Results 1 to 4 of 4

Thread: plz Help me

  1. #1
    Newbie
    Joined
    Oct 2009
    Posts
    8

    plz Help me

    what is y?
    $\displaystyle y^(5)+2y=0$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Dec 2008
    From
    Scotland
    Posts
    901
    Quote Originally Posted by iloveyou7 View Post
    what is y?
    $\displaystyle y^(5)+2y=0$

    I'm not sure I understand. Is this the equation:

    $\displaystyle y^5 + 2y = 0 $?

    You've posted this in the differential equation section, and this isn't a differential equation. Did you mean:

    $\displaystyle \frac{d^5y}{dx^5} + 2y = 0 $?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2009
    Posts
    8
    Quote Originally Posted by Mush View Post
    I'm not sure I understand. Is this the equation:

    $\displaystyle y^5 + 2y = 0 $?

    You've posted this in the differential equation section, and this isn't a differential equation. Did you mean:

    $\displaystyle \frac{d^5y}{dx^5} + 2y = 0 $?
    yes, i mean
    $\displaystyle \frac{d^5y}{dx^5} + 2y = 0 $
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    Quote Originally Posted by iloveyou7 View Post
    what is y?
    $\displaystyle y^{(5)}+2y=0$
    Use a trial solution $\displaystyle e^{\lambda t}$ , this gives you a charateristic equation:

    $\displaystyle \lambda^5+2=0$

    so $\displaystyle \lambda=(-2)^{1/5}$ that is the general solution is:

    $\displaystyle y(t)=Ae^{\lambda_1t}+Be^{\lambda_2t}+Ce^{\lambda_3 t}+De^{\lambda_4t}+Ee^{\lambda_5t}$

    where the $\displaystyle \lambda_i $'s are the fith roots of $\displaystyle -2$

    CB
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum