1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    [tex]y(\frac{\pi}{4})=0[/tex]

    2. Relevant equations

    [tex]\frac{dy}{dx}+Q(x)*y=F(x)[/tex]



    3. The attempt at a solution

    y=u*v
    [tex]u*\frac{dv}{dx}+v*(\frac{du}{dx}+\frac{u}{(1+x^2)})=\frac{\arctan x}{(1+x^2)}[/tex]
    [tex]\frac{du}{dx}+\frac{u}{(1+x^2)}=0[/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]t'=\frac{1}{(1+x^2)}[/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]
    [tex]\frac{1}{e^t}*\frac{dv}{dx}=\frac{t}{(1+x^2)}[/tex]
    [tex]v=\int\frac{t*e^t}{(1+x^2)}\,dx[/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

    [tex]\arctan(\frac{pi}{4})-\frac{1}{(1+\frac{pi^2}{16})}+\frac{C}{e^{\arctan(\frac{pi}{4})}}=0[/tex]

    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
    [tex]C=e^{\arctan(\frac{pi}{4})}*(\frac{1}{(\frac{pi^2}{16}+1)}-\arctan(\frac{pi}{4}))[/tex]
    I have to find rational answer...
     
    Last edited: Nov 8, 2012
  2. jcsd
  3. Nov 8, 2012 #2

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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:

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

    so the integrating factor is:

    [tex]\mu=e^{\arctan(x)}[/tex]

    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]

    or:

    [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:

    [tex]u=e^{\arctan(x)}[/tex]

    then:

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

    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

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Line integrate find C something wrong. dy/dx+Q(x)*y=F(x)
  1. Find dy/dx if f(x,y) = 0 (Replies: 11)

Loading...