# Thread: find two power solutions...

1. ## 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.

$\displaystyle y''-(x+1)y'-y=0$

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

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

im not sure how my book solved this for c2 and

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

whatever help that could be provided would be appreciated...

2. Where is the "="?
Don't you need all the indeces of summation to align?

3. Originally Posted by TheChaz
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

4. Originally Posted by slapmaxwell1
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.

$\displaystyle y''-(x+1)y'-y=0$

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

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

im not sure how my book solved this for c2 and

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

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

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

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

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

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

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

5. Originally Posted by chisigma
Let suppose that the DE...

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

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

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

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

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

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

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

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
can you repost your response? for whatever reason it is not showing up. thanks alot.

6. i dont know why but i keep seeing this latex error message on the board? please repost your message thanks alot.

7. Originally Posted by slapmaxwell1
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

$\displaystyle \chi$ $\displaystyle \sigma$

8. Originally Posted by chisigma
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

$\displaystyle \chi$ $\displaystyle \sigma$
do you know when the latex will be up and running?

9. Originally Posted by slapmaxwell1
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

$\displaystyle \chi$ $\displaystyle \sigma$

10. Originally Posted by chisigma
... 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

$\displaystyle \chi$ $\displaystyle \sigma$