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Differential equation problem

  1. Mar 31, 2012 #1
    1. The problem statement, all variables and given/known data

    http://img546.imageshack.us/img546/1860/dequestion.jpg [Broken]

    3. The attempt at a solution

    My main problem is with the second part where it says find a solution to the DE that satisfies the initial condition x(π) = 2. I found the general solution to be

    [itex]x(t) = \frac{1}{9 t} \ sin \ 3t - \frac{1}{3} \ cos \ 3t + \frac{c}{t}[/itex]

    So when we substitute in we get

    [itex]x(\pi) = 2 = \frac{1}{9 \pi} \ sin \ 3 \pi - \frac{1}{3} \ cos \ 3 \pi + \frac{c}{\pi}[/itex]

    Now I need to solve for the constant C. So my question is: Do I need to calculate this using radians or degrees? I mean, should I have my calculators settings on degrees or radians? :confused:

    By the way this is how I solved the DE:

    [itex]\frac{dx}{dt} + \frac{x}{t} = sin \ 3t[/itex]

    using the integrating factor

    [itex]\mu (t) = e^{\int \frac{1}{t} dt}= k \ e^{\ln |t|} = t[/itex]

    [itex]t \frac{dx}{dt} + t \frac{x}{t}= t(sin \ 3t)[/itex]

    [itex]\int \frac{d}{dt} tx = \int t \ sin \ 3t[/itex]

    Using integration by parts for the RHS

    [itex]tx = \frac{1}{9} \ sin \ 3t - \frac{t}{3} \ cos \ 3t + k[/itex]

    [itex]\therefore \ x(t) = \frac{1}{9 t} \ sin \ 3t - \frac{1}{3} \ cos \ 3t + \frac{c}{t}[/itex]

    Is this correct? I think I solved it correctly, but I would still appreciate it if anyone could let me know if there are any mistakes. I'm a bit unsure because most of the DE solutions I've seen don't have a variable in the denominator like the 1/9t term.
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Mar 31, 2012 #2


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    hi roam! :wink:

    yes, your general solution is fine :smile:
    you always use radians

    always always always! :biggrin:

    (though you shouldn't need a calculator for cos3π and sin3π :wink:)
  4. Mar 31, 2012 #3

    Ray Vickson

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    Always use radians in calculus. The formulas become quite messy if you use degrees. For example, if we measure the angle x in degrees we have (d/dx) sin(x) = (π/180)*cos(x) and (d/dx) cos(x) = -(π/180)*sin(x).

    Last edited by a moderator: May 5, 2017
  5. Apr 3, 2012 #4
    Ah I see! Thanks a lot tiny tim and Ray for your responses, I get it now!
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