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Math Help - please help me with this question

  1. #1
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    please help me with this question

    i need help with the following question attached.

    thanks
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  2. #2
    MHF Contributor chisigma's Avatar
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    The equation is...

    y^{'}-2\cdot y=-5\cdot x (1)

    ... and its caratteristic equation...

    a-2=0

    ... has solution a=2. So the general integral of 'incomplete equation' is...

    y_{i}(x)= c\cdot e^{2x} (2)

    It is aesy to verify that a particular integral of (1) is...

    y_{p}(x)= -\frac{5}{2}\cdot x -\frac{5}{4} (3)

    ... so that the general integral of (1) is...

    y(x)= c\cdot e^{2x} -\frac{5}{2}\cdot x -\frac{5}{4} (4)

    If we insert the 'initial condition' we obtain...

    y(x)= \frac{17}{4}\cdot e^{2x} -\frac{5}{2}\cdot x -\frac{5}{4} (5)

    Kind regards

    \chi \sigma
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  3. #3
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    Hello, cooltowns!

    It's fairly straight forward . . . Where is your difficulty?


    Solve: . \frac{dy}{dx} - 2y \:=\:5x,\quad y(0) = 3

    Integrating factor: . I \:=\:e^{\int \text{-}2\,dx} \:=\:e^{\text{-}2x}

    We have: . e^{-2x}\frac{dy}{dx} - 2e^{-2x} y \;=\;5xe^{-2x}

    . . . . . . . \frac{d}{dx}\left(e^{2x}y\right) \:=\:5xe^{-2x}

    Integrate: . e^{2x}y \;=\;5\!\!\int\!\! xe^{-2x}dx


    . . By parts: . \begin{array}{ccccccc}u &=& x & & dv &=& e^{-2x}dx \\ \\[-4mm] du &=& dx & & v &=&-\frac{1}{2}e^{-2x} \end{array}


    We have: . e^{-2x}y \;=\;5\bigg[\text{-}\tfrac{1}{2}xe^{-2x} + \tfrac{1}{2}\!\!\int e^{-2x}dx\bigg] \;=\;5\bigg[\text{-}\tfrac{1}{2}xe^{-2x} -\tfrac{1}{4}e^{-2x}\bigg] + C

    . . . . . . . e^{-2x}y \;=\;-\tfrac{5}{2}xe^{-2x} - \tfrac{5}{4}e^{-2x} + C


    Multiply by e^{2x}\!:\quad y \;=\;-\tfrac{5}{2}x - \tfrac{5}{4} + Ce^{2x}

    Since y(0)=3\!:\quad3 \;=\;-\tfrac{5}{2}(0) - \tfrac{5}{4} + Ce^0 \quad\Rightarrow\quad C \:=\:\frac{17}{4}


    Therefore: . y \;=\;-\frac{5}{2}x - \frac{5}{4} + \frac{17}{4}e^{2x}

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  4. #4
    Junior Member
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    i haven't done this type of differential equation before i didn't know you have use intergrating factors.
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