Results 1 to 3 of 3

Math Help - Variation of Parameters

  1. #1
    Junior Member
    Joined
    Aug 2009
    Posts
    38

    Variation of Parameters

    Solve using Variation of Parameters

    1. y" + y = tan(x)
    auxiliary equation
    m^2 + 1 = 0
    Yc = c*cos(x) + c*sin(x)

    Y1 = cos(x)
    Y2 = sin(x)
    Wronskian = 1
    f(x) = tan(x)

    u1' = -sin(x)tan(x)
    u1 = ? Not sure how to integrate this...

    u2' = cos(x)tan(x) = sin(x)
    u2 =  -cos(x)

    I'd appreciate it if someone could check over what I've done so far and then tell me how to integrate -sin(x)tan(x).


    2.
    y" - 16y = 2e^4^x
    auxiliary equation
    m^2 - 16 = 0
    Yc = ce^-^4^x + ce^4^x

    Y1 = e^-4^x
    Y2 = e^4^x
    Wronskian = 8
    f(x) = 2e^4^x

    u1' = -(1/4)e^8^x
    u1 = -(1/32)e^8^x

    u2' = 1/4
    u2 = (1/4)x

    Yp = (1/4)xe^4^x - (1/32)e^4^x
    so Y = ce^-^4^x + ce^4^x + (1/4)xe^4^x - (1/32)e^4^x

    Not sure if I did something wrong because when I used undetermined coefficients to solve the problem, I only got Yp = (1/4)x*e^4^x...?

    Thanks in advance.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by Naples View Post
    Solve using Variation of Parameters

    1. y" + y = tan(x)
    auxiliary equation
    m^2 + 1 = 0
    Yc = c*cos(x) + c*sin(x)

    Y1 = cos(x)
    Y2 = sin(x)
    Wronskian = 1
    f(x) = tan(x)

    u1' = -sin(x)tan(x)
    u1 = ? Not sure how to integrate this...

    u2' = cos(x)tan(x) = sin(x)
    u2 =  -cos(x)

    I'd appreciate it if someone could check over what I've done so far and then tell me how to integrate -sin(x)tan(x).
    Spoiler:
    -\sin x\tan x = -\dfrac{\sin^2x}{\cos x} = \cos x-\sec x



    2.
    y" - 16y = 2e^4^x
    auxiliary equation
    m^2 - 16 = 0
    Yc = ce^-^4^x + ce^4^x

    Y1 = e^-4^x
    Y2 = e^4^x
    Wronskian = 8
    f(x) = 2e^4^x

    u1' = -(1/4)e^8^x
    u1 = -(1/32)e^8^x

    u2' = 1/4
    u2 = (1/4)x

    Yp = (1/4)xe^4^x - (1/32)e^4^x
    so Y = ce^-^4^x + ce^4^x + (1/4)xe^4^x - (1/32)e^4^x

    Not sure if I did something wrong because when I used undetermined coefficients to solve the problem, I only got Yp = (1/4)x*e^4^x...?

    Thanks in advance.
    It looks good. Note that we can say that ce^{4x}-\frac{1}{32}e^{4x} \sim ke^{4x};\,\,k=c-\frac{1}{32}, so it doesn't really add "new" information.

    I hope this makes sense.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,886
    Thanks
    325
    Awards
    1
    Quote Originally Posted by Naples View Post
    1. y" + y = tan(x)
    auxiliary equation
    m^2 + 1 = 0
    Yc = c*cos(x) + c*sin(x)
    Just a quick little comment that has nothing to do with variation of parameters. The homogeneous solution to this equation is
    Yc = c_1*cos(x) + c_2*sin(x)

    ie the arbitrary constants are not the same. You did this in your second example as well.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Variation of Parameters
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: January 14th 2012, 12:01 PM
  2. Variation of Parameters
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: February 11th 2010, 04:16 PM
  3. variation of parameters
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: November 24th 2009, 01:27 PM
  4. Variation of Parameters
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 4th 2007, 08:46 PM
  5. Variation of Parameters
    Posted in the Calculus Forum
    Replies: 1
    Last Post: July 4th 2007, 08:44 PM

Search Tags


/mathhelpforum @mathhelpforum