Results 1 to 5 of 5

Math Help - Eliminating the arbitrary constant

  1. #1
    Senior Member
    Joined
    Jul 2006
    From
    Shabu City
    Posts
    381

    Eliminating the arbitrary constant

    im completely lost with this differential equation problem in w/c you eliminate its arbitrary constants
    y = c_{1}x^2 + c_{2}e^{2x}


    derivating
    y' = 2c_{1}x + 2c_{2}e^{2x}

    y'' = 2c_{1} +4c_{2}e^{2x}

    i just do not know what to equate can you please help me w/ this

    answer:
    x(1-x)y'' + (2x^2 - 1)y' - 2(2x - 1)y = 0
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor kalagota's Avatar
    Joined
    Oct 2007
    From
    Taguig City, Philippines
    Posts
    1,026
    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    im completely lost with this differential equation problem in w/c you eliminate its arbitrary constants
    y = c_{1}x^2 + c_{2}e^{2x}


    derivating
    y' = 2c_{1}x + 2c_{2}e^{2x}

    y'' = 2c_{1} +4c_{2}e^{2x}

    i just do not know what to equate can you please help me w/ this

    answer:
    x(1-x)y'' + (2x^2 - 1)y' - 2(2x - 1)y = 0
    kuya, what is the equation?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jul 2006
    From
    Shabu City
    Posts
    381
    Quote Originally Posted by kalagota View Post
    kuya, what is the equation?
    y = c_{1}x^2 + c_{2}e^{2x}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,962
    Thanks
    349
    Awards
    1
    Quote Originally Posted by ^_^Engineer_Adam^_^ View Post
    im completely lost with this differential equation problem in w/c you eliminate its arbitrary constants
    y = c_{1}x^2 + c_{2}e^{2x}


    derivating
    y' = 2c_{1}x + 2c_{2}e^{2x}

    y'' = 2c_{1} +4c_{2}e^{2x}

    i just do not know what to equate can you please help me w/ this

    answer:
    x(1-x)y'' + (2x^2 - 1)y' - 2(2x - 1)y = 0
    You have your two derivatives.

    Solve the first derivative for c_1:
    c_{1} = \frac{y' - 2c_{2}e^{2x}}{2x}

    Plug this value for c_1 into your solution and second derivative:
    y = \left ( \frac{y' - 2c_{2}e^{2x}}{2x} \right ) x^2 + c_{2}e^{2x}

    y'' = 2 \left ( \frac{y' - 2c_{2}e^{2x}}{2x} \right ) + 4c_{2}e^{2x}

    Now solve your solution for c_2 and plug that into your second derivative. Now do a lot of algebra.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Jul 2006
    From
    Shabu City
    Posts
    381
    thank you
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. No solution for an arbitrary constant.
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: April 1st 2011, 04:41 AM
  2. Eliminating the arbitrary constant
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: February 11th 2011, 09:21 AM
  3. Question involving arbitrary constant
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: July 10th 2009, 12:28 PM
  4. Replies: 0
    Last Post: March 5th 2009, 06:31 PM
  5. Eliminating the arbitrary constants
    Posted in the Calculus Forum
    Replies: 0
    Last Post: February 4th 2009, 06:17 PM

Search Tags


/mathhelpforum @mathhelpforum