Internal problem ID [6053]
Internal file name [OUTPUT/5301_Sunday_June_05_2022_03_33_12_PM_94404250/index.tex]
Book: An introduction to Ordinary Differential Equations. Earl A. Coddington. Dover. NY
1961
Section: Chapter 4. Linear equations with Regular Singular Points. Page 166
Problem number: 1(i).
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
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {2 x^{2} y^{\prime \prime }+\left (x^{2}+5 x \right ) y^{\prime }+\left (x^{2}-2\right ) y=0} \] With the expansion point for the power series method at \(x = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ 2 x^{2} y^{\prime \prime }+\left (x^{2}+5 x \right ) y^{\prime }+\left (x^{2}-2\right ) y = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} y^{\prime \prime }+p(x) y^{\prime } + q(x) y &=0 \end {align*}
Where \begin {align*} p(x) &= \frac {x +5}{2 x}\\ q(x) &= \frac {x^{2}-2}{2 x^{2}}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ 2 x^{2} y^{\prime \prime }+\left (x^{2}+5 x \right ) y^{\prime }+\left (x^{2}-2\right ) y = 0 \] Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} \] Then \begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} 2 x^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (x^{2}+5 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (x^{2}-2\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}5 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r +2} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 a_{n} x^{n +r}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r +2} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} x^{n +r} \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}5 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+5 x^{n +r} a_{n} \left (n +r \right )-2 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 2 x^{r} a_{0} r \left (-1+r \right )+5 x^{r} a_{0} r -2 a_{0} x^{r} = 0 \] Or \[ \left (2 x^{r} r \left (-1+r \right )+5 x^{r} r -2 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (2 r^{2}+3 r -2\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 2 r^{2}+3 r -2 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {1}{2}}\\ r_2 &= -2 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (2 r^{2}+3 r -2\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {1}{2}}, -2\right ]\).
Since \(r_1 - r_2 = {\frac {5}{2}}\) is not an integer, then we can construct two linearly independent solutions \begin {align*} y_{1}\left (x \right ) &= x^{r_{1}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= x^{r_{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +\frac {1}{2}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -2} \end {align*}
We start by finding \(y_{1}\left (x \right )\). 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 \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = -\frac {r}{2 r^{2}+7 r +3} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} 2 a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n -1} \left (n +r -1\right )+5 a_{n} \left (n +r \right )+a_{n -2}-2 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {n a_{n -1}+r a_{n -1}+a_{n -2}-a_{n -1}}{2 n^{2}+4 n r +2 r^{2}+3 n +3 r -2}\tag {4} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{n} = \frac {-2 n a_{n -1}-2 a_{n -2}+a_{n -1}}{4 n^{2}+10 n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {1}{2}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r}{2 r^{2}+7 r +3}\) | \(-{\frac {1}{14}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {-r^{2}-6 r -3}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{2}=-{\frac {25}{504}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r}{2 r^{2}+7 r +3}\) | \(-{\frac {1}{14}}\) |
| \(a_{2}\) | \(\frac {-r^{2}-6 r -3}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right )}\) | \(-{\frac {25}{504}}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {3 r^{3}+19 r^{2}+27 r +6}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{3}={\frac {197}{33264}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r}{2 r^{2}+7 r +3}\) | \(-{\frac {1}{14}}\) |
| \(a_{2}\) | \(\frac {-r^{2}-6 r -3}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right )}\) | \(-{\frac {25}{504}}\) |
| \(a_{3}\) | \(\frac {3 r^{3}+19 r^{2}+27 r +6}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right )}\) | \(\frac {197}{33264}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {-r^{4}-r^{3}+37 r^{2}+108 r +57}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right ) \left (2 r^{2}+19 r +42\right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{4}={\frac {1921}{3459456}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r}{2 r^{2}+7 r +3}\) | \(-{\frac {1}{14}}\) |
| \(a_{2}\) | \(\frac {-r^{2}-6 r -3}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right )}\) | \(-{\frac {25}{504}}\) |
| \(a_{3}\) | \(\frac {3 r^{3}+19 r^{2}+27 r +6}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right )}\) | \(\frac {197}{33264}\) |
| \(a_{4}\) | \(\frac {-r^{4}-r^{3}+37 r^{2}+108 r +57}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right ) \left (2 r^{2}+19 r +42\right )}\) | \(\frac {1921}{3459456}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-5 r^{5}-90 r^{4}-574 r^{3}-1579 r^{2}-1737 r -480}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right ) \left (2 r^{2}+19 r +42\right ) \left (2 r^{2}+23 r +63\right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{5}=-{\frac {11653}{103783680}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r}{2 r^{2}+7 r +3}\) | \(-{\frac {1}{14}}\) |
| \(a_{2}\) | \(\frac {-r^{2}-6 r -3}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right )}\) | \(-{\frac {25}{504}}\) |
| \(a_{3}\) | \(\frac {3 r^{3}+19 r^{2}+27 r +6}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right )}\) | \(\frac {197}{33264}\) |
| \(a_{4}\) | \(\frac {-r^{4}-r^{3}+37 r^{2}+108 r +57}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right ) \left (2 r^{2}+19 r +42\right )}\) | \(\frac {1921}{3459456}\) |
| \(a_{5}\) | \(\frac {-5 r^{5}-90 r^{4}-574 r^{3}-1579 r^{2}-1737 r -480}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right ) \left (2 r^{2}+19 r +42\right ) \left (2 r^{2}+23 r +63\right )}\) | \(-{\frac {11653}{103783680}}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {7 r^{6}+140 r^{5}+1036 r^{4}+3445 r^{3}+4703 r^{2}+1050 r -1191}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right ) \left (2 r^{2}+19 r +42\right ) \left (2 r^{2}+23 r +63\right ) \left (2 r^{2}+27 r +88\right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{6}={\frac {12923}{21171870720}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r}{2 r^{2}+7 r +3}\) | \(-{\frac {1}{14}}\) |
| \(a_{2}\) | \(\frac {-r^{2}-6 r -3}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right )}\) | \(-{\frac {25}{504}}\) |
| \(a_{3}\) | \(\frac {3 r^{3}+19 r^{2}+27 r +6}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right )}\) | \(\frac {197}{33264}\) |
| \(a_{4}\) | \(\frac {-r^{4}-r^{3}+37 r^{2}+108 r +57}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right ) \left (2 r^{2}+19 r +42\right )}\) | \(\frac {1921}{3459456}\) |
| \(a_{5}\) | \(\frac {-5 r^{5}-90 r^{4}-574 r^{3}-1579 r^{2}-1737 r -480}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right ) \left (2 r^{2}+19 r +42\right ) \left (2 r^{2}+23 r +63\right )}\) | \(-{\frac {11653}{103783680}}\) |
| \(a_{6}\) | \(\frac {7 r^{6}+140 r^{5}+1036 r^{4}+3445 r^{3}+4703 r^{2}+1050 r -1191}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right ) \left (2 r^{2}+19 r +42\right ) \left (2 r^{2}+23 r +63\right ) \left (2 r^{2}+27 r +88\right )}\) | \(\frac {12923}{21171870720}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=\frac {3 r^{7}+133 r^{6}+2142 r^{5}+16915 r^{4}+71246 r^{3}+157543 r^{2}+160707 r +49386}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right ) \left (2 r^{2}+19 r +42\right ) \left (2 r^{2}+23 r +63\right ) \left (2 r^{2}+27 r +88\right ) \left (2 r^{2}+31 r +117\right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{7}={\frac {917285}{1126343522304}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r}{2 r^{2}+7 r +3}\) | \(-{\frac {1}{14}}\) |
| \(a_{2}\) | \(\frac {-r^{2}-6 r -3}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right )}\) | \(-{\frac {25}{504}}\) |
| \(a_{3}\) | \(\frac {3 r^{3}+19 r^{2}+27 r +6}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right )}\) | \(\frac {197}{33264}\) |
| \(a_{4}\) | \(\frac {-r^{4}-r^{3}+37 r^{2}+108 r +57}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right ) \left (2 r^{2}+19 r +42\right )}\) | \(\frac {1921}{3459456}\) |
| \(a_{5}\) | \(\frac {-5 r^{5}-90 r^{4}-574 r^{3}-1579 r^{2}-1737 r -480}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right ) \left (2 r^{2}+19 r +42\right ) \left (2 r^{2}+23 r +63\right )}\) | \(-{\frac {11653}{103783680}}\) |
| \(a_{6}\) | \(\frac {7 r^{6}+140 r^{5}+1036 r^{4}+3445 r^{3}+4703 r^{2}+1050 r -1191}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right ) \left (2 r^{2}+19 r +42\right ) \left (2 r^{2}+23 r +63\right ) \left (2 r^{2}+27 r +88\right )}\) | \(\frac {12923}{21171870720}\) |
| \(a_{7}\) | \(\frac {3 r^{7}+133 r^{6}+2142 r^{5}+16915 r^{4}+71246 r^{3}+157543 r^{2}+160707 r +49386}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right ) \left (2 r^{2}+19 r +42\right ) \left (2 r^{2}+23 r +63\right ) \left (2 r^{2}+27 r +88\right ) \left (2 r^{2}+31 r +117\right )}\) | \(\frac {917285}{1126343522304}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= \sqrt {x} \left (a_{0}+a_{1} x +a_{2} x^{2}+a_{3} x^{3}+a_{4} x^{4}+a_{5} x^{5}+a_{6} x^{6}+a_{7} x^{7}+a_{8} x^{8}\dots \right ) \\ &= \sqrt {x}\, \left (1-\frac {x}{14}-\frac {25 x^{2}}{504}+\frac {197 x^{3}}{33264}+\frac {1921 x^{4}}{3459456}-\frac {11653 x^{5}}{103783680}+\frac {12923 x^{6}}{21171870720}+\frac {917285 x^{7}}{1126343522304}+O\left (x^{8}\right )\right ) \end {align*}
Now the second solution \(y_{2}\left (x \right )\) 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 \(b_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ b_{1} = -\frac {r}{2 r^{2}+7 r +3} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} 2 b_{n} \left (n +r \right ) \left (n +r -1\right )+b_{n -1} \left (n +r -1\right )+5 b_{n} \left (n +r \right )+b_{n -2}-2 b_{n} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {n b_{n -1}+r b_{n -1}+b_{n -2}-b_{n -1}}{2 n^{2}+4 n r +2 r^{2}+3 n +3 r -2}\tag {4} \] Which for the root \(r = -2\) becomes \[ b_{n} = \frac {-n b_{n -1}-b_{n -2}+3 b_{n -1}}{n \left (2 n -5\right )}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -2\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {r}{2 r^{2}+7 r +3}\) | \(-{\frac {2}{3}}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {-r^{2}-6 r -3}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right )} \] Which for the root \(r = -2\) becomes \[ b_{2}={\frac {5}{6}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {r}{2 r^{2}+7 r +3}\) | \(-{\frac {2}{3}}\) |
| \(b_{2}\) | \(\frac {-r^{2}-6 r -3}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right )}\) | \(\frac {5}{6}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {3 r^{3}+19 r^{2}+27 r +6}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right )} \] Which for the root \(r = -2\) becomes \[ b_{3}={\frac {2}{9}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {r}{2 r^{2}+7 r +3}\) | \(-{\frac {2}{3}}\) |
| \(b_{2}\) | \(\frac {-r^{2}-6 r -3}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right )}\) | \(\frac {5}{6}\) |
| \(b_{3}\) | \(\frac {3 r^{3}+19 r^{2}+27 r +6}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right )}\) | \(\frac {2}{9}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {-r^{4}-r^{3}+37 r^{2}+108 r +57}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right ) \left (2 r^{2}+19 r +42\right )} \] Which for the root \(r = -2\) becomes \[ b_{4}=-{\frac {19}{216}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {r}{2 r^{2}+7 r +3}\) | \(-{\frac {2}{3}}\) |
| \(b_{2}\) | \(\frac {-r^{2}-6 r -3}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right )}\) | \(\frac {5}{6}\) |
| \(b_{3}\) | \(\frac {3 r^{3}+19 r^{2}+27 r +6}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right )}\) | \(\frac {2}{9}\) |
| \(b_{4}\) | \(\frac {-r^{4}-r^{3}+37 r^{2}+108 r +57}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right ) \left (2 r^{2}+19 r +42\right )}\) | \(-{\frac {19}{216}}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {-5 r^{5}-90 r^{4}-574 r^{3}-1579 r^{2}-1737 r -480}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right ) \left (2 r^{2}+19 r +42\right ) \left (2 r^{2}+23 r +63\right )} \] Which for the root \(r = -2\) becomes \[ b_{5}=-{\frac {1}{540}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {r}{2 r^{2}+7 r +3}\) | \(-{\frac {2}{3}}\) |
| \(b_{2}\) | \(\frac {-r^{2}-6 r -3}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right )}\) | \(\frac {5}{6}\) |
| \(b_{3}\) | \(\frac {3 r^{3}+19 r^{2}+27 r +6}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right )}\) | \(\frac {2}{9}\) |
| \(b_{4}\) | \(\frac {-r^{4}-r^{3}+37 r^{2}+108 r +57}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right ) \left (2 r^{2}+19 r +42\right )}\) | \(-{\frac {19}{216}}\) |
| \(b_{5}\) | \(\frac {-5 r^{5}-90 r^{4}-574 r^{3}-1579 r^{2}-1737 r -480}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right ) \left (2 r^{2}+19 r +42\right ) \left (2 r^{2}+23 r +63\right )}\) | \(-{\frac {1}{540}}\) |
For \(n = 6\), using the above recursive equation gives \[ b_{6}=\frac {7 r^{6}+140 r^{5}+1036 r^{4}+3445 r^{3}+4703 r^{2}+1050 r -1191}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right ) \left (2 r^{2}+19 r +42\right ) \left (2 r^{2}+23 r +63\right ) \left (2 r^{2}+27 r +88\right )} \] Which for the root \(r = -2\) becomes \[ b_{6}={\frac {101}{45360}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {r}{2 r^{2}+7 r +3}\) | \(-{\frac {2}{3}}\) |
| \(b_{2}\) | \(\frac {-r^{2}-6 r -3}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right )}\) | \(\frac {5}{6}\) |
| \(b_{3}\) | \(\frac {3 r^{3}+19 r^{2}+27 r +6}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right )}\) | \(\frac {2}{9}\) |
| \(b_{4}\) | \(\frac {-r^{4}-r^{3}+37 r^{2}+108 r +57}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right ) \left (2 r^{2}+19 r +42\right )}\) | \(-{\frac {19}{216}}\) |
| \(b_{5}\) | \(\frac {-5 r^{5}-90 r^{4}-574 r^{3}-1579 r^{2}-1737 r -480}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right ) \left (2 r^{2}+19 r +42\right ) \left (2 r^{2}+23 r +63\right )}\) | \(-{\frac {1}{540}}\) |
| \(b_{6}\) | \(\frac {7 r^{6}+140 r^{5}+1036 r^{4}+3445 r^{3}+4703 r^{2}+1050 r -1191}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right ) \left (2 r^{2}+19 r +42\right ) \left (2 r^{2}+23 r +63\right ) \left (2 r^{2}+27 r +88\right )}\) | \(\frac {101}{45360}\) |
For \(n = 7\), using the above recursive equation gives \[ b_{7}=\frac {3 r^{7}+133 r^{6}+2142 r^{5}+16915 r^{4}+71246 r^{3}+157543 r^{2}+160707 r +49386}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right ) \left (2 r^{2}+19 r +42\right ) \left (2 r^{2}+23 r +63\right ) \left (2 r^{2}+27 r +88\right ) \left (2 r^{2}+31 r +117\right )} \] Which for the root \(r = -2\) becomes \[ b_{7}=-{\frac {4}{35721}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {r}{2 r^{2}+7 r +3}\) | \(-{\frac {2}{3}}\) |
| \(b_{2}\) | \(\frac {-r^{2}-6 r -3}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right )}\) | \(\frac {5}{6}\) |
| \(b_{3}\) | \(\frac {3 r^{3}+19 r^{2}+27 r +6}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right )}\) | \(\frac {2}{9}\) |
| \(b_{4}\) | \(\frac {-r^{4}-r^{3}+37 r^{2}+108 r +57}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right ) \left (2 r^{2}+19 r +42\right )}\) | \(-{\frac {19}{216}}\) |
| \(b_{5}\) | \(\frac {-5 r^{5}-90 r^{4}-574 r^{3}-1579 r^{2}-1737 r -480}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right ) \left (2 r^{2}+19 r +42\right ) \left (2 r^{2}+23 r +63\right )}\) | \(-{\frac {1}{540}}\) |
| \(b_{6}\) | \(\frac {7 r^{6}+140 r^{5}+1036 r^{4}+3445 r^{3}+4703 r^{2}+1050 r -1191}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right ) \left (2 r^{2}+19 r +42\right ) \left (2 r^{2}+23 r +63\right ) \left (2 r^{2}+27 r +88\right )}\) | \(\frac {101}{45360}\) |
| \(b_{7}\) | \(\frac {3 r^{7}+133 r^{6}+2142 r^{5}+16915 r^{4}+71246 r^{3}+157543 r^{2}+160707 r +49386}{\left (2 r^{2}+7 r +3\right ) \left (2 r^{2}+11 r +12\right ) \left (2 r^{2}+15 r +25\right ) \left (2 r^{2}+19 r +42\right ) \left (2 r^{2}+23 r +63\right ) \left (2 r^{2}+27 r +88\right ) \left (2 r^{2}+31 r +117\right )}\) | \(-{\frac {4}{35721}}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= \sqrt {x} \left (b_{0}+b_{1} x +b_{2} x^{2}+b_{3} x^{3}+b_{4} x^{4}+b_{5} x^{5}+b_{6} x^{6}+b_{7} x^{7}+b_{8} x^{8}\dots \right ) \\ &= \frac {1-\frac {2 x}{3}+\frac {5 x^{2}}{6}+\frac {2 x^{3}}{9}-\frac {19 x^{4}}{216}-\frac {x^{5}}{540}+\frac {101 x^{6}}{45360}-\frac {4 x^{7}}{35721}+O\left (x^{8}\right )}{x^{2}} \end {align*}
Therefore the homogeneous solution is \begin{align*} y_h(x) &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= c_{1} \sqrt {x}\, \left (1-\frac {x}{14}-\frac {25 x^{2}}{504}+\frac {197 x^{3}}{33264}+\frac {1921 x^{4}}{3459456}-\frac {11653 x^{5}}{103783680}+\frac {12923 x^{6}}{21171870720}+\frac {917285 x^{7}}{1126343522304}+O\left (x^{8}\right )\right ) + \frac {c_{2} \left (1-\frac {2 x}{3}+\frac {5 x^{2}}{6}+\frac {2 x^{3}}{9}-\frac {19 x^{4}}{216}-\frac {x^{5}}{540}+\frac {101 x^{6}}{45360}-\frac {4 x^{7}}{35721}+O\left (x^{8}\right )\right )}{x^{2}} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \sqrt {x}\, \left (1-\frac {x}{14}-\frac {25 x^{2}}{504}+\frac {197 x^{3}}{33264}+\frac {1921 x^{4}}{3459456}-\frac {11653 x^{5}}{103783680}+\frac {12923 x^{6}}{21171870720}+\frac {917285 x^{7}}{1126343522304}+O\left (x^{8}\right )\right )+\frac {c_{2} \left (1-\frac {2 x}{3}+\frac {5 x^{2}}{6}+\frac {2 x^{3}}{9}-\frac {19 x^{4}}{216}-\frac {x^{5}}{540}+\frac {101 x^{6}}{45360}-\frac {4 x^{7}}{35721}+O\left (x^{8}\right )\right )}{x^{2}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \sqrt {x}\, \left (1-\frac {x}{14}-\frac {25 x^{2}}{504}+\frac {197 x^{3}}{33264}+\frac {1921 x^{4}}{3459456}-\frac {11653 x^{5}}{103783680}+\frac {12923 x^{6}}{21171870720}+\frac {917285 x^{7}}{1126343522304}+O\left (x^{8}\right )\right )+\frac {c_{2} \left (1-\frac {2 x}{3}+\frac {5 x^{2}}{6}+\frac {2 x^{3}}{9}-\frac {19 x^{4}}{216}-\frac {x^{5}}{540}+\frac {101 x^{6}}{45360}-\frac {4 x^{7}}{35721}+O\left (x^{8}\right )\right )}{x^{2}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} \sqrt {x}\, \left (1-\frac {x}{14}-\frac {25 x^{2}}{504}+\frac {197 x^{3}}{33264}+\frac {1921 x^{4}}{3459456}-\frac {11653 x^{5}}{103783680}+\frac {12923 x^{6}}{21171870720}+\frac {917285 x^{7}}{1126343522304}+O\left (x^{8}\right )\right )+\frac {c_{2} \left (1-\frac {2 x}{3}+\frac {5 x^{2}}{6}+\frac {2 x^{3}}{9}-\frac {19 x^{4}}{216}-\frac {x^{5}}{540}+\frac {101 x^{6}}{45360}-\frac {4 x^{7}}{35721}+O\left (x^{8}\right )\right )}{x^{2}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 x^{2} y^{\prime \prime }+\left (x^{2}+5 x \right ) y^{\prime }+\left (x^{2}-2\right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-\frac {\left (x^{2}-2\right ) y}{2 x^{2}}-\frac {\left (x +5\right ) y^{\prime }}{2 x} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime }+\frac {\left (x +5\right ) y^{\prime }}{2 x}+\frac {\left (x^{2}-2\right ) y}{2 x^{2}}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {x +5}{2 x}, P_{3}\left (x \right )=\frac {x^{2}-2}{2 x^{2}}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=\frac {5}{2} \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=-1 \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 2 x^{2} y^{\prime \prime }+x \left (x +5\right ) y^{\prime }+\left (x^{2}-2\right ) y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{2}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (2+r \right ) \left (-1+2 r \right ) x^{r}+\left (a_{1} \left (3+r \right ) \left (1+2 r \right )+a_{0} r \right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (k +r +2\right ) \left (2 k +2 r -1\right )+a_{k -1} \left (k +r -1\right )+a_{k -2}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (2+r \right ) \left (-1+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-2, \frac {1}{2}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (3+r \right ) \left (1+2 r \right )+a_{0} r =0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=-\frac {a_{0} r}{2 r^{2}+7 r +3} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 2 \left (k +r -\frac {1}{2}\right ) \left (k +r +2\right ) a_{k}+a_{k -1} k +a_{k -1} r +a_{k -2}-a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & 2 \left (k +\frac {3}{2}+r \right ) \left (k +4+r \right ) a_{k +2}+a_{k +1} \left (k +2\right )+a_{k +1} r +a_{k}-a_{k +1}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {k a_{k +1}+a_{k +1} r +a_{k}+a_{k +1}}{\left (2 k +3+2 r \right ) \left (k +4+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-2 \\ {} & {} & a_{k +2}=-\frac {k a_{k +1}+a_{k}-a_{k +1}}{\left (2 k -1\right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -2}, a_{k +2}=-\frac {k a_{k +1}+a_{k}-a_{k +1}}{\left (2 k -1\right ) \left (k +2\right )}, a_{1}=-\frac {2 a_{0}}{3}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & a_{k +2}=-\frac {k a_{k +1}+a_{k}+\frac {3}{2} a_{k +1}}{\left (2 k +4\right ) \left (k +\frac {9}{2}\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}}, a_{k +2}=-\frac {k a_{k +1}+a_{k}+\frac {3}{2} a_{k +1}}{\left (2 k +4\right ) \left (k +\frac {9}{2}\right )}, a_{1}=-\frac {a_{0}}{14}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -2}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {1}{2}}\right ), a_{k +2}=-\frac {k a_{k +1}+a_{k}-a_{k +1}}{\left (2 k -1\right ) \left (k +2\right )}, a_{1}=-\frac {2 a_{0}}{3}, b_{k +2}=-\frac {k b_{k +1}+b_{k}+\frac {3}{2} b_{k +1}}{\left (2 k +4\right ) \left (k +\frac {9}{2}\right )}, b_{1}=-\frac {b_{0}}{14}\right ] \end {array} \]
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 -> elliptic -> Legendre -> Whittaker -> hyper3: Equivalence to 1F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 1F1 ODE <- Whittaker successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 55
Order:=8; dsolve(2*x^2*diff(y(x),x$2)+(5*x+x^2)*diff(y(x),x)+(x^2-2)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{2} x^{\frac {5}{2}} \left (1-\frac {1}{14} x -\frac {25}{504} x^{2}+\frac {197}{33264} x^{3}+\frac {1921}{3459456} x^{4}-\frac {11653}{103783680} x^{5}+\frac {12923}{21171870720} x^{6}+\frac {917285}{1126343522304} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+c_{1} \left (1-\frac {2}{3} x +\frac {5}{6} x^{2}+\frac {2}{9} x^{3}-\frac {19}{216} x^{4}-\frac {1}{540} x^{5}+\frac {101}{45360} x^{6}-\frac {4}{35721} x^{7}+\operatorname {O}\left (x^{8}\right )\right )}{x^{2}} \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 116
AsymptoticDSolveValue[2*x^2*y''[x]+(5*x+x^2)*y'[x]+(x^2-2)*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_1 \sqrt {x} \left (\frac {917285 x^7}{1126343522304}+\frac {12923 x^6}{21171870720}-\frac {11653 x^5}{103783680}+\frac {1921 x^4}{3459456}+\frac {197 x^3}{33264}-\frac {25 x^2}{504}-\frac {x}{14}+1\right )+\frac {c_2 \left (-\frac {4 x^7}{35721}+\frac {101 x^6}{45360}-\frac {x^5}{540}-\frac {19 x^4}{216}+\frac {2 x^3}{9}+\frac {5 x^2}{6}-\frac {2 x}{3}+1\right )}{x^2} \]