 

soroban

Well-known member
Feb 2, 2012
409
Hello, rsoy!

$\displaystyle\int \frac{y}{x^2-y^2}\,dx$

Since $y$ is treated as a constant $b$, we have: .$\displaystyle b\int\frac{dx}{x^2-b^2}$


There is a standard integration formula: .$\displaystyle \int \frac{du}{u^2-a^2} \:=\:\frac{1}{2a}\ln\left|\frac{u-a}{u+a}\right|+C $


Therefore: .$\displaystyle b\left(\frac{1}{2b}\right)\ln\left|\frac{x-b}{x+b}\right|+C \;=\;\frac{1}{2}\ln\left|\frac{x-y}{x+y}\right|+C$
 

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
The OP has not specified if this is actually the case...
I believe your earlier suspicion that this arose in the process of solving an ODE was correct, i.e., that $y$ is actually dependent upon $x$.