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    Find the general solution of the differential equation
     \frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 10y = 0

    a) Find the constants k an p such that
     ke^{px}

    is a particular integral of the differential equation

     \frac{d^2y}{dx^2} + 2\frac{dy}{dx} + 10y = 54e^{2x}

    and hence find the solution of this differential equation for which y= 0 and dy/dx = 3 at x = 0.

    I've done all of this and I get the answer as:

     y=e^{-x}(Acos3x+Bsin3x) + 3e^{2x}

    and then I start substituting in and differentiating. My issue is that I get A = -3 and B = -2 whereas the book has them without the negative signs

    (in fact the book has  y=e^{-x}(Acos3x+Bsin3x) + 3e^{3x}

    as an answer, but I presumed that the e^3x at the end was a typo.

    am I going horrendously wrong. Have I missed something, or is the book wrong?
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    I also presume the e^3x is a typo.

    When x = 0, A + 3 = 0, right? So A = -3...
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    That's exactly what I'm getting. perhaps they factorise with -1 and forgot to put a minus outside the bracket.

    So I'm right then and the book's wrong?

    Score: Me 2 (so far): Book 1,536
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    Not a bad little total.
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    Can I get another answer check pls (and you're going to love this one!!!!)

     \frac{e^{-(n+1)\frac{\pi}{2}-\frac{1}{3}arctan3}sin((n+1)\pi + arctan3)}{e^{-(n)\frac{\pi}{2}-\frac{1}{3}arctan3}sin((n)\pi + arctan3)}

    I reckon I can get all the sine bits to cancel out using the addition rules, and then I'm left with the e bits.

    I think I then get

     \frac {e^{\frac{\pi}{3}} \times e^{\frac{-n\pi}{3}}\times e^{\frac{-1}{3}arctan3}}{e^{\frac{-n\pi}{3}}\times e^{\frac{-1}{3}arctan3}}

    i.e. it's e^-pi/3
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    (Original post by studienka)
    Can I get another answer check pls (and you're going to love this one!!!!)

     \frac{e^{-(n+1)\frac{\pi}{2}-\frac{1}{3}arctan3}sin((n+1)\pi + arctan3)}{e^{-(n)\frac{\pi}{2}-\frac{1}{3}arctan3}sin((n)\pi + arctan3)}
    Hmm, let's see. I'm gonna call that expression P, purely a random letter choice because it was the first letter I saw on my whiteboard. :p:

    sin(a) = -sin(a+pi), so:
    \displaystyle P =  \frac{-\exp [-(n+1)\frac{\pi}{2}-\frac{1}{3}\tan^{-1} 3]}{\exp [-(n)\frac{\pi}{2}-\frac{1}{3}\tan^{-1} 3]}

    \displaystyle =  \frac{-\exp [-(n+1)\frac{\pi}{2}] / \exp [\frac{1}{3}\tan^{-1} 3]}{\exp [-(n)\frac{\pi}{2}] / \exp [\frac{1}{3}\tan^{-1} 3]}

    \displaystyle =  \frac{-\exp [-(n+1)\frac{\pi}{2}]}{\exp [-(n)\frac{\pi}{2}]}

    \displaystyle =  \frac{-\exp [-\frac{n\pi}{2} - \frac{\pi}{2}]}{\exp [-\frac{n\pi}{2}]}

    \displaystyle =  -1/\exp \tfrac{\pi}{2} = -e^{-\pi /2}

    Might be wrong.
    • Thread Starter
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    You're right,

    for some reason I switch from Pi/2 to Pi/3 after my first line, so my workings are all correct, it just ends up as:

    -e^-pi/2

    I ignored the negative sign as the question involved moduli.

    thnx for the check though.
 
 
 
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