Solving Equation: ln(y)(x^4 - y^4) + ln(x)(y^4 - x^4) + (y^2 - x^2)^2 = 0

[CIMP]

Hello All,

Could you please indicate me how to resolve this equation:

ln(y) (x^4 - y^4)+ ln(x) (y^4 - x^4) + (y^2 - x^2)^2 = 0

I am really struggling with it...

Thanks in advance,

C:)

 

[Mentallic]

What exactly do you mean by solving it? What do you need to do?

 

[CIMP]

Mentallic,

Thanks a lot for your reply.
I mean, I want to know the value of "X" when "Y=5".

Thanks in advance,

 

[chiro]

Mentallic,

Thanks a lot for your reply.
I mean, I want to know the value of "X" when "Y=5".

Thanks in advance,

Hey CIMP and welcome to the forums.

Do you know about root finding algorithms?

 

[Mentallic]

Mentallic,

Thanks a lot for your reply.
I mean, I want to know the value of "X" when "Y=5".

Thanks in advance,

Then what you're looking for is to solve explicitly for x. However, you can't do that with the use of elementary functions (the usual - addition, multiplication, square roots, powers, trig, exponentials). You will need to find a numeric approximation to the answer.

There are trivial solutions however such as y=x and y=-x, but notice that with logarithms present in the equation, this means you need to restrict x and y to be more than zero, so only y=x, x>0 is a trivial solution.

So, if y=5 then x=5 as well, but that's not the only solution. The others can be found by numeric methods.

 

[Jamma]

How about this, to get going:

x^4-y^4=-(y^4-x^4)

So you could write (ln(y)-ln(x))(x^4-y^4)+(y^2-x^2)^2=0
ln(y/x)(x^4-y^4)+(y^2-x^2)^2=0
ln(y/x)(x^4-y^4)=-(y^2-x^2)^2
ln(y/x)(x^2-y^2)(x^2+y^2)=-(y^2-x^2)^2
ln(y/x)(x^2+y^2)=(y^2-x^2)
ln(y/x)=(y^2-x^2)/R^2

Where R is the distance from the origin.

I don't know if this helps at all- it may give you a better idea of what the graph looks like though

 

[CIMP]

Hello Mentallic,
Thank you very much for your help. I really could advance more with your explanation. I made a plot as suggested by Jamma, and I could see that the function when X<5 it tend to -infinit and if X>5 it also tends to +infinit.

Hello Jama,
Thank you very much for your advice. It is vey useful to make a plot of the function before going to more sophisticated things!

Hello Chiro,
I use Maxima and Matlab and it was the first time that I was looking how to extract the root algoritms with them, but it was quite harder to use it... In fact, I pass all the day of yesterday dealing with that. Then I give up and I ploted the function and it was easy easier to see that it has just one solution when X=5. By lucky!
But I need to learn how to use the software...

Thanks everybody. I am so happy to be in this forum. Thansk again for your time and help...

CIMP

 

[Mentallic]

I could see that the function when X<5 it tend to -infinit and if X>5 it also tends to +infinit.

Not quite. Even though ln(y/x) is real for x,y<0, the original equation was of the form ln(y)-ln(x) in which case x,y<0 is NOT valid. You need to keep in mind that whatever transformations you do, you only consider the domain of the original function, not your manipulated function.

 

[Char. Limit]

Then what you're looking for is to solve explicitly for x. However, you can't do that with the use of elementary functions (the usual - addition, multiplication, square roots, powers, trig, exponentials). You will need to find a numeric approximation to the answer.

There are trivial solutions however such as y=x and y=-x, but notice that with logarithms present in the equation, this means you need to restrict x and y to be more than zero, so only y=x, x>0 is a trivial solution.

So, if y=5 then x=5 as well, but that's not the only solution. The others can be found by numeric methods.

Actually, I'm going to have to disagree with you here. It looks to me as if x=5 IS the only solution corresponding to y=5.

 

[Mentallic]

Actually, I'm going to have to disagree with you here. It looks to me as if x=5 IS the only solution corresponding to y=5.

Ahh yes they're in fact double roots. Thanks for spotting it!

$$\ln{y}(x^4-y^4)+\ln{x}(y^4-x^4)+(y^2-x^2)^2=0$$

$$(y^2-x^2)\left(\ln{\frac{1}{y}}(x^2+y^2)+\ln{x}(x^2+y^2)+y^2-x^2\right)=0$$

The first factor corresponds to the roots of $y=\pm x$ thus only $y=x, x>0$ to satisfy the domain, and the second factor is

$$\ln{\frac{x}{y}}(x^2+y^2)=x^2-y^2$$

Which gives us the same roots as the other factor, since ln(1)=0.

 

[CIMP]

Yes!
Thanks!

"Mathematics is the language of nature..."

 

[Aleta374]

thanks, i hat the same question

 

[Jamma]

So we've just solved all your homework for you guys, pfft.

 

[Mentallic]

CIMP, in the future, please post your questions in the homework help section.