Internal problem ID [2952]
Internal file name [OUTPUT/2444_Sunday_June_05_2022_03_11_32_AM_19801593/index.tex]
Book: Differential equations and linear algebra, Stephen W. Goode and Scott A Annin. Fourth
edition, 2015
Section: Chapter 11, Series Solutions to Linear Differential Equations. Exercises for 11.5. page
771
Problem number: 8.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Repeated root"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x^{2} \left (x^{2}+1\right ) y^{\prime \prime }+7 y^{\prime } x \,{\mathrm e}^{x}+9 \left (1+\tan \left (x \right )\right ) y=0} \] 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. \[ \left (x^{4}+x^{2}\right ) y^{\prime \prime }+7 y^{\prime } x \,{\mathrm e}^{x}+\left (9 \tan \left (x \right )+9\right ) y = 0 \] 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 {7 \,{\mathrm e}^{x}}{x \left (x^{2}+1\right )}\\ q(x) &= \frac {9 \tan \left (x \right )+9}{x^{2} \left (x^{2}+1\right )}\\ \end {align*}
| \(p(x)=\frac {7 \,{\mathrm e}^{x}}{x \left (x^{2}+1\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -i\) | \(\text {``regular''}\) |
| \(x = i\) | \(\text {``regular''}\) |
| \(x = \infty \) | \(\text {``regular''}\) |
| \(q(x)=\frac {9 \tan \left (x \right )+9}{x^{2} \left (x^{2}+1\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -i\) | \(\text {``regular''}\) |
| \(x = i\) | \(\text {``regular''}\) |
| \(x = \frac {1}{2} \pi +Z \pi \) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \(\left [0, -i, i, \infty , \frac {1}{2} \pi +Z \pi \right ]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} \left (x^{2}+1\right ) y^{\prime \prime }+7 y^{\prime } x \,{\mathrm e}^{x}+\left (9 \tan \left (x \right )+9\right ) y = 0 \] 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 (x^{2}+1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+7 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right ) x \,{\mathrm e}^{x}+\left (9 \tan \left (x \right )+9\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} Expanding \(7 x \,{\mathrm e}^{x}\) as Taylor series around \(x=0\) and keeping only the first \(7\) terms gives \begin {align*} 7 x \,{\mathrm e}^{x} &= 7 x +7 x^{2}+\frac {7}{2} x^{3}+\frac {7}{6} x^{4}+\frac {7}{24} x^{5}+\frac {7}{120} x^{6}+\frac {7}{720} x^{7} + \dots \\ &= 7 x +7 x^{2}+\frac {7}{2} x^{3}+\frac {7}{6} x^{4}+\frac {7}{24} x^{5}+\frac {7}{120} x^{6}+\frac {7}{720} x^{7} \end {align*}
Expanding \(9 \tan \left (x \right )+9\) as Taylor series around \(x=0\) and keeping only the first \(7\) terms gives \begin {align*} 9 \tan \left (x \right )+9 &= 9+9 x +3 x^{3}+\frac {6}{5} x^{5}+\frac {17}{35} x^{7} + \dots \\ &= 9+9 x +3 x^{3}+\frac {6}{5} x^{5}+\frac {17}{35} x^{7} \end {align*}
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 )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {7 x^{n +r +6} a_{n} \left (n +r \right )}{720}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {7 x^{n +r +5} a_{n} \left (n +r \right )}{120}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {7 x^{n +r +4} a_{n} \left (n +r \right )}{24}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {7 x^{n +r +3} a_{n} \left (n +r \right )}{6}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {7 x^{n +r +2} a_{n} \left (n +r \right )}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}7 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}7 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}9 a_{n} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}9 x^{1+n +r} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r +3} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {6 x^{n +r +5} a_{n}}{5}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {17 x^{n +r +7} a_{n}}{35}\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 }}\frac {7 x^{n +r +6} a_{n} \left (n +r \right )}{720} &= \moverset {\infty }{\munderset {n =6}{\sum }}\frac {7 a_{n -6} \left (n -6+r \right ) x^{n +r}}{720} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {7 x^{n +r +5} a_{n} \left (n +r \right )}{120} &= \moverset {\infty }{\munderset {n =5}{\sum }}\frac {7 a_{n -5} \left (n -5+r \right ) x^{n +r}}{120} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {7 x^{n +r +4} a_{n} \left (n +r \right )}{24} &= \moverset {\infty }{\munderset {n =4}{\sum }}\frac {7 a_{n -4} \left (n -4+r \right ) x^{n +r}}{24} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {7 x^{n +r +3} a_{n} \left (n +r \right )}{6} &= \moverset {\infty }{\munderset {n =3}{\sum }}\frac {7 a_{n -3} \left (n -3+r \right ) x^{n +r}}{6} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {7 x^{n +r +2} a_{n} \left (n +r \right )}{2} &= \moverset {\infty }{\munderset {n =2}{\sum }}\frac {7 a_{n -2} \left (n +r -2\right ) x^{n +r}}{2} \\ \moverset {\infty }{\munderset {n =0}{\sum }}7 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}7 a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}9 x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}9 a_{n -1} x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r +3} a_{n} &= \moverset {\infty }{\munderset {n =3}{\sum }}3 a_{n -3} x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {6 x^{n +r +5} a_{n}}{5} &= \moverset {\infty }{\munderset {n =5}{\sum }}\frac {6 a_{n -5} x^{n +r}}{5} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\frac {17 x^{n +r +7} a_{n}}{35} &= \moverset {\infty }{\munderset {n =7}{\sum }}\frac {17 a_{n -7} x^{n +r}}{35} \\ \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 )+\left (\moverset {\infty }{\munderset {n =6}{\sum }}\frac {7 a_{n -6} \left (n -6+r \right ) x^{n +r}}{720}\right )+\left (\moverset {\infty }{\munderset {n =5}{\sum }}\frac {7 a_{n -5} \left (n -5+r \right ) x^{n +r}}{120}\right )+\left (\moverset {\infty }{\munderset {n =4}{\sum }}\frac {7 a_{n -4} \left (n -4+r \right ) x^{n +r}}{24}\right )+\left (\moverset {\infty }{\munderset {n =3}{\sum }}\frac {7 a_{n -3} \left (n -3+r \right ) x^{n +r}}{6}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}\frac {7 a_{n -2} \left (n +r -2\right ) x^{n +r}}{2}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}7 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}7 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}9 a_{n} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}9 a_{n -1} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =3}{\sum }}3 a_{n -3} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =5}{\sum }}\frac {6 a_{n -5} x^{n +r}}{5}\right )+\left (\moverset {\infty }{\munderset {n =7}{\sum }}\frac {17 a_{n -7} x^{n +r}}{35}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+7 x^{n +r} a_{n} \left (n +r \right )+9 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )+7 x^{r} a_{0} r +9 a_{0} x^{r} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )+7 x^{r} r +9 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r +3\right )^{2} x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ \left (r +3\right )^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= -3\\ r_2 &= -3 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r +3\right )^{2} x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([-3, -3]\).
Since the root of the indicial equation is repeated, then we can construct two linearly independent solutions. The first solution has the form \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\tag {1A} \end {align*}
Now the second solution \(y_{2}\) is found using \begin {align*} y_{2}\left (x \right ) &= y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +r}\right )\tag {1B} \end {align*}
Then the general solution will be \[ y = c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \] In Eq (1B) the sum starts from 1 and not zero. In Eq (1A), \(a_{0}\) is never zero, and is arbitrary and is typically taken as \(a_{0} = 1\), and \(\{c_{1}, c_{2}\}\) are two arbitray constants of integration which can be found from initial conditions. Using the value of the indicial root found earlier, \(r = -3\), Eqs (1A,1B) become \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n -3}\\ y_{2}\left (x \right ) &= y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n -3}\right ) \end {align*}
We start by finding the first solution \(y_{1}\left (x \right )\). 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 \[ a_{1} = \frac {-7 r -9}{\left (r +4\right )^{2}} \] Substituting \(n = 2\) in Eq. (2B) gives \[ a_{2} = \frac {-2 r^{4}-21 r^{3}+26 r^{2}+270 r +288}{2 \left (r +4\right )^{2} \left (r +5\right )^{2}} \] Substituting \(n = 3\) in Eq. (2B) gives \[ a_{3} = \frac {77 r^{5}+1098 r^{4}+3602 r^{3}-1998 r^{2}-22318 r -22347}{6 \left (r +4\right )^{2} \left (r +5\right )^{2} \left (6+r \right )^{2}} \] Substituting \(n = 4\) in Eq. (2B) gives \[ a_{4} = \frac {24 r^{8}+689 r^{7}+5132 r^{6}-7139 r^{5}-200328 r^{4}-605868 r^{3}+73708 r^{2}+2422164 r +2371896}{24 \left (r +4\right )^{2} \left (r +5\right )^{2} \left (6+r \right )^{2} \left (r +7\right )^{2}} \] Substituting \(n = 5\) in Eq. (2B) gives \[ a_{5} = \frac {-2247 r^{9}-79622 r^{8}-1065025 r^{7}-6575402 r^{6}-16492594 r^{5}+7514875 r^{4}+98942906 r^{3}+50216229 r^{2}-289883610 r -327734910}{120 \left (r +4\right )^{2} \left (r +5\right )^{2} \left (6+r \right )^{2} \left (r +7\right )^{2} \left (r +8\right )^{2}} \] Substituting \(n = 6\) in Eq. (2B) gives \[ a_{6} = \frac {-720 r^{12}-39547 r^{11}-792562 r^{10}-6305240 r^{9}+6536464 r^{8}+490611417 r^{7}+3743792400 r^{6}+12951080410 r^{5}+18943551934 r^{4}-1275211246 r^{3}-21124918552 r^{2}+17754474300 r +38965417200}{720 \left (r +4\right )^{2} \left (r +5\right )^{2} \left (6+r \right )^{2} \left (r +7\right )^{2} \left (r +8\right )^{2} \left (r +9\right )^{2}} \] For \(7\le n\) 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 )+\frac {7 a_{n -6} \left (n -6+r \right )}{720}+\frac {7 a_{n -5} \left (n -5+r \right )}{120}+\frac {7 a_{n -4} \left (n -4+r \right )}{24}+\frac {7 a_{n -3} \left (n -3+r \right )}{6}+\frac {7 a_{n -2} \left (n +r -2\right )}{2}+7 a_{n -1} \left (n +r -1\right )+7 a_{n} \left (n +r \right )+9 a_{n}+9 a_{n -1}+3 a_{n -3}+\frac {6 a_{n -5}}{5}+\frac {17 a_{n -7}}{35} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {5040 n^{2} a_{n -2}+10080 n r a_{n -2}+5040 r^{2} a_{n -2}+49 n a_{n -6}+294 n a_{n -5}+1470 n a_{n -4}+5880 n a_{n -3}-7560 n a_{n -2}+35280 n a_{n -1}+49 r a_{n -6}+294 r a_{n -5}+1470 r a_{n -4}+5880 r a_{n -3}-7560 r a_{n -2}+35280 r a_{n -1}+2448 a_{n -7}-294 a_{n -6}+4578 a_{n -5}-5880 a_{n -4}-2520 a_{n -3}-5040 a_{n -2}+10080 a_{n -1}}{5040 \left (n^{2}+2 n r +r^{2}+6 n +6 r +9\right )}\tag {4} \] Which for the root \(r = -3\) becomes \[ a_{n} = \frac {-5040 n^{2} a_{n -2}+\left (-49 a_{n -6}-294 a_{n -5}-1470 a_{n -4}-5880 a_{n -3}+37800 a_{n -2}-35280 a_{n -1}\right ) n -2448 a_{n -7}+441 a_{n -6}-3696 a_{n -5}+10290 a_{n -4}+20160 a_{n -3}-63000 a_{n -2}+95760 a_{n -1}}{5040 n^{2}}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = -3\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-7 r -9}{\left (r +4\right )^{2}}\) | \(12\) |
| \(a_{2}\) | \(\frac {-2 r^{4}-21 r^{3}+26 r^{2}+270 r +288}{2 \left (r +4\right )^{2} \left (r +5\right )^{2}}\) | \(\frac {117}{8}\) |
| \(a_{3}\) | \(\frac {77 r^{5}+1098 r^{4}+3602 r^{3}-1998 r^{2}-22318 r -22347}{6 \left (r +4\right )^{2} \left (r +5\right )^{2} \left (6+r \right )^{2}}\) | \(-{\frac {67}{36}}\) |
| \(a_{4}\) | \(\frac {24 r^{8}+689 r^{7}+5132 r^{6}-7139 r^{5}-200328 r^{4}-605868 r^{3}+73708 r^{2}+2422164 r +2371896}{24 \left (r +4\right )^{2} \left (r +5\right )^{2} \left (6+r \right )^{2} \left (r +7\right )^{2}}\) | \(\frac {505}{256}\) |
| \(a_{5}\) | \(\frac {-2247 r^{9}-79622 r^{8}-1065025 r^{7}-6575402 r^{6}-16492594 r^{5}+7514875 r^{4}+98942906 r^{3}+50216229 r^{2}-289883610 r -327734910}{120 \left (r +4\right )^{2} \left (r +5\right )^{2} \left (6+r \right )^{2} \left (r +7\right )^{2} \left (r +8\right )^{2}}\) | \(-{\frac {262}{125}}\) |
| \(a_{6}\) | \(\frac {-720 r^{12}-39547 r^{11}-792562 r^{10}-6305240 r^{9}+6536464 r^{8}+490611417 r^{7}+3743792400 r^{6}+12951080410 r^{5}+18943551934 r^{4}-1275211246 r^{3}-21124918552 r^{2}+17754474300 r +38965417200}{720 \left (r +4\right )^{2} \left (r +5\right )^{2} \left (6+r \right )^{2} \left (r +7\right )^{2} \left (r +8\right )^{2} \left (r +9\right )^{2}}\) | \(\frac {2443637}{2304000}\) |
Using the above table, then the first solution \(y_{1}\left (x \right )\) is \begin{align*} y_{1}\left (x \right )&= \frac {1}{x^{3}} \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}\dots \right ) \\ &= \frac {1+12 x +\frac {117 x^{2}}{8}-\frac {67 x^{3}}{36}+\frac {505 x^{4}}{256}-\frac {262 x^{5}}{125}+\frac {2443637 x^{6}}{2304000}+O\left (x^{7}\right )}{x^{3}} \\ \end{align*} Now the second solution is found. The second solution is given by \[ y_{2}\left (x \right ) = y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +r}\right ) \] Where \(b_{n}\) is found using \[ b_{n} = \frac {d}{d r}a_{n ,r} \] And the above is then evaluated at \(r = -3\). The above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table
| \(n\) | \(b_{n ,r}\) | \(a_{n}\) | \(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) | \(b_{n}\left (r =-3\right )\) |
| \(b_{0}\) | \(1\) | \(1\) | N/A since \(b_{n}\) starts from 1 | N/A |
| \(b_{1}\) | \(\frac {-7 r -9}{\left (r +4\right )^{2}}\) | \(12\) | \(\frac {7 r -10}{\left (r +4\right )^{3}}\) | \(-31\) |
| \(b_{2}\) | \(\frac {-2 r^{4}-21 r^{3}+26 r^{2}+270 r +288}{2 \left (r +4\right )^{2} \left (r +5\right )^{2}}\) | \(\frac {117}{8}\) | \(\frac {-15 r^{4}-401 r^{3}-2070 r^{2}-2542 r +216}{2 \left (r +4\right )^{3} \left (r +5\right )^{3}}\) | \(-{\frac {147}{2}}\) |
| \(b_{3}\) | \(\frac {77 r^{5}+1098 r^{4}+3602 r^{3}-1998 r^{2}-22318 r -22347}{6 \left (r +4\right )^{2} \left (r +5\right )^{2} \left (6+r \right )^{2}}\) | \(-{\frac {67}{36}}\) | \(\frac {-77 r^{7}-1041 r^{6}+6288 r^{5}+162666 r^{4}+965118 r^{3}+2435112 r^{2}+2512832 r +629196}{6 \left (r +4\right )^{3} \left (r +5\right )^{3} \left (6+r \right )^{3}}\) | \(\frac {37}{8}\) |
| \(b_{4}\) | \(\frac {24 r^{8}+689 r^{7}+5132 r^{6}-7139 r^{5}-200328 r^{4}-605868 r^{3}+73708 r^{2}+2422164 r +2371896}{24 \left (r +4\right )^{2} \left (r +5\right )^{2} \left (6+r \right )^{2} \left (r +7\right )^{2}}\) | \(\frac {505}{256}\) | \(\frac {367 r^{10}+22078 r^{9}+483282 r^{8}+5154816 r^{7}+27714075 r^{6}+51746274 r^{5}-200593408 r^{4}-1371446536 r^{3}-3140579700 r^{2}-3119788728 r -991921536}{24 \left (r +4\right )^{3} \left (r +5\right )^{3} \left (6+r \right )^{3} \left (r +7\right )^{3}}\) | \(-{\frac {44803}{4608}}\) |
| \(b_{5}\) | \(\frac {-2247 r^{9}-79622 r^{8}-1065025 r^{7}-6575402 r^{6}-16492594 r^{5}+7514875 r^{4}+98942906 r^{3}+50216229 r^{2}-289883610 r -327734910}{120 \left (r +4\right )^{2} \left (r +5\right )^{2} \left (6+r \right )^{2} \left (r +7\right )^{2} \left (r +8\right )^{2}}\) | \(-{\frac {262}{125}}\) | \(\frac {2247 r^{13}+91834 r^{12}+802020 r^{11}-21535712 r^{10}-653860121 r^{9}-8150098608 r^{8}-58894700472 r^{7}-261153930514 r^{6}-671020308468 r^{5}-676780277800 r^{4}+1175421071354 r^{3}+4492941561360 r^{2}+5111619350400 r +1948094750880}{120 \left (r +4\right )^{3} \left (r +5\right )^{3} \left (6+r \right )^{3} \left (r +7\right )^{3} \left (r +8\right )^{3}}\) | \(\frac {5057587}{480000}\) |
| \(b_{6}\) | \(\frac {-720 r^{12}-39547 r^{11}-792562 r^{10}-6305240 r^{9}+6536464 r^{8}+490611417 r^{7}+3743792400 r^{6}+12951080410 r^{5}+18943551934 r^{4}-1275211246 r^{3}-21124918552 r^{2}+17754474300 r +38965417200}{720 \left (r +4\right )^{2} \left (r +5\right )^{2} \left (6+r \right )^{2} \left (r +7\right )^{2} \left (r +8\right )^{2} \left (r +9\right )^{2}}\) | \(\frac {2443637}{2304000}\) | \(\frac {-16613 r^{16}-1757209 r^{15}-77979705 r^{14}-1953565011 r^{13}-30831383043 r^{12}-318268874085 r^{11}-2095326321071 r^{10}-7268248040053 r^{9}+6393794924526 r^{8}+219888860709348 r^{7}+1238615020779714 r^{6}+3790795160732550 r^{5}+6581172050593312 r^{4}+4882069838530340 r^{3}-2771187212166840 r^{2}-7454518221589920 r -3618892518566400}{720 \left (r +4\right )^{3} \left (r +5\right )^{3} \left (6+r \right )^{3} \left (r +7\right )^{3} \left (r +8\right )^{3} \left (r +9\right )^{3}}\) | \(-{\frac {3797765581}{622080000}}\) |
The above table gives all values of \(b_{n}\) needed. Hence the second solution is \begin{align*} y_{2}\left (x \right )&=y_{1}\left (x \right ) \ln \left (x \right )+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}\dots \\ &= \frac {\left (1+12 x +\frac {117 x^{2}}{8}-\frac {67 x^{3}}{36}+\frac {505 x^{4}}{256}-\frac {262 x^{5}}{125}+\frac {2443637 x^{6}}{2304000}+O\left (x^{7}\right )\right ) \ln \left (x \right )}{x^{3}}+\frac {-31 x -\frac {147 x^{2}}{2}+\frac {37 x^{3}}{8}-\frac {44803 x^{4}}{4608}+\frac {5057587 x^{5}}{480000}-\frac {3797765581 x^{6}}{622080000}+O\left (x^{7}\right )}{x^{3}} \\ \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+12 x +\frac {117 x^{2}}{8}-\frac {67 x^{3}}{36}+\frac {505 x^{4}}{256}-\frac {262 x^{5}}{125}+\frac {2443637 x^{6}}{2304000}+O\left (x^{7}\right )\right )}{x^{3}} + c_{2} \left (\frac {\left (1+12 x +\frac {117 x^{2}}{8}-\frac {67 x^{3}}{36}+\frac {505 x^{4}}{256}-\frac {262 x^{5}}{125}+\frac {2443637 x^{6}}{2304000}+O\left (x^{7}\right )\right ) \ln \left (x \right )}{x^{3}}+\frac {-31 x -\frac {147 x^{2}}{2}+\frac {37 x^{3}}{8}-\frac {44803 x^{4}}{4608}+\frac {5057587 x^{5}}{480000}-\frac {3797765581 x^{6}}{622080000}+O\left (x^{7}\right )}{x^{3}}\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= \frac {c_{1} \left (1+12 x +\frac {117 x^{2}}{8}-\frac {67 x^{3}}{36}+\frac {505 x^{4}}{256}-\frac {262 x^{5}}{125}+\frac {2443637 x^{6}}{2304000}+O\left (x^{7}\right )\right )}{x^{3}}+c_{2} \left (\frac {\left (1+12 x +\frac {117 x^{2}}{8}-\frac {67 x^{3}}{36}+\frac {505 x^{4}}{256}-\frac {262 x^{5}}{125}+\frac {2443637 x^{6}}{2304000}+O\left (x^{7}\right )\right ) \ln \left (x \right )}{x^{3}}+\frac {-31 x -\frac {147 x^{2}}{2}+\frac {37 x^{3}}{8}-\frac {44803 x^{4}}{4608}+\frac {5057587 x^{5}}{480000}-\frac {3797765581 x^{6}}{622080000}+O\left (x^{7}\right )}{x^{3}}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1} \left (1+12 x +\frac {117 x^{2}}{8}-\frac {67 x^{3}}{36}+\frac {505 x^{4}}{256}-\frac {262 x^{5}}{125}+\frac {2443637 x^{6}}{2304000}+O\left (x^{7}\right )\right )}{x^{3}}+c_{2} \left (\frac {\left (1+12 x +\frac {117 x^{2}}{8}-\frac {67 x^{3}}{36}+\frac {505 x^{4}}{256}-\frac {262 x^{5}}{125}+\frac {2443637 x^{6}}{2304000}+O\left (x^{7}\right )\right ) \ln \left (x \right )}{x^{3}}+\frac {-31 x -\frac {147 x^{2}}{2}+\frac {37 x^{3}}{8}-\frac {44803 x^{4}}{4608}+\frac {5057587 x^{5}}{480000}-\frac {3797765581 x^{6}}{622080000}+O\left (x^{7}\right )}{x^{3}}\right ) \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1} \left (1+12 x +\frac {117 x^{2}}{8}-\frac {67 x^{3}}{36}+\frac {505 x^{4}}{256}-\frac {262 x^{5}}{125}+\frac {2443637 x^{6}}{2304000}+O\left (x^{7}\right )\right )}{x^{3}}+c_{2} \left (\frac {\left (1+12 x +\frac {117 x^{2}}{8}-\frac {67 x^{3}}{36}+\frac {505 x^{4}}{256}-\frac {262 x^{5}}{125}+\frac {2443637 x^{6}}{2304000}+O\left (x^{7}\right )\right ) \ln \left (x \right )}{x^{3}}+\frac {-31 x -\frac {147 x^{2}}{2}+\frac {37 x^{3}}{8}-\frac {44803 x^{4}}{4608}+\frac {5057587 x^{5}}{480000}-\frac {3797765581 x^{6}}{622080000}+O\left (x^{7}\right )}{x^{3}}\right ) \] Verified OK.
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 75
Order:=7; dsolve(x^2*(x^2+1)*diff(y(x),x$2)+7*x*exp(x)*diff(y(x),x)+9*(1+tan(x))*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {\left (c_{2} \ln \left (x \right )+c_{1} \right ) \left (1+12 x +\frac {117}{8} x^{2}-\frac {67}{36} x^{3}+\frac {505}{256} x^{4}-\frac {262}{125} x^{5}+\frac {2443637}{2304000} x^{6}+\operatorname {O}\left (x^{7}\right )\right )+\left (\left (-31\right ) x -\frac {147}{2} x^{2}+\frac {37}{8} x^{3}-\frac {44803}{4608} x^{4}+\frac {5057587}{480000} x^{5}-\frac {3797765581}{622080000} x^{6}+\operatorname {O}\left (x^{7}\right )\right ) c_{2}}{x^{3}} \]
✓ Solution by Mathematica
Time used: 0.013 (sec). Leaf size: 143
AsymptoticDSolveValue[x^2*(x^2+1)*y''[x]+7*x*Exp[x]*y'[x]+9*(1+Tan[x])*y[x]==0,y[x],{x,0,6}]
\[ y(x)\to \frac {c_1 \left (\frac {2443637 x^6}{2304000}-\frac {262 x^5}{125}+\frac {505 x^4}{256}-\frac {67 x^3}{36}+\frac {117 x^2}{8}+12 x+1\right )}{x^3}+c_2 \left (\frac {-\frac {3797765581 x^6}{622080000}+\frac {5057587 x^5}{480000}-\frac {44803 x^4}{4608}+\frac {37 x^3}{8}-\frac {147 x^2}{2}-31 x}{x^3}+\frac {\left (\frac {2443637 x^6}{2304000}-\frac {262 x^5}{125}+\frac {505 x^4}{256}-\frac {67 x^3}{36}+\frac {117 x^2}{8}+12 x+1\right ) \log (x)}{x^3}\right ) \]