y"' + y'' + 2y' + 2y = 0?

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  • 4 weeks ago

    y"'+y"+2y'+2y=0

    The auxiliary equation is

    m^3+m^2+2m+2=0

    =>

    (m+1)m^2+2(m+1)=0

    =>

    (m+1)(m^2+2)=0

    =>

    m=-1, m=+/-sqr(2)i

    =>

    the general solution of the d.e. is

    y=Ae^(-x)+Bcos[sqr(2)x]+Csin[sqr(2)x],

    where A, B & C are constants.

  • 4 weeks ago

    y"' + y'' + 2y' + 2y = 0

    r^3 + r^2 + 2r + 2y(x) = 0

  • 4 weeks ago

    y’'' + y'' + 2y' + 2y = 0

    r³ + r² + 2r + 2 = 0

    ( r + 1 )( r² + 2 ) = 0

    r = -1, r = ±i√2

    y = c₁ e^(-x) + c₂ cos(x√2) + c₃ sin(x√2)

     

  • Ian H
    Lv 7
    4 weeks ago

    y"' + y'' + 2y' + 2y = 0,

    Speculate that y may be of type y = e^rx

    e^rx[r^3 + r^2 + 2 + 2] = 0

    (r + 1)(r^2 + 2) = 0

    r = - 1, or r = ±√(2)i

    y = Ae^-x + Pe^√(2)i + Qe^-√(2)i which can also be represented by

    y = Ae^-x + Bsin[√(2)x] + Ccos[√(2)x]

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  • 4 weeks ago

    For this type of ODE, you can make teh guess that y = e^(ax) where a is a to be determined constant.  Substituting into your ODE

    a^3 + a^2 +2a +2 = 0 -- we have a cubic equation to solve for three values of a

    Group terms  (a^3 +a^2) + (2a + 2) = a^2*(a +1) + 2*(a +1) = (a^2 +2)(a +1) = 0

    so  a = +/-sqrt(2), a = -1

    and you have three terms in the solution to the ODE

    y = A*e^(-sqrt(2)*x) + B*e^(sqrt(2)*x) + C*e^(-x)  where A, B, and C will be determined by initial and/or boundary conditions

  • rotchm
    Lv 7
    4 weeks ago

    Hints: What type of DE is this? What techniques did you see? What if you let

    y ~ e^(ax) or a poly, or some other type you can try?

  • david
    Lv 7
    4 weeks ago

    y = 0                 

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