Internal problem ID [6998]
Internal file name [OUTPUT/6241_Thursday_August_18_2022_07_11_22_AM_33243587/index.tex]
Book: Elementary differential equations. Rainville, Bedient, Bedient. Prentice Hall. NJ. 8th
edition. 1997.
Section: CHAPTER 18. Power series solutions. 18.11 Many-Term Recurrence Relations. Exercises
page 391
Problem number: 4.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference is integer"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x^{2} y^{\prime \prime }+x \left (1+x \right ) y^{\prime }-\left (6 x^{2}-3 x +1\right ) y=0} \] 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. \[ x^{2} y^{\prime \prime }+\left (x^{2}+x \right ) y^{\prime }+\left (-6 x^{2}+3 x -1\right ) y = 0 \] 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 {1+x}{x}\\ q(x) &= -\frac {6 x^{2}-3 x +1}{x^{2}}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} y^{\prime \prime }+\left (x^{2}+x \right ) y^{\prime }+\left (-6 x^{2}+3 x -1\right ) y = 0 \] 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} \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}+x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (-6 x^{2}+3 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 }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 x^{n +r +2} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{1+n +r} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-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^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 x^{n +r +2} a_{n}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-6 a_{n -2} x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}3 x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}3 a_{n -1} x^{n +r} \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-6 a_{n -2} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}3 a_{n -1} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+x^{n +r} a_{n} \left (n +r \right )-a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )+x^{r} a_{0} r -a_{0} x^{r} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )+x^{r} r -x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r^{2}-1\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r^{2}-1 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 1\\ r_2 &= -1 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r^{2}-1\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([1, -1]\).
Since \(r_1 - r_2 = 2\) is 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 ) &= C y_{1}\left (x \right ) \ln \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 ) &= x \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\frac {\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}}{x} \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{1+n}\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -1}\right ) \end {align*}
Where \(C\) above can be zero. We start by finding \(y_{1}\). 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 \[ a_{1} = \frac {-3-r}{r \left (r +2\right )} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n -1} \left (n +r -1\right )+a_{n} \left (n +r \right )-6 a_{n -2}+3 a_{n -1}-a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {n a_{n -1}+r a_{n -1}-6 a_{n -2}+2 a_{n -1}}{n^{2}+2 n r +r^{2}-1}\tag {4} \] Which for the root \(r = 1\) becomes \[ a_{n} = \frac {-n a_{n -1}+6 a_{n -2}-3 a_{n -1}}{n \left (n +2\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 1\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-r}{r \left (r +2\right )}\) | \(-{\frac {4}{3}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {7 r +12}{\left (r +3\right ) r \left (r +2\right )} \] Which for the root \(r = 1\) becomes \[ a_{2}={\frac {19}{12}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-r}{r \left (r +2\right )}\) | \(-{\frac {4}{3}}\) |
| \(a_{2}\) | \(\frac {7 r +12}{\left (r +3\right ) r \left (r +2\right )}\) | \(\frac {19}{12}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-57-13 r}{\left (r +4\right ) \left (r +2\right ) \left (r +3\right ) r} \] Which for the root \(r = 1\) becomes \[ a_{3}=-{\frac {7}{6}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-r}{r \left (r +2\right )}\) | \(-{\frac {4}{3}}\) |
| \(a_{2}\) | \(\frac {7 r +12}{\left (r +3\right ) r \left (r +2\right )}\) | \(\frac {19}{12}\) |
| \(a_{3}\) | \(\frac {-57-13 r}{\left (r +4\right ) \left (r +2\right ) \left (r +3\right ) r}\) | \(-{\frac {7}{6}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {55 r +210}{\left (r +5\right ) \left (r +3\right ) r \left (r +2\right ) \left (r +4\right )} \] Which for the root \(r = 1\) becomes \[ a_{4}={\frac {53}{72}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-r}{r \left (r +2\right )}\) | \(-{\frac {4}{3}}\) |
| \(a_{2}\) | \(\frac {7 r +12}{\left (r +3\right ) r \left (r +2\right )}\) | \(\frac {19}{12}\) |
| \(a_{3}\) | \(\frac {-57-13 r}{\left (r +4\right ) \left (r +2\right ) \left (r +3\right ) r}\) | \(-{\frac {7}{6}}\) |
| \(a_{4}\) | \(\frac {55 r +210}{\left (r +5\right ) \left (r +3\right ) r \left (r +2\right ) \left (r +4\right )}\) | \(\frac {53}{72}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-795-133 r}{\left (r +6\right ) \left (r +4\right ) \left (r +2\right ) r \left (r +3\right ) \left (r +5\right )} \] Which for the root \(r = 1\) becomes \[ a_{5}=-{\frac {116}{315}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-r}{r \left (r +2\right )}\) | \(-{\frac {4}{3}}\) |
| \(a_{2}\) | \(\frac {7 r +12}{\left (r +3\right ) r \left (r +2\right )}\) | \(\frac {19}{12}\) |
| \(a_{3}\) | \(\frac {-57-13 r}{\left (r +4\right ) \left (r +2\right ) \left (r +3\right ) r}\) | \(-{\frac {7}{6}}\) |
| \(a_{4}\) | \(\frac {55 r +210}{\left (r +5\right ) \left (r +3\right ) r \left (r +2\right ) \left (r +4\right )}\) | \(\frac {53}{72}\) |
| \(a_{5}\) | \(\frac {-795-133 r}{\left (r +6\right ) \left (r +4\right ) \left (r +2\right ) r \left (r +3\right ) \left (r +5\right )}\) | \(-{\frac {116}{315}}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {463 r +2784}{\left (r +7\right ) \left (r +5\right ) \left (r +3\right ) r \left (r +2\right ) \left (r +4\right ) \left (r +6\right )} \] Which for the root \(r = 1\) becomes \[ a_{6}={\frac {3247}{20160}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-r}{r \left (r +2\right )}\) | \(-{\frac {4}{3}}\) |
| \(a_{2}\) | \(\frac {7 r +12}{\left (r +3\right ) r \left (r +2\right )}\) | \(\frac {19}{12}\) |
| \(a_{3}\) | \(\frac {-57-13 r}{\left (r +4\right ) \left (r +2\right ) \left (r +3\right ) r}\) | \(-{\frac {7}{6}}\) |
| \(a_{4}\) | \(\frac {55 r +210}{\left (r +5\right ) \left (r +3\right ) r \left (r +2\right ) \left (r +4\right )}\) | \(\frac {53}{72}\) |
| \(a_{5}\) | \(\frac {-795-133 r}{\left (r +6\right ) \left (r +4\right ) \left (r +2\right ) r \left (r +3\right ) \left (r +5\right )}\) | \(-{\frac {116}{315}}\) |
| \(a_{6}\) | \(\frac {463 r +2784}{\left (r +7\right ) \left (r +5\right ) \left (r +3\right ) r \left (r +2\right ) \left (r +4\right ) \left (r +6\right )}\) | \(\frac {3247}{20160}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=\frac {-9741-1261 r}{\left (8+r \right ) \left (r +6\right ) \left (r +4\right ) \left (r +2\right ) r \left (r +3\right ) \left (r +5\right ) \left (r +7\right )} \] Which for the root \(r = 1\) becomes \[ a_{7}=-{\frac {5501}{90720}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-r}{r \left (r +2\right )}\) | \(-{\frac {4}{3}}\) |
| \(a_{2}\) | \(\frac {7 r +12}{\left (r +3\right ) r \left (r +2\right )}\) | \(\frac {19}{12}\) |
| \(a_{3}\) | \(\frac {-57-13 r}{\left (r +4\right ) \left (r +2\right ) \left (r +3\right ) r}\) | \(-{\frac {7}{6}}\) |
| \(a_{4}\) | \(\frac {55 r +210}{\left (r +5\right ) \left (r +3\right ) r \left (r +2\right ) \left (r +4\right )}\) | \(\frac {53}{72}\) |
| \(a_{5}\) | \(\frac {-795-133 r}{\left (r +6\right ) \left (r +4\right ) \left (r +2\right ) r \left (r +3\right ) \left (r +5\right )}\) | \(-{\frac {116}{315}}\) |
| \(a_{6}\) | \(\frac {463 r +2784}{\left (r +7\right ) \left (r +5\right ) \left (r +3\right ) r \left (r +2\right ) \left (r +4\right ) \left (r +6\right )}\) | \(\frac {3247}{20160}\) |
| \(a_{7}\) | \(\frac {-9741-1261 r}{\left (8+r \right ) \left (r +6\right ) \left (r +4\right ) \left (r +2\right ) r \left (r +3\right ) \left (r +5\right ) \left (r +7\right )}\) | \(-{\frac {5501}{90720}}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= 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}+a_{7} x^{7}+a_{8} x^{8}\dots \right ) \\ &= x \left (1-\frac {4 x}{3}+\frac {19 x^{2}}{12}-\frac {7 x^{3}}{6}+\frac {53 x^{4}}{72}-\frac {116 x^{5}}{315}+\frac {3247 x^{6}}{20160}-\frac {5501 x^{7}}{90720}+O\left (x^{8}\right )\right ) \end {align*}
Now the second solution \(y_{2}\left (x \right )\) is found. Let \[ r_{1}-r_{2} = N \] Where \(N\) is positive integer which is the difference between the two roots. \(r_{1}\) is taken as the larger root. Hence for this problem we have \(N=2\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{2}\left (r \right )\). If this limit exists, then \(C = 0\), else we need to keep the log term and \(C \neq 0\). The above table shows that \begin {align*} a_N &= a_{2} \\ &= \frac {7 r +12}{\left (r +3\right ) r \left (r +2\right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {7 r +12}{\left (r +3\right ) r \left (r +2\right )}&= \lim _{r\rightarrow -1}\frac {7 r +12}{\left (r +3\right ) r \left (r +2\right )}\\ &= -{\frac {5}{2}} \end {align*}
The limit is \(-{\frac {5}{2}}\). Since the limit exists then the log term is not needed and we can set \(C = 0\). Therefore the second solution has the form \begin {align*} y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r}\\ &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -1} \end {align*}
Eq (3) derived above is 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(3) gives \[ b_{1} = -\frac {r +3}{r \left (r +2\right )} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{4} b_{n} \left (n +r \right ) \left (n +r -1\right )+b_{n -1} \left (n +r -1\right )+b_{n} \left (n +r \right )-6 b_{n -2}+3 b_{n -1}-b_{n} = 0 \end{equation} Which for for the root \(r = -1\) becomes \begin{equation} \tag{4A} b_{n} \left (n -1\right ) \left (n -2\right )+b_{n -1} \left (n -2\right )+b_{n} \left (n -1\right )-6 b_{n -2}+3 b_{n -1}-b_{n} = 0 \end{equation} Solving for \(b_{n}\) from the recursive equation (4) gives \[ b_{n} = -\frac {n b_{n -1}+r b_{n -1}-6 b_{n -2}+2 b_{n -1}}{n^{2}+2 n r +r^{2}-1}\tag {5} \] Which for the root \(r = -1\) becomes \[ b_{n} = -\frac {n b_{n -1}-6 b_{n -2}+b_{n -1}}{n^{2}-2 n}\tag {6} \] 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.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-3-r}{r \left (r +2\right )}\) | \(2\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {7 r +12}{\left (r +3\right ) r \left (r +2\right )} \] Which for the root \(r = -1\) becomes \[ b_{2}=-{\frac {5}{2}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-3-r}{r \left (r +2\right )}\) | \(2\) |
| \(b_{2}\) | \(\frac {7 r +12}{\left (r +3\right ) r \left (r +2\right )}\) | \(-{\frac {5}{2}}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=-\frac {13 r +57}{\left (r^{2}+6 r +8\right ) \left (r +3\right ) r} \] Which for the root \(r = -1\) becomes \[ b_{3}={\frac {22}{3}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-3-r}{r \left (r +2\right )}\) | \(2\) |
| \(b_{2}\) | \(\frac {7 r +12}{\left (r +3\right ) r \left (r +2\right )}\) | \(-{\frac {5}{2}}\) |
| \(b_{3}\) | \(\frac {-57-13 r}{\left (r +4\right ) \left (r +2\right ) \left (r +3\right ) r}\) | \(\frac {22}{3}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {55 r +210}{r \left (r +2\right ) \left (r +4\right ) \left (r^{2}+8 r +15\right )} \] Which for the root \(r = -1\) becomes \[ b_{4}=-{\frac {155}{24}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-3-r}{r \left (r +2\right )}\) | \(2\) |
| \(b_{2}\) | \(\frac {7 r +12}{\left (r +3\right ) r \left (r +2\right )}\) | \(-{\frac {5}{2}}\) |
| \(b_{3}\) | \(\frac {-57-13 r}{\left (r +4\right ) \left (r +2\right ) \left (r +3\right ) r}\) | \(\frac {22}{3}\) |
| \(b_{4}\) | \(\frac {55 r +210}{\left (r +5\right ) \left (r +3\right ) r \left (r +2\right ) \left (r +4\right )}\) | \(-{\frac {155}{24}}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-\frac {133 r +795}{\left (r^{2}+10 r +24\right ) \left (r +2\right ) r \left (r +3\right ) \left (r +5\right )} \] Which for the root \(r = -1\) becomes \[ b_{5}={\frac {331}{60}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-3-r}{r \left (r +2\right )}\) | \(2\) |
| \(b_{2}\) | \(\frac {7 r +12}{\left (r +3\right ) r \left (r +2\right )}\) | \(-{\frac {5}{2}}\) |
| \(b_{3}\) | \(\frac {-57-13 r}{\left (r +4\right ) \left (r +2\right ) \left (r +3\right ) r}\) | \(\frac {22}{3}\) |
| \(b_{4}\) | \(\frac {55 r +210}{\left (r +5\right ) \left (r +3\right ) r \left (r +2\right ) \left (r +4\right )}\) | \(-{\frac {155}{24}}\) |
| \(b_{5}\) | \(\frac {-795-133 r}{\left (r +6\right ) \left (r +4\right ) \left (r +2\right ) r \left (r +3\right ) \left (r +5\right )}\) | \(\frac {331}{60}\) |
For \(n = 6\), using the above recursive equation gives \[ b_{6}=\frac {463 r +2784}{\left (r^{2}+12 r +35\right ) \left (r +3\right ) r \left (r +2\right ) \left (r +4\right ) \left (r +6\right )} \] Which for the root \(r = -1\) becomes \[ b_{6}=-{\frac {2321}{720}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-3-r}{r \left (r +2\right )}\) | \(2\) |
| \(b_{2}\) | \(\frac {7 r +12}{\left (r +3\right ) r \left (r +2\right )}\) | \(-{\frac {5}{2}}\) |
| \(b_{3}\) | \(\frac {-57-13 r}{\left (r +4\right ) \left (r +2\right ) \left (r +3\right ) r}\) | \(\frac {22}{3}\) |
| \(b_{4}\) | \(\frac {55 r +210}{\left (r +5\right ) \left (r +3\right ) r \left (r +2\right ) \left (r +4\right )}\) | \(-{\frac {155}{24}}\) |
| \(b_{5}\) | \(\frac {-795-133 r}{\left (r +6\right ) \left (r +4\right ) \left (r +2\right ) r \left (r +3\right ) \left (r +5\right )}\) | \(\frac {331}{60}\) |
| \(b_{6}\) | \(\frac {463 r +2784}{\left (r +7\right ) \left (r +5\right ) \left (r +3\right ) r \left (r +2\right ) \left (r +4\right ) \left (r +6\right )}\) | \(-{\frac {2321}{720}}\) |
For \(n = 7\), using the above recursive equation gives \[ b_{7}=-\frac {1261 r +9741}{\left (r +4\right ) \left (r +2\right ) r \left (r +3\right ) \left (r +5\right ) \left (r +7\right ) \left (r^{2}+14 r +48\right )} \] Which for the root \(r = -1\) becomes \[ b_{7}={\frac {106}{63}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-3-r}{r \left (r +2\right )}\) | \(2\) |
| \(b_{2}\) | \(\frac {7 r +12}{\left (r +3\right ) r \left (r +2\right )}\) | \(-{\frac {5}{2}}\) |
| \(b_{3}\) | \(\frac {-57-13 r}{\left (r +4\right ) \left (r +2\right ) \left (r +3\right ) r}\) | \(\frac {22}{3}\) |
| \(b_{4}\) | \(\frac {55 r +210}{\left (r +5\right ) \left (r +3\right ) r \left (r +2\right ) \left (r +4\right )}\) | \(-{\frac {155}{24}}\) |
| \(b_{5}\) | \(\frac {-795-133 r}{\left (r +6\right ) \left (r +4\right ) \left (r +2\right ) r \left (r +3\right ) \left (r +5\right )}\) | \(\frac {331}{60}\) |
| \(b_{6}\) | \(\frac {463 r +2784}{\left (r +7\right ) \left (r +5\right ) \left (r +3\right ) r \left (r +2\right ) \left (r +4\right ) \left (r +6\right )}\) | \(-{\frac {2321}{720}}\) |
| \(b_{7}\) | \(\frac {-9741-1261 r}{\left (8+r \right ) \left (r +6\right ) \left (r +4\right ) \left (r +2\right ) r \left (r +3\right ) \left (r +5\right ) \left (r +7\right )}\) | \(\frac {106}{63}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= 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}+b_{7} x^{7}+b_{8} x^{8}\dots \right ) \\ &= \frac {1+2 x -\frac {5 x^{2}}{2}+\frac {22 x^{3}}{3}-\frac {155 x^{4}}{24}+\frac {331 x^{5}}{60}-\frac {2321 x^{6}}{720}+\frac {106 x^{7}}{63}+O\left (x^{8}\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 ) \\ &= c_{1} x \left (1-\frac {4 x}{3}+\frac {19 x^{2}}{12}-\frac {7 x^{3}}{6}+\frac {53 x^{4}}{72}-\frac {116 x^{5}}{315}+\frac {3247 x^{6}}{20160}-\frac {5501 x^{7}}{90720}+O\left (x^{8}\right )\right ) + \frac {c_{2} \left (1+2 x -\frac {5 x^{2}}{2}+\frac {22 x^{3}}{3}-\frac {155 x^{4}}{24}+\frac {331 x^{5}}{60}-\frac {2321 x^{6}}{720}+\frac {106 x^{7}}{63}+O\left (x^{8}\right )\right )}{x} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x \left (1-\frac {4 x}{3}+\frac {19 x^{2}}{12}-\frac {7 x^{3}}{6}+\frac {53 x^{4}}{72}-\frac {116 x^{5}}{315}+\frac {3247 x^{6}}{20160}-\frac {5501 x^{7}}{90720}+O\left (x^{8}\right )\right )+\frac {c_{2} \left (1+2 x -\frac {5 x^{2}}{2}+\frac {22 x^{3}}{3}-\frac {155 x^{4}}{24}+\frac {331 x^{5}}{60}-\frac {2321 x^{6}}{720}+\frac {106 x^{7}}{63}+O\left (x^{8}\right )\right )}{x} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x \left (1-\frac {4 x}{3}+\frac {19 x^{2}}{12}-\frac {7 x^{3}}{6}+\frac {53 x^{4}}{72}-\frac {116 x^{5}}{315}+\frac {3247 x^{6}}{20160}-\frac {5501 x^{7}}{90720}+O\left (x^{8}\right )\right )+\frac {c_{2} \left (1+2 x -\frac {5 x^{2}}{2}+\frac {22 x^{3}}{3}-\frac {155 x^{4}}{24}+\frac {331 x^{5}}{60}-\frac {2321 x^{6}}{720}+\frac {106 x^{7}}{63}+O\left (x^{8}\right )\right )}{x} \\ \end{align*}
Verification of solutions
\[ y = c_{1} x \left (1-\frac {4 x}{3}+\frac {19 x^{2}}{12}-\frac {7 x^{3}}{6}+\frac {53 x^{4}}{72}-\frac {116 x^{5}}{315}+\frac {3247 x^{6}}{20160}-\frac {5501 x^{7}}{90720}+O\left (x^{8}\right )\right )+\frac {c_{2} \left (1+2 x -\frac {5 x^{2}}{2}+\frac {22 x^{3}}{3}-\frac {155 x^{4}}{24}+\frac {331 x^{5}}{60}-\frac {2321 x^{6}}{720}+\frac {106 x^{7}}{63}+O\left (x^{8}\right )\right )}{x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime } x^{2}+\left (x^{2}+x \right ) y^{\prime }+\left (-6 x^{2}+3 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 (6 x^{2}-3 x +1\right ) y}{x^{2}}-\frac {\left (1+x \right ) y^{\prime }}{x} \\ \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 (1+x \right ) y^{\prime }}{x}-\frac {\left (6 x^{2}-3 x +1\right ) y}{x^{2}}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {1+x}{x}, P_{3}\left (x \right )=-\frac {6 x^{2}-3 x +1}{x^{2}}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=1 \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=-1 \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & y^{\prime \prime } x^{2}+x \left (1+x \right ) y^{\prime }+\left (-6 x^{2}+3 x -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^{2}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (1+r \right ) \left (-1+r \right ) x^{r}+\left (a_{1} \left (2+r \right ) r +a_{0} \left (3+r \right )\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (k +r +1\right ) \left (k +r -1\right )+a_{k -1} \left (k +2+r \right )-6 a_{k -2}\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+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-1, 1\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (2+r \right ) r +a_{0} \left (3+r \right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=-\frac {a_{0} \left (3+r \right )}{r \left (2+r \right )} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (k +r +1\right ) \left (k +r -1\right )+a_{k -1} \left (k +2+r \right )-6 a_{k -2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +2} \left (k +r +3\right ) \left (k +r +1\right )+a_{k +1} \left (k +4+r \right )-6 a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {k a_{k +1}+r a_{k +1}-6 a_{k}+4 a_{k +1}}{\left (k +r +3\right ) \left (k +r +1\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1 \\ {} & {} & a_{k +2}=-\frac {k a_{k +1}-6 a_{k}+3 a_{k +1}}{\left (k +2\right ) k} \\ \bullet & {} & \textrm {Series not valid for}\hspace {3pt} r =-1\hspace {3pt}\textrm {, division by}\hspace {3pt} 0\hspace {3pt}\textrm {in the recursion relation at}\hspace {3pt} k =0 \\ {} & {} & a_{k +2}=-\frac {k a_{k +1}-6 a_{k}+3 a_{k +1}}{\left (k +2\right ) k} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +2}=-\frac {k a_{k +1}-6 a_{k}+5 a_{k +1}}{\left (k +4\right ) \left (k +2\right )} \\ \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 a_{k +1}-6 a_{k}+5 a_{k +1}}{\left (k +4\right ) \left (k +2\right )}, a_{1}=-\frac {4 a_{0}}{3}\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) Reducible group (found another exponential solution) <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 53
Order:=8; dsolve(x^2*diff(y(x),x$2)+x*(1+x)*diff(y(x),x)-(1-3*x+6*x^2)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x \left (1-\frac {4}{3} x +\frac {19}{12} x^{2}-\frac {7}{6} x^{3}+\frac {53}{72} x^{4}-\frac {116}{315} x^{5}+\frac {3247}{20160} x^{6}-\frac {5501}{90720} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+\frac {c_{2} \left (-2-4 x +5 x^{2}-\frac {44}{3} x^{3}+\frac {155}{12} x^{4}-\frac {331}{30} x^{5}+\frac {2321}{360} x^{6}-\frac {212}{63} x^{7}+\operatorname {O}\left (x^{8}\right )\right )}{x} \]
✓ Solution by Mathematica
Time used: 0.149 (sec). Leaf size: 92
AsymptoticDSolveValue[x^2*y''[x]+x*(1+x)*y'[x]-(1-3*x+6*x^2)*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_1 \left (-\frac {2321 x^5}{720}+\frac {331 x^4}{60}-\frac {155 x^3}{24}+\frac {22 x^2}{3}-\frac {5 x}{2}+\frac {1}{x}+2\right )+c_2 \left (\frac {3247 x^7}{20160}-\frac {116 x^6}{315}+\frac {53 x^5}{72}-\frac {7 x^4}{6}+\frac {19 x^3}{12}-\frac {4 x^2}{3}+x\right ) \]