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Method of reduction

  1. Apr 21, 2010 #1
    find the general solution of (cos x)y''-y'+y = 0

    L[y1] = 0
    L[y2] = 0

    L[y1] x y2 : (cos x)y1'' y2 - y1'y2 + y1y2 = 0 ...(i)
    L[y2] x y1 : (cos x)y2'' y1 - y2'y1 + y1y2 = 0 ...(ii)

    (i) -(ii) : (cos x)(y2''y1 - y1''y2) - (y2'y1 - y1'y2) = 0

    W = | y1 y2 | = y1y2' - y2y1'
    | y1' y2'|

    W' = y1'y2'+y1y2''-y2'y1' - y2y1''
    = y1y2'' - y2y1''

    (cos x) W' - W = 0
    W' - (1/cos x) W = 0

    miu(x) exp(-integration of (1/cos x dx) = exp(- ln |cos x|)
    = 1/cos x

    integration of d (W.(1/cos x)) = 0 x integration of (1/cos x) dx
    W/cos x = c, c= constant
    W = c cos x

    Since W = y1y2' - y2y1' ..(*)

    let y1 = x^r y1' = r(x^(r-1))


    thus insert y1 and y2 in (*) : W = x^r(y2')-(rx^(r-1)(y2))
    = x^r (y2' - (1/x)y2)
    = x^r(y2' -(1/x)y2) = c cos x

    ===> y1' - (1/x) y2 = (c/(x^r) x cos x) ...(**)

    miu(x) = exp (- integration of (1/x) dx) = 1/x

    miu(x) x (**) : integration of (y x (1/x)) = integration of (r/(x^r+1) x cos x) dx


    i get stuck here... how can i integrate the above function since it involves 3 variables - x,y and r... huhuhu... please help me...

    is there any other way to solve this problem rather than reduction method??? anyone?:cry:
     
    Last edited: Apr 22, 2010
  2. jcsd
  3. Apr 21, 2010 #2

    LCKurtz

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    Dunno if I can help you or not because it isn't clear what you are asking. You titled the thread "method of reduction". I would normally take that to mean reducing the order of the DE by using a known solution. But that doesn't seem to be what you are doing since you didn't indicate you have one solution. Could you be a bit more clear about what you are trying to do?
     
  4. Apr 21, 2010 #3
    find the general solution of (cos x)y''-y'+y = 0

    this i so the question, so in my opinion, the method of reduction is the best method to solve the question. is there any other way that much more easier than this????
     
  5. Apr 22, 2010 #4
    another way of my attempt is

    [tex](cos x)y''-y'+y = 0[/tex]

    [tex]y''- \frac{ y'}{cos x}+ \frac{y}{cos x} = 0[/tex]

    The roots of the characteristic equation are the solutions to this problem.

    [tex]\lambda^2 - \frac{ \lambda }{cos x} + \frac{1}{cos x} = 0[/tex]

    If the roots don't present in nice form, like in the case of this equation, you can put them into the quadratic equation.

    [tex]\frac{ secx +/- \sqrt{ sec^2 x - 4secx } }{2}[/tex]

    Just taking the posative root for now

    [tex]\frac{ secx + \sqrt{ secx } \sqrt{ secx - 4 } }{2}[/tex]

    Well....I'm not sure how to simplify this but this is one of our solutions (e to the power of this). :confused:
     
    Last edited: Apr 22, 2010
  6. Apr 22, 2010 #5

    LCKurtz

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    Stop right there. This is not a constant coefficient differential equation, and you can't solve it with the characteristic equation method used for constant coefficient differential equations.

    Also, use [tex] tags, not [math] tags to display mathematics. You can alway preview your post to see if it displays correctly.

    As far as solving your DE goes, I don't know how to solve it. Maybe someone else will see a way.
     
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