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Thread: 2nd order diff eq

  1. #1
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    2nd order diff eq

    im having problems again with the constant at the end!! just need a bit of help trying to figure it out..

    $\displaystyle
    \begin{gathered}
    \frac{{d^2 y}}
    {{dx^2 }} - 8\frac{{dy}}
    {{dx}} + 7y = 0 \hfill \\
    \hfill \\
    (\lambda - 1)(\lambda - 7) \hfill \\
    \lambda = 1,\lambda = 7 \hfill \\
    \hfill \\
    y(x) = Ae^x + Be^7 ^x \hfill \\
    \hfill \\
    0 = y(0) = A + B \hfill \\
    \hfill \\
    5 = y'(0) = ... \hfill \\
    \end{gathered}
    $

    Its just this last part with the constant here I'm strugging with can someone help finish this off..

    I know the answer it should be,

    $\displaystyle
    y = \frac{5}
    {6}(e^7 ^x - e^x ))
    $
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  2. #2
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    Quote Originally Posted by dankelly07 View Post
    im having problems again with the constant at the end!! just need a bit of help trying to figure it out..

    $\displaystyle

    \frac{d^2 y}{dx^2 } - 8 \frac{dy}{dx} + 7y = 0

    $

    $\displaystyle
    (\lambda - 1)(\lambda - 7) = 0
    $

    $\displaystyle \lambda = 1 , \lambda = 7
    $

    $\displaystyle y(x) = Ae^x + Be^{7x}$


    $\displaystyle 0 = A + B
    $

    $\displaystyle 5 = y'(0)
    $.....

    Its just this last part with the constant here I'm strugging with can someone help finish this off..

    I know the answer it should be,

    $\displaystyle
    y = \frac{5}
    {6}(e^{7x} - e^x )
    $

    find a better program to output your latex, whatever you just used sucked!

    okay ow we can all read your question

    firsly you must find out what $\displaystyle \frac{dy}{dx}$ is

    $\displaystyle \frac{dy}{dx} = Ae^x + 7Be^{7x} $

    then sub in the given values

    $\displaystyle 5 = A + 7B \ \ \ (1)$
    $\displaystyle 0 = A + B \ \ \ \ (2) $

    and the rest is easy.


    Bobak
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  3. #3
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    Thanks for your help, any suggestions on a better latex program? that was the first time i used it...
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  4. #4
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    Well I did all the corrections by hand on that. If your going to post here frequently it may be worth picking up a few common inputs such as \frac \int

    Look on the Latex Help forum of the website http://www.mathhelpforum.com/math-help/latex-help/ there are many posts about programs.

    Bobak
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  5. #5
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    similar question

    ok now i have another problem...

    say for example I have a general solution of

    y(x) = Aexp{-4x}+Bex{-x}

    with y(0)=1 and y(1)=0

    I get

    1 = A + B
    0 = Aexp{-4}+Bexp{-x}

    so if Iam right how do I procced from here?
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  6. #6
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    Quote Originally Posted by dankelly07 View Post
    ok now i have another problem...

    say for example I have a general solution of

    y(x) = Aexp{-4x}+Bex{-x}

    with y(0)=1 and y(1)=0

    I get

    1 = A + B
    0 = A exp{-4}+B exp{-1} Mr F says: Note the correction.

    so if Iam right how do I procced from here?
    Solve the equations simultaneously for A and B. $\displaystyle e^{-4}$ and $\displaystyle e^{-1}$ are just numbers ......
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  7. #7
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    thanks, I just want to check,
    so now I get

    -exp{-1}=-exp{-1}
    Aexp{-4}=0

    so

    Aexp{-3} = 0

    A= exp{-3}

    B=-exp{-5} ???

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  8. #8
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    Quote Originally Posted by dankelly07 View Post
    thanks, I just want to check,
    so now I get

    -exp{-1}=-exp{-1}
    Aexp{-4}=0

    so

    Aexp{-3} = 0

    A= exp{-3}

    B=-exp{-5} ???

    Sorry but you're working makes absolutely no sense.

    And did you check your answers? Do they satisfy A + B = 1?

    If I asked you to solve

    1 = A + B

    0 = 2A + 3B

    how would you go about doing it .......?
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  9. #9
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    I would minus 3 from the first equation to get rid of the B to get

    -A = 1


    A + B = 1
    2A + 3B = 0

    A = -1
    B = 1

    ......
    my maths history isnt gr8 as u can probably tell...
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  10. #10
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    Quote Originally Posted by dankelly07 View Post
    I would minus 3 from the first equation to get rid of the B to get

    -A = 1


    A + B = 1
    2A + 3B = 0

    A = -1
    B = 1

    ......
    my maths history isnt gr8 as u can probably tell...
    If this is a true representation of your level of algebraic skill, your prospects for success in a subject covering differential equations are very bleak unless you embark on an extensive revision program that covers basic algebra.
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  11. #11
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    well I guess thats what I get for studying Maths at uni without studying maths at college... I studied music at college, which doesnt really help..lol thanks anyways
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