Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Second order linear DE

  1. Jan 21, 2009 #1

    s7b

    User Avatar

    Hi,

    I'm having problems solving this equation;

    y'' -2y' +3y =0 y(0)=-1 , y'(0)=(root 2) -1

    I found the auxiliary equation r^2 -2r +3 = 0
    and since b^2 -4ac is less than zero this the case where r1 and r2 are complex numbers.

    This is as far as I get without getting stuck.

    Please help me out if you know this.
    Thanks :)
     
  2. jcsd
  3. Jan 21, 2009 #2

    rock.freak667

    User Avatar
    Homework Helper

    For ay''+by'+cy=0

    the auxiliary equation is ar2+br+c=0. If b2-4ac <0, such that the roots are in the form [itex]r=M \pm Ni[/itex]

    then y=eMx(Acos(Nx)+Bsin(Nx))
     
  4. Jan 21, 2009 #3

    s7b

    User Avatar

    To find R1 and R2 do you just use the quadratic formula?

    So r1= 1+i
    and r2=1-i
     
  5. Jan 21, 2009 #4

    rock.freak667

    User Avatar
    Homework Helper

    yep..so [itex]r= 1 \pm i[/itex] i.e. M=1 and N=1
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Second order linear DE
  1. Second order DE (Replies: 6)

  2. First order Linear DE (Replies: 2)

  3. 2nd order Linear DE (Replies: 1)

  4. DE of second order (Replies: 5)

Loading...