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Math Help - General Solution

  1. #1
    Junior Member
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    General Solution

    Can someone help me with this please

    D^2Y
    +2DY +2Y=SinX
    DX^2 DX

    Find the G.S. given that the CF is e^-x(C1Cosx+C2SinX)
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  2. #2
    Super Member

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    Lexington, MA (USA)
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    Hello, Mathsnewbie!

    What methods have you been taught?
    And there is a typo . . .


    \frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 2y \:=\:\sin x

    Find the G.S. given that the CF is: e^{-x}(C_1\cos{\color{red}2}x+C_2\sin{\color{red}2}x)
    Undetermined Coefficients


    Let: y \:=\:A\sin x + B\cos x \quad\Rightarrow\quad 2y \:=\:2A\sin x + 2B\cos x .[1]

    . . \frac{dy}{dx} \:=\: A\cos x -B\sin x\quad\Rightarrow\quad 2\frac{dy}{dx} \:=\:\text{-}2B\sin x + 2A\cos x .[2]

    . . \frac{d^2y}{dx^2} \:=\:\text{-}A\sin x - B\cos x .[3]


    Add [1], [2], [3]: . \frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 2y \:=\:(A-2B)\sin x + (2A+B)\cos x \;=\;\sin x

    And we have: . \begin{array}{ccc}A - 2B &=& 1 \\ 2A + B &=& 0 \end{array}

    Solve the system: . A \,=\,\tfrac{1}{5},\;B \,=\,\text{-}\tfrac{2}{5}

    . . Hence, the particular solution is: . y \:=\:\tfrac{1}{5}\sin x - \tfrac{2}{5}\cos x


    The General Solution is: . y \;=\;e^{-x}\left(C_1\cos2x + C_2\sin2x\right) + \tfrac{1}{5}\sin x - \tfrac{2}{5}\cos x

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