Results 1 to 6 of 6

Math Help - Ordinary differential equations

  1. #1
    Newbie
    Joined
    Jan 2009
    Posts
    4

    Ordinary differential equations

    I want to solve the differential equation:

    dy/dx=sin(x) + y.tan(x)

    Using the integrating factor method I rearrange in the required form of :

    dy/dx + g(x)y = h(x) to give:

    dy/dx - tan(x).y = sin(x)

    The integrating factor p(x) = exp[Sg(x) dx] where in this case
    g(x) = -tan(x) (S represents integration !)

    But the integral of tan(x) depends on whether x is negative or positive, and since -pi/2 < x < pi/2 I'm left in a bit of a pickle as to how to proceed! If I use the two integrals for the positive and negative values and integrate between limits things get a bit messy, and I grind to a halt.

    If I use the modulus sign for tan (x) and push on then I obtain the integrating factor of p(x)= cos(x).

    On multiplying through the differential equation by p(x), the R.H.S. then becomes Cos(x)Sin(x). Following the proceedure for the I.F. method, I first write the l.h.s. as d/dx[p(x)y] and then integrate both sides. I now have the problem of trying to integrate Cos(x)Sin(x) and again grind to a halt.

    The final answer for the initial condition y(0)= 1/2 :
    y = sec x -1/2 cos x but it's the integrating factor I'm after to get there.

    The general solution before applying the initial conditions is:

    y = C sec x -1/2 Cos x

    I have tried to go in reverse from the G.S. but didn't get far!

    I apologise if it seems I'm coming on here only to take and not give to the forum, I 've had to re-register after last year, not that I was the most prolific/helpfull poster anyway.

    Any help would be much appreciated.

    Cheers.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hello,

    I now have the problem of trying to integrate Cos(x)Sin(x) and again grind to a halt.
    \int \cos(x) \sin(x) ~dx=\frac 12 \int \sin(2x) ~ dx



    I think that your problem with the arctan is not very important, because of the integration constant. I'm sorry I can only help for the point above, because I don't know the other methods
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2009
    Posts
    4
    Quote Originally Posted by Moo View Post
    Hello,


    \int \cos(x) \sin(x) ~dx=\frac 12 \int \sin(2x) ~ dx


    Thanks for the reply, but how have you gone about integrating Cos(x)Sin(x) ? Is it something you recognise? A standard integral? or (more likely??) a trig identity? Which part of maths do I have to go back and re-re-re-re-revise again the knowledge is falling out as fast as I can get it in!

    Cheers.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by Troutfisch View Post
    Thanks for the reply, but how have you gone about integrating Cos(x)Sin(x) ? Is it something you recognise? A standard integral? or (more likely??) a trig identity? Which part of maths do I have to go back and re-re-re-re-revise again the knowledge is falling out as fast as I can get it in!

    Cheers.
    It's the double angle formula :
    \sin(2x)=2\sin(x)\cos(x)

    Or you can just recall that cos is the derivative of sine, and foresee a substitution : t=\sin(x) (hence dt=cos(x) dx and the integral becomes \int t ~dt)


    One will give -\cos(2x)+c, the other one will give \frac{\sin^2(x)}{2}+c, which are in fact equivalent thanks to the identity \cos(2x)=1-2\sin^2(x)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Jan 2009
    Posts
    4
    Quote Originally Posted by Moo View Post
    It's the double angle formula :
    \sin(2x)=2\sin(x)\cos(x)
    Yo're almost a star bear with me while I make sure I've got this.

    Pushing on from my first post, I take the modulus of tan(x), so the integrating factor becomes Cos (x). Multiplying through the d.e. by the Cos(x) and following the necessary procedure I'm left with:

    Cos(x).y = S Cos(x)Sin(x) dx

    Using the D.A. formulae, then:

    Cos(x)y = 1/2 S sin(2x)dx (remember S is integrate)

    On integrating:

    Cos(x)y = -1/4 Cos(2x) +C (C=constant of integration).

    Rearrange for y:

    y = C.Sec(x) -1/4 Cos(2x)Sec(x)

    But this does not equate to the given general solution of:

    y= C.Sec(x) - 1/2 Cos(x) .......or are they equivalent?

    But!1! if I now apply the initial condition of y(0) = 1/2 I obtain the particular solution of:

    y=3/4 Sec(x) - 1/4 Cos(2x)Sec(x) compared to the given solution of :

    y = sec(x) - 1/2 Cos(x)

    Hmmm, maybe my G.S. isn't correct after all!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Jan 2009
    Posts
    4
    Quote Originally Posted by Moo View Post
    It's the double angle formula :
    \sin(2x)=2\sin(x)\cos(x)

    Or you can just recall that cos is the derivative of sine, and foresee a substitution : t=\sin(x) (hence dt=cos(x) dx and the integral becomes \int t ~dt)


    One will give -\cos(2x)+c, the other one will give \frac{\sin^2(x)}{2}+c, which are in fact equivalent thanks to the identity \cos(2x)=1-2\sin^2(x)
    It's taken me some time (a day!), but all the info I needed was there and I've crunched it out in the end.

    Many thanks.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Ordinary differential equations
    Posted in the Differential Equations Forum
    Replies: 7
    Last Post: April 30th 2011, 03:16 PM
  2. Ordinary Differential Equations
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: October 24th 2010, 07:15 AM
  3. Ordinary differential equations
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: October 8th 2010, 06:31 PM
  4. Ordinary differential equations....
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: July 22nd 2009, 10:13 AM
  5. Ordinary differential equations
    Posted in the Calculus Forum
    Replies: 6
    Last Post: September 24th 2006, 07:35 PM

Search Tags


/mathhelpforum @mathhelpforum