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Math Help - Homog Lin... #2

  1. #1
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    Homog Lin... #2

    2.) Find the general sol'n of hte following Dif EQ: \frac{d^5y}{dx^5}-2\frac{d^4y}{dx^4} + 17\frac{d^3y}{dx^3} = 0.

    Work:

    So,  m^5 - 2m^4 + 17m^3 = 0

    Factor: m^3*(m^2 - 2m + 17)

    Thus, m = 0.

    So, the general solution is just y = c_1 + c_2 since e^0 = 1 just cancels out.
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  2. #2
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    Hello, Ideasman!

    Your algebra is off . . .


    2) Find the general sol'n of: . \frac{d^5y}{dx^5}-2\frac{d^4y}{dx^4} + 17\frac{d^3y}{dx^3} \:= \:0

    We have: . m^5 - 2m^4 + 17m^3\:=\:0\quad\Rightarrow\quad m^3(m^2-2m + 17) \:=\:0


    We have two equations to solve:

    . . m^3\,=\,0\quad\Rightarrow\quad m\,=\,0\quad\Rightarrow\quad y \:=\:C_1

    . . m^2-2m + 17\:=\:0\quad\Rightarrow\quad m \:=\:\frac{2 \pm\sqrt{-64}}{2} \:=\:1 \pm 4i \quad\Rightarrow\quad y \:= . e^x\left(C_2\cos4x + C_3\sin4x\right)


    General solution: . y \;=\;C_1 + e^x\left(C_2\cos4x + C_3\sin4x\right)

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, Ideasman!

    Your algebra is off . . .



    We have: . m^5 - 2m^4 + 17m^3\:=\:0\quad\Rightarrow\quad m^3(m^2-2m + 17) \:=\:0


    We have two equations to solve:

    . . m^3\,=\,0\quad\Rightarrow\quad m\,=\,0\quad\Rightarrow\quad y \:=\:C_1

    . . m^2-2m + 17\:=\:0\quad\Rightarrow\quad m \:=\:\frac{2 \pm\sqrt{-64}}{2} \:=\:1 \pm 4i \quad\Rightarrow\quad y \:= . e^x\left(C_2\cos4x + C_3\sin4x\right)


    General solution: . y \;=\;C_1 + e^x\left(C_2\cos4x + C_3\sin4x\right)


    Thank you Soroban. I see how I was careless in computing the root for m^2 - 2m + 17!
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