Internal problem ID [6496]
Internal file name [OUTPUT/5744_Sunday_June_05_2022_03_52_28_PM_85151222/index.tex]
Book: Differential Equations: Theory, Technique, and Practice by George Simmons, Steven
Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 4. Power Series Solutions and Special Functions. Problems for review and
discovert. (B) Challenge Problems . Page 194
Problem number: 1(a).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Repeated root"
Maple gives the following as the ode type
[[_Emden, _Fowler]]
\[ \boxed {x^{3} y^{\prime \prime }+x^{2} y^{\prime }+y=0} \] With the expansion point for the power series method at \(x = \infty \).
Since expansion is around \(\infty \), then the independent variable \(x\) is replaced by \(\frac {1}{t}\) and the expansion is made around \(t = 0\) and after solving, the solution is changed back to \(x\) using \(x = \frac {1}{t}\). Changing variables results in the new ode \begin {align*} -\frac {-\left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right ) t^{2}-2 t \left (\frac {d}{d t}y \left (t \right )\right )}{t}-\frac {d}{d t}y \left (t \right )+y \left (t \right ) = 0 \end {align*}
The transformed ODE is now solved. The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right ) t +y \left (t \right )+\frac {d}{d t}y \left (t \right ) = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} \frac {d^{2}}{d t^{2}}y \left (t \right )+p(t) \frac {d}{d t}y \left (t \right ) + q(t) y \left (t \right ) &=0 \end {align*}
Where \begin {align*} p(t) &= \frac {1}{t}\\ q(t) &= \frac {1}{t}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0]\)
Irregular singular points : \([\infty ]\)
Since \(t = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right ) t +y \left (t \right )+\frac {d}{d t}y \left (t \right ) = 0 \] Let the solution be represented as Frobenius power series of the form \[ y \left (t \right ) = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r} \] Then \begin{align*} \frac {d}{d t}y \left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1} \\ \frac {d^{2}}{d t^{2}}y \left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} t^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} t^{n +r -2}\right ) t +\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r}\right ) = 0 \end{equation} The next step is to make all powers of \(t\) be \(n +r -1\) in each summation term. Going over each summation term above with power of \(t\) in it which is not already \(t^{n +r -1}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} t^{n +r -1} \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(t\) are the same and equal to \(n +r -1\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} t^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ t^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+\left (n +r \right ) a_{n} t^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ t^{-1+r} a_{0} r \left (-1+r \right )+r a_{0} t^{-1+r} = 0 \] Or \[ \left (t^{-1+r} r \left (-1+r \right )+r \,t^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ t^{-1+r} r^{2} = 0 \] Since the above is true for all \(t\) then the indicial equation becomes \[ r^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 0\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ t^{-1+r} r^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([0, 0]\).
Since the root of the indicial equation is repeated, then we can construct two linearly independent solutions. The first solution has the form \begin {align*} y_{1}\left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r}\tag {1A} \end {align*}
Now the second solution \(y_{2}\) is found using \begin {align*} y_{2}\left (t \right ) &= y_{1}\left (t \right ) \ln \left (t \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} t^{n +r}\right )\tag {1B} \end {align*}
Then the general solution will be \[ y \left (t \right ) = c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right ) \] In Eq (1B) the sum starts from 1 and not zero. In Eq (1A), \(a_{0}\) is never zero, and is arbitrary and is typically taken as \(a_{0} = 1\), and \(\{c_{1}, c_{2}\}\) are two arbitray constants of integration which can be found from initial conditions. We start by finding the first solution \(y_{1}\left (t \right )\). Eq (2B) derived above is now used to find all \(a_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n} \left (n +r \right )+a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -1}}{n^{2}+2 n r +r^{2}}\tag {4} \] Which for the root \(r = 0\) becomes \[ a_{n} = -\frac {a_{n -1}}{n^{2}}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 0\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ a_{1}=-\frac {1}{\left (r +1\right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{1}=-1 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{\left (r +1\right )^{2}}\) | \(-1\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{2}={\frac {1}{4}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{\left (r +1\right )^{2}}\) | \(-1\) |
| \(a_{2}\) | \(\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2}}\) | \(\frac {1}{4}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{3}=-{\frac {1}{36}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{\left (r +1\right )^{2}}\) | \(-1\) |
| \(a_{2}\) | \(\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2}}\) | \(\frac {1}{4}\) |
| \(a_{3}\) | \(-\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\) | \(-{\frac {1}{36}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{4}={\frac {1}{576}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{\left (r +1\right )^{2}}\) | \(-1\) |
| \(a_{2}\) | \(\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2}}\) | \(\frac {1}{4}\) |
| \(a_{3}\) | \(-\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\) | \(-{\frac {1}{36}}\) |
| \(a_{4}\) | \(\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(\frac {1}{576}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{5}=-{\frac {1}{14400}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{\left (r +1\right )^{2}}\) | \(-1\) |
| \(a_{2}\) | \(\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2}}\) | \(\frac {1}{4}\) |
| \(a_{3}\) | \(-\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\) | \(-{\frac {1}{36}}\) |
| \(a_{4}\) | \(\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(\frac {1}{576}\) |
| \(a_{5}\) | \(-\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\) | \(-{\frac {1}{14400}}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2} \left (r +6\right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{6}={\frac {1}{518400}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{\left (r +1\right )^{2}}\) | \(-1\) |
| \(a_{2}\) | \(\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2}}\) | \(\frac {1}{4}\) |
| \(a_{3}\) | \(-\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\) | \(-{\frac {1}{36}}\) |
| \(a_{4}\) | \(\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(\frac {1}{576}\) |
| \(a_{5}\) | \(-\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\) | \(-{\frac {1}{14400}}\) |
| \(a_{6}\) | \(\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2} \left (r +6\right )^{2}}\) | \(\frac {1}{518400}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=-\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2} \left (r +6\right )^{2} \left (7+r \right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{7}=-{\frac {1}{25401600}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{\left (r +1\right )^{2}}\) | \(-1\) |
| \(a_{2}\) | \(\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2}}\) | \(\frac {1}{4}\) |
| \(a_{3}\) | \(-\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\) | \(-{\frac {1}{36}}\) |
| \(a_{4}\) | \(\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(\frac {1}{576}\) |
| \(a_{5}\) | \(-\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\) | \(-{\frac {1}{14400}}\) |
| \(a_{6}\) | \(\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2} \left (r +6\right )^{2}}\) | \(\frac {1}{518400}\) |
| \(a_{7}\) | \(-\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2} \left (r +6\right )^{2} \left (7+r \right )^{2}}\) | \(-{\frac {1}{25401600}}\) |
Using the above table, then the first solution \(y_{1}\left (t \right )\) becomes \begin{align*} y_{1}\left (t \right )&= a_{0}+a_{1} t +a_{2} t^{2}+a_{3} t^{3}+a_{4} t^{4}+a_{5} t^{5}+a_{6} t^{6}+a_{7} t^{7}+a_{8} t^{8}\dots \\ &= -t +1+\frac {t^{2}}{4}-\frac {t^{3}}{36}+\frac {t^{4}}{576}-\frac {t^{5}}{14400}+\frac {t^{6}}{518400}-\frac {t^{7}}{25401600}+O\left (t^{8}\right ) \\ \end{align*} Now the second solution is found. The second solution is given by \[ y_{2}\left (t \right ) = y_{1}\left (t \right ) \ln \left (t \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} t^{n +r}\right ) \] Where \(b_{n}\) is found using \[ b_{n} = \frac {d}{d r}a_{n ,r} \] And the above is then evaluated at \(r = 0\). The above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table
| \(n\) | \(b_{n ,r}\) | \(a_{n}\) | \(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) | \(b_{n}\left (r =0\right )\) |
| \(b_{0}\) | \(1\) | \(1\) | N/A since \(b_{n}\) starts from 1 | N/A |
| \(b_{1}\) | \(-\frac {1}{\left (r +1\right )^{2}}\) | \(-1\) | \(\frac {2}{\left (r +1\right )^{3}}\) | \(2\) |
| \(b_{2}\) | \(\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2}}\) | \(\frac {1}{4}\) | \(\frac {-4 r -6}{\left (r +1\right )^{3} \left (r +2\right )^{3}}\) | \(-{\frac {3}{4}}\) |
| \(b_{3}\) | \(-\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\) | \(-{\frac {1}{36}}\) | \(\frac {6 r^{2}+24 r +22}{\left (r +1\right )^{3} \left (r +2\right )^{3} \left (r +3\right )^{3}}\) | \(\frac {11}{108}\) |
| \(b_{4}\) | \(\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(\frac {1}{576}\) | \(\frac {-8 r^{3}-60 r^{2}-140 r -100}{\left (r +1\right )^{3} \left (r +2\right )^{3} \left (r +3\right )^{3} \left (r +4\right )^{3}}\) | \(-{\frac {25}{3456}}\) |
| \(b_{5}\) | \(-\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\) | \(-{\frac {1}{14400}}\) | \(\frac {10 r^{4}+120 r^{3}+510 r^{2}+900 r +548}{\left (r +1\right )^{3} \left (r +2\right )^{3} \left (r +3\right )^{3} \left (r +4\right )^{3} \left (r +5\right )^{3}}\) | \(\frac {137}{432000}\) |
| \(b_{6}\) | \(\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2} \left (r +6\right )^{2}}\) | \(\frac {1}{518400}\) | \(\frac {-12 r^{5}-210 r^{4}-1400 r^{3}-4410 r^{2}-6496 r -3528}{\left (r +1\right )^{3} \left (r +2\right )^{3} \left (r +3\right )^{3} \left (r +4\right )^{3} \left (r +5\right )^{3} \left (r +6\right )^{3}}\) | \(-{\frac {49}{5184000}}\) |
| \(b_{7}\) | \(-\frac {1}{\left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2} \left (r +6\right )^{2} \left (7+r \right )^{2}}\) | \(-{\frac {1}{25401600}}\) | \(\frac {14 r^{6}+336 r^{5}+3220 r^{4}+15680 r^{3}+40614 r^{2}+52528 r +26136}{\left (r +1\right )^{3} \left (r +2\right )^{3} \left (r +3\right )^{3} \left (r +4\right )^{3} \left (r +5\right )^{3} \left (r +6\right )^{3} \left (7+r \right )^{3}}\) | \(\frac {121}{592704000}\) |
The above table gives all values of \(b_{n}\) needed. Hence the second solution is \begin{align*} y_{2}\left (t \right )&=y_{1}\left (t \right ) \ln \left (t \right )+b_{0}+b_{1} t +b_{2} t^{2}+b_{3} t^{3}+b_{4} t^{4}+b_{5} t^{5}+b_{6} t^{6}+b_{7} t^{7}+b_{8} t^{8}\dots \\ &= \left (-t +1+\frac {t^{2}}{4}-\frac {t^{3}}{36}+\frac {t^{4}}{576}-\frac {t^{5}}{14400}+\frac {t^{6}}{518400}-\frac {t^{7}}{25401600}+O\left (t^{8}\right )\right ) \ln \left (t \right )+2 t -\frac {3 t^{2}}{4}+\frac {11 t^{3}}{108}-\frac {25 t^{4}}{3456}+\frac {137 t^{5}}{432000}-\frac {49 t^{6}}{5184000}+\frac {121 t^{7}}{592704000}+O\left (t^{8}\right ) \\ \end{align*} Therefore the homogeneous solution is \begin{align*} y_h(t) &= c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right ) \\ &= c_{1} \left (-t +1+\frac {t^{2}}{4}-\frac {t^{3}}{36}+\frac {t^{4}}{576}-\frac {t^{5}}{14400}+\frac {t^{6}}{518400}-\frac {t^{7}}{25401600}+O\left (t^{8}\right )\right ) + c_{2} \left (\left (-t +1+\frac {t^{2}}{4}-\frac {t^{3}}{36}+\frac {t^{4}}{576}-\frac {t^{5}}{14400}+\frac {t^{6}}{518400}-\frac {t^{7}}{25401600}+O\left (t^{8}\right )\right ) \ln \left (t \right )+2 t -\frac {3 t^{2}}{4}+\frac {11 t^{3}}{108}-\frac {25 t^{4}}{3456}+\frac {137 t^{5}}{432000}-\frac {49 t^{6}}{5184000}+\frac {121 t^{7}}{592704000}+O\left (t^{8}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y \left (t \right ) &= y_h \\ &= c_{1} \left (-t +1+\frac {t^{2}}{4}-\frac {t^{3}}{36}+\frac {t^{4}}{576}-\frac {t^{5}}{14400}+\frac {t^{6}}{518400}-\frac {t^{7}}{25401600}+O\left (t^{8}\right )\right )+c_{2} \left (\left (-t +1+\frac {t^{2}}{4}-\frac {t^{3}}{36}+\frac {t^{4}}{576}-\frac {t^{5}}{14400}+\frac {t^{6}}{518400}-\frac {t^{7}}{25401600}+O\left (t^{8}\right )\right ) \ln \left (t \right )+2 t -\frac {3 t^{2}}{4}+\frac {11 t^{3}}{108}-\frac {25 t^{4}}{3456}+\frac {137 t^{5}}{432000}-\frac {49 t^{6}}{5184000}+\frac {121 t^{7}}{592704000}+O\left (t^{8}\right )\right ) \\ \end{align*} Replacing \(t\) by \(\frac {1}{x}\) gives \begin {align*} y = c_{1} \left (-\frac {1}{x}+1+\frac {1}{4 x^{2}}-\frac {1}{36 x^{3}}+\frac {1}{576 x^{4}}-\frac {1}{14400 x^{5}}+\frac {1}{518400 x^{6}}-\frac {1}{25401600 x^{7}}+O\left (\frac {1}{x^{8}}\right )\right )+c_{2} \left (\left (-\frac {1}{x}+1+\frac {1}{4 x^{2}}-\frac {1}{36 x^{3}}+\frac {1}{576 x^{4}}-\frac {1}{14400 x^{5}}+\frac {1}{518400 x^{6}}-\frac {1}{25401600 x^{7}}+O\left (\frac {1}{x^{8}}\right )\right ) \ln \left (\frac {1}{x}\right )+\frac {2}{x}-\frac {3}{4 x^{2}}+\frac {11}{108 x^{3}}-\frac {25}{3456 x^{4}}+\frac {137}{432000 x^{5}}-\frac {49}{5184000 x^{6}}+\frac {121}{592704000 x^{7}}+O\left (\frac {1}{x^{8}}\right )\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (-\frac {1}{x}+1+\frac {1}{4 x^{2}}-\frac {1}{36 x^{3}}+\frac {1}{576 x^{4}}-\frac {1}{14400 x^{5}}+\frac {1}{518400 x^{6}}-\frac {1}{25401600 x^{7}}+O\left (\frac {1}{x^{8}}\right )\right )+c_{2} \left (\left (-\frac {1}{x}+1+\frac {1}{4 x^{2}}-\frac {1}{36 x^{3}}+\frac {1}{576 x^{4}}-\frac {1}{14400 x^{5}}+\frac {1}{518400 x^{6}}-\frac {1}{25401600 x^{7}}+O\left (\frac {1}{x^{8}}\right )\right ) \ln \left (\frac {1}{x}\right )+\frac {2}{x}-\frac {3}{4 x^{2}}+\frac {11}{108 x^{3}}-\frac {25}{3456 x^{4}}+\frac {137}{432000 x^{5}}-\frac {49}{5184000 x^{6}}+\frac {121}{592704000 x^{7}}+O\left (\frac {1}{x^{8}}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (-\frac {1}{x}+1+\frac {1}{4 x^{2}}-\frac {1}{36 x^{3}}+\frac {1}{576 x^{4}}-\frac {1}{14400 x^{5}}+\frac {1}{518400 x^{6}}-\frac {1}{25401600 x^{7}}+O\left (\frac {1}{x^{8}}\right )\right )+c_{2} \left (\left (-\frac {1}{x}+1+\frac {1}{4 x^{2}}-\frac {1}{36 x^{3}}+\frac {1}{576 x^{4}}-\frac {1}{14400 x^{5}}+\frac {1}{518400 x^{6}}-\frac {1}{25401600 x^{7}}+O\left (\frac {1}{x^{8}}\right )\right ) \ln \left (\frac {1}{x}\right )+\frac {2}{x}-\frac {3}{4 x^{2}}+\frac {11}{108 x^{3}}-\frac {25}{3456 x^{4}}+\frac {137}{432000 x^{5}}-\frac {49}{5184000 x^{6}}+\frac {121}{592704000 x^{7}}+O\left (\frac {1}{x^{8}}\right )\right ) \] Verified OK.
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel <- Bessel successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 207
Order:=8; dsolve(x^3*diff(y(x),x$2)+x^2*diff(y(x),x)+y(x)=0,y(x),type='series',x=infinity);
\[ y \left (x \right ) = \left (1-\frac {\left (x -\operatorname {Infinity} \right )^{2}}{2 \operatorname {Infinity}^{3}}+\frac {2 \left (x -\operatorname {Infinity} \right )^{3}}{3 \operatorname {Infinity}^{4}}+\frac {\left (-18 \operatorname {Infinity} +1\right ) \left (x -\operatorname {Infinity} \right )^{4}}{24 \operatorname {Infinity}^{6}}+\frac {\left (96 \operatorname {Infinity} -14\right ) \left (x -\operatorname {Infinity} \right )^{5}}{120 \operatorname {Infinity}^{7}}+\frac {\left (-600 \operatorname {Infinity}^{2}+156 \operatorname {Infinity} -1\right ) \left (x -\operatorname {Infinity} \right )^{6}}{720 \operatorname {Infinity}^{9}}+\frac {\left (4320 \operatorname {Infinity}^{2}-1692 \operatorname {Infinity} +30\right ) \left (x -\operatorname {Infinity} \right )^{7}}{5040 \operatorname {Infinity}^{10}}\right ) y \left (\operatorname {Infinity} \right )+\left (x -\operatorname {Infinity} -\frac {\left (x -\operatorname {Infinity} \right )^{2}}{2 \operatorname {Infinity}}+\frac {\left (2 \operatorname {Infinity}^{2}-\operatorname {Infinity} \right ) \left (x -\operatorname {Infinity} \right )^{3}}{6 \operatorname {Infinity}^{4}}-\frac {\left (\operatorname {Infinity} -\frac {4}{3}\right ) \left (x -\operatorname {Infinity} \right )^{4}}{4 \operatorname {Infinity}^{4}}+\frac {\left (24 \operatorname {Infinity}^{3}-58 \operatorname {Infinity}^{2}+\operatorname {Infinity} \right ) \left (x -\operatorname {Infinity} \right )^{5}}{120 \operatorname {Infinity}^{7}}+\frac {\left (-120 \operatorname {Infinity}^{4}+444 \operatorname {Infinity}^{3}-21 \operatorname {Infinity}^{2}\right ) \left (x -\operatorname {Infinity} \right )^{6}}{720 \operatorname {Infinity}^{9}}+\frac {\left (720 \operatorname {Infinity}^{4}-3708 \operatorname {Infinity}^{3}+324 \operatorname {Infinity}^{2}-\operatorname {Infinity} \right ) \left (x -\operatorname {Infinity} \right )^{7}}{5040 \operatorname {Infinity}^{10}}\right ) D\left (y \right )\left (\operatorname {Infinity} \right )+O\left (x^{8}\right ) \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 171
AsymptoticDSolveValue[x^3*y''[x]+x^2*y'[x]+y[x]==0,y[x],{x,infinity,7}]
\[ y(x)\to c_1 \left (-\frac {1}{25401600 x^7}+\frac {1}{518400 x^6}-\frac {1}{14400 x^5}+\frac {1}{576 x^4}-\frac {1}{36 x^3}+\frac {1}{4 x^2}-\frac {1}{x}+1\right )+c_2 \left (\frac {121}{592704000 x^7}+\frac {\log (x)}{25401600 x^7}-\frac {49}{5184000 x^6}-\frac {\log (x)}{518400 x^6}+\frac {137}{432000 x^5}+\frac {\log (x)}{14400 x^5}-\frac {25}{3456 x^4}-\frac {\log (x)}{576 x^4}+\frac {11}{108 x^3}+\frac {\log (x)}{36 x^3}-\frac {3}{4 x^2}-\frac {\log (x)}{4 x^2}+\frac {2}{x}+\frac {\log (x)}{x}-\log (x)\right ) \]