Results 1 to 4 of 4

Math Help - Homog Lin... #3

  1. #1
    Member
    Joined
    Sep 2006
    Posts
    221

    Homog Lin... #3

    3.) Solve the following Dif EQ, given the indicated initial conditions:

    \frac{d^4y}{dt^4} - y = 0, y(0) = y'(0) = y''(0) = 0, y'''(0) = 1

    Work:

    So m^4 - 3m^3 + 3m^2 - m = 0

    m*(m-1)*(m^2 + m + 1)

    Thus, m = 0 or 1.

    So, the general sol'n would be:

    y = c_1 + c_2e^{t}

    So, with the conditions:

    0 = c_1 + c_2e^(0)

    So c_1 = -c_2

    Then, would I find the 3rd derivative? Not really sure.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Sep 2006
    Posts
    221
    Quote Originally Posted by Ideasman View Post
    3.) Solve the following Dif EQ, given the indicated initial conditions:

    \frac{d^4y}{dt^4} - y = 0, y(0) = y'(0) = y''(0) = 0, y'''(0) = 1

    Work:

    So m^4 - 3m^3 + 3m^2 - m = 0

    m*(m-1)*(m^2 + m + 1)

    Thus, m = 0 or 1.

    So, the general sol'n would be:

    y = c_1 + c_2e^{t}

    So, with the conditions:

    0 = c_1 + c_2e^(0)

    So c_1 = -c_2

    Then, would I find the 3rd derivative? Not really sure.
    Hmm, I am definitely going without much sleep. Sorry, it should be m^4 - m = 0. But, it turns out it's the same exact answer. That is, factor:

    m*(m-1)*(m^2 + m + 1)...

    etc etc, just not sure how to proceed after.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Sep 2006
    Posts
    221
    3.) Solve the following Dif EQ, given the indicated initial conditions:

    \frac{d^4y}{dt^4} - y = 0, y(0) = y'(0) = y''(0) = 0, y'''(0) = 1

    Okay, my second attempt.. I def. messed up the first time.

    The auxiliary equation is m^4 - 1 = 0.

    We get: (m - 1)*(m + 1)*(m^2 + 1) = 0

    Thus, m = 1, -1 , i, -i

    So, y = c_1e^t + c_2e^(-t) + c_3e^t(sin(x) + c_4e^tcos(x)

    Is this right? I think it is. So, if this is right, then I can find [tex]y', y'', y''' and solve the equations. However, wouldn't it be hard trying to solve the equations for c_1, c_2, c_3, c_4... it just feels as there is too much unknown..
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,663
    Thanks
    298
    Awards
    1
    Quote Originally Posted by Ideasman View Post
    3.) Solve the following Dif EQ, given the indicated initial conditions:

    \frac{d^4y}{dt^4} - y = 0, y(0) = y'(0) = y''(0) = 0, y'''(0) = 1

    Okay, my second attempt.. I def. messed up the first time.

    The auxiliary equation is m^4 - 1 = 0.

    We get: (m - 1)*(m + 1)*(m^2 + 1) = 0

    Thus, m = 1, -1 , i, -i

    So, y = c_1e^t + c_2e^(-t) + c_3e^t(sin(x) + c_4e^tcos(x)

    Is this right? I think it is. So, if this is right, then I can find [tex]y', y'', y''' and solve the equations. However, wouldn't it be hard trying to solve the equations for c_1, c_2, c_3, c_4... it just feels as there is too much unknown..
    Aside from a wild set of parenthesis and a LaTeX coding error, this is a valid way to express the solution. You will have to solve some simultaneous equations to get the values of the constants, so you still have some work ahead of you.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. 2nd Order, Homog., Reduction of Order
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: November 27th 2011, 06:36 AM
  2. Using Power Series to solve Non-Homog. DE
    Posted in the Differential Equations Forum
    Replies: 1
    Last Post: March 24th 2010, 03:16 AM
  3. Homog Lin... #2
    Posted in the Calculus Forum
    Replies: 2
    Last Post: October 4th 2007, 02:42 PM
  4. Homog. Lin. Eq's with Constant Coeff.
    Posted in the Calculus Forum
    Replies: 3
    Last Post: October 4th 2007, 11:51 AM
  5. Homog diff eqn
    Posted in the Calculus Forum
    Replies: 0
    Last Post: December 3rd 2006, 05:50 PM

Search Tags


/mathhelpforum @mathhelpforum