Internal problem ID [5515]
Internal file name [OUTPUT/4763_Sunday_June_05_2022_03_05_10_PM_89001034/index.tex]
Book: Notes on Diffy Qs. Differential Equations for Engineers. By by Jiri Lebl, 2013.
Section: Chapter 7. POWER SERIES METHODS. 7.3.2 The method of Frobenius. Exercises.
page 300
Problem number: 7.3.3.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Complex roots"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x^{2} y^{\prime \prime }+y^{\prime } x +\left (1+x \right ) y=0} \] 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. \[ x^{2} y^{\prime \prime }+y^{\prime } x +\left (1+x \right ) y = 0 \] 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 {1}{x}\\ q(x) &= \frac {1+x}{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 \[ x^{2} y^{\prime \prime }+y^{\prime } x +\left (1+x \right ) y = 0 \] 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} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right ) x^{2}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right ) x +\left (1+x \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 }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n}\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} &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} 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 }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+x^{n +r} a_{n} \left (n +r \right )+a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )+x^{r} a_{0} r +a_{0} x^{r} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )+x^{r} r +x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r^{2}+1\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r^{2}+1 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= i\\ r_2 &= -i \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r^{2}+1\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([i, -i]\).
Since the roots are complex conjugates, then two linearly independent solutions can be constructed using \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 +i}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -i} \end {align*}
\(y_{1}\left (x \right )\) is found first. 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}+a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -1}}{n^{2}+2 n r +r^{2}+1}\tag {4} \] Which for the root \(r = i\) becomes \[ a_{n} = -\frac {a_{n -1}}{n \left (2 i+n \right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = i\) 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 \[ a_{1}=-\frac {1}{r^{2}+2 r +2} \] Which for the root \(r = i\) becomes \[ a_{1}=-\frac {1}{5}+\frac {2 i}{5} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{r^{2}+2 r +2}\) | \(-\frac {1}{5}+\frac {2 i}{5}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {1}{\left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right )} \] Which for the root \(r = i\) becomes \[ a_{2}=-\frac {1}{40}-\frac {3 i}{40} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{r^{2}+2 r +2}\) | \(-\frac {1}{5}+\frac {2 i}{5}\) |
| \(a_{2}\) | \(\frac {1}{\left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right )}\) | \(-\frac {1}{40}-\frac {3 i}{40}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {1}{\left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right )} \] Which for the root \(r = i\) becomes \[ a_{3}=\frac {3}{520}+\frac {7 i}{1560} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{r^{2}+2 r +2}\) | \(-\frac {1}{5}+\frac {2 i}{5}\) |
| \(a_{2}\) | \(\frac {1}{\left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right )}\) | \(-\frac {1}{40}-\frac {3 i}{40}\) |
| \(a_{3}\) | \(-\frac {1}{\left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right )}\) | \(\frac {3}{520}+\frac {7 i}{1560}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {1}{\left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right )} \] Which for the root \(r = i\) becomes \[ a_{4}=-\frac {1}{2496}-\frac {i}{12480} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{r^{2}+2 r +2}\) | \(-\frac {1}{5}+\frac {2 i}{5}\) |
| \(a_{2}\) | \(\frac {1}{\left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right )}\) | \(-\frac {1}{40}-\frac {3 i}{40}\) |
| \(a_{3}\) | \(-\frac {1}{\left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right )}\) | \(\frac {3}{520}+\frac {7 i}{1560}\) |
| \(a_{4}\) | \(\frac {1}{\left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right )}\) | \(-\frac {1}{2496}-\frac {i}{12480}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {1}{\left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right ) \left (r^{2}+10 r +26\right )} \] Which for the root \(r = i\) becomes \[ a_{5}=\frac {9}{603200}-\frac {i}{361920} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{r^{2}+2 r +2}\) | \(-\frac {1}{5}+\frac {2 i}{5}\) |
| \(a_{2}\) | \(\frac {1}{\left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right )}\) | \(-\frac {1}{40}-\frac {3 i}{40}\) |
| \(a_{3}\) | \(-\frac {1}{\left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right )}\) | \(\frac {3}{520}+\frac {7 i}{1560}\) |
| \(a_{4}\) | \(\frac {1}{\left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right )}\) | \(-\frac {1}{2496}-\frac {i}{12480}\) |
| \(a_{5}\) | \(-\frac {1}{\left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right ) \left (r^{2}+10 r +26\right )}\) | \(\frac {9}{603200}-\frac {i}{361920}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin{align*} y_{1}\left (x \right )&= x^{i} \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}\dots \right ) \\ &= x^{i} \left (1+\left (-\frac {1}{5}+\frac {2 i}{5}\right ) x +\left (-\frac {1}{40}-\frac {3 i}{40}\right ) x^{2}+\left (\frac {3}{520}+\frac {7 i}{1560}\right ) x^{3}+\left (-\frac {1}{2496}-\frac {i}{12480}\right ) x^{4}+\left (\frac {9}{603200}-\frac {i}{361920}\right ) x^{5}+O\left (x^{6}\right )\right ) \\ \end{align*} The second solution \(y_{2}\left (x \right )\) is found by taking the complex conjugate of \(y_{1}\left (x \right )\) which gives \[ y_{2}\left (x \right )= x^{-i} \left (1+\left (-\frac {1}{5}-\frac {2 i}{5}\right ) x +\left (-\frac {1}{40}+\frac {3 i}{40}\right ) x^{2}+\left (\frac {3}{520}-\frac {7 i}{1560}\right ) x^{3}+\left (-\frac {1}{2496}+\frac {i}{12480}\right ) x^{4}+\left (\frac {9}{603200}+\frac {i}{361920}\right ) x^{5}+O\left (x^{6}\right )\right ) \] 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} x^{i} \left (1+\left (-\frac {1}{5}+\frac {2 i}{5}\right ) x +\left (-\frac {1}{40}-\frac {3 i}{40}\right ) x^{2}+\left (\frac {3}{520}+\frac {7 i}{1560}\right ) x^{3}+\left (-\frac {1}{2496}-\frac {i}{12480}\right ) x^{4}+\left (\frac {9}{603200}-\frac {i}{361920}\right ) x^{5}+O\left (x^{6}\right )\right ) + c_{2} x^{-i} \left (1+\left (-\frac {1}{5}-\frac {2 i}{5}\right ) x +\left (-\frac {1}{40}+\frac {3 i}{40}\right ) x^{2}+\left (\frac {3}{520}-\frac {7 i}{1560}\right ) x^{3}+\left (-\frac {1}{2496}+\frac {i}{12480}\right ) x^{4}+\left (\frac {9}{603200}+\frac {i}{361920}\right ) x^{5}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{i} \left (1+\left (-\frac {1}{5}+\frac {2 i}{5}\right ) x +\left (-\frac {1}{40}-\frac {3 i}{40}\right ) x^{2}+\left (\frac {3}{520}+\frac {7 i}{1560}\right ) x^{3}+\left (-\frac {1}{2496}-\frac {i}{12480}\right ) x^{4}+\left (\frac {9}{603200}-\frac {i}{361920}\right ) x^{5}+O\left (x^{6}\right )\right )+c_{2} x^{-i} \left (1+\left (-\frac {1}{5}-\frac {2 i}{5}\right ) x +\left (-\frac {1}{40}+\frac {3 i}{40}\right ) x^{2}+\left (\frac {3}{520}-\frac {7 i}{1560}\right ) x^{3}+\left (-\frac {1}{2496}+\frac {i}{12480}\right ) x^{4}+\left (\frac {9}{603200}+\frac {i}{361920}\right ) x^{5}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{i} \left (1+\left (-\frac {1}{5}+\frac {2 i}{5}\right ) x +\left (-\frac {1}{40}-\frac {3 i}{40}\right ) x^{2}+\left (\frac {3}{520}+\frac {7 i}{1560}\right ) x^{3}+\left (-\frac {1}{2496}-\frac {i}{12480}\right ) x^{4}+\left (\frac {9}{603200}-\frac {i}{361920}\right ) x^{5}+O\left (x^{6}\right )\right )+c_{2} x^{-i} \left (1+\left (-\frac {1}{5}-\frac {2 i}{5}\right ) x +\left (-\frac {1}{40}+\frac {3 i}{40}\right ) x^{2}+\left (\frac {3}{520}-\frac {7 i}{1560}\right ) x^{3}+\left (-\frac {1}{2496}+\frac {i}{12480}\right ) x^{4}+\left (\frac {9}{603200}+\frac {i}{361920}\right ) x^{5}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{i} \left (1+\left (-\frac {1}{5}+\frac {2 i}{5}\right ) x +\left (-\frac {1}{40}-\frac {3 i}{40}\right ) x^{2}+\left (\frac {3}{520}+\frac {7 i}{1560}\right ) x^{3}+\left (-\frac {1}{2496}-\frac {i}{12480}\right ) x^{4}+\left (\frac {9}{603200}-\frac {i}{361920}\right ) x^{5}+O\left (x^{6}\right )\right )+c_{2} x^{-i} \left (1+\left (-\frac {1}{5}-\frac {2 i}{5}\right ) x +\left (-\frac {1}{40}+\frac {3 i}{40}\right ) x^{2}+\left (\frac {3}{520}-\frac {7 i}{1560}\right ) x^{3}+\left (-\frac {1}{2496}+\frac {i}{12480}\right ) x^{4}+\left (\frac {9}{603200}+\frac {i}{361920}\right ) x^{5}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) x^{2}+y^{\prime } x +\left (1+x \right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=-\frac {y^{\prime }}{x}-\frac {\left (1+x \right ) y}{x^{2}} \\ \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} \\ {} & {} & \frac {d}{d x}y^{\prime }+\frac {y^{\prime }}{x}+\frac {\left (1+x \right ) y}{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 {1}{x}, P_{3}\left (x \right )=\frac {1+x}{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}}}=1 \\ {} & \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} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) x^{2}+y^{\prime } x +\left (1+x \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..1 \\ {} & {} & 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 \cdot y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{2}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\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 (r^{2}+1\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (k^{2}+2 k r +r^{2}+1\right )+a_{k -1}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r^{2}+1=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{\mathrm {-I}, \mathrm {I}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (k^{2}+2 k r +r^{2}+1\right )+a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +1} \left (\left (k +1\right )^{2}+2 \left (k +1\right ) r +r^{2}+1\right )+a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {a_{k}}{k^{2}+2 k r +r^{2}+2 k +2 r +2} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\mathrm {-I} \\ {} & {} & a_{k +1}=-\frac {a_{k}}{k^{2}-2 \,\mathrm {I} k +1-2 \,\mathrm {I}+2 k} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\mathrm {-I} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\mathrm {I}}, a_{k +1}=-\frac {a_{k}}{k^{2}-2 \,\mathrm {I} k +1-2 \,\mathrm {I}+2 k}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\mathrm {I} \\ {} & {} & a_{k +1}=-\frac {a_{k}}{k^{2}+2 \,\mathrm {I} k +1+2 \,\mathrm {I}+2 k} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\mathrm {I} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\mathrm {I}}, a_{k +1}=-\frac {a_{k}}{k^{2}+2 \,\mathrm {I} k +1+2 \,\mathrm {I}+2 k}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\mathrm {I}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\mathrm {I}}\right ), a_{k +1}=-\frac {a_{k}}{k^{2}-2 \,\mathrm {I} k +1-2 \,\mathrm {I}+2 k}, b_{k +1}=-\frac {b_{k}}{k^{2}+2 \,\mathrm {I} k +1+2 \,\mathrm {I}+2 k}\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 <- Bessel successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 69
Order:=6; dsolve(x^2*diff(y(x),x$2)+x*diff(y(x),x)+(1+x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{-i} \left (1+\left (-\frac {1}{5}-\frac {2 i}{5}\right ) x +\left (-\frac {1}{40}+\frac {3 i}{40}\right ) x^{2}+\left (\frac {3}{520}-\frac {7 i}{1560}\right ) x^{3}+\left (-\frac {1}{2496}+\frac {i}{12480}\right ) x^{4}+\left (\frac {9}{603200}+\frac {i}{361920}\right ) x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} x^{i} \left (1+\left (-\frac {1}{5}+\frac {2 i}{5}\right ) x +\left (-\frac {1}{40}-\frac {3 i}{40}\right ) x^{2}+\left (\frac {3}{520}+\frac {7 i}{1560}\right ) x^{3}+\left (-\frac {1}{2496}-\frac {i}{12480}\right ) x^{4}+\left (\frac {9}{603200}-\frac {i}{361920}\right ) x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.01 (sec). Leaf size: 90
AsymptoticDSolveValue[x^2*y''[x]+x*y'[x]+(1+x)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to \left (\frac {1}{12480}+\frac {i}{2496}\right ) c_2 x^{-i} \left (i x^4-(8+16 i) x^3+(168+96 i) x^2-(1056-288 i) x+(480-2400 i)\right )-\left (\frac {1}{2496}+\frac {i}{12480}\right ) c_1 x^i \left (x^4-(16+8 i) x^3+(96+168 i) x^2+(288-1056 i) x-(2400-480 i)\right ) \]