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ODE example

  1. Nov 20, 2011 #1
    this example was in a book I bought ( maths methods for physics, Mathews and Walker)

    dy/dx + sqrt( (1-y^2) / (1-x^2) ) = 0

    dy/(1-y^2) + dx/(1-x^2) = 0

    sin -1 y + sin -1 x = c [1]

    or, taking the sine of both sides

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




    My response/confusion with this.....

    [1] Ok so i'm fine with this, standard integral

    [2] ???Where on earth did this come from
    just by taking the sin of the equation
    above.

    Please help if you can.
     
  2. jcsd
  3. Nov 20, 2011 #2

    lurflurf

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    use from elementary trigonometry
    [tex]\sin(x+y)=\sin(x)\cos(y)+\cos(x)\sin(y)[/tex]
    and
    [tex]\sin( \cos^{-1}(x))=\sqrt{1-x^2}[/tex]
     
  4. Nov 20, 2011 #3

    HallsofIvy

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    Because [itex]sin^2(y)+ cos^2(y)= 1[/itex], [itex]sin(y)= \sqrt{1- cos^2(y)}[/itex] and so if x= cos(y) (so that [itex]y= cos^{-1}(x))[/itex], then [itex]sin(cos^{-1}(x))= sin(y)= \sqrt{1- cos^2(y)}= \sqrt{1- x^2}[/itex].

    Another way of seeing this is to think of [itex]cos^{-1}(x)[/itex] as representing the angle in a right triangle that has cosine equal to x. Since "cosine= near side over hypotenuse" and x= x/1, that right triangle can have "near side" of length x and hypotenuse 1. By the Pythagorean theorem, the third side, the "opposite side" to the angle has length [itex]\sqrt{1- x^2}[/itex] and so the sine of that angle, [itex]sin(cos^{-1}(x)[/itex], is [itex]\sqrt{1- x^2}/1= \sqrt{1- x^2}[/itex].
     
  5. Nov 20, 2011 #4
    Thanks folks for your insight really appreciate it.
     
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