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Differential Equations

  1. Feb 13, 2010 #1
    1. The problem statement, all variables and given/known data

    Im having trouble with "initial value" problems. I cant seem to grasp the "way" around it. any help is greatly appreciated.

    [tex]\sqrt{}1-y^2[/tex] dx - [tex]\sqrt{}1-x^2[/tex] dy = 0
    y(0) = sqr(3) / 2


    2. Relevant equations

    solve by separation of variables and inital value.

    3. The attempt at a solution

    rearranging
    i get dy / [tex]\sqrt{}1-y^2[/tex] = dx / [tex]\sqrt{}1-x^2[/tex]
    integrating
    arcsin(y) = arcsin(x) + c

    y = sin (arcsin (x) + c))
    y (o) = sqr3/2

    c = pi/3

    therefore y = sin(arcsin x + pi/3)

    the back of the book says:

    y=.5x + sqr(3)/2 ([tex]\sqrt{}1-x^2[/tex]

    ====
    i was reading the book but theres only 1 example about initial value problems.
    so If anyone can give me "tips" on how to solve this problems that'd be great
     
  2. jcsd
  3. Feb 13, 2010 #2

    Dick

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    Your answer is correct, but so is the book's. Use sin(a+b)=sin(a)cos(b)+cos(a)sin(b) to show this.
     
  4. Feb 14, 2010 #3
    Thank you.

    Another question regarding my homework, same topic (differential equations)

    Im doing "substitution" where you have to identify that M(x,y) and N(x,y) have the same degree by doing M(tx,ty) and N(tx,ty) and such. . . substitute by u= x/y or u=y/x

    anyways I dont see any "usefulness" in this process. . .


    1. Problem:
    (x-y)dx + xdy = 0

    M = x - y
    N = x

    u=x/y
    x=uy
    dx = ydu + udy

    subsitute:

    (uy - y)(ydu + udy) + uydy = 0 (now everything is more complicated. . .yay?)
    multiply trough?
    or what do I do next?
    y^2 udu + u^2 ydy - y^2 du = 0

    and now what?

    thnkz again
     
  5. Feb 14, 2010 #4

    Dick

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    Divide the equation through by xdx. (1-y/x)+dy/dx=0. Set u=y/x and figure out what dy/dx is in terms of du/dx. Eliminate the y's. Not the x's.
     
  6. Feb 14, 2010 #5
    are there any specific "tips" on like how to work around them? from what i saw there its good to get a dy/dx somewhere (like there by dividing everything by xdx) and then doing the substitution. . .
    but how do you know if its better to use u=y/x or v=x/y ?
     
  7. Feb 14, 2010 #6

    Dick

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    Depends what you want the dependent variable to be. If you want to find y as a function of x, then I suggest u=y/x. If you want x as a function of y then x/y is probably better. Really, just practicing with different combinations will tell you what's better. That's how I learned.
     
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