Expert: Paul Klarreich - 10/24/2006

Question
please tell me what im doing wrong.

general solution of separable equation:
dx/dy = (x-x^3)/(y-y^3)

dx/dy = g(x)/h(y)

h(y)dx/dy=g(x)

integral h(y(x))(y`(x)) dx = integral g(x) dx
where y=y(x) dy=y`(x) dx

integral h(y) dy = integral g(x) dx

integral (y-y^3) dy = integral (x-x^3) dx

y^2/2 - y^4/4 = x^2/2 - x^4/4 + C

4(y^2) - 2(y^4) = 4(x^2) - 2(x^4) + C

here im stuck. i should solve for y


Answer
Questioner:  Daniel
Category:  Calculus
 
Subject:  separable equations
Question:  please tell me what im doing wrong.

general solution of separable equation:
dx/dy = (x-x^3)/(y-y^3)

dx/dy = g(x)/h(y)

h(y)dx/dy=g(x)

integral h(y(x))(y`(x)) dx = integral g(x) dx
where y=y(x) dy=y`(x) dx

integral h(y) dy = integral g(x) dx

integral (y-y^3) dy = integral (x-x^3) dx

y^2/2 - y^4/4 = x^2/2 - x^4/4 + C

4(y^2) - 2(y^4) = 4(x^2) - 2(x^4) + C

here im stuck. i should solve for y
........................................
Hi, Daniel,

Are you sure the left side is  dx/dy?  not dy/dx?  You seem to have gotten the separation upside down.

I think you just separate the variables as they are:
WARNING: THE FOLLOWING DISCUSSION MAY CONTAIN FRACTIONS AND OTHER MATERIAL INAPPROPRIATE FOR CERTAIN COMPUTING SYSTEMS.  BE SURE TO VIEW IT IN A FIXED-SIZE FONT, SUCH AS COURIER.


dx   x - x^3
-- = -------
dy   y - y^3

  dx        dy
------- = -------
x - x^3   y - y^3

[Not  (x - x^3) dx, I think.]

Now integrate each side separately.  Fortunately, they are the same, so you only have to do the work once.

  1            1
------- = ---------------
x - x^3   x(1 - x)(1 + x)

Looks like a partial fractions example.

     1            A      B        C
--------------- = --- + ------ + ----- =
x(1 - x)(1 + x)    x    1 - x    1 + x

A(1 - x^2) + Bx(1 + x) + Cx(1 - x)
---------------------------------- =
         x(1 - x)(1 + x)

A - Ax^2 + Bx + Bx^2 + Cx - Cx^2           1
---------------------------------- = ---------------
   x(1 - x)(1 + x)                  x(1 - x)(1 + x)

So the equations to solve are:

A          = 1    <<<  constants
    B + C = 0    <<<  x - terms
-A + B - C = 0    <<<  x^2 - terms.

-A + 2B = 0  -->  2B = A,  B = 1/2
B + C = 0  -->  C = -1/2

Almost ready:  We have to integrate:

1     1/2     -1/2
--- + ------ + -----
x    1 - x    1 + x
That gives:

ln x - 1/2 ln(1 - x) - 1/2 ln(1 + x) =

ln x - 1/2 [ ln(1 - x) + ln(1 + x)] =

ln x - 1/2 ln[(1 - x)(1 + x)] =

ln x - 1/2 ln[1 - x^2] =

ln x - ln[ sqrt(1 - x^2)] =

       x
ln[ ----------- ]
   sqrt(1-x^2)

That's it for the x-part.  The y-part looks the same, but we will have a constant.]


       x                  y
ln[ ----------- ] = ln[-----------] + c0
   sqrt(1-x^2)        sqrt(1-y^2)


       x                  y
ln[ ----------- ] - ln[-----------] = c0
   sqrt(1-x^2)        sqrt(1-y^2)

More logarithm tricks:
        x       sqrt(1-y^2)
ln [ ----------- -----------] = c0
    sqrt(1-x^2)      y


    x       sqrt(1-y^2)
----------- -----------  = C1,  which is just  e^c0
sqrt(1-x^2)      y

  x^2    1-y^2
------- -------  = C2,  which is just  C1^2
 1-x^2    y^2

And I guess that's where you stop.  I.E. You DON'T bother solving for y.  You got the original equation in IMPLICIT form, so you can leave the answer that way.