Internal problem ID [14007]
Internal file name [OUTPUT/13179_Friday_February_23_2024_06_58_24_AM_73898812/index.tex]
Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell.
second edition. CRC Press. FL, USA. 2020
Section: Chapter 36. The big theorem on the the Frobenius method. Additional Exercises. page
739
Problem number: 36.2 (d).
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]]
\[ \boxed {x^{2} \left (-x^{2}+2\right ) y^{\prime \prime }+\left (4 x^{2}+5 x \right ) y^{\prime }+y \left (x^{2}+1\right )=0} \] 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. \[ \left (-x^{4}+2 x^{2}\right ) y^{\prime \prime }+\left (4 x^{2}+5 x \right ) y^{\prime }+y \left (x^{2}+1\right ) = 0 \] 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 {4 x +5}{x \left (x^{2}-2\right )}\\ q(x) &= -\frac {x^{2}+1}{x^{2} \left (x^{2}-2\right )}\\ \end {align*}
| \(p(x)=-\frac {4 x +5}{x \left (x^{2}-2\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = \sqrt {2}\) | \(\text {``regular''}\) |
| \(x = -\sqrt {2}\) | \(\text {``regular''}\) |
| \(q(x)=-\frac {x^{2}+1}{x^{2} \left (x^{2}-2\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = \sqrt {2}\) | \(\text {``regular''}\) |
| \(x = -\sqrt {2}\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \(\left [0, \sqrt {2}, -\sqrt {2}, \infty \right ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ -y^{\prime \prime } x^{2} \left (x^{2}-2\right )+\left (4 x^{2}+5 x \right ) y^{\prime }+y \left (x^{2}+1\right ) = 0 \] 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 (x^{2}-2\right )+\left (4 x^{2}+5 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) \left (x^{2}+1\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\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 }}4 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 )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}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 }}\left (-x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}4 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}4 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} \moverset {\infty }{\munderset {n =2}{\sum }}\left (-a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r}\right )+\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 }}4 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 )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+5 x^{n +r} a_{n} \left (n +r \right )+a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 2 x^{r} a_{0} r \left (-1+r \right )+5 x^{r} a_{0} r +a_{0} x^{r} = 0 \] Or \[ \left (2 x^{r} r \left (-1+r \right )+5 x^{r} r +x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (2 r^{2}+3 r +1\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 2 r^{2}+3 r +1 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= -{\frac {1}{2}}\\ r_2 &= -1 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (2 r^{2}+3 r +1\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [-{\frac {1}{2}}, -1\right ]\).
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 (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 -1} \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 \[ a_{1} = -\frac {4 r}{2 r^{2}+7 r +6} \] 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 )+2 a_{n} \left (n +r \right ) \left (n +r -1\right )+4 a_{n -1} \left (n +r -1\right )+5 a_{n} \left (n +r \right )+a_{n -2}+a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {n^{2} a_{n -2}+2 n r a_{n -2}+r^{2} a_{n -2}-5 n a_{n -2}-4 n a_{n -1}-5 r a_{n -2}-4 r a_{n -1}+5 a_{n -2}+4 a_{n -1}}{2 n^{2}+4 n r +2 r^{2}+3 n +3 r +1}\tag {4} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ a_{n} = \frac {4 n^{2} a_{n -2}+\left (-24 a_{n -2}-16 a_{n -1}\right ) n +31 a_{n -2}+24 a_{n -1}}{8 n^{2}+4 n}\tag {5} \] 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.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {4 r}{2 r^{2}+7 r +6}\) | \(\frac {2}{3}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {2 r^{4}+5 r^{3}+13 r^{2}+3 r -6}{\left (2 r^{2}+7 r +6\right ) \left (2 r^{2}+11 r +15\right )} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ a_{2}=-{\frac {19}{120}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {4 r}{2 r^{2}+7 r +6}\) | \(\frac {2}{3}\) |
| \(a_{2}\) | \(\frac {2 r^{4}+5 r^{3}+13 r^{2}+3 r -6}{\left (2 r^{2}+7 r +6\right ) \left (2 r^{2}+11 r +15\right )}\) | \(-{\frac {19}{120}}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-16 r^{5}-88 r^{4}-188 r^{3}-132 r^{2}+60 r +48}{\left (2 r^{2}+7 r +6\right ) \left (2 r^{2}+11 r +15\right ) \left (2 r^{2}+15 r +28\right )} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ a_{3}={\frac {1}{180}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {4 r}{2 r^{2}+7 r +6}\) | \(\frac {2}{3}\) |
| \(a_{2}\) | \(\frac {2 r^{4}+5 r^{3}+13 r^{2}+3 r -6}{\left (2 r^{2}+7 r +6\right ) \left (2 r^{2}+11 r +15\right )}\) | \(-{\frac {19}{120}}\) |
| \(a_{3}\) | \(\frac {-16 r^{5}-88 r^{4}-188 r^{3}-132 r^{2}+60 r +48}{\left (2 r^{2}+7 r +6\right ) \left (2 r^{2}+11 r +15\right ) \left (2 r^{2}+15 r +28\right )}\) | \(\frac {1}{180}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {4 r^{8}+52 r^{7}+345 r^{6}+1396 r^{5}+3385 r^{4}+4310 r^{3}+1555 r^{2}-1422 r -744}{\left (2 r^{2}+7 r +6\right ) \left (2 r^{2}+11 r +15\right ) \left (2 r^{2}+15 r +28\right ) \left (2 r^{2}+19 r +45\right )} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ a_{4}=-{\frac {23}{51840}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {4 r}{2 r^{2}+7 r +6}\) | \(\frac {2}{3}\) |
| \(a_{2}\) | \(\frac {2 r^{4}+5 r^{3}+13 r^{2}+3 r -6}{\left (2 r^{2}+7 r +6\right ) \left (2 r^{2}+11 r +15\right )}\) | \(-{\frac {19}{120}}\) |
| \(a_{3}\) | \(\frac {-16 r^{5}-88 r^{4}-188 r^{3}-132 r^{2}+60 r +48}{\left (2 r^{2}+7 r +6\right ) \left (2 r^{2}+11 r +15\right ) \left (2 r^{2}+15 r +28\right )}\) | \(\frac {1}{180}\) |
| \(a_{4}\) | \(\frac {4 r^{8}+52 r^{7}+345 r^{6}+1396 r^{5}+3385 r^{4}+4310 r^{3}+1555 r^{2}-1422 r -744}{\left (2 r^{2}+7 r +6\right ) \left (2 r^{2}+11 r +15\right ) \left (2 r^{2}+15 r +28\right ) \left (2 r^{2}+19 r +45\right )}\) | \(-{\frac {23}{51840}}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {4 \left (12 r^{9}+228 r^{8}+1885 r^{7}+8785 r^{6}+24886 r^{5}+42331 r^{4}+37332 r^{3}+5623 r^{2}-13647 r -5676\right )}{\left (2 r^{2}+7 r +6\right ) \left (2 r^{2}+11 r +15\right ) \left (2 r^{2}+15 r +28\right ) \left (2 r^{2}+19 r +45\right ) \left (2 r^{2}+23 r +66\right )} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ a_{5}={\frac {557}{1425600}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {4 r}{2 r^{2}+7 r +6}\) | \(\frac {2}{3}\) |
| \(a_{2}\) | \(\frac {2 r^{4}+5 r^{3}+13 r^{2}+3 r -6}{\left (2 r^{2}+7 r +6\right ) \left (2 r^{2}+11 r +15\right )}\) | \(-{\frac {19}{120}}\) |
| \(a_{3}\) | \(\frac {-16 r^{5}-88 r^{4}-188 r^{3}-132 r^{2}+60 r +48}{\left (2 r^{2}+7 r +6\right ) \left (2 r^{2}+11 r +15\right ) \left (2 r^{2}+15 r +28\right )}\) | \(\frac {1}{180}\) |
| \(a_{4}\) | \(\frac {4 r^{8}+52 r^{7}+345 r^{6}+1396 r^{5}+3385 r^{4}+4310 r^{3}+1555 r^{2}-1422 r -744}{\left (2 r^{2}+7 r +6\right ) \left (2 r^{2}+11 r +15\right ) \left (2 r^{2}+15 r +28\right ) \left (2 r^{2}+19 r +45\right )}\) | \(-{\frac {23}{51840}}\) |
| \(a_{5}\) | \(-\frac {4 \left (12 r^{9}+228 r^{8}+1885 r^{7}+8785 r^{6}+24886 r^{5}+42331 r^{4}+37332 r^{3}+5623 r^{2}-13647 r -5676\right )}{\left (2 r^{2}+7 r +6\right ) \left (2 r^{2}+11 r +15\right ) \left (2 r^{2}+15 r +28\right ) \left (2 r^{2}+19 r +45\right ) \left (2 r^{2}+23 r +66\right )}\) | \(\frac {557}{1425600}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= \frac {1}{\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}\dots \right ) \\ &= \frac {1+\frac {2 x}{3}-\frac {19 x^{2}}{120}+\frac {x^{3}}{180}-\frac {23 x^{4}}{51840}+\frac {557 x^{5}}{1425600}+O\left (x^{6}\right )}{\sqrt {x}} \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 \[ b_{1} = -\frac {4 r}{2 r^{2}+7 r +6} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} -b_{n -2} \left (n +r -2\right ) \left (n -3+r \right )+2 b_{n} \left (n +r \right ) \left (n +r -1\right )+4 b_{n -1} \left (n +r -1\right )+5 b_{n} \left (n +r \right )+b_{n -2}+b_{n} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = \frac {n^{2} b_{n -2}+2 n r b_{n -2}+r^{2} b_{n -2}-5 n b_{n -2}-4 n b_{n -1}-5 r b_{n -2}-4 r b_{n -1}+5 b_{n -2}+4 b_{n -1}}{2 n^{2}+4 n r +2 r^{2}+3 n +3 r +1}\tag {4} \] Which for the root \(r = -1\) becomes \[ b_{n} = \frac {n^{2} b_{n -2}+\left (-7 b_{n -2}-4 b_{n -1}\right ) n +11 b_{n -2}+8 b_{n -1}}{2 n^{2}-n}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -1\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {4 r}{2 r^{2}+7 r +6}\) | \(4\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {2 r^{4}+5 r^{3}+13 r^{2}+3 r -6}{\left (2 r^{2}+7 r +6\right ) \left (2 r^{2}+11 r +15\right )} \] Which for the root \(r = -1\) becomes \[ b_{2}={\frac {1}{6}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {4 r}{2 r^{2}+7 r +6}\) | \(4\) |
| \(b_{2}\) | \(\frac {2 r^{4}+5 r^{3}+13 r^{2}+3 r -6}{\left (2 r^{2}+7 r +6\right ) \left (2 r^{2}+11 r +15\right )}\) | \(\frac {1}{6}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {-16 r^{5}-88 r^{4}-188 r^{3}-132 r^{2}+60 r +48}{\left (2 r^{2}+7 r +6\right ) \left (2 r^{2}+11 r +15\right ) \left (2 r^{2}+15 r +28\right )} \] Which for the root \(r = -1\) becomes \[ b_{3}=-{\frac {14}{45}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {4 r}{2 r^{2}+7 r +6}\) | \(4\) |
| \(b_{2}\) | \(\frac {2 r^{4}+5 r^{3}+13 r^{2}+3 r -6}{\left (2 r^{2}+7 r +6\right ) \left (2 r^{2}+11 r +15\right )}\) | \(\frac {1}{6}\) |
| \(b_{3}\) | \(\frac {-16 r^{5}-88 r^{4}-188 r^{3}-132 r^{2}+60 r +48}{\left (2 r^{2}+7 r +6\right ) \left (2 r^{2}+11 r +15\right ) \left (2 r^{2}+15 r +28\right )}\) | \(-{\frac {14}{45}}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {4 r^{8}+52 r^{7}+345 r^{6}+1396 r^{5}+3385 r^{4}+4310 r^{3}+1555 r^{2}-1422 r -744}{\left (2 r^{2}+7 r +6\right ) \left (2 r^{2}+11 r +15\right ) \left (2 r^{2}+15 r +28\right ) \left (2 r^{2}+19 r +45\right )} \] Which for the root \(r = -1\) becomes \[ b_{4}={\frac {209}{2520}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {4 r}{2 r^{2}+7 r +6}\) | \(4\) |
| \(b_{2}\) | \(\frac {2 r^{4}+5 r^{3}+13 r^{2}+3 r -6}{\left (2 r^{2}+7 r +6\right ) \left (2 r^{2}+11 r +15\right )}\) | \(\frac {1}{6}\) |
| \(b_{3}\) | \(\frac {-16 r^{5}-88 r^{4}-188 r^{3}-132 r^{2}+60 r +48}{\left (2 r^{2}+7 r +6\right ) \left (2 r^{2}+11 r +15\right ) \left (2 r^{2}+15 r +28\right )}\) | \(-{\frac {14}{45}}\) |
| \(b_{4}\) | \(\frac {4 r^{8}+52 r^{7}+345 r^{6}+1396 r^{5}+3385 r^{4}+4310 r^{3}+1555 r^{2}-1422 r -744}{\left (2 r^{2}+7 r +6\right ) \left (2 r^{2}+11 r +15\right ) \left (2 r^{2}+15 r +28\right ) \left (2 r^{2}+19 r +45\right )}\) | \(\frac {209}{2520}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-\frac {4 \left (12 r^{9}+228 r^{8}+1885 r^{7}+8785 r^{6}+24886 r^{5}+42331 r^{4}+37332 r^{3}+5623 r^{2}-13647 r -5676\right )}{\left (2 r^{2}+7 r +6\right ) \left (2 r^{2}+11 r +15\right ) \left (2 r^{2}+15 r +28\right ) \left (2 r^{2}+19 r +45\right ) \left (2 r^{2}+23 r +66\right )} \] Which for the root \(r = -1\) becomes \[ b_{5}=-{\frac {823}{28350}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {4 r}{2 r^{2}+7 r +6}\) | \(4\) |
| \(b_{2}\) | \(\frac {2 r^{4}+5 r^{3}+13 r^{2}+3 r -6}{\left (2 r^{2}+7 r +6\right ) \left (2 r^{2}+11 r +15\right )}\) | \(\frac {1}{6}\) |
| \(b_{3}\) | \(\frac {-16 r^{5}-88 r^{4}-188 r^{3}-132 r^{2}+60 r +48}{\left (2 r^{2}+7 r +6\right ) \left (2 r^{2}+11 r +15\right ) \left (2 r^{2}+15 r +28\right )}\) | \(-{\frac {14}{45}}\) |
| \(b_{4}\) | \(\frac {4 r^{8}+52 r^{7}+345 r^{6}+1396 r^{5}+3385 r^{4}+4310 r^{3}+1555 r^{2}-1422 r -744}{\left (2 r^{2}+7 r +6\right ) \left (2 r^{2}+11 r +15\right ) \left (2 r^{2}+15 r +28\right ) \left (2 r^{2}+19 r +45\right )}\) | \(\frac {209}{2520}\) |
| \(b_{5}\) | \(-\frac {4 \left (12 r^{9}+228 r^{8}+1885 r^{7}+8785 r^{6}+24886 r^{5}+42331 r^{4}+37332 r^{3}+5623 r^{2}-13647 r -5676\right )}{\left (2 r^{2}+7 r +6\right ) \left (2 r^{2}+11 r +15\right ) \left (2 r^{2}+15 r +28\right ) \left (2 r^{2}+19 r +45\right ) \left (2 r^{2}+23 r +66\right )}\) | \(-{\frac {823}{28350}}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= \frac {1}{\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}\dots \right ) \\ &= \frac {1+4 x +\frac {x^{2}}{6}-\frac {14 x^{3}}{45}+\frac {209 x^{4}}{2520}-\frac {823 x^{5}}{28350}+O\left (x^{6}\right )}{x} \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 ) \\ &= \frac {c_{1} \left (1+\frac {2 x}{3}-\frac {19 x^{2}}{120}+\frac {x^{3}}{180}-\frac {23 x^{4}}{51840}+\frac {557 x^{5}}{1425600}+O\left (x^{6}\right )\right )}{\sqrt {x}} + \frac {c_{2} \left (1+4 x +\frac {x^{2}}{6}-\frac {14 x^{3}}{45}+\frac {209 x^{4}}{2520}-\frac {823 x^{5}}{28350}+O\left (x^{6}\right )\right )}{x} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= \frac {c_{1} \left (1+\frac {2 x}{3}-\frac {19 x^{2}}{120}+\frac {x^{3}}{180}-\frac {23 x^{4}}{51840}+\frac {557 x^{5}}{1425600}+O\left (x^{6}\right )\right )}{\sqrt {x}}+\frac {c_{2} \left (1+4 x +\frac {x^{2}}{6}-\frac {14 x^{3}}{45}+\frac {209 x^{4}}{2520}-\frac {823 x^{5}}{28350}+O\left (x^{6}\right )\right )}{x} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1} \left (1+\frac {2 x}{3}-\frac {19 x^{2}}{120}+\frac {x^{3}}{180}-\frac {23 x^{4}}{51840}+\frac {557 x^{5}}{1425600}+O\left (x^{6}\right )\right )}{\sqrt {x}}+\frac {c_{2} \left (1+4 x +\frac {x^{2}}{6}-\frac {14 x^{3}}{45}+\frac {209 x^{4}}{2520}-\frac {823 x^{5}}{28350}+O\left (x^{6}\right )\right )}{x} \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1} \left (1+\frac {2 x}{3}-\frac {19 x^{2}}{120}+\frac {x^{3}}{180}-\frac {23 x^{4}}{51840}+\frac {557 x^{5}}{1425600}+O\left (x^{6}\right )\right )}{\sqrt {x}}+\frac {c_{2} \left (1+4 x +\frac {x^{2}}{6}-\frac {14 x^{3}}{45}+\frac {209 x^{4}}{2520}-\frac {823 x^{5}}{28350}+O\left (x^{6}\right )\right )}{x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -y^{\prime \prime } x^{2} \left (x^{2}-2\right )+\left (4 x^{2}+5 x \right ) y^{\prime }+y \left (x^{2}+1\right )=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 {y \left (x^{2}+1\right )}{x^{2} \left (x^{2}-2\right )}+\frac {\left (4 x +5\right ) y^{\prime }}{x \left (x^{2}-2\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 {\left (4 x +5\right ) y^{\prime }}{x \left (x^{2}-2\right )}-\frac {y \left (x^{2}+1\right )}{x^{2} \left (x^{2}-2\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 {4 x +5}{x \left (x^{2}-2\right )}, P_{3}\left (x \right )=-\frac {x^{2}+1}{x^{2} \left (x^{2}-2\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}}}=\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}}}=\frac {1}{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} \\ {} & {} & y^{\prime \prime } x^{2} \left (x^{2}-2\right )-x \left (4 x +5\right ) y^{\prime }+\left (-x^{2}-1\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^{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 (1+r \right ) \left (1+2 r \right ) x^{r}+\left (-a_{1} \left (2+r \right ) \left (3+2 r \right )-4 a_{0} r \right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (-a_{k} \left (k +r +1\right ) \left (2 k +2 r +1\right )-4 a_{k -1} \left (k +r -1\right )+a_{k -2} \left (\left (k -2\right )^{2}+2 \left (k -2\right ) r +r^{2}-k +1-r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -\left (1+r \right ) \left (1+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-1, -\frac {1}{2}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & -a_{1} \left (2+r \right ) \left (3+2 r \right )-4 a_{0} r =0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=-\frac {4 a_{0} r}{2 r^{2}+7 r +6} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (-2 a_{k}+a_{k -2}\right ) k^{2}+\left (\left (-4 a_{k}+2 a_{k -2}\right ) r -3 a_{k}-5 a_{k -2}-4 a_{k -1}\right ) k +\left (-2 a_{k}+a_{k -2}\right ) r^{2}+\left (-3 a_{k}-5 a_{k -2}-4 a_{k -1}\right ) r -a_{k}+5 a_{k -2}+4 a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \left (-2 a_{k +2}+a_{k}\right ) \left (k +2\right )^{2}+\left (\left (-4 a_{k +2}+2 a_{k}\right ) r -3 a_{k +2}-5 a_{k}-4 a_{k +1}\right ) \left (k +2\right )+\left (-2 a_{k +2}+a_{k}\right ) r^{2}+\left (-3 a_{k +2}-5 a_{k}-4 a_{k +1}\right ) r -a_{k +2}+5 a_{k}+4 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}-4 k a_{k +1}-r a_{k}-4 r a_{k +1}-a_{k}-4 a_{k +1}}{2 k^{2}+4 k r +2 r^{2}+11 k +11 r +15} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1 \\ {} & {} & a_{k +2}=\frac {k^{2} a_{k}-3 k a_{k}-4 k a_{k +1}+a_{k}}{2 k^{2}+7 k +6} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -1}, a_{k +2}=\frac {k^{2} a_{k}-3 k a_{k}-4 k a_{k +1}+a_{k}}{2 k^{2}+7 k +6}, a_{1}=4 a_{0}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {1}{2} \\ {} & {} & a_{k +2}=\frac {k^{2} a_{k}-2 k a_{k}-4 k a_{k +1}-\frac {1}{4} a_{k}-2 a_{k +1}}{2 k^{2}+9 k +10} \\ \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^{2} a_{k}-2 k a_{k}-4 k a_{k +1}-\frac {1}{4} a_{k}-2 a_{k +1}}{2 k^{2}+9 k +10}, a_{1}=\frac {2 a_{0}}{3}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k -\frac {1}{2}}\right ), a_{k +2}=\frac {k^{2} a_{k}-3 k a_{k}-4 k a_{k +1}+a_{k}}{2 k^{2}+7 k +6}, a_{1}=4 a_{0}, b_{k +2}=\frac {k^{2} b_{k}-2 k b_{k}-4 k b_{k +1}-\frac {1}{4} b_{k}-2 b_{k +1}}{2 k^{2}+9 k +10}, b_{1}=\frac {2 b_{0}}{3}\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: 47
Order:=6; dsolve(x^2*(2-x^2)*diff(y(x),x$2)+(5*x+4*x^2)*diff(y(x),x)+(1+x^2)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{1} \left (1+4 x +\frac {1}{6} x^{2}-\frac {14}{45} x^{3}+\frac {209}{2520} x^{4}-\frac {823}{28350} x^{5}+\operatorname {O}\left (x^{6}\right )\right )}{x}+\frac {c_{2} \left (1+\frac {2}{3} x -\frac {19}{120} x^{2}+\frac {1}{180} x^{3}-\frac {23}{51840} x^{4}+\frac {557}{1425600} x^{5}+\operatorname {O}\left (x^{6}\right )\right )}{\sqrt {x}} \]
✓ Solution by Mathematica
Time used: 0.007 (sec). Leaf size: 86
AsymptoticDSolveValue[x^2*(2-x^2)*y''[x]+(5*x+4*x^2)*y'[x]+(1+x^2)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to \frac {c_1 \left (\frac {557 x^5}{1425600}-\frac {23 x^4}{51840}+\frac {x^3}{180}-\frac {19 x^2}{120}+\frac {2 x}{3}+1\right )}{\sqrt {x}}+\frac {c_2 \left (-\frac {823 x^5}{28350}+\frac {209 x^4}{2520}-\frac {14 x^3}{45}+\frac {x^2}{6}+4 x+1\right )}{x} \]