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1+x^2 dy/dx = 1+y^2, differential equation! a little Help here!

  1. Aug 31, 2010 #1
    1. The problem statement, all variables and given/known data

    1+x^2 dy/dx = 1+y^2



    2. Relevant equations



    3. The attempt at a solution

    if I clean this up a little, I would get

    1/ (1+x^2) dx = 1/ (1+y^2) dy

    correct?

    since the integral of 1+x^2 is arctanX, I get

    arctanX + C = arctanY + C.

    And I don't know what to do from here. Please help!
     
  2. jcsd
  3. Aug 31, 2010 #2

    hunt_mat

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    You have:
    [tex]
    \tan^{-1}y=\tan^{-1}x+\tan^{-1}(d)
    [/tex]
    Take tan of both sides and use the double angle formula for tan.
     
  4. Sep 1, 2010 #3

    HallsofIvy

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    First, you don't need both "C"s. And you especially should not write the same symbol, "C", for both- that makes it look like they will cancel. Since the constant of integration may not be the same on both sides, you shold write "arctan(x)+ C1= arctan(y)+ C2" (also, do NOT use "x" and "X" interchangeably- they are different symbols).

    Now, if you like you can write arctan(y)= arctan(x)+ (C1- C2) and since C1 and C2 are unknown constants, C1- C2 can be any constant. Just call it "C": C= C1- C2. Then you have arctan(y)= arctan(x)+ C.

    If you want to solve for y, just take the tangent of both sides:
    y= tan(arctan(x)+ C)
    Caution- this is NOT the same as tan(arctan(x))+ arctan(C)!

    What is true is that
    [tex]tan(A+ B)= \frac{tan(A)+ tan(B)}{1+ tan(A)tan(B)}[/tex]
    with A= arctan(x) and B= C, tan(A)= x so
    [tex]y= tan(arctan(x)+ C)= \frac{x+ tan(C)}{1+ tan(C)x}[/tex]

    and, again, since C is an arbitrary constant and the range of tangent is all real numbers, tan(C) is an arbitrary constant. letting tan(C)= C',
    [tex]y= \frac{x+ C'}{1+ C'x}[/tex]
     
  5. Sep 1, 2010 #4

    ehild

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    Sorry, HallsofIvy, it is not true. The correct formula is:

    [tex]tan(A+ B)= \frac{tan(A)+ tan(B)}{1- tan(A)tan(B)}[/tex]

    By the way, the original equation

    1+x^2 dy/dx = 1+y^2

    is equivalent to x^2 dy/dx = y^2 :wink: with the solution

    [tex]y=\frac{x}{1+cx}[/tex]:


    ehild
     
    Last edited: Sep 2, 2010
  6. Sep 2, 2010 #5

    Hurkyl

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    The joys of sloppy use of parentheses. :frown:

    It looks like the OP meant to have a pair of parentheses around 1+x2, but I suppose we should let him clarify rather than assuming one way or another.
     
  7. Sep 2, 2010 #6

    ehild

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    Yes, Hurkyl, but I answered to HallsofIvy, not to the OP. This custom of not using parenthesis is quite frequent on the Forum and I think we should do something against it, instead of reading minds.

    ehild
     
  8. Sep 2, 2010 #7

    HallsofIvy

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    Thanks. I keep getting that wrong!

    By the way, the original equation

    Unfortunately people on this board tend to be so sloppy with parentheses I just automatically assumed that [itex](1+ x^2) dy/dx= 1+ y^2[/itex] was intended, but yes, the correct interpretation of what was written is as you say.
     
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