Results 1 to 4 of 4

Math Help - Almost there...

  1. #1
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823

    Almost there...

    Hey, I got most of the way through this reduction of order problem, but then my brain gave out right at the end.

    Given the equation y''-4y'+4y=0 and the solution y_1=e^{2x}, write the general solution.

    I make the substitution y_2(x)=u(x)y_1(x) and do all the proper work and get here:

    e^{2x}u''=0 and my brain gives out. I know this is easy, How do I find u?
    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 VonNemo19 View Post
    Hey, I got most of the way through this reduction of order problem, but then my brain gave out right at the end.

    Given the equation y''-4y'+4y=0 and the solution y_1=e^{2x}, write the general solution.

    I make the substitution y_2(x)=u(x)y_1(x) and do all the proper work and get here:

    e^{2x}u''=0 and my brain gives out. I know this is easy, How do I find u?
    Well since e^{2x} \ne 0,\forall x divide both sides by it to get

    u''=0 \iff \frac{d}{dx}u'=0 \iff u'=c \iff u=cx+d now choose c=1,d=0 to get u=x
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,404
    Thanks
    1293
    Of course, if you solve the Characteristic equation and realise you get a repeated root \displaystyle m = 2, then you immediately know that the solution of the DE is \displaystyle y = C_1e^{2x} + C_2x\,e^{2x}...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by Prove It View Post
    Of course, if you solve the Characteristic equation and realise you get a repeated root \displaystyle m = 2, then you immediately know that the solution of the DE is \displaystyle y = C_1e^{2x} + C_2x\,e^{2x}...
    Right but I think the point of the questions is to see how to get a 2nd linerly inpendant vector when you have a repeated root to the charateristic equation. Without this derivation how would we know what the 2nd vector is? Plus a great gereralization of this is variation of parameters!
    Follow Math Help Forum on Facebook and Google+


/mathhelpforum @mathhelpforum