Results 1 to 5 of 5

Thread: second order differential equation

  1. #1
    Newbie
    Joined
    Oct 2009
    Posts
    23

    second order differential equation

    solve y''-4y'+13y=0 if y(0)= -1 and y'(0)= 2.

    from my lecture notes im confused about this step
    general solution (alpha=2, beta=3)
    y=Ae^[(2+3i)x] + Be^[(2-3i)x]
    y=C(a small 1 at the corner)(e^2x)cos3x + C(a small 2 at the corner)(e^2x)sin3x

    how to get from the first step to the second step? someone please explain to me thanks =)
    Follow Math Help Forum on Facebook and Google+

  2. #2

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    Quote Originally Posted by yen yen View Post
    solve y''-4y'+13y=0 if y(0)= -1 and y'(0)= 2.

    from my lecture notes im confused about this step
    general solution (alpha=2, beta=3)
    y=Ae^[(2+3i)x] + Be^[(2-3i)x]
    y=C(a small 1 at the corner)(e^2x)cos3x + C(a small 2 at the corner)(e^2x)sin3x

    how to get from the first step to the second step? someone please explain to me thanks =)
    You start with:

    $\displaystyle y=Ae^{(2+3i)x} + Be^{(2-3i)x}$

    which may be written:

    $\displaystyle y=Ae^{2x}e^{3ix} + Be^{2x}e^{-3ix}$

    Now expand the two expontials with imaginary powers into $\displaystyle \sin $'s and $\displaystyle \cos $'s and rearrange and re-lable the constanants.

    (that is use: $\displaystyle e^{3ix}=\cos(3x)+i \sin(3x)$ and $\displaystyle e^{-3ix}=\cos(3x)-i \sin(3x)$ )

    CB
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Oct 2009
    Posts
    23
    Quote Originally Posted by CaptainBlack View Post

    Now expand the two expontials with imaginary powers into $\displaystyle \sin $'s and $\displaystyle \cos $'s and rearrange and re-lable the constanants.

    (that is use: $\displaystyle e^{3ix}=\cos(3x)+i \sin(3x)$ and $\displaystyle e^{-3ix}=\cos(3x)-i \sin(3x)$ )

    CB
    i get that but how do you get to
    y=C(a small 1 at the corner)(e^2x)cos3x + C(a small 2 at the corner)(e^2x)sin3x

    wouldnt u get
    y=A(e^2x)(cos3x + i sin3x) +B(e^2x)(cos3x - i sin3x)???
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    5
    Quote Originally Posted by yen yen View Post
    i get that but how do you get to
    y=C(a small 1 at the corner)(e^2x)cos3x + C(a small 2 at the corner)(e^2x)sin3x

    wouldnt u get
    y=A(e^2x)(cos3x + i sin3x) +B(e^2x)(cos3x - i sin3x)???
    Now collect the terms with cos and sign together to get:

    $\displaystyle
    y=(A+B)e^{2x}\cos(3x)+(A-B)e^{2x} i \sin(3x)
    $

    and put $\displaystyle C_1=(A+B)$ and $\displaystyle C_2=i(A-B)$

    CB
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: Sep 21st 2011, 03:37 PM
  2. First order differential equation
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: Nov 3rd 2010, 05:29 PM
  3. Second Order Differential Equation
    Posted in the Differential Equations Forum
    Replies: 3
    Last Post: Nov 2nd 2009, 02:08 PM
  4. First order differential equation
    Posted in the Differential Equations Forum
    Replies: 0
    Last Post: Feb 7th 2009, 08:06 PM
  5. Replies: 2
    Last Post: Nov 25th 2008, 09:29 PM

Search Tags


/mathhelpforum @mathhelpforum