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Find value

  1. Jul 19, 2006 #1
    i try to solce the initial value problem question and got f(x,y) = x^2 -x +y^2 +y +c and y(0)=2 is given

    how do i find c??
    is it just make the eqn x^2 -x +y^2 +y +c = 0 and subs y=2 and x=0??

    pls help

    thanx
     
  2. jcsd
  3. Jul 19, 2006 #2

    berkeman

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    Staff: Mentor

    I think you need more info. To find c, you would need to know f(0,0), x(0) and y(0). Are those available?
     
  4. Jul 19, 2006 #3
    the eqn is (x-1)dx + (y+1)dy=0 , y(1)=0
    just only this
     
  5. Jul 19, 2006 #4

    berkeman

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    I don't understand the problem. Could you please post the full text of the problem? Is it an integration or differentiation or what? Are you taking partial derivatives?
     
  6. Jul 19, 2006 #5

    berkeman

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    And what do you mean by both y(1)=0 and y(0)=2? What is the argument to y? Time? Too many things are getting mixed up here.
     
  7. Jul 19, 2006 #6
    Are the following differential equations exact? Solve the IVP:

    (x-1)dx + (y+1)dy=0 , y(1)=0
     
  8. Jul 19, 2006 #7

    berkeman

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    Sorry, I'm not familiar with that form of differential equation. I'll see if I can find somebody to help.
     
  9. Jul 19, 2006 #8
    okok........thanx
     
  10. Jul 19, 2006 #9

    HallsofIvy

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    Of course it's exact: [itex]\frac{\partial (x-1)}{\partial y}= \frac{\partial (y+1)}{\partial x}= 0[/itex]. In fact, it's separable.
    (x-1)dx+ (y+1)dy= 0 can be integrated directly to get
    [tex]\frac{1}{2}x^2- x+ \frac{1}{2}y^2+ y+ C= 0[/itex]
    Now put x= 1, y= 0 to find C.
     
  11. Jul 20, 2006 #10
    thanx............
     
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