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    to find the general solution of a linear second order differential equation, i find the auxiliary roots.

    y" + 2y' + y = 0

    so i get 1+i and 1-i.

    i have two auxiliary rules:

    (i) If the auxiliary quadratic equation has pure imaginary roots +/-ni arising from the m^2 + n^2 = 0. The general solution is

    y = Acos nx + Bsin nx

    Where A and B are constants and n is of the element of real numbers.



    (ii) If the auxiliary quadratic equation has complex conjugate roots p +/- iq where p,q are of the real set of numbes. (condition b^2 < 4ac) the general solution is

    y = e^{px} ( Acos qx + Bsin qx )


    Where A and B are constants.




    ok so teacher has used (ii) to get e^x (Acosx + Bsinx)- is this because of the b^2<4ac? like, if it was = or > would i definitely use (i)?

    plus what's the difference between +-ni and p+-iq?

    and what is m^2 + n^2 = 0 and where has it come from?!

    thanks
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    (Original post by furryvision)
    ok so teacher has used (ii) to get e^x (Acosx + Bsinx)- is this because of the b^2<4ac? like, if it was = or > would i definitely use (i)?
    No - if b^2 >= 4ac, your roots would be real, so there'd be no 'complex conjugate' nonsense going on because there'd be no imagnary part to your roots. You use (ii) in all cases; (i) is just what you get from (ii) when p = 0 (and if you look at the quadratic formula, you'll see that's equivalent to b = 0).

    (Original post by furryvision)
    plus what's the difference between +-ni and p+-iq?
    I'm sure you can answer that. Is 2+3i of the form p +/- iq? Is it of the form +/- ni? (What's probably confusing you is that anything of the form +/- ni is also of the form p +/- iq for p = 0. But you'll notice that if you put p = 0 into (ii), you just get the same as in (i).)

    (Original post by furryvision)
    and what is m^2 + n^2 = 0 and where has it come from?!
    If y'' + n2y = 0, the auxiliary equation you need to solve is m^2 + n^2 = 0, giving solutions m = +/- ni.
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    Your complementary function is always of the form Ae^{\alpha x} +Be^{\beta x} (unless \alpha =\beta in which case they're both real and the CF is Ae^{\alpha x} +Bxe^{\alpha x}=e^{\alpha x}(A+Bx) )

    If \alpha\not=\beta and they are complex/imaginary, then use the fact that they will be complex conjugates of each other (lets say that \alpha=u+iv and \beta=u-iv) to get

    Ae^{\alpha x} +Be^{\beta x}\newline =Ae^{ux + ivx} +Be^{ux-ivx}\newline =e^{ux}(Ae^{ivx}+Be^{-ivx})\newline =e^{ux}(A(\cos vx +i\sin vx)+B(\cos vx -i\sin vx))\newline =e^{ux}(P\cos vx + Q\sin vx)

    If the roots are purely imaginary then u=0 giving just P\cos vx + Q\sin vx
 
 
 
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