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How did they isolate y in this seperable equation?

  1. Sep 6, 2013 #1
    Hello -

    This is from a textbook example in the chapter on seperable equations:

    [itex]y^{2}-2y = x^{3}+2x^{2}+2x+3[/itex]

    "We solve for y in terms of x, a simple matter":

    [itex]y = 1\pm \sqrt{x^{3}+2x^{2}+2x+4}[/itex]

    Only, it's not a simple matter for me. I've been unable to figure it out. Some direction would be appreciated! :)
  2. jcsd
  3. Sep 6, 2013 #2

    Simon Bridge

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    Hint: quadratic equation.
  4. Sep 6, 2013 #3
    Yes, I was experimenting with that; I turned it into y(y-2) and (y-1)^2 - 1... but that didn't help me. I could take that -1 and add it to the other side, but then I have (y-1)^2 on the left, and taking the square root of both sides results in the square root of (y-1) on the left, which I can't do anything with... right?
  5. Sep 6, 2013 #4

    Simon Bridge

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    Put the equation in the form ##ay^2+by+c=0## ... c will have a lot of x's in it.
  6. Sep 6, 2013 #5
    I am still confused but I will play with it using your guidance. Also, how can I proceed if y is cubed in the equation? It looks like one of the other problems has that issue...

    Thank you
  7. Sep 6, 2013 #6
    I should solve this?

    [itex]y^{2} - 2y + (x^{3}+2x^{2}+2x+3) = 0[/itex]

    Because I don't know how to solve that... or how to factor the thing even...
  8. Sep 6, 2013 #7


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    Get it into (y-b)^2 = a + RHS and then take square roots and re-arrange.
  9. Sep 6, 2013 #8
    Ah yes, thank you. What can I do about ones that have y^3?
  10. Sep 6, 2013 #9


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    You will have to show an example but if you can separate it, you do the same sort of thing (express as (y+blah)^3, take the cubic root and express it as y = blah).

    If you have trouble getting an explicit solution, you can use differential calculus where you solve for the derivative as a function of x and y and then use a numerical scheme to evaluate the function with some degree of error.
  11. Sep 6, 2013 #10

    Simon Bridge

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    You appear a little lost with your algebra there ... perhaps you are not seeing it because of all the x's where you are used to seeing a single constant?

    Starting from: [itex]y^{2}-2y = x^{3}+2x^{2}+2x+3[/itex]

    You see that this is a quadratic in y ... solving it for y means finding the roots.
    You've done this before.

    Put it in standard form: ##ay^2+by+c=0##
    then ##a=1, b=-2, c=x^{3}+2x^{2}+2x+3## and use the quadratic equation:
    $$y\in\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$... and you get:
    $$y\in\frac{2\pm\sqrt{4+4(x^{3}+2x^{2}+2x+3)}}{2}$$ ... and simplify.

    In your case, the RHS is a nice quadratic in y. So there's a shortcut.

    ##y^2-2y=(y-1)^2-1## so we can rewrite the original equation: $$(y-1)^2-1= x^{3}+2x^{2}+2x+3\\ \Rightarrow (y-1)^2=x^{3}+2x^{2}+2x+4$$ ... take the square-root of both sides and follow your nose.

    This process is called "completing the square". Not all quadratics are nice like this.

    There is also a cubic equation for if y appears to the third power.
    ... you could also try completing the cube. i.e. if you can express it as $$(y+\alpha)^3+\beta = f(x)$$ then you can just as $$y=(f(x)-\beta)^{1/3}-\alpha$$

    More generally you may have ##\sum a_ny^n = f(x)## ... in which case you can get y in terms of f(x) by a combination of guessing and long division - like you learned to find the roots of a polynomial.

    Really generally, a separable equation could end up as f(y)=g(x) ... i.e. the LHS is a function of y alone and the RHS is a function of x alone. How you approach this depends on f and g. There is no general approach.
    Last edited: Sep 7, 2013
  12. Sep 7, 2013 #11
  13. Sep 7, 2013 #12
    Now I understand the general idea behind ones with squared y's... but I am having problems with this:

    y^3 - 3y^2 = ......

    How do i handle that y side? I don't think the cubic equation is involved; it is too complicated for this material. How can I factor out and get rid of a y? With quadratic I did it by completing the square...

  14. Sep 7, 2013 #13

    Simon Bridge

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    eg. $$y^3+3y^2+3y = x^3-1$$ ... notice that this (the LHS) differs from ##(y+1)^3## by a constant so you can rewrite the original equation as: $$(y+1)^3 =x^3 \Rightarrow y+1=x \Rightarrow y=x-1$$

    But if we take your original example:
    ##y^{2}-2y = x^{3}+2x^{2}+2x+3##
    ... and say we wanted to find x in terms of y: start by moving that 3 to the other side:

    ##y^{2}-2y-3 = x^{3}+2x^{2}+2x##
    ... and try to complete the cube in the RHS:

    ##x^{3}+2x^{2}+2x = (x+1)^3 -x^2-x+3##
    ... which doesn't help.

    Instead rewrite as $$x^{3}+2x^{2}+2x+c=0: c=3-y^{2}-2y$$ and work out the roots in terms of c(y) by some other method. This can get very hard to do - at your level, you'll probably not see the hard ones.

    The reason your text just said it was clear is that you are supposed to already know how to solve quadratics.
    Where the LHS is a quadratic, you'll get:

    ##ay^2+by+c = f(x) \Rightarrow y^2+by/a = (f(x)-c)/a## and you can finish by completing the square.
    Last edited: Sep 7, 2013
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