Solving set of non-linear implicit eqns: x^3 - 5((y^2)/x) = 6/y & y^3 - 5((x^2)/y) =

[andrew180]

Solving set of non-linear implicit eqns: x^3 - 5((y^2)/x) = 6/y & y^3 - 5((x^2)/y) =

Hey guys,

I hope this is the right place. I'm an mech engineer who has been out of school and working for a few years now but I'm trying to re-study some old math subjects I've since forgotten. One is numerical methods and I'm stuck on solving these higher order set of equations. For example this one.

x^3 - 5((y^2)/x) = 6/y
y^3 - 5((x^2)/y) = 6/x

Could anyone help me find some examples or documents to read on how to solve equations like these? Thanks a lot.

 

[Ishuda]

Hey guys,

I hope this is the right place. I'm an mech engineer who has been out of school and working for a few years now but I'm trying to re-study some old math subjects I've since forgotten. One is numerical methods and I'm stuck on solving these higher order set of equations. For example this one.

x^3 - 5((y^2)/x) = 6/y
y^3 - 5((x^2)/y) = 6/x

Could anyone help me find some examples or documents to read on how to solve equations like these? Thanks a lot.

Just as a point reference, these two equations are essentially the same formulas. Generally, in a situation such as this I try to manipulate the equation into something 'familiar' So, taking the top equation, multiplying both sides by -xy we are left with a cubic in y:
5 y³ - x⁴ y + 6x = 0
So, I might look at solutions for roots of cubics [there are closed form solutions]

But to get back to your question, you might also look at
https://archive.org/search.php?query=numerical methods
Don't know if it is still out there or new editions are available but I always kept a copy of 'Numerical Methods that Usually Work' by Forman S. Acton.


EDIT: Dumb me! For some reason I was thinking solve each equation independently. Off to the corner I go.

 

[stapel]

x^3 - 5((y^2)/x) = 6/y
y^3 - 5((x^2)/y) = 6/x

To be clear, I'm taking the above as meaning the following:

. . . . .\(\displaystyle x^3\, -\, 5\left(\dfrac{y^2}{x}\right)\, =\, \dfrac{6}{y}\)

. . . . .\(\displaystyle y^3\, -\, 5\left(\dfrac{x^2}{y}\right)\, =\, \dfrac{6}{x}\)

(Rational expressions are hard to format, though you did great with the grouping symbols. Thank you! I'm just typing it out in "nice" formatting so that other folks like me, losing our eyesight, can "see" it better.)

Could anyone help me find some examples or documents to read on how to solve equations like these?

A lot of the "method" for this generally boils down to (1) using intelligent substitution and (2) hoping you don't see things bigger than quadratics. (here) So... ouch.

My first step would be to multiply through on each equation, to clear the denominators. This will get me as far as:

. . . . .\(\displaystyle x^4y\, -\, 5y^3\, =\, 6xy\)

. . . . .\(\displaystyle xy^4\, -\, 5x^3\, =\, 6xy\)

Edit: The above equations are incorrect. The corrected equations are:

. . . . .\(\displaystyle \color{purple}{x^4y\, -\, 5y^3\, =\, 6x}\)

. . . . .\(\displaystyle \color{purple}{xy^4\, -\, 5x^3\, =\, 6y}\)

Which means that the right-hand sides of the equations are no longer equal, so the rest doesn't follow. Bummer....

However, we can make them equal:

. . . . .\(\displaystyle x^4y^2\, -\, 5y^4\, =\, 6xy\)

. . . . .\(\displaystyle x^2y^4\, -\, 5x^4\, =\, 6xy\)

Then we get:

. . . . .\(\displaystyle x^4y^2\, -\, 5y^4\, =\, x^2y^4\, -\, 5x^4\)

. . . . .\(\displaystyle x^4y^2\, -\, x^2y^4\, +\, 5x^4\, -\, 5y^4\, =\, 0\)

. . . . .\(\displaystyle x^2y^2\, (x^2\, -\, y^2)\, +\, 5(x^2\, -\, y^2)(x^2\, +\, y^2)\, =\, 0\)

. . . . .\(\displaystyle (x^2\, -\, y^2)(x^2y^2\, +\, 5x^2\, +\, 5y^2)\, =\, 0\)

. . . . .\(\displaystyle (x\, -\, y)(x\, +\, y)(5x^2\, +\, x^2y^2\, +\, 5y^2)\, =\, 0\)

The solvable factors tell us that x = y and x = -y. But surely there are restrictions on this?

However, I've ground to a halt. Sorry.

 

[andrew180]

Just as a point reference, these two equations are essentially the same formulas. Generally, in a situation such as this I try to manipulate the equation into something 'familiar' So, taking the top equation, multiplying both sides by -xy we are left with a cubic in y:
5 y³ - x⁴ y + 6x = 0
So, I might look at solutions for roots of cubics [there are closed form solutions]

But to get back to your question, you might also look at
https://archive.org/search.php?query=numerical methods
Don't know if it is still out there or new editions are available but I always kept a copy of 'Numerical Methods that Usually Work' by Forman S. Acton.


EDIT: Dumb me! For some reason I was thinking solve each equation independently. Off to the corner I go.

Cool thanks for the response! Helped out.

 

[andrew180]

Just as a point reference, these two equations are essentially the same formulas. Generally, in a situation such as this I try to manipulate the equation into something 'familiar' So, taking the top equation, multiplying both sides by -xy we are left with a cubic in y:
5 y³ - x⁴ y + 6x = 0
So, I might look at solutions for roots of cubics [there are closed form solutions]

But to get back to your question, you might also look at
https://archive.org/search.php?query=numerical methods
Don't know if it is still out there or new editions are available but I always kept a copy of 'Numerical Methods that Usually Work' by Forman S. Acton.


EDIT: Dumb me! For some reason I was thinking solve each equation independently. Off to the corner I go.

This is great thanks a lot!

 

[andrew180]

To be clear, I'm taking the above as meaning the following:

. . . . .\(\displaystyle x^3\, -\, 5\left(\dfrac{y^2}{x}\right)\, =\, \dfrac{6}{y}\)

. . . . .\(\displaystyle y^3\, -\, 5\left(\dfrac{x^2}{y}\right)\, =\, \dfrac{6}{x}\)

(Rational expressions are hard to format, though you did great with the grouping symbols. Thank you! I'm just typing it out in "nice" formatting so that other folks like me, losing our eyesight, can "see" it better.)


A lot of the "method" for this generally boils down to (1) using intelligent substitution and (2) hoping you don't see things bigger than quadratics. (here) So... ouch.

My first step would be to multiply through on each equation, to clear the denominators. This will get me as far as:

. . . . .\(\displaystyle x^4y\, -\, 5y^3\, =\, 6xy\)

. . . . .\(\displaystyle xy^4\, -\, 5x^3\, =\, 6xy\)

Edit: The above equations are incorrect. The corrected equations are:

. . . . .\(\displaystyle \color{purple}{x^4y\, -\, 5y^3\, =\, 6x}\)

. . . . .\(\displaystyle \color{purple}{xy^4\, -\, 5x^3\, =\, 6y}\)

Which means that the right-hand sides of the equations are no longer equal, so the rest doesn't follow. Bummer....

However, we can make them equal:

. . . . .\(\displaystyle x^4y^2\, -\, 5y^4\, =\, 6xy\)

. . . . .\(\displaystyle x^2y^4\, -\, 5x^4\, =\, 6xy\)

Then we get:

. . . . .\(\displaystyle x^4y^2\, -\, 5y^4\, =\, x^2y^4\, -\, 5x^4\)

. . . . .\(\displaystyle x^4y^2\, -\, x^2y^4\, +\, 5x^4\, -\, 5y^4\, =\, 0\)

. . . . .\(\displaystyle x^2y^2\, (x^2\, -\, y^2)\, +\, 5(x^2\, -\, y^2)(x^2\, +\, y^2)\, =\, 0\)

. . . . .\(\displaystyle (x^2\, -\, y^2)(x^2y^2\, +\, 5x^2\, +\, 5y^2)\, =\, 0\)

. . . . .\(\displaystyle (x\, -\, y)(x\, +\, y)(5x^2\, +\, x^2y^2\, +\, 5y^2)\, =\, 0\)

The solvable factors tell us that x = y and x = -y. But surely there are restrictions on this?

However, I've ground to a halt. Sorry.

Cool thanks a lot. Helped me out!