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Trouble solving this ode

  1. Nov 9, 2011 #1
    I am having trouble solving this ode, I wasn't sure if this should go under calculus section or differential equations section, but I figured since were given this in calculus 1 it belongs here. We just recently started learning integral calculus and I don't know a whole lot about differential equations. How would you solve this?

    dx/dt = 2000-500x/100

    the answer is supposed to be X(t). Thank you in advance.
  2. jcsd
  3. Nov 9, 2011 #2
    Get all the x's to the dx/dt side, then integrate both sides. When I say integrate, I mean integrate the side with the x's in terms of x and integrate the other in terms of t. That will leave with something that you can use simple algebraic manipulation to get it into a explicit form.
  4. Nov 9, 2011 #3
    When i tried that i got
    dx/dt - 5x = 20
    dx - 5x = 20 dt
    ∫dx - 5x = ∫20 dt
    (5x^2)/2 = 20t
    5x^2 = 40t
    x^2 = 8
    x = 2.828

    am I on the right track here?
  5. Nov 9, 2011 #4
    *x = 2.828t
  6. Nov 9, 2011 #5
    Oh, I forgot to mention that you have to get some function of x to be multiplied by dx. In other words, dx/x is okay, but dx+x isn't. More generally, f(x)dx is okay, f(x)+dx isn't.

    [tex]\int \frac{dx}{400-x}=\int 5 dt[/tex]
    and so on...
  7. Nov 9, 2011 #6
    how would i do the integral of the left side?
  8. Nov 9, 2011 #7


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    Homework Helper

    Have you heard of a function called log(x)?
  9. Nov 9, 2011 #8
    actually no
  10. Nov 10, 2011 #9
    dx/dt = 2000-500x/100
    Integrating it,-ln(4-x)=5t+k(integrating constant)
    (ln(x) is one of the basic functions in maths)
  11. Nov 10, 2011 #10
    [tex]\int \frac{dx}{400-x}[/tex] is a specific case of [tex]\int \frac{dx}{x}[/tex]. Given [tex]\int \frac{dx}{x} = ln(x)[/tex], you can solve it using substitution.

    BTW, omkar13 is close but wrong. The correct answer can be found on Wolfram Alpha, if you want proof: http://www.wolframalpha.com/input/?i=dx/dt+=+2000-500x/100.
  12. Nov 10, 2011 #11


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    Science Advisor

    This is just bad algebra. If you start with dx/dt- 5x= 20 and multiply both sides by dt, you get dx- 5xdt= 20dt. However, the best thing to do is to go back to dx/dt= 5x+ 20, multiply both sides by dt and divide both sides by 5x+ 20: dx/(5x+20)= dt. Now let u= 5x+ 20 so that du= 5dx or dx= (1/5)du. The equation becomes (1/5) du/u= dt. Integrat both sides, then go back to x.

    No, "dx- 5x" (which was wrong anyway) is NOT -5xdx

    Pretty much every thing you did, both algebra and Calculus, is wrong.
  13. Nov 11, 2011 #12
    dude we just started on the integral in class you don't have to be douchebag I was just asking for help
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