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Math Help - General solution step

  1. #1
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    General solution step

    I'm attepmting to solve a second order partial differential equation using the seperation of variable method, but have gotten up to where i have reduced to ordinary differential equations and has been so long since i did them i'm confused.


    If i have say;

    X'' - X/(k^2) = 0

    And i try a solution of the form X = e^{(\lambda x)}

    hence X' = \lambda e^{(\lambda x)}

    and X'' = \lambda^2 e^{(\lambda x)}

    when substituted in gives;

     \lambda^2 e^{(\lambda x)} - \frac{e^{(\lambda x)}}{k^2} = 0

    hence;  \lambda = + (1/k) or - 1/k

    is this correct up to here? and the following is where i'm confused,

    this seems to me to be real distinct solutions hence general solution should be;

    X(x) =  A.e^{\frac{x}{k}} +B.e^{\frac{-x}{k}}

    but i remember the lecturer saying that this should be a trig or hyperbolic solution, as there is no complex roots i'm gussing hyperbolics?

    Any help would be greatly appreciated, i'm very hazy on this stuff,

    Cheers.
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  2. #2
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    Quote Originally Posted by monster View Post
    I'm attepmting to solve a second order partial differential equation using the seperation of variable method, but have gotten up to where i have reduced to ordinary differential equations and has been so long since i did them i'm confused.


    If i have say;

    X'' - X/(k^2) = 0

    And i try a solution of the form X = e^{(\lambda x)}

    hence X' = \lambda e^{(\lambda x)}

    and X'' = \lambda^2 e^{(\lambda x)}

    when substituted in gives;

     \lambda^2 e^{(\lambda x)} - \frac{e^{(\lambda x)}}{k^2} = 0

    hence;  \lambda = + (1/k) or - 1/k

    is this correct up to here? and the following is where i'm confused,

    this seems to me to be real distinct solutions hence general solution should be;

    X(x) =  A.e^{\frac{x}{k}} +B.e^{\frac{-x}{k}}

    but i remember the lecturer saying that this should be a trig or hyperbolic solution, as there is no complex roots i'm gussing hyperbolics?

    Any help would be greatly appreciated, i'm very hazy on this stuff,

    Cheers.
    I would agree with your solution if I knew that k was real. But it might be complex.

    If it was complex then you could write

    k = \alpha + i\beta.

    Therefore \frac{x}{k} = \frac{x}{\alpha + i\beta} = \frac{x(\alpha - i\beta)}{\alpha^2 + \beta^2} = \frac{\alpha x - i\beta x}{\alpha^2 + \beta^2}.

    Now you can use Euler's Formula to clean it up.
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  3. #3
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    yes k is a real number such that k > 0,

    in my notes there is an example where;

    X'' + (k^2) X = 0

    When using trial solution  e^{\lambda x}

    gets to  \lambda = (+-) ki

    so general solution is;

    X =  a. e^{kxi} + b. e^{-kxi}

    which goes to;

    X = Acos(kx) + Bsin(kx)
    which i don't understand as;

    1/2( (e^ix) + (e^-ix) ) = cos(x) so how is the solution a combination of cos and sin ?
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  4. #4
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    Quote Originally Posted by monster View Post
    yes k is a real number such that k > 0,

    in my notes there is an example where;

    X'' + (k^2) X = 0

    When using trial solution  e^{\lambda x}

    gets to  \lambda = (+-) ki

    so general solution is;

    X =  a. e^{kxi} + b. e^{-kxi}

    which goes to;

    X = Acos(kx) + Bsin(kx)
    which i don't understand as;

    1/2( (e^ix) + (e^-ix) ) = cos(x) so how is the solution a combination of cos and sin ?
    Because e^{i\theta} = \cos{\theta} + i\sin{\theta} - this is Euler's Formula.
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  5. #5
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    Quote Originally Posted by monster View Post
    yes k is a real number such that k > 0,

    in my notes there is an example where;

    X'' + (k^2) X = 0

    When using trial solution  e^{\lambda x}

    gets to  \lambda = (+-) ki

    so general solution is;

    X =  a. e^{kxi} + b. e^{-kxi}

    which goes to;

    X = Acos(kx) + Bsin(kx)
    which i don't understand as;

    1/2( (e^ix) + (e^-ix) ) = cos(x) so how is the solution a combination of cos and sin ?
    You have both e^{kxi} and =ke^{-kxi}.

    Yes, cos(x)= \frac{e^{ix}+ e^{-ix}}{2}= cos(x) and you should also know that sin(x)= \frac{e^{ix}- e^{-ix}}{2i}.

    Multiplying the second equation by i, i sin(x)= \frac{e^{ix}- e^{-ix}}{2} and adding that to the first, cos(x)+ i sin(x)= e^{ix}, the formula Prove It gave. Subtracting that from the first gives cos(x)- i sin(x)= e^{-ix}.

    From that, ae^{ix}+ be^{-ix}= a(cos(x)+ i sin(x))+ b(cos(x)- i sin(x)) = acos(x)+ ai sin(x)+ bcos(x)- bi sin(x) = (a+ b)cos(x)+ i(a- b)sin(x)= A cos(x)+ B sin(x) with A= a+ b and B= a- b. If a and b were equal, that would be only a "cos(x)" term. It is because they are not equal that you have both cos(x) and sin(x).
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