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Math Help - find two power solutions...

  1. #1
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    find two power solutions...

    ok i was working this problem and im not sure how to finish it...well i kinda have an idea..so here is the problem.

    y''-(x+1)y'-y=0

    and this is what i have so far....

    \sum_{k=0}^{\infty }(k+2)(k+1)c_{k+2}x^{k}-\sum_{k=1}^{\infty}kc_{k}x^{k}-\sum_{k=0}^{\infty}(k+1)c_{k+1}x^{k}-\sum_{k=0}^{\infty}c_{k}x^{k}

    im not sure how my book solved this for c2 and

    c_{k+2}=\frac{c_{k+1}+c_{k}}{k+2}

    whatever help that could be provided would be appreciated...
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  2. #2
    Super Member TheChaz's Avatar
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    Where is the "="?
    Don't you need all the indeces of summation to align?
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  3. #3
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    Quote Originally Posted by TheChaz View Post
    Where is the "="?
    Don't you need all the indeces of summation to align?
    your right, it all equals 0, sorry i forgot to include that. thanks
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  4. #4
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by slapmaxwell1 View Post
    ok i was working this problem and im not sure how to finish it...well i kinda have an idea..so here is the problem.

    y''-(x+1)y'-y=0

    and this is what i have so far....

    \sum_{k=0}^{\infty }(k+2)(k+1)c_{k+2}x^{k}-\sum_{k=1}^{\infty}kc_{k}x^{k}-\sum_{k=0}^{\infty}(k+1)c_{k+1}x^{k}-\sum_{k=0}^{\infty}c_{k}x^{k}

    im not sure how my book solved this for c2 and

    c_{k+2}=\frac{c_{k+1}+c_{k}}{k+2}

    whatever help that could be provided would be appreciated...
    Let suppose that the DE...

    \displaystyle y^{''} - (1+x)\ y^{'} - y = 0\ , \ y(0)=a_{0}\ , y^{'}(0)=a_{1} (1)

    ... has the solution of the form...

    \displaystyle y(x)= \sum_{n=0}^{\infty} a_{n}\ x^{n} = \sum_{n=0}^{\infty} \frac{y^{(n)} (0)}{n!}\ x^{n} (2)

    The values of a_{0} and a_{1} are given by the 'initial conditions' and the successive a_{n} are found from (1) and (2) as follows...

    \displaystyle y^{''}= \{(1+x)\ y^{'} + y\}_{x=0} = a_{0} + a_{1} \implies a_{2}= \frac{a_{0}+a_{1}}{2}

    \displaystyle y^{(3)}= \{(1+x)\ y^{''} + 2 y^{'}\}_{x=0} = a_{0} + 3 a_{1} \implies a_{3}= \frac{a_{0}+3 a_{1}}{6}

    \displaystyle y^{(4)}= \{(1+x)\ y^{(3)} + 3 y^{''}\}_{x=0} = 4 a_{0} + 6 a_{1} \implies a_{4}= \frac{4 a_{0}+6 a_{1}}{24}

    ... and so one so that is \displaystyle a_{n}= \frac{a_{n-1}+a_{n-2}}{n} exactly as in your book...

    Kind regards

    \chi \sigma
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  5. #5
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    Quote Originally Posted by chisigma View Post
    Let suppose that the DE...

    \displaystyle y^{''} - (1+x)\ y^{'} - y = 0\ , \ y(0)=a_{0}\ , y^{'}(0)=a_{1} (1)

    ... has the solution of the form...

    \displaystyle y(x)= \sum_{n=0}^{\infty} a_{n}\ x^{n} = \sum_{n=0}^{\infty} \frac{y^{(n)} (0)}{n!}\ x^{n} (2)

    The values of a_{0} and a_{1} are given by the 'initial conditions' and the successive a_{n} are found from (1) and (2) as follows...

    \displaystyle y^{''}= \{(1+x)\ y^{'} + y\}_{x=0} = a_{0} + a_{1} \implies a_{2}= \frac{a_{0}+a_{1}}{2}

    \displaystyle y^{(3)}= \{(1+x)\ y^{''} + 2 y^{'}\}_{x=0} = a_{0} + 3 a_{1} \implies a_{3}= \frac{a_{0}+3 a_{1}}{6}

    \displaystyle y^{(4)}= \{(1+x)\ y^{(3)} + 3 y^{''}\}_{x=0} = 4 a_{0} + 6 a_{1} \implies a_{4}= \frac{4 a_{0}+6 a_{1}}{24}

    ... and so one so that is \displaystyle a_{n}= \frac{a_{n-1}+a_{n-2}}{n} exactly as in your book...

    Kind regards

    \chi \sigma
    can you repost your response? for whatever reason it is not showing up. thanks alot.
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  6. #6
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    i dont know why but i keep seeing this latex error message on the board? please repost your message thanks alot.
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  7. #7
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by slapmaxwell1 View Post
    can you repost your response? for whatever reason it is not showing up. thanks alot.
    Sorry but at the moment the Latex is 'knocket out'... You can go to my post and press the 'Replay With Quote' button and see [in Latex language..] what I posted...

    Kind regards

    \chi \sigma
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  8. #8
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    Quote Originally Posted by chisigma View Post
    Sorry but at the moment the Latex is 'knocket out'... You can go to my post and press the 'Replay With Quote' button and see [in Latex language..] what I posted...



    Kind regards

    \chi \sigma
    do you know when the latex will be up and running?
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  9. #9
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by slapmaxwell1 View Post
    do you know when the latex will be up and running?
    ... no, I don't... all informations about that are from mederators or administrators... regarding Your specific problem we can have, if You like, further discussion in...

    Art of Problem Solving

    ...

    Kind regards

    \chi \sigma
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  10. #10
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    Quote Originally Posted by chisigma View Post
    ... no, I don't... all informations about that are from mederators or administrators... regarding Your specific problem we can have, if You like, further discussion in...

    Art of Problem Solving

    ...

    Kind regards

    \chi \sigma
    thanks for your help!

    do you know of a good website that could explain how add, subtract and reduce summations? or power series? i think the various sigma notations in my problems is throwing me off a bit.
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