Results 1 to 3 of 3

Math Help - method of reduction

  1. #1
    Junior Member
    Joined
    Jan 2010
    Posts
    49

    Exclamation method of reduction

    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member AllanCuz's Avatar
    Joined
    Apr 2010
    From
    Canada
    Posts
    384
    Thanks
    4
    Quote Originally Posted by bobey View Post
    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?
    I'm not entirely sure why you are introducing another variable. Reduction of order typically doesn't require this. However, there is a another way for homogenious DE equations.

    (cos x)y''-y'+y = 0

    y''- \frac{ y'}{cos x}+ \frac{y}{cos x} = 0

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

     \lambda^2 - \frac{ \lambda }{cos x} + \frac{1}{cos x} = 0

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

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

    Just taking the posative root for now

     \frac{ secx + \sqrt{ secx } \sqrt{ secx - 4 } }{2}

    Well....I'm not sure how to simplify this but this is one of our solutions (e to the power of this).

    Reduction of order might be best in this case. But to be honest, I'm not a fan of the way your notation it's kinda hard to read. Also, reduction of order problems typically already give you one solution so that you can find the other. Are you asked to find in general the reduction of order for this quesstion (i.e. both y1 and y2 without being given either of them)?

    I also don't recognize how you're doing this. You multiply by Y1 and Y2 and then subtract the equations? Generally, reduction of order is

    let

     y_2 = u(x) y_1

    Then we differentiate here and sub back into our equation. We can then find a linear representation for this equation and find a solution.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2010
    Posts
    49

    Exclamation

    Quote Originally Posted by AllanCuz View Post
    I'm not entirely sure why you are introducing another variable. Reduction of order typically doesn't require this. However, there is a another way for homogenious DE equations.

    (cos x)y''-y'+y = 0

    y''- \frac{ y'}{cos x}+ \frac{y}{cos x} = 0

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

     \lambda^2 - \frac{ \lambda }{cos x} + \frac{1}{cos x} = 0

    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.


    i hope someone can help me... plz3x
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Reduction method of integration
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 2nd 2010, 10:05 PM
  2. method of reduction of order!
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: March 11th 2010, 04:25 AM
  3. Reduction Method
    Posted in the Calculus Forum
    Replies: 4
    Last Post: February 10th 2010, 09:37 AM
  4. Integration - Reduction Method
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 9th 2010, 01:13 AM
  5. Reduction Method Question
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 22nd 2009, 06:35 AM

Search Tags


/mathhelpforum @mathhelpforum