I'm trying to solve the following equation in terms of y.

x=1-[3(1-y)^(2/3)]+2-2y

can't seem to solve it, have had a look at non-linear second order differential equations but still can't seem to work it out.

Can anyone help???

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- October 16th 2005, 08:20 PMslug000tricky equation
I'm trying to solve the following equation in terms of y.

x=1-[3(1-y)^(2/3)]+2-2y

can't seem to solve it, have had a look at non-linear second order differential equations but still can't seem to work it out.

Can anyone help??? - October 16th 2005, 10:25 PMrgep
As written it isn't a differential equation as there is no differentiation involved! Have you posted it correctly?

- October 18th 2005, 12:31 AMslug000
It is posted correctly, I thought I might be able to solve by using a differential equation (didn't mean it was one). Basically I just want to get a function in terms of y (y=...). When I tried to solve it algebraically I got a messy looking equation and still couldn't seem to get y by itself (I got a polynomial function of y's i.e y^2, y^3...). I don't know if it is possible to get y by itself. Basically I want to graph x against y so thats why I want a function in terms of y.

- October 18th 2005, 01:56 AMticbol
Tricky question?

x=1-[3(1-y)^(2/3)]+2-2y --------(i)

Simplifying that,

x = 3 -2y -3(1-y)^2 -----(ii)

So you only want to solve for "y" in terms of "x" so that you could graph the the equation in the x,y plane?

Why don't you out-trick the original equation then?

In (i) or (ii), swap the x and y,

y =1 -[3(1-x)^(2/3)] +2 -2y --------(ia)

Or,

y = 3 -2x -3(1-x)^2 -----(iia)

1.) Now graph (iia) the usual way.

2.) Then rotate the graph clockwise by 90 dergrees.

The "positive y" now of (iia) will become the real/usual positive x.

But the "positive x" of (iia) will become the real/usual negative y.

3.) So, reverse or flip or rotate this new graph in (2.) about the new, or now real/usual, x-axis by 180 degrees.

The "positive y" now of (iia) will remain as the real/usual positive x.

And the "positive x" of (iia) will become the real/usual positive y.

4.) There now is the graph of x=1-[3(1-y)^(2/3)]+2-2y --------(i). - October 18th 2005, 05:21 PMslug000
Thanks, sounds like a good way to go about it. The only problem I have now is how to solve such a thing in Microsoft Excel. Basically I need to plot the function of y in excel, as y is a set of data and x is a set of data.

- October 19th 2005, 02:16 AMticbol
Sorry, I am not an expert on Excel. I do not even know how to use any graphing calculator or any Math applications like MathLab, MathCad, etc..

I thought you only know how to graph an equation when it is in the form y = ......, that'sa why I suggested what I wrote before.

If you know how to use Excel, you cannot graph the original equation as it is? - October 19th 2005, 07:18 PMslug000
Thanks anyway ticbol, I might still be able to put your idea to use. If anyone has any ideas about how I might graph such a function using excel, i'm all ears.

- October 19th 2005, 08:12 PMMathGuru
Take a look at this and let me know if it works for you