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    This differential equation:
    L[x(t)] = 0 where  L = (\frac{d}{dt} + 2)(\frac{d^2}{dt^2} + 3)
    (\frac{dx}{dt} + 2x)(\frac{d^2x}{dt^2} + 3x) = 0

    Characteristic Equation:
    (m+2)(m^2+3)=0
    m^3 + 2m^2 + 3m + 6 = 0

    Thus:
    m_1 = -2

m_2 = +3i

m_3 = -3i

    General Solution:
    x(t) = Ae^{-2t} + e^{0t}[Bcos3t + Csin(3t)]
    x(t) = Ae^{-2t} + Bcos(3t) + Csin(3t)

    I don't have the marking scheme and I wanted to double-check if this is right.
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    (Original post by Alpha-Omega)
    This differential equation:
    L[x(t)] = 0 where  L = (\frac{d}{dt} + 2)(\frac{d^2}{dt^2} + 3)
    (\frac{dx}{dt} + 2x)(\frac{d^2x}{dt^2} + 3x) = 0

    Characteristic Equation:
    (m+2)(m^2+3)=0
    m^3 + 2m^2 + 3m + 6 = 0

    Thus:
    m_1 = -2

m_2 = +3i

m_3 = -3i

    General Solution:
    x(t) = Ae^{-2t} + e^{0t}[Bcos3t + Csin(3t)]
    x(t) = Ae^{-2t} + Bcos(3t) + Csin(3t)

    I don't have the marking scheme and I wanted to double-check if this is right.
    I don't think this right
    Consider that the equation will be 0 when
    1.
    \frac{dx}{dt}+2x=0
    2.
    \frac{d^2x}{dt^2}+3x=0
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    (Original post by ztibor)
    I don't think this right
    Consider that the equation will be 0 when
    1.
    \frac{dx}{dt}+2x=0
    2.
    \frac{d^2x}{dt^2}+3x=0
    Hmm... do you know how to get the correct solution?
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    (Original post by Alpha-Omega)
    Hmm... do you know how to get the correct solution?
    It depends on L
    A simple multiplication of two operators L=L_1\cdot L_2
    as you wrote in the first equation
    or one linear operator meaning
    L=\left(\frac{d}{dt}+2\right )\left(\frac{d^2}{dt^2}+3\right )=\frac{d^3}{dt^3}+2\frac{d^2}{d  t^2}+3\frac{d}{dt}+6
    as you wrote with characteristic polinomial
    I think to this
    Then your method is correct with that
    m_2=\sqrt{3}\cdot i
    m_3=-\sqrt{3}\cdot i
    For simple multiplication you will get same base solutions
    but then the real and complex solutions will be separate solutions so you can not get the linear combination of them (only for the two complex conjugate to get real functions) because the initial equation is not linear.
    1.
    x_1(t)=A\cdot e^{-2t}
    2.
    THe base solutions
    \left {u_1=e^{\sqrt{3}it},u_2=e^{-\sqrt{3}it}\right }
    so for real
    x_{2a}=\frac{B}{2}(u_1+u_2)=B \cdot cosh (i\cdot \sqrt{3}t)=B\cdot \cos(\sqrt{3}t)
    x_{2b}=\frac{C}{2i}(u_1-u_2)=\frac{C}{i}\cdot sinh (i\cdot \sqrt{3}t)=C\cdot \sin(\sqrt{3}t)
    So another solution for x(t)
    x_2(t)=B\cdot \cos(\sqrt{3}t)+C\cdot \sin(\sqrt{3}t)
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    (Original post by ztibor)
    It depends on L
    A simple multiplication of two operators L=L_1\cdot L_2
    as you wrote in the first equation
    or one linear operator meaning
    L=\left(\frac{d}{dt}+2\right )\left(\frac{d^2}{dt^2}+3\right )=\frac{d^3}{dt^3}+2\frac{d^2}{d  t^2}+3\frac{d}{dt}+6
    as you wrote with characteristic polinomial
    I think to this
    Then your method is correct with that
    m_2=\sqrt{3}\cdot i
    m_3=-\sqrt{3}\cdot i
    For simple multiplication you will get same base solutions
    but then the real and complex solutions will be separate solutions so you can not get the linear combination of them (only for the two complex conjugate to get real functions) because the initial equation is not linear.
    1.
    x_1(t)=A\cdot e^{-2t}
    2.
    THe base solutions
    \left {u_1=e^{\sqrt{3}it},u_2=e^{-\sqrt{3}it}\right }
    so for real
    x_{2a}=\frac{B}{2}(u_1+u_2)=B \cdot cosh (i\cdot \sqrt{3}t)=B\cdot \cos(\sqrt{3}t)
    x_{2b}=\frac{C}{2i}(u_1-u_2)=\frac{C}{i}\cdot sinh (i\cdot \sqrt{3}t)=C\cdot \sin(\sqrt{3}t)
    So another solution for x(t)
    x_2(t)=B\cdot \cos(\sqrt{3}t)+C\cdot \sin(\sqrt{3}t)
    Thanks for the answer.

    Although, I don't fully understand why do the real and complex solutions have to be separated if the initial equation is not linear. Could explain this a bit further or give me a link to your reference. Cheers
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    (Original post by Alpha-Omega)
    Thanks for the answer.

    Although, I don't fully understand why do the real and complex solutions have to be separated if the initial equation is not linear. Could explain this a bit further or give me a link to your reference. Cheers

    In first half of my answer I assume L is one linear operator of
    \displaystyle L=\left (\frac{d}{dt}+2\right )\left (\frac{d^2}{dt^2}+3\right )=\frac{d}{dt}\left (\frac{d^2}{dt^2}\right )+2\frac{d^2}{dt^2}+3\frac{d}{dt  }+6
    so
    \displaystyle L[x(t)]=0
    does not mean the following product as you wrote
    \displaystyle \left (\frac{dx}{dt}+2x\right )\left (\frac{d^2x}{dt^2}+3x\right )=0
    but
    \displaystyle \frac{d^3x}{dt^3}+2\frac{d^2x}{d  t^2}+3\frac{dx}{dt}+6x=0 equation
    I wrore that you have the correct solution for this except two m values;
    \displaystyle m_1=-2
    \displaystyle m_2=\sqrt{3}\cdot i
    \displaystyle m_3=-\sqrt{3}\cdot i
    So the base solutions are \displaystyle {e^{-2t},e^{-\sqrt{3}it},e^{\sqrt{3}it}}
    from this the general solution in Complex is the linear combination of base solutions (only if the equation is linear and it is)
    \displaystyle x(t)=Ae^{-2t}+Be^{\sqrt{3}it}+Ce^{\sqrt{3}  it}
    or functions with real variables (with further linear combinations as i wrote)
    \displaystyle x(t)=Ae^{-2t}+B\cos (\sqrt{3}t)+C\sin (\sqrt{3}t)

    At the second half of my answer I assumed that \displaystyle L=L_1\cdot L_2
    where
    \displaystyle L_1=\left (\frac{d}{dt}+2\right)
    and
    \displaystyle L_2=\left (\frac{d^2}{dt^2}+3\right)
    so
    \displaystyle L[x(t)]=0 which means
    \displaystyle \left (\frac{dx}{dt}+2x\right )\left (\frac{d^2x}{dt^2}+3x\right )=0
    That is
    \displaystyle \left (\frac{dx}{dt}+2x\right )=0
    OR
    \displaystyle \left (\frac{d^2x}{dt^2}+3x\right )=0
    Which means to solve the two equations separately and not simoultaneously
    because
    L_1[(x(t)]=0 involves L(x)=0
    and L_2[x(t)]=0 involves that too.
    So there is two solutions for x(t)
    from the 1st equation
    \displaystyle x_1(t)=A\cdot e^{-2t}
    from the 2nd
    x_2(t)=B\cdot e^{\sqrt{3}it}+C\cdot e^{-\sqrt{3}it}
    or for real functions
    x_2(t)=B\cdot \cos (\sqrt{3}t)+C\cdot \sin (\sqrt{3}t)

    http://en.wikipedia.org/wiki/Charact...ion_(calculus) (Characteristic equation)
    http://en.wikipedia.org/wiki/Superposition_principle (Superposition)
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    (Original post by ztibor)
    In first half of my answer I assume L is one linear operator of
    \displaystyle L=\left (\frac{d}{dt}+2\right )\left (\frac{d^2}{dt^2}+3\right )=\frac{d}{dt}\left (\frac{d^2}{dt^2}\right )+2\frac{d^2}{dt^2}+3\frac{d}{dt  }+6
    so
    \displaystyle L[x(t)]=0
    does not mean the following product as you wrote
    \displaystyle \left (\frac{dx}{dt}+2x\right )\left (\frac{d^2x}{dt^2}+3x\right )=0
    but
    \displaystyle \frac{d^3x}{dt^3}+2\frac{d^2x}{d  t^2}+3\frac{dx}{dt}+6x=0 equation
    I wrore that you have the correct solution for this except two m values;
    \displaystyle m_1=-2
    \displaystyle m_2=\sqrt{3}\cdot i
    \displaystyle m_3=-\sqrt{3}\cdot i
    So the base solutions are \displaystyle {e^{-2t},e^{-\sqrt{3}it},e^{\sqrt{3}it}}
    from this the general solution in Complex is the linear combination of base solutions (only if the equation is linear and it is)
    \displaystyle x(t)=Ae^{-2t}+Be^{\sqrt{3}it}+Ce^{\sqrt{3}  it}
    or in real (with further linear combinations as i wrote)
    \displaystyle x(t)=Ae^{-2t}+B\cos (\sqrt{3}t)+C\sin (\sqrt{3}t)

    At the second half of my answer I assumed that \displaystyle L=L_1\cdot L_2
    where
    \displaystyle L_1=\left (\frac{d}{dt}+2\right)
    and
    \displaystyle L_2=\left (\frac{d^2}{dt^2}+3\right)
    so
    \displaystyle L[x(t)]=0 which means
    \displaystyle \left (\frac{dx}{dt}+2x\right )\left (\frac{d^2x}{dt^2}+3x\right )=0
    That is
    \displaystyle \left (\frac{dx}{dt}+2x\right )=0
    OR
    \displaystyle \left (\frac{d^2x}{dt^2}+3x\right )=0
    Which means to solve the two equations separately and not simoultaneously
    because
    L_1[(x(t)]=0 involves L(x)=0
    and L_2[x(t)]=0 involves that too.
    So there is two solutions for x(t)
    from the 1st equation
    \displaystyle x_1(t)=A\cdot e^{-2t}
    from the 2nd
    x_2(t)=B\cdot e^{\sqrt{3}it}+C\cdot e^{-\sqrt{3}it}
    or for real functions
    x_2(t)=B\cdot \cos (\sqrt{3}t)+C\cdot \sin (\sqrt{3}t)

    I have no queation or answer books or links just my knowledge and skills
    Substituting back the solution into the original equation it works, and it is enough to prove that this was a good solution.
    Cheers, that makes sense.
 
 
 
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