Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies.
Following the comments in the hints, we see that is a regular singular point. Therefore, we should look for series solutions of the form
![{\displaystyle y=\sum _{n=0}^{\infty }a_{n}x^{n+r}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/6750cc01074af9207e2e561368508aba20e2f296)
We have that
![{\displaystyle y=\sum _{n=0}^{\infty }a_{n}x^{n+r},\;y'=\sum _{\color {red}n=0}^{\infty }a_{n}(n+r)x^{n+r-1},\;y''=\sum _{\color {red}n=0}^{\infty }a_{n}(n+r)(n+r-1)x^{n+r-2}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/3a47cec9f667033c77e0b60cae01b692e1d01498)
(Please note that the both of the sums for and start at .)
Substitute these into the ODE
![{\displaystyle 2x^{2}y''+(3x+x^{2})y'-y=0,}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/99fbd50d27da8439624520ae8c7c5a393161abf7)
we get that
![{\displaystyle \sum _{n=0}^{\infty }2a_{n}(n+r)(n+r-1)x^{n+r}+\sum _{n=0}^{\infty }3a_{n}(n+r)x^{n+r}{\color {blue}+\sum _{n=0}^{\infty }a_{n}(n+r)x^{n+r+1}}-\sum _{n=0}^{\infty }a_{n}x^{n+r}=0.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/ee91cfc3834708ce22b766324ae714fd3f3d70a8)
We should shift the index of so the order of matches with the other terms. To do so, we should replace every with . Then
![{\displaystyle {\color {blue}\sum _{n=0}^{\infty }a_{n}(n+r)x^{n+r+1}}=\sum _{n={\color {red}1}}^{\infty }a_{n{\color {red}-1}}(n+r{\color {red}-1})x^{\color {red}n+r}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/9c7b455dfb1277bbd6c65a2b91e626728b5a601b)
Then we have
![{\displaystyle \sum _{n=0}^{\infty }2a_{n}(n+r)(n+r-1)x^{n+r}+\sum _{n=0}^{\infty }3a_{n}(n+r)x^{n+r}{\color {red}+\sum _{n=1}^{\infty }a_{n-1}(n+r-1)x^{n+r}}-\sum _{n=0}^{\infty }a_{n}x^{n+r}=0}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/eaf32a533b712f4fcf65758b0cc951223010e637)
which, after separating the terms, can be rewritten as
![{\displaystyle [2a_{0}r(r-1)+3a_{0}r-a_{0}]\;x^{r}+\sum _{n=1}^{\infty }\;{\color {OliveGreen}[2a_{n}(n+r)(n+r-1)+3a_{n}(n+r)+a_{n-1}(n+r-1)-a_{n}]}\;x^{n+r}=0.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/d81ad188a3907ba9b6e8809f44a30ea1f6b55a8f)
Look at the above equation, because the right hand side is 0, the coefficient of the term must be 0 as well.
Hence,
![{\displaystyle [2a_{0}r(r-1)+3a_{0}r-a_{0}]=a_{0}(2r(r-1)+3r-1)=0.\,}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/9a86520e361d43323a31f10553c838862d0099af)
Now, if , then all the subsequent will be zero as well and we only get the trivial solution .
(Why? You can verify that yourself after we finish the question. Set and you will see the other terms will be forced to be 0 by the recursion relation.)
In other words, we must have
![{\displaystyle 2r(r-1)+3r-1=2r^{2}+r-1=(2r-1)(r+1)=0\,}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/3b68a0130ad21927cd596a3401b434aedb1ec551)
giving us two roots and .
(Note: In the above, is known as the indicial equation.)
By the same reason as , we have that
![{\displaystyle {\color {OliveGreen}[2a_{n}(n+r)(n+r-1)+3a_{n}(n+r)+a_{n-1}(n+r-1)-a_{n}]}=0\;\;{\text{for each}}\;\;n\geq 1.\,}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/3d56cb33a97231496a3969a340a4e4e4305a8eee)
If we Isolate , we get the recurrence relation
![{\displaystyle a_{n}={\frac {-(n+r-1)a_{n-1}}{2(n+r)(n+r-1)+3(n+r)-1}}\;\;{\text{for}}\;\;n\geq 1.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/758da7b7dc3804161a6c134410c1f437e79501f9)
After simplifying and factoring the expression in the denominator, the above expression can be rewritten as
![{\displaystyle a_{n}={\frac {-(n+r-1)a_{n-1}}{2(n+r)^{2}+(n+r)-1}}={\frac {-(n+r-1)a_{n-1}}{(2(n+r)-1)((n+r)+1)}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/f91b7799a0d92247ba57adc31d5d2b900981f223)
which may be easier to do calculations with.
Now recall that from the indicial equation, we found and .
In the case , .
Note that the above equation does not satisfy due to the term, so the expression corresponding to is not what the question is asking for.
In the case , .
Note that the above equation satisfies . In other words, this is the one the question is asking for.
It remains to find and in terms of .
To find , substitue and into the recurrence relation. We get that
![{\displaystyle a_{1}={\frac {-(1+{\frac {1}{2}}-1)a_{0}}{(2(1+{\frac {1}{2}})-1)(1+{\frac {1}{2}}+1)}}={\frac {-1}{10}}a_{0}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/3d836589b9cc89eb765a64c54343b4fd2f09e77a)
To find , substitue and into the recurrence relation. We get that
![{\displaystyle a_{2}={\frac {-(2+{\frac {1}{2}}-1)a_{1}}{(2(2+{\frac {1}{2}})-1)(2+{\frac {1}{2}}+1)}}={\frac {-3}{28}}a_{1}={\frac {-3}{28}}({\frac {-1}{10}}a_{0})={\frac {3}{280}}a_{0}.}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/fdfb47f41d77b334623ba852e1ed4042c8bad9a8)
Therefore, we have
![{\displaystyle y(x)=a_{0}x^{\frac {1}{2}}+a_{1}x^{\frac {3}{2}}+a_{2}x^{\frac {5}{2}}+\cdots =a_{0}x^{\frac {1}{2}}-{\frac {1}{10}}a_{0}x^{\frac {3}{2}}+{\frac {3}{280}}a_{0}x^{\frac {5}{2}}+\cdots }](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/a729c7701556e4f62874ef9f21132ee9373b0782)
|