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Trouble solving this ode

  1. Nov 9, 2011 #1
    1. The problem statement, all variables and given/known data

    dx/dt = 2000-500x/100

    Solve this linear ODE using integration. You should get a function of t, x(t). This is the "analytical solution". Use the differential equation above, separate the variables, and then integrate to find x(t). Find the integration constant and simplify your final result.

    2. Relevant equations

    dx/dt = 2000-500x/100

    3. The attempt at a solution

    I tried doing this but do no think its correct

    dx-5x = 20dt then I would integrate that.
     
  2. jcsd
  3. Nov 9, 2011 #2

    Ray Vickson

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    As written, you DE is dx/dt = 2000 - 5x, so letting y = x-400 we have dy/dt = -5y, which is easy to solve.

    However, perhaps you meant dx/dt = (2000 - 500x)/100 = 20 - 5x. If that is what you meant, that is what you should have written. Use brackets.

    RGV
     
  4. Nov 9, 2011 #3
    Sorry it was supposed to be dx/dt = (2000-500x)/100
     
  5. Nov 9, 2011 #4
    can you help me solve it?
     
  6. Nov 9, 2011 #5

    SammyS

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    (2000-500x)/100 = 20 - 5x, for one thing.

    The ODE is separable.

    [itex]\displaystyle\frac{dx}{20 - 5x}=dt\,.[/itex]

    Now, integrate both sides.
     
  7. Nov 9, 2011 #6
    how do I integrate with dx in the numerator?
     
  8. Nov 9, 2011 #7

    Ray Vickson

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    The nature of you questions has me wondering: what is your situation? Are you in a course that is far above your background level? Are you using a textbook that does not have any of this material in it? You are asking introductory questions that you should have seen discussed before. You can't learn the material from an on-line homework assistance site.

    RGV
     
  9. Nov 9, 2011 #8
    I am taking my first calculus course and haven't come across this before
     
  10. Nov 9, 2011 #9
    Why wouldn't you integrate with dx in the numerator?

    [tex]\int \frac{1}{20-5x}dx = \int dt[/tex]

    [itex]\frac{dx}{20-5x} = \frac{1}{20-5x}dx[/itex] after all right?
     
  11. Nov 9, 2011 #10

    SammyS

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    [itex]\displaystyle\int\frac{dx}{20 - 5x}=dt[/itex]

    Let u = 20 - 5x . → du =-5dx
     
  12. Nov 9, 2011 #11
    We just started the integral so I don't know all of the integrating rules yet.
     
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