Internal problem ID [6484]
Internal file name [OUTPUT/5732_Sunday_June_05_2022_03_49_30_PM_54862276/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. (A) Drill Exercises . Page 194
Problem number: 2(a).
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]]
\[ \boxed {\left (x^{2}+1\right ) x^{2} y^{\prime \prime }-y^{\prime } x +\left (x +2\right ) y=0} \] 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. \[ \left (x^{4}+x^{2}\right ) y^{\prime \prime }-y^{\prime } x +\left (x +2\right ) y = 0 \] 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 {1}{\left (x^{2}+1\right ) x}\\ q(x) &= \frac {x +2}{x^{2} \left (x^{2}+1\right )}\\ \end {align*}
| \(p(x)=-\frac {1}{\left (x^{2}+1\right ) x}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -i\) | \(\text {``regular''}\) |
| \(x = i\) | \(\text {``regular''}\) |
| \(q(x)=\frac {x +2}{x^{2} \left (x^{2}+1\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -i\) | \(\text {``regular''}\) |
| \(x = i\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, -i, i, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ \left (x^{2}+1\right ) x^{2} y^{\prime \prime }-y^{\prime } x +\left (x +2\right ) y = 0 \] 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} 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}+1\right )-\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right ) x +\left (x +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 }}x^{n +r +2} 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 ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}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^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r} \\ \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 =2}{\sum }}a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-x^{n +r} a_{n} \left (n +r \right )+2 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )-x^{r} a_{0} r +2 a_{0} x^{r} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )-x^{r} r +2 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r^{2}-2 r +2\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r^{2}-2 r +2 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 1+i\\ r_2 &= 1-i \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r^{2}-2 r +2\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([1+i, 1-i]\).
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 +1+i}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +1-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 \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = -\frac {1}{r^{2}+1} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} a_{n -2} \left (n +r -2\right ) \left (n -3+r \right )+a_{n} \left (n +r \right ) \left (n +r -1\right )-a_{n} \left (n +r \right )+a_{n -1}+2 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {n^{2} a_{n -2}+2 n r a_{n -2}+r^{2} a_{n -2}-5 n a_{n -2}-5 r a_{n -2}+6 a_{n -2}+a_{n -1}}{n^{2}+2 n r +r^{2}-2 n -2 r +2}\tag {4} \] Which for the root \(r = 1+i\) becomes \[ a_{n} = \frac {\left (-n^{2}+\left (3-2 i\right ) n -1+3 i\right ) a_{n -2}-a_{n -1}}{n \left (n +2 i\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 1+i\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{r^{2}+1}\) | \(-\frac {1}{5}+\frac {2 i}{5}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {-r^{4}+r^{3}-r^{2}+r +1}{\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right )} \] Which for the root \(r = 1+i\) becomes \[ a_{2}=-\frac {1}{40}-\frac {13 i}{40} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{r^{2}+1}\) | \(-\frac {1}{5}+\frac {2 i}{5}\) |
| \(a_{2}\) | \(\frac {-r^{4}+r^{3}-r^{2}+r +1}{\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right )}\) | \(-\frac {1}{40}-\frac {13 i}{40}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {2 r^{4}+2 r^{3}+5 r^{2}+r -1}{\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right )} \] Which for the root \(r = 1+i\) becomes \[ a_{3}=\frac {71}{520}-\frac {17 i}{520} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{r^{2}+1}\) | \(-\frac {1}{5}+\frac {2 i}{5}\) |
| \(a_{2}\) | \(\frac {-r^{4}+r^{3}-r^{2}+r +1}{\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right )}\) | \(-\frac {1}{40}-\frac {13 i}{40}\) |
| \(a_{3}\) | \(\frac {2 r^{4}+2 r^{3}+5 r^{2}+r -1}{\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right )}\) | \(\frac {71}{520}-\frac {17 i}{520}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {r^{8}+6 r^{7}+13 r^{6}+10 r^{5}-4 r^{4}-15 r^{3}-37 r^{2}-34 r -9}{\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right )} \] Which for the root \(r = 1+i\) becomes \[ a_{4}=-\frac {31}{832}+\frac {541 i}{4160} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{r^{2}+1}\) | \(-\frac {1}{5}+\frac {2 i}{5}\) |
| \(a_{2}\) | \(\frac {-r^{4}+r^{3}-r^{2}+r +1}{\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right )}\) | \(-\frac {1}{40}-\frac {13 i}{40}\) |
| \(a_{3}\) | \(\frac {2 r^{4}+2 r^{3}+5 r^{2}+r -1}{\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right )}\) | \(\frac {71}{520}-\frac {17 i}{520}\) |
| \(a_{4}\) | \(\frac {r^{8}+6 r^{7}+13 r^{6}+10 r^{5}-4 r^{4}-15 r^{3}-37 r^{2}-34 r -9}{\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right )}\) | \(-\frac {31}{832}+\frac {541 i}{4160}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-3 r^{8}-30 r^{7}-132 r^{6}-330 r^{5}-528 r^{4}-570 r^{3}-303 r^{2}+60 r +69}{\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right )} \] Which for the root \(r = 1+i\) becomes \[ a_{5}=-\frac {1423}{20800}-\frac {7 i}{4160} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{r^{2}+1}\) | \(-\frac {1}{5}+\frac {2 i}{5}\) |
| \(a_{2}\) | \(\frac {-r^{4}+r^{3}-r^{2}+r +1}{\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right )}\) | \(-\frac {1}{40}-\frac {13 i}{40}\) |
| \(a_{3}\) | \(\frac {2 r^{4}+2 r^{3}+5 r^{2}+r -1}{\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right )}\) | \(\frac {71}{520}-\frac {17 i}{520}\) |
| \(a_{4}\) | \(\frac {r^{8}+6 r^{7}+13 r^{6}+10 r^{5}-4 r^{4}-15 r^{3}-37 r^{2}-34 r -9}{\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right )}\) | \(-\frac {31}{832}+\frac {541 i}{4160}\) |
| \(a_{5}\) | \(\frac {-3 r^{8}-30 r^{7}-132 r^{6}-330 r^{5}-528 r^{4}-570 r^{3}-303 r^{2}+60 r +69}{\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right )}\) | \(-\frac {1423}{20800}-\frac {7 i}{4160}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {-r^{12}-21 r^{11}-188 r^{10}-930 r^{9}-2742 r^{8}-4764 r^{7}-4068 r^{6}+1014 r^{5}+8233 r^{4}+14610 r^{3}+15926 r^{2}+8811 r +1767}{\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right ) \left (r^{2}+10 r +26\right )} \] Which for the root \(r = 1+i\) becomes \[ a_{6}=\frac {12849}{416000}-\frac {10853 i}{156000} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{r^{2}+1}\) | \(-\frac {1}{5}+\frac {2 i}{5}\) |
| \(a_{2}\) | \(\frac {-r^{4}+r^{3}-r^{2}+r +1}{\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right )}\) | \(-\frac {1}{40}-\frac {13 i}{40}\) |
| \(a_{3}\) | \(\frac {2 r^{4}+2 r^{3}+5 r^{2}+r -1}{\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right )}\) | \(\frac {71}{520}-\frac {17 i}{520}\) |
| \(a_{4}\) | \(\frac {r^{8}+6 r^{7}+13 r^{6}+10 r^{5}-4 r^{4}-15 r^{3}-37 r^{2}-34 r -9}{\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right )}\) | \(-\frac {31}{832}+\frac {541 i}{4160}\) |
| \(a_{5}\) | \(\frac {-3 r^{8}-30 r^{7}-132 r^{6}-330 r^{5}-528 r^{4}-570 r^{3}-303 r^{2}+60 r +69}{\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right )}\) | \(-\frac {1423}{20800}-\frac {7 i}{4160}\) |
| \(a_{6}\) | \(\frac {-r^{12}-21 r^{11}-188 r^{10}-930 r^{9}-2742 r^{8}-4764 r^{7}-4068 r^{6}+1014 r^{5}+8233 r^{4}+14610 r^{3}+15926 r^{2}+8811 r +1767}{\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right ) \left (r^{2}+10 r +26\right )}\) | \(\frac {12849}{416000}-\frac {10853 i}{156000}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=\frac {4 r^{12}+108 r^{11}+1298 r^{10}+9150 r^{9}+42072 r^{8}+133134 r^{7}+298869 r^{6}+482955 r^{5}+553706 r^{4}+403821 r^{3}+106210 r^{2}-69957 r -37647}{\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right ) \left (r^{2}+10 r +26\right ) \left (r^{2}+12 r +37\right )} \] Which for the root \(r = 1+i\) becomes \[ a_{7}=\frac {209609}{5088000}+\frac {106907 i}{17808000} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{r^{2}+1}\) | \(-\frac {1}{5}+\frac {2 i}{5}\) |
| \(a_{2}\) | \(\frac {-r^{4}+r^{3}-r^{2}+r +1}{\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right )}\) | \(-\frac {1}{40}-\frac {13 i}{40}\) |
| \(a_{3}\) | \(\frac {2 r^{4}+2 r^{3}+5 r^{2}+r -1}{\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right )}\) | \(\frac {71}{520}-\frac {17 i}{520}\) |
| \(a_{4}\) | \(\frac {r^{8}+6 r^{7}+13 r^{6}+10 r^{5}-4 r^{4}-15 r^{3}-37 r^{2}-34 r -9}{\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right )}\) | \(-\frac {31}{832}+\frac {541 i}{4160}\) |
| \(a_{5}\) | \(\frac {-3 r^{8}-30 r^{7}-132 r^{6}-330 r^{5}-528 r^{4}-570 r^{3}-303 r^{2}+60 r +69}{\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right )}\) | \(-\frac {1423}{20800}-\frac {7 i}{4160}\) |
| \(a_{6}\) | \(\frac {-r^{12}-21 r^{11}-188 r^{10}-930 r^{9}-2742 r^{8}-4764 r^{7}-4068 r^{6}+1014 r^{5}+8233 r^{4}+14610 r^{3}+15926 r^{2}+8811 r +1767}{\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right ) \left (r^{2}+10 r +26\right )}\) | \(\frac {12849}{416000}-\frac {10853 i}{156000}\) |
| \(a_{7}\) | \(\frac {4 r^{12}+108 r^{11}+1298 r^{10}+9150 r^{9}+42072 r^{8}+133134 r^{7}+298869 r^{6}+482955 r^{5}+553706 r^{4}+403821 r^{3}+106210 r^{2}-69957 r -37647}{\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right ) \left (r^{2}+10 r +26\right ) \left (r^{2}+12 r +37\right )}\) | \(\frac {209609}{5088000}+\frac {106907 i}{17808000}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin{align*} y_{1}\left (x \right )&= x^{1+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}+a_{7} x^{7}+a_{8} x^{8}\dots \right ) \\ &= x^{1+i} \left (1+\left (-\frac {1}{5}+\frac {2 i}{5}\right ) x +\left (-\frac {1}{40}-\frac {13 i}{40}\right ) x^{2}+\left (\frac {71}{520}-\frac {17 i}{520}\right ) x^{3}+\left (-\frac {31}{832}+\frac {541 i}{4160}\right ) x^{4}+\left (-\frac {1423}{20800}-\frac {7 i}{4160}\right ) x^{5}+\left (\frac {12849}{416000}-\frac {10853 i}{156000}\right ) x^{6}+\left (\frac {209609}{5088000}+\frac {106907 i}{17808000}\right ) x^{7}+O\left (x^{8}\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 \[ y_{2}\left (x \right )= x^{1-i} \left (1+\left (-\frac {1}{5}-\frac {2 i}{5}\right ) x +\left (-\frac {1}{40}+\frac {13 i}{40}\right ) x^{2}+\left (\frac {71}{520}+\frac {17 i}{520}\right ) x^{3}+\left (-\frac {31}{832}-\frac {541 i}{4160}\right ) x^{4}+\left (-\frac {1423}{20800}+\frac {7 i}{4160}\right ) x^{5}+\left (\frac {12849}{416000}+\frac {10853 i}{156000}\right ) x^{6}+\left (\frac {209609}{5088000}-\frac {106907 i}{17808000}\right ) x^{7}+O\left (x^{8}\right )\right ) \] 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^{1+i} \left (1+\left (-\frac {1}{5}+\frac {2 i}{5}\right ) x +\left (-\frac {1}{40}-\frac {13 i}{40}\right ) x^{2}+\left (\frac {71}{520}-\frac {17 i}{520}\right ) x^{3}+\left (-\frac {31}{832}+\frac {541 i}{4160}\right ) x^{4}+\left (-\frac {1423}{20800}-\frac {7 i}{4160}\right ) x^{5}+\left (\frac {12849}{416000}-\frac {10853 i}{156000}\right ) x^{6}+\left (\frac {209609}{5088000}+\frac {106907 i}{17808000}\right ) x^{7}+O\left (x^{8}\right )\right ) + c_{2} x^{1-i} \left (1+\left (-\frac {1}{5}-\frac {2 i}{5}\right ) x +\left (-\frac {1}{40}+\frac {13 i}{40}\right ) x^{2}+\left (\frac {71}{520}+\frac {17 i}{520}\right ) x^{3}+\left (-\frac {31}{832}-\frac {541 i}{4160}\right ) x^{4}+\left (-\frac {1423}{20800}+\frac {7 i}{4160}\right ) x^{5}+\left (\frac {12849}{416000}+\frac {10853 i}{156000}\right ) x^{6}+\left (\frac {209609}{5088000}-\frac {106907 i}{17808000}\right ) x^{7}+O\left (x^{8}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{1+i} \left (1+\left (-\frac {1}{5}+\frac {2 i}{5}\right ) x +\left (-\frac {1}{40}-\frac {13 i}{40}\right ) x^{2}+\left (\frac {71}{520}-\frac {17 i}{520}\right ) x^{3}+\left (-\frac {31}{832}+\frac {541 i}{4160}\right ) x^{4}+\left (-\frac {1423}{20800}-\frac {7 i}{4160}\right ) x^{5}+\left (\frac {12849}{416000}-\frac {10853 i}{156000}\right ) x^{6}+\left (\frac {209609}{5088000}+\frac {106907 i}{17808000}\right ) x^{7}+O\left (x^{8}\right )\right )+c_{2} x^{1-i} \left (1+\left (-\frac {1}{5}-\frac {2 i}{5}\right ) x +\left (-\frac {1}{40}+\frac {13 i}{40}\right ) x^{2}+\left (\frac {71}{520}+\frac {17 i}{520}\right ) x^{3}+\left (-\frac {31}{832}-\frac {541 i}{4160}\right ) x^{4}+\left (-\frac {1423}{20800}+\frac {7 i}{4160}\right ) x^{5}+\left (\frac {12849}{416000}+\frac {10853 i}{156000}\right ) x^{6}+\left (\frac {209609}{5088000}-\frac {106907 i}{17808000}\right ) x^{7}+O\left (x^{8}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{1+i} \left (1+\left (-\frac {1}{5}+\frac {2 i}{5}\right ) x +\left (-\frac {1}{40}-\frac {13 i}{40}\right ) x^{2}+\left (\frac {71}{520}-\frac {17 i}{520}\right ) x^{3}+\left (-\frac {31}{832}+\frac {541 i}{4160}\right ) x^{4}+\left (-\frac {1423}{20800}-\frac {7 i}{4160}\right ) x^{5}+\left (\frac {12849}{416000}-\frac {10853 i}{156000}\right ) x^{6}+\left (\frac {209609}{5088000}+\frac {106907 i}{17808000}\right ) x^{7}+O\left (x^{8}\right )\right )+c_{2} x^{1-i} \left (1+\left (-\frac {1}{5}-\frac {2 i}{5}\right ) x +\left (-\frac {1}{40}+\frac {13 i}{40}\right ) x^{2}+\left (\frac {71}{520}+\frac {17 i}{520}\right ) x^{3}+\left (-\frac {31}{832}-\frac {541 i}{4160}\right ) x^{4}+\left (-\frac {1423}{20800}+\frac {7 i}{4160}\right ) x^{5}+\left (\frac {12849}{416000}+\frac {10853 i}{156000}\right ) x^{6}+\left (\frac {209609}{5088000}-\frac {106907 i}{17808000}\right ) x^{7}+O\left (x^{8}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{1+i} \left (1+\left (-\frac {1}{5}+\frac {2 i}{5}\right ) x +\left (-\frac {1}{40}-\frac {13 i}{40}\right ) x^{2}+\left (\frac {71}{520}-\frac {17 i}{520}\right ) x^{3}+\left (-\frac {31}{832}+\frac {541 i}{4160}\right ) x^{4}+\left (-\frac {1423}{20800}-\frac {7 i}{4160}\right ) x^{5}+\left (\frac {12849}{416000}-\frac {10853 i}{156000}\right ) x^{6}+\left (\frac {209609}{5088000}+\frac {106907 i}{17808000}\right ) x^{7}+O\left (x^{8}\right )\right )+c_{2} x^{1-i} \left (1+\left (-\frac {1}{5}-\frac {2 i}{5}\right ) x +\left (-\frac {1}{40}+\frac {13 i}{40}\right ) x^{2}+\left (\frac {71}{520}+\frac {17 i}{520}\right ) x^{3}+\left (-\frac {31}{832}-\frac {541 i}{4160}\right ) x^{4}+\left (-\frac {1423}{20800}+\frac {7 i}{4160}\right ) x^{5}+\left (\frac {12849}{416000}+\frac {10853 i}{156000}\right ) x^{6}+\left (\frac {209609}{5088000}-\frac {106907 i}{17808000}\right ) x^{7}+O\left (x^{8}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime \prime } \left (x^{2}+1\right )-y^{\prime } x +\left (x +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\right ) y}{x^{2} \left (x^{2}+1\right )}+\frac {y^{\prime }}{x \left (x^{2}+1\right )} \\ \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 {y^{\prime }}{x \left (x^{2}+1\right )}+\frac {\left (x +2\right ) y}{x^{2} \left (x^{2}+1\right )}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{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}{\left (x^{2}+1\right ) x}, P_{3}\left (x \right )=\frac {x +2}{x^{2} \left (x^{2}+1\right )}\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}}}=2 \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime \prime } \left (x^{2}+1\right )-y^{\prime } x +\left (x +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..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^{m}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..4 \\ {} & {} & x^{m}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & x^{m}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (r^{2}-2 r +2\right ) x^{r}+\left (\left (r^{2}+1\right ) a_{1}+a_{0}\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (k^{2}+2 k r +r^{2}-2 k -2 r +2\right )+a_{k -1}+a_{k -2} \left (k -2+r \right ) \left (k -3+r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r^{2}-2 r +2=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{1-\mathrm {I}, 1+\mathrm {I}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & \left (r^{2}+1\right ) a_{1}+a_{0}=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=-\frac {a_{0}}{r^{2}+1} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k -2} \left (k -2+r \right ) \left (k -3+r \right )+\left (k^{2}+\left (2 r -2\right ) k +r^{2}-2 r +2\right ) a_{k}+a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k} \left (k +r \right ) \left (k +r -1\right )+\left (\left (k +2\right )^{2}+\left (2 r -2\right ) \left (k +2\right )+r^{2}-2 r +2\right ) a_{k +2}+a_{k +1}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}+2 k r a_{k}+r^{2} a_{k}-k a_{k}-r a_{k}+a_{k +1}}{k^{2}+2 k r +r^{2}+2 k +2 r +2} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1-\mathrm {I} \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}+\left (2-2 \,\mathrm {I}\right ) k a_{k}-\left (1+\mathrm {I}\right ) a_{k}-k a_{k}+a_{k +1}}{k^{2}+\left (2-2 \,\mathrm {I}\right ) k +4-4 \,\mathrm {I}+2 k} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1-\mathrm {I} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1-\mathrm {I}}, a_{k +2}=-\frac {k^{2} a_{k}+\left (2-2 \,\mathrm {I}\right ) k a_{k}-\left (1+\mathrm {I}\right ) a_{k}-k a_{k}+a_{k +1}}{k^{2}+\left (2-2 \,\mathrm {I}\right ) k +4-4 \,\mathrm {I}+2 k}, a_{1}=\left (-\frac {1}{5}-\frac {2 \,\mathrm {I}}{5}\right ) a_{0}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1+\mathrm {I} \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}+\left (2+2 \,\mathrm {I}\right ) k a_{k}+\left (-1+\mathrm {I}\right ) a_{k}-k a_{k}+a_{k +1}}{k^{2}+\left (2+2 \,\mathrm {I}\right ) k +4+4 \,\mathrm {I}+2 k} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1+\mathrm {I} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1+\mathrm {I}}, a_{k +2}=-\frac {k^{2} a_{k}+\left (2+2 \,\mathrm {I}\right ) k a_{k}+\left (-1+\mathrm {I}\right ) a_{k}-k a_{k}+a_{k +1}}{k^{2}+\left (2+2 \,\mathrm {I}\right ) k +4+4 \,\mathrm {I}+2 k}, a_{1}=\left (-\frac {1}{5}+\frac {2 \,\mathrm {I}}{5}\right ) a_{0}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1-\mathrm {I}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +1+\mathrm {I}}\right ), a_{k +2}=-\frac {k^{2} a_{k}+\left (2-2 \,\mathrm {I}\right ) k a_{k}-\left (1+\mathrm {I}\right ) a_{k}-k a_{k}+a_{k +1}}{k^{2}+\left (2-2 \,\mathrm {I}\right ) k +4-4 \,\mathrm {I}+2 k}, a_{1}=\left (-\frac {1}{5}-\frac {2 \,\mathrm {I}}{5}\right ) a_{0}, b_{k +2}=-\frac {k^{2} b_{k}+\left (2+2 \,\mathrm {I}\right ) k b_{k}+\left (-1+\mathrm {I}\right ) b_{k}-k b_{k}+b_{k +1}}{k^{2}+\left (2+2 \,\mathrm {I}\right ) k +4+4 \,\mathrm {I}+2 k}, b_{1}=\left (-\frac {1}{5}+\frac {2 \,\mathrm {I}}{5}\right ) b_{0}\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 -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius trying a solution in terms of MeijerG functions -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius <- Heun successful: received ODE is equivalent to the HeunG ODE, case a <> 0, e <> 0, g <> 0, c = 0 `
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 87
Order:=8; dsolve((x^2+1)*x^2*diff(y(x),x$2)-x*diff(y(x),x)+(2+x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{1-i} \left (1+\left (-\frac {1}{5}-\frac {2 i}{5}\right ) x +\left (-\frac {1}{40}+\frac {13 i}{40}\right ) x^{2}+\left (\frac {71}{520}+\frac {17 i}{520}\right ) x^{3}+\left (-\frac {31}{832}-\frac {541 i}{4160}\right ) x^{4}+\left (-\frac {1423}{20800}+\frac {7 i}{4160}\right ) x^{5}+\left (\frac {12849}{416000}+\frac {10853 i}{156000}\right ) x^{6}+\left (\frac {209609}{5088000}-\frac {106907 i}{17808000}\right ) x^{7}+\operatorname {O}\left (x^{8}\right )\right )+c_{2} x^{1+i} \left (1+\left (-\frac {1}{5}+\frac {2 i}{5}\right ) x +\left (-\frac {1}{40}-\frac {13 i}{40}\right ) x^{2}+\left (\frac {71}{520}-\frac {17 i}{520}\right ) x^{3}+\left (-\frac {31}{832}+\frac {541 i}{4160}\right ) x^{4}+\left (-\frac {1423}{20800}-\frac {7 i}{4160}\right ) x^{5}+\left (\frac {12849}{416000}-\frac {10853 i}{156000}\right ) x^{6}+\left (\frac {209609}{5088000}+\frac {106907 i}{17808000}\right ) x^{7}+\operatorname {O}\left (x^{8}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.055 (sec). Leaf size: 122
AsymptoticDSolveValue[(x^2+1)*x^2*y''[x]-x*y'[x]+(2+x)*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to \left (\frac {1}{156000}+\frac {i}{1248000}\right ) c_2 x^{1-i} \left ((6080+10093 i) x^6-(10476-1572 i) x^5-(8220+19260 i) x^4+(21600+2400 i) x^3+(2400+50400 i) x^2-(38400+57600 i) x+(153600-19200 i)\right )-\left (\frac {1}{1248000}+\frac {i}{156000}\right ) c_1 x^{1+i} \left ((10093+6080 i) x^6+(1572-10476 i) x^5-(19260+8220 i) x^4+(2400+21600 i) x^3+(50400+2400 i) x^2-(57600+38400 i) x-(19200-153600 i)\right ) \]