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Math Help - Derivative of General Solution

  1. #1
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    Derivative of General Solution

    this should be simple but I can't get it to come out right

    i have the general solution as x(t)= C1e^(-t)cos4.36t + C2e^(-t)sin4.36t

    can someone get the first deriv of x(t)? C1 and C2 are what I need to solve for to find the particular solution
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  2. #2
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    is your general solution correct??
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  3. #3
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    Quote Originally Posted by johan View Post
    is your general solution correct??
    yeah i'm pretty sure..the roots to my equation were M1=-1+4.36i and M2=-1-4.36i
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  4. #4
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    Quote Originally Posted by chief27 View Post
    this should be simple but I can't get it to come out right

    i have the general solution as x(t)= C1e^(-t)cos4.36t + C2e^(-t)sin4.36t

    can someone get the first deriv of x(t)? C1 and C2 are what I need to solve for to find the particular solution
    Maybe you should post the whole question ....
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    Maybe you should post the whole question ....
    The DE is y"+2y'+20x = 0 initial conditions are x(0)=1 and x'(0)=0

    my characteristic equation is m^2+2m+20=0


    use initial conditions to determine the particular solution
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  6. #6
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    Second order differential equation

    Hello chief27
    Quote Originally Posted by chief27 View Post
    The DE is y"+2y'+20x = 0 initial conditions are x(0)=1 and x'(0)=0

    my characteristic equation is m^2+2m+20=0


    use initial conditions to determine the particular solution
    I agree with your general solution. I've left the term as \sqrt{19} rather than 4.36.

    x = e^{-t}(C_1\cos\sqrt{19}t+C_2\sin\sqrt{19}t)

    \Rightarrow x(0) = C_1 = 1

    \Rightarrow x = e^{-t}(\cos\sqrt{19}t+C_2\sin\sqrt{19}t)

    \Rightarrow \dot{x} = e^{-t}(-\sqrt{19}\sin\sqrt{19}t + \sqrt{19}C_2\cos\sqrt{19}t) - e^{-t}(\cos\sqrt{19}t+C_2\sin\sqrt{19}t)

     \Rightarrow \dot{x}(0) = \sqrt{19}C_2 - 1 = 0

    \Rightarrow C_2 = \frac{1}{\sqrt{19}}

    \Rightarrow x = e^{-t}\left(\cos\sqrt{19}t+\frac{1}{\sqrt{19}}\sin\sqr  t{19}t\right)

    Grandad
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