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Homework Help: Linear First order ODE problem

  1. Oct 26, 2008 #1
    1. The problem statement, all variables and given/known data

    x[tex](\theta)[/tex]= x tanθ−cosθ


    2. Relevant equations

    The integrating factor is (note it should be negative tan, but the latex won't display it);

    [tex]I= \int -tan \theta\ d \theta[/tex]

    The solution is given by;

    [tex]x \ e^{I}=-\int \ e^{I}*cos\theta\ d \theta[/tex]

    3. The attempt at a solution

    [tex]I=\int \tan \theta\ d \theta[/tex]
    [tex]=-(-\ln|cos \theta |)[/tex]

    So, [tex]e^{I}=cos \theta[/tex]

    Therefore,

    [tex]x*cos \theta = \int \ cos^{2} \theta d \theta[/tex]

    Using the half angle formulae gives,

    [tex]x*cos \theta =\frac{1}{2} \int \cos(2 \theta) +1[/tex]

    Which, after integration and simplification gives,

    [tex]x=-\frac{1}{2}*(sin \theta + \theta\ sec\theta)[/tex]

    When I check my answer I end up with zero,

    [tex]\frac{dx}{d \theta}=-\frac{1}{2}*(cos \theta \ +\ sec\theta \ +\ \theta*sec \theta \tan \theta)[/tex]

    [tex]x*tan \theta=-\frac{1}{2}*(sin \theta * tan \theta + \theta* sec \theta* tan \theta)[/tex]

    [tex]\frac{dx}{d \theta}-x*tan \theta=\frac{1}{2}*[(\frac{cos^{2}\theta\ {-1}}{cos \theta}-\theta* sec \theta* tan \theta)-(-\frac{sin^{2} \theta}{cos \theta}-\theta* sec \theta* tan \theta)][/tex]

    [tex]=\frac{sin^{2} \theta \ - \ (1-cos^{2} \theta)}{2* cos \theta}=0[/tex]

    Please could someone go through my calculation and see where my mistake is?
     
    Last edited: Oct 26, 2008
  2. jcsd
  3. Oct 26, 2008 #2

    HallsofIvy

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    Science Advisor

    I have corrected that: you had a "\" before the "-".

    Haven't you forgotten the "-" on the right side?

    After integration, this is x cos(theta)= 1/2(theta)- (1/4) sin(2\theta)

    You appear to have forgotten the "-" on the "- cos(theta)" to begin, with as well as the "2" in the 2 theta, and, finally, the theta from integrating "1".

     
  4. Oct 26, 2008 #3
    Hi HallsofIvy,

    I've always struggled with minus signs, I find them really confusing! :redface:

    When I simplified I used [tex]\frac{1}{2}sin 2\theta=sin\theta*cos\theta[/tex].

    Popping in that minus and the missing theta gave me the correct solution, I think.

    [tex]x=-\frac{1}{2}sin\theta + \theta*cos\theta[/tex]

    To check;

    [tex]\frac{dx}{d\theta}=-\frac{1}{2}*(cos\theta + sec\theta + \theta*sec\theta*tan\theta)[/tex]

    [tex]x*tan\theta=-\frac{1}{2}*(\frac{sin^{2}\theta}{cos\theta} + \frac{\theta*sin\theta}{cos^{2}\theta})[/tex]

    [tex]\frac{dx}{d\theta}-x*tan\theta=-\frac{1}{2}*(cos\theta + sec\theta + \theta*sec\theta*tan\theta - \frac{sin^{2}\theta}{cos\theta} -\frac{\theta*sin\theta}{cos^{2}\theta})[/tex]

    [tex]=-\frac{1}{2}*[\frac{cos^{2}\theta + 1 -sin^{2}\theta}{cos\theta}][/tex]

    [tex]=-\frac{1}{2}*[\frac{2*cos^{2}\theta -1 + 1}{cos\theta}]=-cos\theta[/tex]

    Does this look reasonable to you?

    Thanks for your help, I was going round in circles with that problem!
     
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