Internal problem ID [5468]
Internal file name [OUTPUT/4959_Wednesday_February_14_2024_02_06_00_AM_35920087/index.tex]
Book: Schaums Outline. Theory and problems of Differential Equations, 1st edition. Frank Ayres.
McGraw Hill 1952
Section: Chapter 26. Integration in series (singular points). Supplemetary problems. Page
218
Problem number: 20.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference not integer"
Maple gives the following as the ode type
[[_Emden, _Fowler]]
\[ \boxed {2 x^{3} y^{\prime \prime }+x^{2} y^{\prime }+y=0} \] 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 {2 \left (-\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 )\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. \[ 2 \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right ) t +3 \frac {d}{d t}y \left (t \right )+y \left (t \right ) = 0 \] 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 {3}{2 t}\\ q(t) &= \frac {1}{2 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 \[ 2 \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right ) t +3 \frac {d}{d t}y \left (t \right )+y \left (t \right ) = 0 \] 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} 2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} t^{n +r -2}\right ) t +3 \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} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}2 t^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 \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 }}2 t^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 \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 \[ 2 t^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+3 \left (n +r \right ) a_{n} t^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ 2 t^{-1+r} a_{0} r \left (-1+r \right )+3 r a_{0} t^{-1+r} = 0 \] Or \[ \left (2 t^{-1+r} r \left (-1+r \right )+3 r \,t^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (2 r^{2}+r \right ) t^{-1+r} = 0 \] Since the above is true for all \(t\) then the indicial equation becomes \[ 2 r^{2}+r = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 0\\ r_2 &= -{\frac {1}{2}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (2 r^{2}+r \right ) t^{-1+r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [0, -{\frac {1}{2}}\right ]\).
Since \(r_1 - r_2 = {\frac {1}{2}}\) is not an integer, then we can construct two linearly independent solutions \begin {align*} y_{1}\left (t \right ) &= t^{r_{1}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\right )\\ y_{2}\left (t \right ) &= t^{r_{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n}\right ) \end {align*}
Or \begin {align*} y_{1}\left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n}\\ y_{2}\left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} t^{n -\frac {1}{2}} \end {align*}
We start by finding \(y_{1}\left (t \right )\). 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} 2 a_{n} \left (n +r \right ) \left (n +r -1\right )+3 a_{n} \left (n +r \right )+a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -1}}{2 n^{2}+4 n r +2 r^{2}+n +r}\tag {4} \] Which for the root \(r = 0\) becomes \[ a_{n} = -\frac {a_{n -1}}{2 n^{2}+n}\tag {5} \] 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 \[ a_{1}=-\frac {1}{2 r^{2}+5 r +3} \] Which for the root \(r = 0\) becomes \[ a_{1}=-{\frac {1}{3}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{2 r^{2}+5 r +3}\) | \(-{\frac {1}{3}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {1}{4 r^{4}+28 r^{3}+71 r^{2}+77 r +30} \] Which for the root \(r = 0\) becomes \[ a_{2}={\frac {1}{30}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{2 r^{2}+5 r +3}\) | \(-{\frac {1}{3}}\) |
| \(a_{2}\) | \(\frac {1}{4 r^{4}+28 r^{3}+71 r^{2}+77 r +30}\) | \(\frac {1}{30}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {1}{8 r^{6}+108 r^{5}+590 r^{4}+1665 r^{3}+2552 r^{2}+2007 r +630} \] Which for the root \(r = 0\) becomes \[ a_{3}=-{\frac {1}{630}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{2 r^{2}+5 r +3}\) | \(-{\frac {1}{3}}\) |
| \(a_{2}\) | \(\frac {1}{4 r^{4}+28 r^{3}+71 r^{2}+77 r +30}\) | \(\frac {1}{30}\) |
| \(a_{3}\) | \(-\frac {1}{8 r^{6}+108 r^{5}+590 r^{4}+1665 r^{3}+2552 r^{2}+2007 r +630}\) | \(-{\frac {1}{630}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {1}{16 r^{8}+352 r^{7}+3304 r^{6}+17248 r^{5}+54649 r^{4}+107338 r^{3}+127251 r^{2}+82962 r +22680} \] Which for the root \(r = 0\) becomes \[ a_{4}={\frac {1}{22680}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{2 r^{2}+5 r +3}\) | \(-{\frac {1}{3}}\) |
| \(a_{2}\) | \(\frac {1}{4 r^{4}+28 r^{3}+71 r^{2}+77 r +30}\) | \(\frac {1}{30}\) |
| \(a_{3}\) | \(-\frac {1}{8 r^{6}+108 r^{5}+590 r^{4}+1665 r^{3}+2552 r^{2}+2007 r +630}\) | \(-{\frac {1}{630}}\) |
| \(a_{4}\) | \(\frac {1}{16 r^{8}+352 r^{7}+3304 r^{6}+17248 r^{5}+54649 r^{4}+107338 r^{3}+127251 r^{2}+82962 r +22680}\) | \(\frac {1}{22680}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {1}{32 r^{10}+1040 r^{9}+14880 r^{8}+123240 r^{7}+653226 r^{6}+2310945 r^{5}+5514295 r^{4}+8741785 r^{3}+8786367 r^{2}+5039190 r +1247400} \] Which for the root \(r = 0\) becomes \[ a_{5}=-{\frac {1}{1247400}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{2 r^{2}+5 r +3}\) | \(-{\frac {1}{3}}\) |
| \(a_{2}\) | \(\frac {1}{4 r^{4}+28 r^{3}+71 r^{2}+77 r +30}\) | \(\frac {1}{30}\) |
| \(a_{3}\) | \(-\frac {1}{8 r^{6}+108 r^{5}+590 r^{4}+1665 r^{3}+2552 r^{2}+2007 r +630}\) | \(-{\frac {1}{630}}\) |
| \(a_{4}\) | \(\frac {1}{16 r^{8}+352 r^{7}+3304 r^{6}+17248 r^{5}+54649 r^{4}+107338 r^{3}+127251 r^{2}+82962 r +22680}\) | \(\frac {1}{22680}\) |
| \(a_{5}\) | \(-\frac {1}{32 r^{10}+1040 r^{9}+14880 r^{8}+123240 r^{7}+653226 r^{6}+2310945 r^{5}+5514295 r^{4}+8741785 r^{3}+8786367 r^{2}+5039190 r +1247400}\) | \(-{\frac {1}{1247400}}\) |
Using the above table, then the solution \(y_{1}\left (t \right )\) is \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}\dots \\ &= 1-\frac {t}{3}+\frac {t^{2}}{30}-\frac {t^{3}}{630}+\frac {t^{4}}{22680}-\frac {t^{5}}{1247400}+O\left (t^{6}\right ) \end {align*}
Now the second solution \(y_{2}\left (t \right )\) is found. Eq (2B) derived above is now used to find all \(b_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(b_{0}\) is arbitrary and taken as \(b_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} 2 b_{n} \left (n +r \right ) \left (n +r -1\right )+3 \left (n +r \right ) b_{n}+b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {b_{n -1}}{2 n^{2}+4 n r +2 r^{2}+n +r}\tag {4} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ b_{n} = -\frac {b_{n -1}}{n \left (2 n -1\right )}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -{\frac {1}{2}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ b_{1}=-\frac {1}{2 r^{2}+5 r +3} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ b_{1}=-1 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {1}{2 r^{2}+5 r +3}\) | \(-1\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {1}{4 r^{4}+28 r^{3}+71 r^{2}+77 r +30} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ b_{2}={\frac {1}{6}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {1}{2 r^{2}+5 r +3}\) | \(-1\) |
| \(b_{2}\) | \(\frac {1}{4 r^{4}+28 r^{3}+71 r^{2}+77 r +30}\) | \(\frac {1}{6}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=-\frac {1}{8 r^{6}+108 r^{5}+590 r^{4}+1665 r^{3}+2552 r^{2}+2007 r +630} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ b_{3}=-{\frac {1}{90}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {1}{2 r^{2}+5 r +3}\) | \(-1\) |
| \(b_{2}\) | \(\frac {1}{4 r^{4}+28 r^{3}+71 r^{2}+77 r +30}\) | \(\frac {1}{6}\) |
| \(b_{3}\) | \(-\frac {1}{8 r^{6}+108 r^{5}+590 r^{4}+1665 r^{3}+2552 r^{2}+2007 r +630}\) | \(-{\frac {1}{90}}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {1}{16 r^{8}+352 r^{7}+3304 r^{6}+17248 r^{5}+54649 r^{4}+107338 r^{3}+127251 r^{2}+82962 r +22680} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ b_{4}={\frac {1}{2520}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {1}{2 r^{2}+5 r +3}\) | \(-1\) |
| \(b_{2}\) | \(\frac {1}{4 r^{4}+28 r^{3}+71 r^{2}+77 r +30}\) | \(\frac {1}{6}\) |
| \(b_{3}\) | \(-\frac {1}{8 r^{6}+108 r^{5}+590 r^{4}+1665 r^{3}+2552 r^{2}+2007 r +630}\) | \(-{\frac {1}{90}}\) |
| \(b_{4}\) | \(\frac {1}{16 r^{8}+352 r^{7}+3304 r^{6}+17248 r^{5}+54649 r^{4}+107338 r^{3}+127251 r^{2}+82962 r +22680}\) | \(\frac {1}{2520}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-\frac {1}{32 r^{10}+1040 r^{9}+14880 r^{8}+123240 r^{7}+653226 r^{6}+2310945 r^{5}+5514295 r^{4}+8741785 r^{3}+8786367 r^{2}+5039190 r +1247400} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ b_{5}=-{\frac {1}{113400}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {1}{2 r^{2}+5 r +3}\) | \(-1\) |
| \(b_{2}\) | \(\frac {1}{4 r^{4}+28 r^{3}+71 r^{2}+77 r +30}\) | \(\frac {1}{6}\) |
| \(b_{3}\) | \(-\frac {1}{8 r^{6}+108 r^{5}+590 r^{4}+1665 r^{3}+2552 r^{2}+2007 r +630}\) | \(-{\frac {1}{90}}\) |
| \(b_{4}\) | \(\frac {1}{16 r^{8}+352 r^{7}+3304 r^{6}+17248 r^{5}+54649 r^{4}+107338 r^{3}+127251 r^{2}+82962 r +22680}\) | \(\frac {1}{2520}\) |
| \(b_{5}\) | \(-\frac {1}{32 r^{10}+1040 r^{9}+14880 r^{8}+123240 r^{7}+653226 r^{6}+2310945 r^{5}+5514295 r^{4}+8741785 r^{3}+8786367 r^{2}+5039190 r +1247400}\) | \(-{\frac {1}{113400}}\) |
Using the above table, then the solution \(y_{2}\left (t \right )\) is \begin {align*} y_{2}\left (t \right )&= 1 \left (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}\dots \right ) \\ &= \frac {1-t +\frac {t^{2}}{6}-\frac {t^{3}}{90}+\frac {t^{4}}{2520}-\frac {t^{5}}{113400}+O\left (t^{6}\right )}{\sqrt {t}} \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 (1-\frac {t}{3}+\frac {t^{2}}{30}-\frac {t^{3}}{630}+\frac {t^{4}}{22680}-\frac {t^{5}}{1247400}+O\left (t^{6}\right )\right ) + \frac {c_{2} \left (1-t +\frac {t^{2}}{6}-\frac {t^{3}}{90}+\frac {t^{4}}{2520}-\frac {t^{5}}{113400}+O\left (t^{6}\right )\right )}{\sqrt {t}} \\ \end{align*} Hence the final solution is \begin{align*} y \left (t \right ) &= y_h \\ &= c_{1} \left (1-\frac {t}{3}+\frac {t^{2}}{30}-\frac {t^{3}}{630}+\frac {t^{4}}{22680}-\frac {t^{5}}{1247400}+O\left (t^{6}\right )\right )+\frac {c_{2} \left (1-t +\frac {t^{2}}{6}-\frac {t^{3}}{90}+\frac {t^{4}}{2520}-\frac {t^{5}}{113400}+O\left (t^{6}\right )\right )}{\sqrt {t}} \\ \end{align*} Replacing \(t\) by \(\frac {1}{x}\) gives \begin {align*} y = c_{1} \left (1-\frac {1}{3 x}+\frac {1}{30 x^{2}}-\frac {1}{630 x^{3}}+\frac {1}{22680 x^{4}}-\frac {1}{1247400 x^{5}}+O\left (\frac {1}{x^{6}}\right )\right )+\frac {c_{2} \left (1-\frac {1}{x}+\frac {1}{6 x^{2}}-\frac {1}{90 x^{3}}+\frac {1}{2520 x^{4}}-\frac {1}{113400 x^{5}}+O\left (\frac {1}{x^{6}}\right )\right )}{\sqrt {\frac {1}{x}}} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (1-\frac {1}{3 x}+\frac {1}{30 x^{2}}-\frac {1}{630 x^{3}}+\frac {1}{22680 x^{4}}-\frac {1}{1247400 x^{5}}+O\left (\frac {1}{x^{6}}\right )\right )+\frac {c_{2} \left (1-\frac {1}{x}+\frac {1}{6 x^{2}}-\frac {1}{90 x^{3}}+\frac {1}{2520 x^{4}}-\frac {1}{113400 x^{5}}+O\left (\frac {1}{x^{6}}\right )\right )}{\sqrt {\frac {1}{x}}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (1-\frac {1}{3 x}+\frac {1}{30 x^{2}}-\frac {1}{630 x^{3}}+\frac {1}{22680 x^{4}}-\frac {1}{1247400 x^{5}}+O\left (\frac {1}{x^{6}}\right )\right )+\frac {c_{2} \left (1-\frac {1}{x}+\frac {1}{6 x^{2}}-\frac {1}{90 x^{3}}+\frac {1}{2520 x^{4}}-\frac {1}{113400 x^{5}}+O\left (\frac {1}{x^{6}}\right )\right )}{\sqrt {\frac {1}{x}}} \] Verified OK.
Maple trace Kovacic algorithm successful
`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 A Liouvillian solution exists Group is reducible or imprimitive <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 117
Order:=6; dsolve(2*x^3*diff(y(x),x$2)+x^2*diff(y(x),x)+y(x)=0,y(x),type='series',x=infinity);
\[ y \left (x \right ) = \left (1-\frac {\left (x -\operatorname {Infinity} \right )^{2}}{4 \operatorname {Infinity}^{3}}+\frac {7 \left (x -\operatorname {Infinity} \right )^{3}}{24 \operatorname {Infinity}^{4}}+\frac {\left (-59 \operatorname {Infinity} +2\right ) \left (x -\operatorname {Infinity} \right )^{4}}{192 \operatorname {Infinity}^{6}}+\frac {\left (605 \operatorname {Infinity} -52\right ) \left (x -\operatorname {Infinity} \right )^{5}}{1920 \operatorname {Infinity}^{7}}\right ) y \left (\operatorname {Infinity} \right )+\left (x -\operatorname {Infinity} -\frac {\left (x -\operatorname {Infinity} \right )^{2}}{4 \operatorname {Infinity}}+\frac {\left (3 \operatorname {Infinity}^{2}-2 \operatorname {Infinity} \right ) \left (x -\operatorname {Infinity} \right )^{3}}{24 \operatorname {Infinity}^{4}}-\frac {5 \left (\operatorname {Infinity} -\frac {28}{15}\right ) \left (x -\operatorname {Infinity} \right )^{4}}{64 \operatorname {Infinity}^{4}}+\frac {\left (105 \operatorname {Infinity}^{3}-370 \operatorname {Infinity}^{2}+4 \operatorname {Infinity} \right ) \left (x -\operatorname {Infinity} \right )^{5}}{1920 \operatorname {Infinity}^{7}}\right ) D\left (y \right )\left (\operatorname {Infinity} \right )+O\left (x^{6}\right ) \]
✓ Solution by Mathematica
Time used: 0.003 (sec). Leaf size: 96
AsymptoticDSolveValue[2*x^3*y''[x]+x^2*y'[x]+y[x]==0,y[x],{x,infinity,5}]
\[ y(x)\to c_2 \left (\frac {1}{6 x^{3/2}}-\frac {1}{90 x^{5/2}}+\frac {1}{2520 x^{7/2}}-\frac {1}{113400 x^{9/2}}+\sqrt {x}-\frac {1}{\sqrt {x}}\right )+c_1 \left (-\frac {1}{1247400 x^5}+\frac {1}{22680 x^4}-\frac {1}{630 x^3}+\frac {1}{30 x^2}-\frac {1}{3 x}+1\right ) \]