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Homework Help: Initial-value problem

  1. Oct 23, 2006 #1
    i have no idea on how to start this problem.


    Find a solution of the initial-value problem

    dy/dx= -y^2 , y(0)=.0625


    so..

    (1/-y^2)dy = dx

    int(1/-y^2)dy=int(dx)

    1/y=x+c
    please if someone can help me start it or explain it to me
     
  2. jcsd
  3. Oct 23, 2006 #2

    radou

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    Note that you have [tex]-\frac{1}{y} = x + C[/tex]. Now all you have to do is write [tex]y = -\frac{1}{x+C}[/tex] and use the initial value.
     
    Last edited: Oct 23, 2006
  4. Oct 23, 2006 #3
    so am i solving for C?
     
  5. Oct 23, 2006 #4

    radou

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    Yes, you are.
     
  6. Oct 23, 2006 #5
    so
    C= (1/y)-x ??

    so C=(1/.0625) if y(0)=.0625 ????
    is this correct?
     
  7. Oct 23, 2006 #6

    radou

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    Yes, C = -16.
     
  8. Oct 23, 2006 #7
    negative???
     
  9. Oct 23, 2006 #8

    radou

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    Yes, since [tex]y(x) = -\frac{1}{x+C} \Rightarrow y(0) = -\frac{1}{C} = 0.0625[/tex]
     
  10. Oct 23, 2006 #9
    but the integral of -1/y^2 dy is 1/y
     
  11. Oct 23, 2006 #10

    radou

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    Yes, you're right. I apologize, that minus sign slipped through somehow. :smile: So, your answer is correct.
     
  12. Oct 23, 2006 #11
    ok but i put the answer c=16 and it was wrong. is there another step?
     
  13. Oct 23, 2006 #12
    ok i got it the answe is y=1/(x+16)

    thank you for your help
     
  14. Oct 23, 2006 #13

    radou

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    Yes, I just wanted to write that the answer is a function, not a constant.
     
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