Internal problem ID [1363]
Internal file name [OUTPUT/1364_Sunday_June_05_2022_02_13_17_AM_2731518/index.tex]
Book: Elementary differential equations with boundary value problems. William F. Trench.
Brooks/Cole 2001
Section: Chapter 7 Series Solutions of Linear Second Equations. 7.6 THE METHOD OF
FROBENIUS II. Exercises 7.6. Page 374
Problem number: 11.
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]]
\[ \boxed {9 x^{2} \left (x +1\right ) y^{\prime \prime }+3 x \left (-x^{2}+11 x +5\right ) y^{\prime }+\left (-7 x^{2}+16 x +1\right ) y=0} \] 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. \[ \left (9 x^{3}+9 x^{2}\right ) y^{\prime \prime }+\left (-3 x^{3}+33 x^{2}+15 x \right ) y^{\prime }+\left (-7 x^{2}+16 x +1\right ) y = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} y^{\prime \prime }+p(x) y^{\prime } + q(x) y &=0 \end {align*}
Where \begin {align*} p(x) &= -\frac {x^{2}-11 x -5}{3 x \left (x +1\right )}\\ q(x) &= -\frac {7 x^{2}-16 x -1}{9 x^{2} \left (x +1\right )}\\ \end {align*}
| \(p(x)=-\frac {x^{2}-11 x -5}{3 x \left (x +1\right )}\) | |
| singularity | type |
| \(x = -1\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
| \(q(x)=-\frac {7 x^{2}-16 x -1}{9 x^{2} \left (x +1\right )}\) | |
| singularity | type |
| \(x = -1\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([-1, 0]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ 9 x^{2} \left (x +1\right ) y^{\prime \prime }+\left (-3 x^{3}+33 x^{2}+15 x \right ) y^{\prime }+\left (-7 x^{2}+16 x +1\right ) y = 0 \] 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} 9 x^{2} \left (x +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 )+\left (-3 x^{3}+33 x^{2}+15 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (-7 x^{2}+16 x +1\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 }}9 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}9 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 x^{n +r +2} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}33 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}15 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-7 x^{n +r +2} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}16 x^{1+n +r} 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 }}9 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}9 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 x^{n +r +2} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-3 a_{n -2} \left (n +r -2\right ) x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}33 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}33 a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-7 x^{n +r +2} a_{n}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-7 a_{n -2} x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}16 x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}16 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 =1}{\sum }}9 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}9 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-3 a_{n -2} \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}33 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}15 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-7 a_{n -2} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}16 a_{n -1} 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 \[ 9 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+15 x^{n +r} a_{n} \left (n +r \right )+a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 9 x^{r} a_{0} r \left (-1+r \right )+15 x^{r} a_{0} r +a_{0} x^{r} = 0 \] Or \[ \left (9 x^{r} r \left (-1+r \right )+15 x^{r} r +x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (3 r +1\right )^{2} x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ \left (3 r +1\right )^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= -{\frac {1}{3}}\\ r_2 &= -{\frac {1}{3}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (3 r +1\right )^{2} x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [-{\frac {1}{3}}, -{\frac {1}{3}}\right ]\).
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, \(r = -{\frac {1}{3}}\), Eqs (1A,1B) become \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n -\frac {1}{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 -\frac {1}{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 \[ a_{1} = -1 \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} 9 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+9 a_{n} \left (n +r \right ) \left (n +r -1\right )-3 a_{n -2} \left (n +r -2\right )+33 a_{n -1} \left (n +r -1\right )+15 a_{n} \left (n +r \right )-7 a_{n -2}+16 a_{n -1}+a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {3 n a_{n -1}+3 r a_{n -1}-a_{n -2}+a_{n -1}}{1+3 n +3 r}\tag {4} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ a_{n} = \frac {-3 n a_{n -1}+a_{n -2}}{3 n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = -{\frac {1}{3}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-1\) | \(-1\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {3 r +8}{7+3 r} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ a_{2}={\frac {7}{6}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-1\) | \(-1\) |
| \(a_{2}\) | \(\frac {3 r +8}{7+3 r}\) | \(\frac {7}{6}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-9 r^{2}-57 r -87}{9 r^{2}+51 r +70} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ a_{3}=-{\frac {23}{18}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-1\) | \(-1\) |
| \(a_{2}\) | \(\frac {3 r +8}{7+3 r}\) | \(\frac {7}{6}\) |
| \(a_{3}\) | \(\frac {-9 r^{2}-57 r -87}{9 r^{2}+51 r +70}\) | \(-{\frac {23}{18}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {27 r^{3}+297 r^{2}+1056 r +1211}{27 r^{3}+270 r^{2}+873 r +910} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ a_{4}={\frac {11}{8}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-1\) | \(-1\) |
| \(a_{2}\) | \(\frac {3 r +8}{7+3 r}\) | \(\frac {7}{6}\) |
| \(a_{3}\) | \(\frac {-9 r^{2}-57 r -87}{9 r^{2}+51 r +70}\) | \(-{\frac {23}{18}}\) |
| \(a_{4}\) | \(\frac {27 r^{3}+297 r^{2}+1056 r +1211}{27 r^{3}+270 r^{2}+873 r +910}\) | \(\frac {11}{8}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-81 r^{4}-1350 r^{3}-8208 r^{2}-21531 r -20507}{\left (16+3 r \right ) \left (9 r^{2}+51 r +70\right ) \left (13+3 r \right )} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ a_{5}=-{\frac {1577}{1080}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-1\) | \(-1\) |
| \(a_{2}\) | \(\frac {3 r +8}{7+3 r}\) | \(\frac {7}{6}\) |
| \(a_{3}\) | \(\frac {-9 r^{2}-57 r -87}{9 r^{2}+51 r +70}\) | \(-{\frac {23}{18}}\) |
| \(a_{4}\) | \(\frac {27 r^{3}+297 r^{2}+1056 r +1211}{27 r^{3}+270 r^{2}+873 r +910}\) | \(\frac {11}{8}\) |
| \(a_{5}\) | \(\frac {-81 r^{4}-1350 r^{3}-8208 r^{2}-21531 r -20507}{\left (16+3 r \right ) \left (9 r^{2}+51 r +70\right ) \left (13+3 r \right )}\) | \(-{\frac {1577}{1080}}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {243 r^{5}+5670 r^{4}+51597 r^{3}+228465 r^{2}+491139 r +409009}{243 r^{5}+5265 r^{4}+44415 r^{3}+181935 r^{2}+360942 r +276640} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ a_{6}={\frac {3319}{2160}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-1\) | \(-1\) |
| \(a_{2}\) | \(\frac {3 r +8}{7+3 r}\) | \(\frac {7}{6}\) |
| \(a_{3}\) | \(\frac {-9 r^{2}-57 r -87}{9 r^{2}+51 r +70}\) | \(-{\frac {23}{18}}\) |
| \(a_{4}\) | \(\frac {27 r^{3}+297 r^{2}+1056 r +1211}{27 r^{3}+270 r^{2}+873 r +910}\) | \(\frac {11}{8}\) |
| \(a_{5}\) | \(\frac {-81 r^{4}-1350 r^{3}-8208 r^{2}-21531 r -20507}{\left (16+3 r \right ) \left (9 r^{2}+51 r +70\right ) \left (13+3 r \right )}\) | \(-{\frac {1577}{1080}}\) |
| \(a_{6}\) | \(\frac {243 r^{5}+5670 r^{4}+51597 r^{3}+228465 r^{2}+491139 r +409009}{243 r^{5}+5265 r^{4}+44415 r^{3}+181935 r^{2}+360942 r +276640}\) | \(\frac {3319}{2160}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=\frac {-729 r^{6}-22599 r^{5}-285120 r^{4}-1870803 r^{3}-6720192 r^{2}-12502695 r -9387831}{\left (22+3 r \right ) \left (13+3 r \right ) \left (9 r^{2}+51 r +70\right ) \left (16+3 r \right ) \left (19+3 r \right )} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ a_{7}=-{\frac {72853}{45360}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-1\) | \(-1\) |
| \(a_{2}\) | \(\frac {3 r +8}{7+3 r}\) | \(\frac {7}{6}\) |
| \(a_{3}\) | \(\frac {-9 r^{2}-57 r -87}{9 r^{2}+51 r +70}\) | \(-{\frac {23}{18}}\) |
| \(a_{4}\) | \(\frac {27 r^{3}+297 r^{2}+1056 r +1211}{27 r^{3}+270 r^{2}+873 r +910}\) | \(\frac {11}{8}\) |
| \(a_{5}\) | \(\frac {-81 r^{4}-1350 r^{3}-8208 r^{2}-21531 r -20507}{\left (16+3 r \right ) \left (9 r^{2}+51 r +70\right ) \left (13+3 r \right )}\) | \(-{\frac {1577}{1080}}\) |
| \(a_{6}\) | \(\frac {243 r^{5}+5670 r^{4}+51597 r^{3}+228465 r^{2}+491139 r +409009}{243 r^{5}+5265 r^{4}+44415 r^{3}+181935 r^{2}+360942 r +276640}\) | \(\frac {3319}{2160}\) |
| \(a_{7}\) | \(\frac {-729 r^{6}-22599 r^{5}-285120 r^{4}-1870803 r^{3}-6720192 r^{2}-12502695 r -9387831}{\left (22+3 r \right ) \left (13+3 r \right ) \left (9 r^{2}+51 r +70\right ) \left (16+3 r \right ) \left (19+3 r \right )}\) | \(-{\frac {72853}{45360}}\) |
Using the above table, then the first solution \(y_{1}\left (x \right )\) is \begin{align*} y_{1}\left (x \right )&= \frac {1}{x^{{1}/{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}+a_{8} x^{8}\dots \right ) \\ &= \frac {1-x +\frac {7 x^{2}}{6}-\frac {23 x^{3}}{18}+\frac {11 x^{4}}{8}-\frac {1577 x^{5}}{1080}+\frac {3319 x^{6}}{2160}-\frac {72853 x^{7}}{45360}+O\left (x^{8}\right )}{x^{{1}/{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 \(r = -{\frac {1}{3}}\). The above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table
| \(n\) | \(b_{n ,r}\) | \(a_{n}\) | \(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) | \(b_{n}\left (r =-\frac {1}{3}\right )\) |
| \(b_{0}\) | \(1\) | \(1\) | N/A since \(b_{n}\) starts from 1 | N/A |
| \(b_{1}\) | \(-1\) | \(-1\) | \(0\) | \(0\) |
| \(b_{2}\) | \(\frac {3 r +8}{7+3 r}\) | \(\frac {7}{6}\) | \(-\frac {3}{\left (7+3 r \right )^{2}}\) | \(-{\frac {1}{12}}\) |
| \(b_{3}\) | \(\frac {-9 r^{2}-57 r -87}{9 r^{2}+51 r +70}\) | \(-{\frac {23}{18}}\) | \(\frac {54 r^{2}+306 r +447}{\left (9 r^{2}+51 r +70\right )^{2}}\) | \(\frac {13}{108}\) |
| \(b_{4}\) | \(\frac {27 r^{3}+297 r^{2}+1056 r +1211}{27 r^{3}+270 r^{2}+873 r +910}\) | \(\frac {11}{8}\) | \(\frac {-729 r^{4}-9882 r^{3}-50220 r^{2}-113400 r -96243}{\left (27 r^{3}+270 r^{2}+873 r +910\right )^{2}}\) | \(-{\frac {131}{864}}\) |
| \(b_{5}\) | \(\frac {-81 r^{4}-1350 r^{3}-8208 r^{2}-21531 r -20507}{\left (16+3 r \right ) \left (9 r^{2}+51 r +70\right ) \left (13+3 r \right )}\) | \(-{\frac {1577}{1080}}\) | \(\frac {8748 r^{6}+205578 r^{5}+2001105 r^{4}+10325232 r^{3}+29787507 r^{2}+45579186 r +28934526}{\left (16+3 r \right )^{2} \left (9 r^{2}+51 r +70\right )^{2} \left (13+3 r \right )^{2}}\) | \(\frac {11449}{64800}\) |
| \(b_{6}\) | \(\frac {243 r^{5}+5670 r^{4}+51597 r^{3}+228465 r^{2}+491139 r +409009}{243 r^{5}+5265 r^{4}+44415 r^{3}+181935 r^{2}+360942 r +276640}\) | \(\frac {3319}{2160}\) | \(\frac {-98415 r^{8}-3490452 r^{7}-53745525 r^{6}-469145034 r^{5}-2538718200 r^{4}-8720362962 r^{3}-18569781900 r^{2}-22420989630 r -11759833518}{\left (243 r^{5}+5265 r^{4}+44415 r^{3}+181935 r^{2}+360942 r +276640\right )^{2}}\) | \(-{\frac {76919}{388800}}\) |
| \(b_{7}\) | \(\frac {-729 r^{6}-22599 r^{5}-285120 r^{4}-1870803 r^{3}-6720192 r^{2}-12502695 r -9387831}{\left (22+3 r \right ) \left (13+3 r \right ) \left (9 r^{2}+51 r +70\right ) \left (16+3 r \right ) \left (19+3 r \right )}\) | \(-{\frac {72853}{45360}}\) | \(\frac {1062882 r^{10}+52553610 r^{9}+1159663311 r^{8}+15034741452 r^{7}+126794500524 r^{6}+726656053590 r^{5}+2865553796322 r^{4}+7677332912286 r^{3}+13374342543807 r^{2}+13682424177432 r +6244921647564}{\left (22+3 r \right )^{2} \left (13+3 r \right )^{2} \left (9 r^{2}+51 r +70\right )^{2} \left (16+3 r \right )^{2} \left (19+3 r \right )^{2}}\) | \(\frac {4118557}{19051200}\) |
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}+b_{8} x^{8}\dots \\ &= \frac {\left (1-x +\frac {7 x^{2}}{6}-\frac {23 x^{3}}{18}+\frac {11 x^{4}}{8}-\frac {1577 x^{5}}{1080}+\frac {3319 x^{6}}{2160}-\frac {72853 x^{7}}{45360}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{x^{{1}/{3}}}+\frac {-\frac {x^{2}}{12}+\frac {13 x^{3}}{108}-\frac {131 x^{4}}{864}+\frac {11449 x^{5}}{64800}-\frac {76919 x^{6}}{388800}+\frac {4118557 x^{7}}{19051200}+O\left (x^{8}\right )}{x^{{1}/{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-x +\frac {7 x^{2}}{6}-\frac {23 x^{3}}{18}+\frac {11 x^{4}}{8}-\frac {1577 x^{5}}{1080}+\frac {3319 x^{6}}{2160}-\frac {72853 x^{7}}{45360}+O\left (x^{8}\right )\right )}{x^{{1}/{3}}} + c_{2} \left (\frac {\left (1-x +\frac {7 x^{2}}{6}-\frac {23 x^{3}}{18}+\frac {11 x^{4}}{8}-\frac {1577 x^{5}}{1080}+\frac {3319 x^{6}}{2160}-\frac {72853 x^{7}}{45360}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{x^{{1}/{3}}}+\frac {-\frac {x^{2}}{12}+\frac {13 x^{3}}{108}-\frac {131 x^{4}}{864}+\frac {11449 x^{5}}{64800}-\frac {76919 x^{6}}{388800}+\frac {4118557 x^{7}}{19051200}+O\left (x^{8}\right )}{x^{{1}/{3}}}\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= \frac {c_{1} \left (1-x +\frac {7 x^{2}}{6}-\frac {23 x^{3}}{18}+\frac {11 x^{4}}{8}-\frac {1577 x^{5}}{1080}+\frac {3319 x^{6}}{2160}-\frac {72853 x^{7}}{45360}+O\left (x^{8}\right )\right )}{x^{{1}/{3}}}+c_{2} \left (\frac {\left (1-x +\frac {7 x^{2}}{6}-\frac {23 x^{3}}{18}+\frac {11 x^{4}}{8}-\frac {1577 x^{5}}{1080}+\frac {3319 x^{6}}{2160}-\frac {72853 x^{7}}{45360}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{x^{{1}/{3}}}+\frac {-\frac {x^{2}}{12}+\frac {13 x^{3}}{108}-\frac {131 x^{4}}{864}+\frac {11449 x^{5}}{64800}-\frac {76919 x^{6}}{388800}+\frac {4118557 x^{7}}{19051200}+O\left (x^{8}\right )}{x^{{1}/{3}}}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1} \left (1-x +\frac {7 x^{2}}{6}-\frac {23 x^{3}}{18}+\frac {11 x^{4}}{8}-\frac {1577 x^{5}}{1080}+\frac {3319 x^{6}}{2160}-\frac {72853 x^{7}}{45360}+O\left (x^{8}\right )\right )}{x^{{1}/{3}}}+c_{2} \left (\frac {\left (1-x +\frac {7 x^{2}}{6}-\frac {23 x^{3}}{18}+\frac {11 x^{4}}{8}-\frac {1577 x^{5}}{1080}+\frac {3319 x^{6}}{2160}-\frac {72853 x^{7}}{45360}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{x^{{1}/{3}}}+\frac {-\frac {x^{2}}{12}+\frac {13 x^{3}}{108}-\frac {131 x^{4}}{864}+\frac {11449 x^{5}}{64800}-\frac {76919 x^{6}}{388800}+\frac {4118557 x^{7}}{19051200}+O\left (x^{8}\right )}{x^{{1}/{3}}}\right ) \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1} \left (1-x +\frac {7 x^{2}}{6}-\frac {23 x^{3}}{18}+\frac {11 x^{4}}{8}-\frac {1577 x^{5}}{1080}+\frac {3319 x^{6}}{2160}-\frac {72853 x^{7}}{45360}+O\left (x^{8}\right )\right )}{x^{{1}/{3}}}+c_{2} \left (\frac {\left (1-x +\frac {7 x^{2}}{6}-\frac {23 x^{3}}{18}+\frac {11 x^{4}}{8}-\frac {1577 x^{5}}{1080}+\frac {3319 x^{6}}{2160}-\frac {72853 x^{7}}{45360}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{x^{{1}/{3}}}+\frac {-\frac {x^{2}}{12}+\frac {13 x^{3}}{108}-\frac {131 x^{4}}{864}+\frac {11449 x^{5}}{64800}-\frac {76919 x^{6}}{388800}+\frac {4118557 x^{7}}{19051200}+O\left (x^{8}\right )}{x^{{1}/{3}}}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 9 x^{2} \left (x +1\right ) y^{\prime \prime }+\left (-3 x^{3}+33 x^{2}+15 x \right ) y^{\prime }+\left (-7 x^{2}+16 x +1\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 (7 x^{2}-16 x -1\right ) y}{9 x^{2} \left (x +1\right )}+\frac {\left (x^{2}-11 x -5\right ) y^{\prime }}{3 x \left (x +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 {\left (x^{2}-11 x -5\right ) y^{\prime }}{3 x \left (x +1\right )}-\frac {\left (7 x^{2}-16 x -1\right ) y}{9 x^{2} \left (x +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 {x^{2}-11 x -5}{3 x \left (x +1\right )}, P_{3}\left (x \right )=-\frac {7 x^{2}-16 x -1}{9 x^{2} \left (x +1\right )}\right ] \\ {} & \circ & \left (x +1\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=\frac {7}{3} \\ {} & \circ & \left (x +1\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=0 \\ {} & \circ & x =-1\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}=-1 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 9 x^{2} \left (x +1\right ) y^{\prime \prime }-3 x \left (x^{2}-11 x -5\right ) y^{\prime }+\left (-7 x^{2}+16 x +1\right ) y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -1\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (9 u^{3}-18 u^{2}+9 u \right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (-3 u^{3}+42 u^{2}-60 u +21\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (-7 u^{2}+30 u -22\right ) y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot y \left (u \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..3 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & 3 a_{0} r \left (4+3 r \right ) u^{-1+r}+\left (3 a_{1} \left (1+r \right ) \left (7+3 r \right )-2 a_{0} \left (9 r^{2}+21 r +11\right )\right ) u^{r}+\left (3 a_{2} \left (2+r \right ) \left (10+3 r \right )-2 a_{1} \left (9 r^{2}+39 r +41\right )+3 a_{0} \left (2+r \right ) \left (5+3 r \right )\right ) u^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (3 a_{k +1} \left (k +1+r \right ) \left (3 k +3 r +7\right )-2 a_{k} \left (9 k^{2}+18 k r +9 r^{2}+21 k +21 r +11\right )+3 a_{k -1} \left (k +1+r \right ) \left (3 k +2+3 r \right )-a_{k -2} \left (3 k +1+3 r \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 3 r \left (4+3 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -\frac {4}{3}\right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} u \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [3 a_{1} \left (1+r \right ) \left (7+3 r \right )-2 a_{0} \left (9 r^{2}+21 r +11\right )=0, 3 a_{2} \left (2+r \right ) \left (10+3 r \right )-2 a_{1} \left (9 r^{2}+39 r +41\right )+3 a_{0} \left (2+r \right ) \left (5+3 r \right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=\frac {2 a_{0} \left (9 r^{2}+21 r +11\right )}{3 \left (3 r^{2}+10 r +7\right )}, a_{2}=\frac {a_{0} \left (243 r^{4}+1593 r^{3}+3699 r^{2}+3567 r +1174\right )}{9 \left (9 r^{4}+78 r^{3}+241 r^{2}+312 r +140\right )}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 9 \left (-2 a_{k}+a_{k -1}+a_{k +1}\right ) k^{2}+3 \left (6 \left (-2 a_{k}+a_{k -1}+a_{k +1}\right ) r -14 a_{k}-a_{k -2}+5 a_{k -1}+10 a_{k +1}\right ) k +9 \left (-2 a_{k}+a_{k -1}+a_{k +1}\right ) r^{2}+3 \left (-14 a_{k}-a_{k -2}+5 a_{k -1}+10 a_{k +1}\right ) r -22 a_{k}-a_{k -2}+6 a_{k -1}+21 a_{k +1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & 9 \left (-2 a_{k +2}+a_{k +1}+a_{k +3}\right ) \left (k +2\right )^{2}+3 \left (6 \left (-2 a_{k +2}+a_{k +1}+a_{k +3}\right ) r -14 a_{k +2}-a_{k}+5 a_{k +1}+10 a_{k +3}\right ) \left (k +2\right )+9 \left (-2 a_{k +2}+a_{k +1}+a_{k +3}\right ) r^{2}+3 \left (-14 a_{k +2}-a_{k}+5 a_{k +1}+10 a_{k +3}\right ) r -22 a_{k +2}-a_{k}+6 a_{k +1}+21 a_{k +3}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=-\frac {9 k^{2} a_{k +1}-18 k^{2} a_{k +2}+18 k r a_{k +1}-36 k r a_{k +2}+9 r^{2} a_{k +1}-18 r^{2} a_{k +2}-3 k a_{k}+51 k a_{k +1}-114 k a_{k +2}-3 r a_{k}+51 r a_{k +1}-114 r a_{k +2}-7 a_{k}+72 a_{k +1}-178 a_{k +2}}{3 \left (3 k^{2}+6 k r +3 r^{2}+22 k +22 r +39\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +3}=-\frac {9 k^{2} a_{k +1}-18 k^{2} a_{k +2}-3 k a_{k}+51 k a_{k +1}-114 k a_{k +2}-7 a_{k}+72 a_{k +1}-178 a_{k +2}}{3 \left (3 k^{2}+22 k +39\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k}, a_{k +3}=-\frac {9 k^{2} a_{k +1}-18 k^{2} a_{k +2}-3 k a_{k}+51 k a_{k +1}-114 k a_{k +2}-7 a_{k}+72 a_{k +1}-178 a_{k +2}}{3 \left (3 k^{2}+22 k +39\right )}, a_{1}=\frac {22 a_{0}}{21}, a_{2}=\frac {587 a_{0}}{630}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k}, a_{k +3}=-\frac {9 k^{2} a_{k +1}-18 k^{2} a_{k +2}-3 k a_{k}+51 k a_{k +1}-114 k a_{k +2}-7 a_{k}+72 a_{k +1}-178 a_{k +2}}{3 \left (3 k^{2}+22 k +39\right )}, a_{1}=\frac {22 a_{0}}{21}, a_{2}=\frac {587 a_{0}}{630}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {4}{3} \\ {} & {} & a_{k +3}=-\frac {9 k^{2} a_{k +1}-18 k^{2} a_{k +2}-3 k a_{k}+27 k a_{k +1}-66 k a_{k +2}-3 a_{k}+20 a_{k +1}-58 a_{k +2}}{3 \left (3 k^{2}+14 k +15\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {4}{3} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k -\frac {4}{3}}, a_{k +3}=-\frac {9 k^{2} a_{k +1}-18 k^{2} a_{k +2}-3 k a_{k}+27 k a_{k +1}-66 k a_{k +2}-3 a_{k}+20 a_{k +1}-58 a_{k +2}}{3 \left (3 k^{2}+14 k +15\right )}, a_{1}=\frac {2 a_{0}}{3}, a_{2}=\frac {7 a_{0}}{18}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k -\frac {4}{3}}, a_{k +3}=-\frac {9 k^{2} a_{k +1}-18 k^{2} a_{k +2}-3 k a_{k}+27 k a_{k +1}-66 k a_{k +2}-3 a_{k}+20 a_{k +1}-58 a_{k +2}}{3 \left (3 k^{2}+14 k +15\right )}, a_{1}=\frac {2 a_{0}}{3}, a_{2}=\frac {7 a_{0}}{18}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (x +1\right )^{k -\frac {4}{3}}\right ), a_{k +3}=-\frac {9 k^{2} a_{1+k}-18 k^{2} a_{k +2}-3 k a_{k}+51 k a_{1+k}-114 k a_{k +2}-7 a_{k}+72 a_{1+k}-178 a_{k +2}}{3 \left (3 k^{2}+22 k +39\right )}, a_{1}=\frac {22 a_{0}}{21}, a_{2}=\frac {587 a_{0}}{630}, b_{k +3}=-\frac {9 k^{2} b_{1+k}-18 k^{2} b_{k +2}-3 k b_{k}+27 k b_{1+k}-66 k b_{k +2}-3 b_{k}+20 b_{1+k}-58 b_{k +2}}{3 \left (3 k^{2}+14 k +15\right )}, b_{1}=\frac {2 b_{0}}{3}, b_{2}=\frac {7 b_{0}}{18}\right ] \end {array} \]
Maple trace Kovacic algorithm successful
`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 A Liouvillian solution exists Reducible group (found an exponential solution) Group is reducible, not completely reducible Solution has integrals. Trying a special function solution free of integrals... -> 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 -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius <- Heun successful: received ODE is equivalent to the HeunC ODE, case a <> 0, e <> 0, c = 0 <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.079 (sec). Leaf size: 79
Order:=8; dsolve(9*x^2*(1+x)*diff(y(x),x$2)+3*x*(5+11*x-x^2)*diff(y(x),x)+(1+16*x-7*x^2)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {\left (c_{2} \ln \left (x \right )+c_{1} \right ) \left (1-x +\frac {7}{6} x^{2}-\frac {23}{18} x^{3}+\frac {11}{8} x^{4}-\frac {1577}{1080} x^{5}+\frac {3319}{2160} x^{6}-\frac {72853}{45360} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+\left (-\frac {1}{12} x^{2}+\frac {13}{108} x^{3}-\frac {131}{864} x^{4}+\frac {11449}{64800} x^{5}-\frac {76919}{388800} x^{6}+\frac {4118557}{19051200} x^{7}+\operatorname {O}\left (x^{8}\right )\right ) c_{2}}{x^{\frac {1}{3}}} \]
✓ Solution by Mathematica
Time used: 0.016 (sec). Leaf size: 167
AsymptoticDSolveValue[9*x^2*(1+x)*y''[x]+3*x*(5+11*x-x^2)*y'[x]+(1+16*x-7*x^2)*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to \frac {c_1 \left (-\frac {72853 x^7}{45360}+\frac {3319 x^6}{2160}-\frac {1577 x^5}{1080}+\frac {11 x^4}{8}-\frac {23 x^3}{18}+\frac {7 x^2}{6}-x+1\right )}{\sqrt [3]{x}}+c_2 \left (\frac {\frac {4118557 x^7}{19051200}-\frac {76919 x^6}{388800}+\frac {11449 x^5}{64800}-\frac {131 x^4}{864}+\frac {13 x^3}{108}-\frac {x^2}{12}}{\sqrt [3]{x}}+\frac {\left (-\frac {72853 x^7}{45360}+\frac {3319 x^6}{2160}-\frac {1577 x^5}{1080}+\frac {11 x^4}{8}-\frac {23 x^3}{18}+\frac {7 x^2}{6}-x+1\right ) \log (x)}{\sqrt [3]{x}}\right ) \]