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Initial-value problem

  1. Oct 22, 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


    please if someone can help me start it or explain it to me
     
  2. jcsd
  3. Oct 22, 2006 #2

    quasar987

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    Do you know what a separable differential equation is?
     
  4. Oct 22, 2006 #3
    you need to first use separation of variables, to get [tex] y(x) [/tex].

    So: [tex] -\frac{dy}{y^{2}} = dx [/tex]

    [tex] \int -\frac{dy}{y^{2}} = \int dx [/tex]
     
  5. Oct 22, 2006 #4
    no we just started covering DE on friday
     
  6. Oct 22, 2006 #5

    quasar987

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    Well they're the type of differential equation that are the easiest to solve. If you've not yet seen how to solve this type of DE, why not just postpone the resolution of the problem til you've learned about it?
     
  7. Oct 22, 2006 #6
    in class all we went over was exponential growth and decay. but i dont remember going over initial-value problems.
     
  8. Oct 22, 2006 #7

    quasar987

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    All an "initial value problem" is is a DE together with an "initial condition", i.e. a value of y(0). So just solve the differential equation, and set y(0)=0.0625. this will specify a concrete value of the intgegration constant. So, start with solving the differential equation like courtrigrad showed you. Can you do that?
     
  9. Oct 22, 2006 #8
    im not sure exactly how to solve the DE. What does dy and dx become?
    do they just simply become x and y?
     
  10. Oct 22, 2006 #9

    quasar987

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    You don't know how to solve an integral?!
     
  11. Oct 22, 2006 #10
    i can solve integrals, but im confused about what the integral of dx becomes. i can solve the integral of -1/y^2 which is 1/y. i dont know what goes on with the integral of dy or dx.
     
  12. Oct 22, 2006 #11

    quasar987

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    [itex]\int dx[/itex] is a notation for [itex]\int 1dx[/itex]. Surely you can solve [itex]\int 1dx[/itex]? It's only a matter of finding a function F(x) such that F'(x) = 1 and adding an "integration constant" C to it.
     
  13. Oct 22, 2006 #12
    oh so the integral of dx is simply x+c. but what is going on with the dy?
     
  14. Oct 22, 2006 #13

    quasar987

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    The dy is only there to say that you must integrate with respect to y.

    [tex] \int -\frac{dy}{y^{2}}[/tex] is a notation for [tex] \int -\frac{1}{y^{2}} \ \ dy[/tex]
     
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