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Homework Help: Line integrate find C something wrong. dy/dx+Q(x)*y=F(x)

  1. Nov 8, 2012 #1
    1. The problem statement, all variables and given/known data

    [tex]\frac{dy}{dx}+\frac{y}{(1+x^2)} = \frac{\arctan x}{(1+x^2)}[/tex] when

    2. Relevant equations


    3. The attempt at a solution

    [tex]u*\frac{dv}{dx}+v*(\frac{du}{dx}+\frac{u}{(1+x^2)})=\frac{\arctan x}{(1+x^2)}[/tex]
    [tex]\int\frac{1}{u}\,du=-\int\frac{1}{(1+x^2)}\,dx [/tex]
    [tex]\ln u=-\arctan x [/tex]
    [tex]t=\arctan x[/tex]
    [tex] u=\frac{1}{e^t} [/tex]
    So let's go back to
    [tex]u*\frac{dv}{dx}+v*(\frac{du}{dx}+\frac{u}{(1+x^2)})=\frac{\arctan x}{(1+x^2)}[/tex]
    Integrating by parts...
    [tex]v=t * e^t-e^t*dt+C[/tex]
    Going back to y=u*v
    [tex]y=t-t'+\frac{C}{e^t}[/tex] - global answer.

    The problem starts now... I need to find C when
    [tex]y(\frac{\pi}{4})=0[/tex] so i get


    And now i am clueless how to find C. Becouse I cannot find arctan pi/4 and there is a problem with pi^2/16+1
    I assume that C=0 becouse graphic of arctan x and 1/(x^2+1) are very close in point 0.79 (pi/4), but i cannot prove it
    I have to find rational answer...
    Last edited: Nov 8, 2012
  2. jcsd
  3. Nov 8, 2012 #2


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    That should be
    $$dt = \frac{1}{(1+x^2)}\,dx$$
    That's wrong. You can't have a lone dt sitting around.
  4. Nov 8, 2012 #3
    [tex] (u*v)'=u'*v+u*v'[/tex]
    I meant it like dt=t'
    I should write ' instead of d, but i am right here i think.
    If u=f(x) and v=e^q(x)
    Then (u*v)'=f'(x)*e^q(x)+f(x)*q'(x)*e^q(x)
    It doesnt solve my problem with Constant
  5. Nov 8, 2012 #4


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    No, that's not right.
  6. Nov 8, 2012 #5
    Anyway if you are right i still dont know what to do with
    arctan pi/4-1+C/e^arctan pi/4=0
    C=e^arctan x - arctan x * e^arctan x
    I have to transform it somehow to get rational answer
  7. Nov 8, 2012 #6
    You're all over the place with that. Try and treat it more carefully. First write it as:


    so the integrating factor is:


    apply that to both sides and integrate to get:

    [tex]\int d\left(y e^{\arctan(x)}\right)=\int e^{\arctan(x)}\frac{\arctan(x)}{1+x^2}dx[/tex]


    [tex]ye^{\arctan(x)}=\int e^{\arctan(x)}\frac{\arctan(x)}{1+x^2}dx[/tex]

    so now you just have to integrate

    [tex]\int e^{\arctan(x)}\frac{\arctan(x)}{1+x^2}dx[/tex]

    You can do that right?
  8. Nov 8, 2012 #7
    well i did that... if you look carefully you can notice that i treat t=arctan(x)
    so basicaly [tex]\int e^t\frac{t}{1+x^2}dx[/tex] equal to [tex]\int e^{\arctan(x)}\frac{\arctan(x)}{1+x^2}dx[/tex]
    and my answer is [tex]v=\arctan(x)*e^{\arctan(x)}-\frac{e^{\arctan(x)}}{x^2+1}+C[/tex] and then i turn back to y=u*v and get[tex]y=\arctan(x)-\frac{1}{x^2+1}+\frac{C}{e^{\arctan(x)}}[/tex] i need to aply that [tex]y(\frac{pi}{4})=0[/tex] and find C, but i am stack.
  9. Nov 8, 2012 #8
    You shouldn't have t's and x's in the same integrand. Start with:

    [tex]\int e^{\arctan(x)} \frac{\arctan(x)}{1+x^2}dx[/tex]

    and let:




    now substitute all that in the integral to get:

    [tex]\int \ln(u)du[/tex]

    see, all u's now. Now integrate that and don't forget the constant of integration.
  10. Nov 8, 2012 #9


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    Which isn't correct. If you differentiate your answer, you do not get the integrand back. From here, it appears your mistake is resulting from your sloppy notation.
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