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Euler Differential Equation

  1. May 21, 2012 #1
    Problem: Solve the initial Value:
    when x=1, y=0
    dy/dx = 1

    2x^2(d^2y/dx^2) + 3x (dy/dx) - 15y = 0

    My attempt:

    x = e^t
    dx/dt = e ^t

    dy/dt = dy/dx * dx/dt
    dy/dt = x*dy/dx

    d^2y/dt^2
    = d/dt(dy/dt)
    = d/dt(x*dy/dx)
    =d/dx(x*dy/dx)*(dx/dt)
    since dx/dt = x
    =(x^2*d^2y/dx^2) + (x*dy/dx)
    =(x^2*d^2y/dx^2) + (dy/dt)
    (x^2*d^2y/dx^2) = (d^2y/dt^2) - (dy/dt)

    From this point, do I simply substitute the Values for the 1st/2nd derivatives in ? i.e.

    2((d^2y/dt^2) - (dy/dt)) + 3(dy/dt) - 15y = 0
    2d^2y/dt^2 - 2dy/dy +3dy/dt - 15y = 0

    2m^2 +3m - 1 = 0

    Then solve for m,

    Then create the solution in the form of :

    Ae^mt + Be^mt = 0 ??

    Is that the right path? Any help is appreciated.

    Cheers.
     
    Last edited: May 21, 2012
  2. jcsd
  3. May 22, 2012 #2

    tiny-tim

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    Hi James! :smile:

    (try using the X2 button just above the Reply box :wink:)
    (you mean = y, not 0 !)

    Yes, that's the method! :smile:

    (i haven't checked your result)
     
  4. May 22, 2012 #3
    Excellent. Sorry for the lack of real looking equations, I am on my phone.
    2m^2 + 3m - 1 = 0
    m = [-3+/- sqrt( 9 + 8 )]/4
    = 1.1231 / 4
    =0.2808
    m = -7.1231/4
    = -1.7808

    Should y be assigned a coefficient?

    Ae^mx + Be^mx = Cy
    Ae^0.2808x + Be^-1.7808x = Cy
    Sub in IC 1
    x = 1
    y = 0

    Ae^0.2808 + Be^-1.7808 = 0

    And IC 2
    dy/dx = 1

    Ae^0.2808/0.2808 -
    Be^-1.7808/1.7808 = 1

    Is that going in the right direction?
     
  5. May 23, 2012 #4
    Excellent! You are thinking like mathematican! Keeping going! gOOD JOB!
     
  6. May 23, 2012 #5

    tiny-tim

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    Hi James! :smile:

    (just got up :zzz:)

    no, it's unnecessary, it makes no difference (and it'll probably lose you a mark in the exam) :redface:

    then fine :smile:, until …
    those factors should be on top :wink:

    hmm … usually exam questions like this factor out nicely :confused:

    let's go back and check …
    ahhh! :biggrin:

    how did you get that? :rolleyes:
     
  7. May 23, 2012 #6
    Cool,

    So the assumption of my quadratic formula should be:

    2m2 + 3m - 15 = 0

    m = (-3±√129)/4
    m = 2.089
    m = -3.589

    So that general assumption is:
    Aemx + Bemx = y
    Ae2.089x + Be-3.589x = y

    IC 1
    y = Ae2.089 + Be-3.589 = 0

    IC2
    dy/dx = 2.089Ae2.089 - 3.589Be-3.589 = 1

    How do I go about finding the A and B co efficients??

    PS

    Thanks for the responses :D
     
  8. May 23, 2012 #7

    tiny-tim

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    is your screen too small? :confused:

    2m2 + m - 15 = 0 :smile:
     
  9. May 23, 2012 #8
    Thanks very much. Officially on a computer now. My screen was far to small, and i was writing out by hand first to see what they'd really look like. :)
    Thanks for being patient.

    2[itex]\frac{d^2y}{dt^2}[/itex] - 2[itex]\frac{dy}{dt}[/itex] +3[itex]\frac{dy}{dt}[/itex] - 15y = 0

    2[itex]\frac{d^2y}{dt^2}[/itex] +[itex]\frac{dy}{dt}[/itex] - 15y = 0


    2m2+ m - 15 = 0

    m = [itex]\frac{-1±√121}{4}[/itex]
    m = 5/2
    m = -3

    y = Aemx + Bemx
    y = Ae[itex]\frac{5x}{2}[/itex] + Be-3x

    IC1 : y(1) = 0
    y = Ae[itex]\frac{5}{2}[/itex] + Be-3 = 0

    IC2 : [itex]\frac{dy}{dx}[/itex] = 1

    [itex]\frac{dy}{dx}[/itex] = [itex]\frac{5}{2}[/itex] Ae[itex]\frac{5}{2}[/itex] - 3Be-3 = 1

    [itex]\frac{d^2y}{dx^2}[/itex] = ?

    Not entirely sure where to go now with substitution..

    Gah, i am getting stuck too often. I am so tired :( heading off to bed, its 4am here. Really appreciate your help tim :D
     
  10. May 23, 2012 #9

    tiny-tim

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    uh-oh, you've lost the plot :yuck: …

    the plot was, put x = et and solve for y against t ! :rolleyes:

    fine until then! :smile:
     
  11. May 24, 2012 #10
  12. May 24, 2012 #11

    tiny-tim

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    y = Ae5t/2 + Be-3t :smile:

    (now convert to x, then solve for the initial conditions)
     
  13. May 24, 2012 #12

    HallsofIvy

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    Yes, that's good.

    But where did you get this? The characteristic equation for 2y''+ y'- 15= 0 is
    [itex]2m^2+ m- 15= (2m- 5)(m+ 3)= 0[/itex] with roots m= 5/2 and m= -3.

    Right path- wrong solution to the equation.

    By the way, while I am a strong advocate of changing the variable to convert the equation to one with constant coefficients, a "short cut" is to "try" a solution to the original equation of the form [itex]x^m[/itex]. Then [itex]y'= mx^{m-1}[/itex], [itex]y''= m(m-1)x^{m-2}[/itex] and your differential equation becomes
    [tex]2x^2(m(m-1)x^{m-2})+ 3x(mx^{m-1})- 15x^m= (2m(m-1)+ 3m- 15)x^m= 0[/itex]
    In order that that be 0 for all x, we must have [itex]2m(m-1)+ 3m- 15= 2m^2+ m- 15= 0[/itex], exactly the same characteristic equation as before. Since that has roots -3 and 5/2, the general solution is
    [tex]y= Ax^{-3}+ Bx^{5/2}[/tex]
     
  14. May 25, 2012 #13
    Ah. Terriffic.

    y = ax-3 + bx[itex]\frac{5}{2}[/itex]

    Apply IC1: y(1) = 0

    y(1) = a(1)-3 + b(1)[itex]\frac{5}{2}[/itex] = 0

    y(1) = a + b

    => a = -b

    Then
    [itex]\frac{dy}{dx}[/itex] = -3ax-3 + [itex]\frac{5}{2}[/itex]bx[itex]\frac{5}{2}[/itex]

    IC2:
    y'(1) = 1
    1 = -3a + [itex]\frac{5}{2}[/itex]b

    Since a = -b

    3b + [itex]\frac{5}{2}[/itex]b = 1

    [itex]\frac{6}{2}[/itex]b + [itex]\frac{5}{2}[/itex]b = 1

    11b = 2

    b = [itex]\frac{2}{11}[/itex]

    a = [itex]\frac{-2}{11}[/itex]

    y(x) = [itex]\frac{-2}{11}[/itex]x-3 + [itex]\frac{2}{11}[/itex]x[itex]\frac{5}{2}[/itex]

    Thanks very much for all your help Tim and HallsofIvy, ps your method seems much faster!
     
    Last edited: May 25, 2012
  15. May 25, 2012 #14

    HallsofIvy

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    You mean a+ b= 0.
    Probably a typo: -3x-4 but fortunately at x= 1, it doesn't matter.

    As long as your characteristic equation has only distinct real roots it is. But if you have multiple roots or complex roots (or a non-homogeneous equation) it may be simpler to use what you have learned for linear d.e s with constant coefficients.
     
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