Results 1 to 6 of 6

Math Help - Euler Differential Equation help.

  1. #1
    Newbie
    Joined
    May 2008
    Posts
    19

    Euler Differential Equation help.

    Hey everyone!

    I have a question:

    2x^2\frac{d^2y}{dx^2} + 3x\frac{dy}{dx} - 15y = 0

    So I let:
    y = x^k
    y' = kx^{k-1}
    y'' = k(k-1)x^{k-2}

    Therefore,

    2x^2k(k-1)x^{k-2} + 3xkx^{k-1} - 15x^k = 0

    So would I be right in saying:

    x^k(2k(k-1) + 3k - 15) = 0

    x^k \ne 0

    2k(k-1) + 3k - 15 = 0

    Therefore, k = 2.5, k = -3

    y = \frac{A}{x^3} + Bx^{\tfrac{5}{2}}

    Would this solution be right?

    Thanks in advance.
    Last edited by Chris L T521; May 10th 2009 at 08:38 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by falconed View Post
    Hey everyone!

    I have a question:

    2x^2\frac{d^2y}{dx^2} + 3x\frac{dy}{dx} - 15y = 0

    So I let:
    y = x^k
    y' = kx^{k-1}
    y'' = k^2x^{k-2}

    Therefore,

    2x^2k^2x^{k-2} + 3xkx^{k-1} - 15x^k = 0

    So would I be right in saying:

    x^k(2k(k-1) + 3k - 15) = 0

    x^k \ne 0

    2k(k-1) + 3k - 15 = 0

    Therefore, k = 2.5, k = -3

    y = \frac{A}{x^3} + Bx^{\tfrac{5}{2}}

    Would this solution be right?

    Thanks in advance.
    I think your signs are backwards

    2k^2-k-15=(2k+5)(k-3)

    So k=3 and k=-\frac{5}{2}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by falconed View Post
    Hey everyone!

    I have a question:

    2x^2\frac{d^2y}{dx^2} + 3x\frac{dy}{dx} - 15y = 0

    So I let:
    y = x^k
    y' = kx^{k-1}
    y'' = k(k-1)x^{k-2}

    Therefore,

    2x^2k(k-1)x^{k-2} + 3xkx^{k-1} - 15x^k = 0

    So would I be right in saying:

    x^k(2k(k-1) + 3k - 15) = 0

    x^k \ne 0

    2k(k-1) + 3k - 15 = 0

    Therefore, k = 2.5, k = -3

    y = \frac{A}{x^3} + Bx^{\tfrac{5}{2}}

    Would this solution be right?

    Thanks in advance.
    I fixed a mistake in your second derivative.

    Yes, that solution you have would be correct.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Chicago, IL
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by TheEmptySet View Post
    I think your signs are backwards

    2k^2-k-15=(2k+5)(k-3)

    So k=3 and k=-\frac{5}{2}
    It should be 2k^2\,{\color{red}+}\,k-15=(2k-5)(k+3)...right?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by Chris L T521 View Post
    It should be 2k^2\,{\color{red}+}\,k-15=(2k-5)(k+3)...right?
    Yes your are correct
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    May 2008
    Posts
    19
    Thank you guys! Much appreciated!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Euler Differential Equation
    Posted in the Differential Equations Forum
    Replies: 5
    Last Post: January 22nd 2009, 09:16 AM
  2. Bernoulli and Euler Differential Equations
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: January 14th 2009, 08:08 AM
  3. Replies: 0
    Last Post: October 27th 2008, 06:34 PM
  4. Numerical solution of differential equations (Euler method)
    Posted in the Differential Equations Forum
    Replies: 2
    Last Post: October 27th 2008, 03:12 PM
  5. Replies: 1
    Last Post: May 31st 2008, 09:03 AM

Search Tags


/mathhelpforum @mathhelpforum