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    i've forgotten what that rule is. has it got a name?

    because i've got a square root containing -4, [equation being 2^2 - 4(1)(2)], and i know i have to put sqrt4i or something along those lines. any help??
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    i^2 = -1

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    Well, both are wrong.
    But i^2=-1.
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    i^2 = -1 and (-i)^2 = -1. i is one of the square roots of -1.
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    What?

    i^2 = -1 \Rightarrow i = \sqrt{-1} (if taking the 'positive' root), and \sqrt{-4} = \sqrt{-1}\sqrt{4}, so your answer is pretty much straightforward from there.

    But if your equation is 2^2 - \sqrt{4}, then that's different to 2^2 - \sqrt{-4} or whatever; i is only for if the minus is inside the root.

    EDIT: Beaten to it three times!
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    yes! atleast i was close! thanks
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    ok sorry confused again! :P

    y'' + 2' + 2 = 0

    (-2 (+-) sqrt(2^2 - 4(1)(2)))/2(1)

    = (-2 (+-) sqrt(-4))/2

    i remember the teacher doing something like:

    (-2 (+-) sqrt(4i))/2 or similar, is that right?
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    (Original post by furryvision)
    ok sorry confused again! :P

    y'' + 2' + 2 = 0

    (-2 (+-) sqrt(2^2 - 4(1)(2)))/2(1)

    = (-2 (+-) sqrt(-4))/2

    i remember the teacher doing something like:

    (-2 (+-) sqrt(4i))/2 or similar, is that right?
    See nuodai's post. Informally, sqrt(-4) = sqrt(4)*sqrt(-1) = sqrt(4) * i. Definitely not sqrt(4i) (because there you've just replaced -1 with i, which we told you was wrong :p:).
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    No, because the i should be outside the root. \dfrac{-2 \pm \sqrt{4}i}{2} = \dfrac{-2 \pm 2i}{2} = 1 \pm i.

    But, by y'' do you mean y² or \dfrac{\mbox{d}^2y}{\mbox{d}x^2}? Because if you mean \dfrac{\mbox{d}^2y}{\mbox{d}x^2}, you're going about this very much the wrong way, although you would need to solve the equation \lambda^2 + 2\lambda + 2 = 0 to get the auxiliary equation so the 1 \pm i bit isn't completely wasted.
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    (Original post by nuodai)
    you would need to solve the equation \lambda^2 + 2\lambda + 2 = 0 to get the auxiliary equation.
    That's what I assumed she was doing.
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    (Original post by nuodai)
    No, because the i should be outside the root. \dfrac{-2 \pm \sqrt{4}i}{2} = \dfrac{-2 \pm 2i}{2} = 1 \pm i.

    But, by y'' do you mean y² or \dfrac{\mbox{d}^2y}{\mbox{d}x^2}? Because if you mean \dfrac{\mbox{d}^2y}{\mbox{d}x^2}, you're going about this very much the wrong way, although you would need to solve the equation \lambda^2 + 2\lambda + 2 = 0 to get the auxiliary equation so the 1 \pm i bit isn't completely wasted.
    thanks, i'm making sense of it all now! y" is the d^2y/dx^2, but it equals cos(x), not 0, i just said that to make the question easier. it says in my notes to find the general solutions first and my teacher got the 1+-i so i understand now. thanks!
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    (Original post by furryvision)
    thanks, i'm making sense of it all now! y" is the d^2y/dx^2, but it equals cos(x), not 0, i just said that to make the question easier. it says in my notes to find the general solutions first and my teacher got the 1+-i so i understand now. thanks!
    Fair enough; in such a case you don't need to do any quadratic equation. To get your solution, you need to:
    1. Let y = A\sin x + B\cos x, and solve for A and B by substituting in and differentiating etc. to give your particular integral
    2. Let \lambda^2 + 2\lambda + 2 = 0 (ignore the cos x here) and solve for \lambda giving two solutions \lambda_1 and \lambda_2, so your complementary function is Ce^{\lambda_1} + De^{\lambda_2}
    3. Sum the complementary function and particular integral
 
 
 
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