Internal problem ID [6976]
Internal file name [OUTPUT/6219_Friday_August_12_2022_11_05_52_PM_96052526/index.tex]
Book: Elementary differential equations. Rainville, Bedient, Bedient. Prentice Hall. NJ. 8th
edition. 1997.
Section: CHAPTER 18. Power series solutions. 18.8 Indicial Equation with Difference of Roots a
Positive Integer: Nonlogarithmic Case. Exercises page 380
Problem number: 12.
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]]
\[ \boxed {x y^{\prime \prime }+\left (3+2 x \right ) y^{\prime }+4 y=0} \] 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. \[ x y^{\prime \prime }+\left (3+2 x \right ) y^{\prime }+4 y = 0 \] 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 {3+2 x}{x}\\ q(x) &= \frac {4}{x}\\ \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 \[ x y^{\prime \prime }+\left (3+2 x \right ) y^{\prime }+4 y = 0 \] 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 \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+2 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+4 \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 -1} 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 )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 a_{n} x^{n +r}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r -1\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r -1}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}4 a_{n} x^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} x^{n +r -1} \\ \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 \(n +r -1\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+3 \left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ x^{-1+r} a_{0} r \left (-1+r \right )+3 r a_{0} x^{-1+r} = 0 \] Or \[ \left (x^{-1+r} r \left (-1+r \right )+3 r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ r \,x^{-1+r} \left (2+r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r \left (2+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 0\\ r_2 &= -2 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ r \,x^{-1+r} \left (2+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([0, -2]\).
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 ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\\ 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^{2}} \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{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 -2}\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 \(a_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} a_{n} \left (n +r \right ) \left (n +r -1\right )+2 a_{n -1} \left (n +r -1\right )+3 a_{n} \left (n +r \right )+4 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {2 a_{n -1} \left (n +r +1\right )}{n^{2}+2 n r +r^{2}+2 n +2 r}\tag {4} \] Which for the root \(r = 0\) becomes \[ a_{n} = -\frac {2 a_{n -1} \left (n +1\right )}{n \left (n +2\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 0\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ a_{1}=\frac {-4-2 r}{r^{2}+4 r +3} \] Which for the root \(r = 0\) becomes \[ a_{1}=-{\frac {4}{3}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-4-2 r}{r^{2}+4 r +3}\) | \(-{\frac {4}{3}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {4}{\left (r +4\right ) \left (r +1\right )} \] Which for the root \(r = 0\) becomes \[ a_{2}=1 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-4-2 r}{r^{2}+4 r +3}\) | \(-{\frac {4}{3}}\) |
| \(a_{2}\) | \(\frac {4}{\left (r +4\right ) \left (r +1\right )}\) | \(1\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {8}{\left (r +1\right ) \left (r +5\right ) \left (3+r \right )} \] Which for the root \(r = 0\) becomes \[ a_{3}=-{\frac {8}{15}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-4-2 r}{r^{2}+4 r +3}\) | \(-{\frac {4}{3}}\) |
| \(a_{2}\) | \(\frac {4}{\left (r +4\right ) \left (r +1\right )}\) | \(1\) |
| \(a_{3}\) | \(-\frac {8}{\left (r +1\right ) \left (r +5\right ) \left (3+r \right )}\) | \(-{\frac {8}{15}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {16}{\left (r +1\right ) \left (3+r \right ) \left (r +6\right ) \left (r +4\right )} \] Which for the root \(r = 0\) becomes \[ a_{4}={\frac {2}{9}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-4-2 r}{r^{2}+4 r +3}\) | \(-{\frac {4}{3}}\) |
| \(a_{2}\) | \(\frac {4}{\left (r +4\right ) \left (r +1\right )}\) | \(1\) |
| \(a_{3}\) | \(-\frac {8}{\left (r +1\right ) \left (r +5\right ) \left (3+r \right )}\) | \(-{\frac {8}{15}}\) |
| \(a_{4}\) | \(\frac {16}{\left (r +1\right ) \left (3+r \right ) \left (r +6\right ) \left (r +4\right )}\) | \(\frac {2}{9}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {32}{\left (r +1\right ) \left (3+r \right ) \left (r +4\right ) \left (r +7\right ) \left (r +5\right )} \] Which for the root \(r = 0\) becomes \[ a_{5}=-{\frac {8}{105}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-4-2 r}{r^{2}+4 r +3}\) | \(-{\frac {4}{3}}\) |
| \(a_{2}\) | \(\frac {4}{\left (r +4\right ) \left (r +1\right )}\) | \(1\) |
| \(a_{3}\) | \(-\frac {8}{\left (r +1\right ) \left (r +5\right ) \left (3+r \right )}\) | \(-{\frac {8}{15}}\) |
| \(a_{4}\) | \(\frac {16}{\left (r +1\right ) \left (3+r \right ) \left (r +6\right ) \left (r +4\right )}\) | \(\frac {2}{9}\) |
| \(a_{5}\) | \(-\frac {32}{\left (r +1\right ) \left (3+r \right ) \left (r +4\right ) \left (r +7\right ) \left (r +5\right )}\) | \(-{\frac {8}{105}}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {64}{\left (r +1\right ) \left (3+r \right ) \left (r +4\right ) \left (r +5\right ) \left (8+r \right ) \left (r +6\right )} \] Which for the root \(r = 0\) becomes \[ a_{6}={\frac {1}{45}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-4-2 r}{r^{2}+4 r +3}\) | \(-{\frac {4}{3}}\) |
| \(a_{2}\) | \(\frac {4}{\left (r +4\right ) \left (r +1\right )}\) | \(1\) |
| \(a_{3}\) | \(-\frac {8}{\left (r +1\right ) \left (r +5\right ) \left (3+r \right )}\) | \(-{\frac {8}{15}}\) |
| \(a_{4}\) | \(\frac {16}{\left (r +1\right ) \left (3+r \right ) \left (r +6\right ) \left (r +4\right )}\) | \(\frac {2}{9}\) |
| \(a_{5}\) | \(-\frac {32}{\left (r +1\right ) \left (3+r \right ) \left (r +4\right ) \left (r +7\right ) \left (r +5\right )}\) | \(-{\frac {8}{105}}\) |
| \(a_{6}\) | \(\frac {64}{\left (r +1\right ) \left (3+r \right ) \left (r +4\right ) \left (r +5\right ) \left (8+r \right ) \left (r +6\right )}\) | \(\frac {1}{45}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=-\frac {128}{\left (r +1\right ) \left (3+r \right ) \left (r +4\right ) \left (r +5\right ) \left (r +6\right ) \left (9+r \right ) \left (r +7\right )} \] Which for the root \(r = 0\) becomes \[ a_{7}=-{\frac {16}{2835}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-4-2 r}{r^{2}+4 r +3}\) | \(-{\frac {4}{3}}\) |
| \(a_{2}\) | \(\frac {4}{\left (r +4\right ) \left (r +1\right )}\) | \(1\) |
| \(a_{3}\) | \(-\frac {8}{\left (r +1\right ) \left (r +5\right ) \left (3+r \right )}\) | \(-{\frac {8}{15}}\) |
| \(a_{4}\) | \(\frac {16}{\left (r +1\right ) \left (3+r \right ) \left (r +6\right ) \left (r +4\right )}\) | \(\frac {2}{9}\) |
| \(a_{5}\) | \(-\frac {32}{\left (r +1\right ) \left (3+r \right ) \left (r +4\right ) \left (r +7\right ) \left (r +5\right )}\) | \(-{\frac {8}{105}}\) |
| \(a_{6}\) | \(\frac {64}{\left (r +1\right ) \left (3+r \right ) \left (r +4\right ) \left (r +5\right ) \left (8+r \right ) \left (r +6\right )}\) | \(\frac {1}{45}\) |
| \(a_{7}\) | \(-\frac {128}{\left (r +1\right ) \left (3+r \right ) \left (r +4\right ) \left (r +5\right ) \left (r +6\right ) \left (9+r \right ) \left (r +7\right )}\) | \(-{\frac {16}{2835}}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= 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 \\ &= 1-\frac {4 x}{3}+x^{2}-\frac {8 x^{3}}{15}+\frac {2 x^{4}}{9}-\frac {8 x^{5}}{105}+\frac {x^{6}}{45}-\frac {16 x^{7}}{2835}+O\left (x^{8}\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 {4}{\left (r +4\right ) \left (r +1\right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {4}{\left (r +4\right ) \left (r +1\right )}&= \lim _{r\rightarrow -2}\frac {4}{\left (r +4\right ) \left (r +1\right )}\\ &= -2 \end {align*}
The limit is \(-2\). 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 -2} \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 \(b_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{4} b_{n} \left (n +r \right ) \left (n +r -1\right )+2 b_{n -1} \left (n +r -1\right )+3 \left (n +r \right ) b_{n}+4 b_{n -1} = 0 \end{equation} Which for for the root \(r = -2\) becomes \begin{equation} \tag{4A} b_{n} \left (n -2\right ) \left (n -3\right )+2 b_{n -1} \left (n -3\right )+3 \left (n -2\right ) b_{n}+4 b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from the recursive equation (4) gives \[ b_{n} = -\frac {2 b_{n -1} \left (n +r +1\right )}{n^{2}+2 n r +r^{2}+2 n +2 r}\tag {5} \] Which for the root \(r = -2\) becomes \[ b_{n} = -\frac {2 b_{n -1} \left (n -1\right )}{n^{2}-2 n}\tag {6} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -2\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ b_{1}=-\frac {2 \left (2+r \right )}{r^{2}+4 r +3} \] Which for the root \(r = -2\) becomes \[ b_{1}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-4-2 r}{r^{2}+4 r +3}\) | \(0\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {4}{\left (r +4\right ) \left (r +1\right )} \] Which for the root \(r = -2\) becomes \[ b_{2}=-2 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-4-2 r}{r^{2}+4 r +3}\) | \(0\) |
| \(b_{2}\) | \(\frac {4}{\left (r +4\right ) \left (r +1\right )}\) | \(-2\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=-\frac {8}{\left (r +1\right ) \left (r^{2}+8 r +15\right )} \] Which for the root \(r = -2\) becomes \[ b_{3}={\frac {8}{3}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-4-2 r}{r^{2}+4 r +3}\) | \(0\) |
| \(b_{2}\) | \(\frac {4}{\left (r +4\right ) \left (r +1\right )}\) | \(-2\) |
| \(b_{3}\) | \(-\frac {8}{\left (r +1\right ) \left (r +5\right ) \left (3+r \right )}\) | \(\frac {8}{3}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {16}{\left (r +1\right ) \left (3+r \right ) \left (r^{2}+10 r +24\right )} \] Which for the root \(r = -2\) becomes \[ b_{4}=-2 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-4-2 r}{r^{2}+4 r +3}\) | \(0\) |
| \(b_{2}\) | \(\frac {4}{\left (r +4\right ) \left (r +1\right )}\) | \(-2\) |
| \(b_{3}\) | \(-\frac {8}{\left (r +1\right ) \left (r +5\right ) \left (3+r \right )}\) | \(\frac {8}{3}\) |
| \(b_{4}\) | \(\frac {16}{\left (r +1\right ) \left (3+r \right ) \left (r +6\right ) \left (r +4\right )}\) | \(-2\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-\frac {32}{\left (r +1\right ) \left (3+r \right ) \left (r +4\right ) \left (r^{2}+12 r +35\right )} \] Which for the root \(r = -2\) becomes \[ b_{5}={\frac {16}{15}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-4-2 r}{r^{2}+4 r +3}\) | \(0\) |
| \(b_{2}\) | \(\frac {4}{\left (r +4\right ) \left (r +1\right )}\) | \(-2\) |
| \(b_{3}\) | \(-\frac {8}{\left (r +1\right ) \left (r +5\right ) \left (3+r \right )}\) | \(\frac {8}{3}\) |
| \(b_{4}\) | \(\frac {16}{\left (r +1\right ) \left (3+r \right ) \left (r +6\right ) \left (r +4\right )}\) | \(-2\) |
| \(b_{5}\) | \(-\frac {32}{\left (r +1\right ) \left (3+r \right ) \left (r +4\right ) \left (r +7\right ) \left (r +5\right )}\) | \(\frac {16}{15}\) |
For \(n = 6\), using the above recursive equation gives \[ b_{6}=\frac {64}{\left (r +1\right ) \left (3+r \right ) \left (r +4\right ) \left (r +5\right ) \left (r^{2}+14 r +48\right )} \] Which for the root \(r = -2\) becomes \[ b_{6}=-{\frac {4}{9}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-4-2 r}{r^{2}+4 r +3}\) | \(0\) |
| \(b_{2}\) | \(\frac {4}{\left (r +4\right ) \left (r +1\right )}\) | \(-2\) |
| \(b_{3}\) | \(-\frac {8}{\left (r +1\right ) \left (r +5\right ) \left (3+r \right )}\) | \(\frac {8}{3}\) |
| \(b_{4}\) | \(\frac {16}{\left (r +1\right ) \left (3+r \right ) \left (r +6\right ) \left (r +4\right )}\) | \(-2\) |
| \(b_{5}\) | \(-\frac {32}{\left (r +1\right ) \left (3+r \right ) \left (r +4\right ) \left (r +7\right ) \left (r +5\right )}\) | \(\frac {16}{15}\) |
| \(b_{6}\) | \(\frac {64}{\left (r +1\right ) \left (3+r \right ) \left (r +4\right ) \left (r +5\right ) \left (8+r \right ) \left (r +6\right )}\) | \(-{\frac {4}{9}}\) |
For \(n = 7\), using the above recursive equation gives \[ b_{7}=-\frac {128}{\left (r +1\right ) \left (3+r \right ) \left (r +4\right ) \left (r +5\right ) \left (r +6\right ) \left (r^{2}+16 r +63\right )} \] Which for the root \(r = -2\) becomes \[ b_{7}={\frac {16}{105}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-4-2 r}{r^{2}+4 r +3}\) | \(0\) |
| \(b_{2}\) | \(\frac {4}{\left (r +4\right ) \left (r +1\right )}\) | \(-2\) |
| \(b_{3}\) | \(-\frac {8}{\left (r +1\right ) \left (r +5\right ) \left (3+r \right )}\) | \(\frac {8}{3}\) |
| \(b_{4}\) | \(\frac {16}{\left (r +1\right ) \left (3+r \right ) \left (r +6\right ) \left (r +4\right )}\) | \(-2\) |
| \(b_{5}\) | \(-\frac {32}{\left (r +1\right ) \left (3+r \right ) \left (r +4\right ) \left (r +7\right ) \left (r +5\right )}\) | \(\frac {16}{15}\) |
| \(b_{6}\) | \(\frac {64}{\left (r +1\right ) \left (3+r \right ) \left (r +4\right ) \left (r +5\right ) \left (8+r \right ) \left (r +6\right )}\) | \(-{\frac {4}{9}}\) |
| \(b_{7}\) | \(-\frac {128}{\left (r +1\right ) \left (3+r \right ) \left (r +4\right ) \left (r +5\right ) \left (r +6\right ) \left (9+r \right ) \left (r +7\right )}\) | \(\frac {16}{105}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= 1 \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^{2}+\frac {8 x^{3}}{3}-2 x^{4}+\frac {16 x^{5}}{15}-\frac {4 x^{6}}{9}+\frac {16 x^{7}}{105}+O\left (x^{8}\right )}{x^{2}} \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} \left (1-\frac {4 x}{3}+x^{2}-\frac {8 x^{3}}{15}+\frac {2 x^{4}}{9}-\frac {8 x^{5}}{105}+\frac {x^{6}}{45}-\frac {16 x^{7}}{2835}+O\left (x^{8}\right )\right ) + \frac {c_{2} \left (1-2 x^{2}+\frac {8 x^{3}}{3}-2 x^{4}+\frac {16 x^{5}}{15}-\frac {4 x^{6}}{9}+\frac {16 x^{7}}{105}+O\left (x^{8}\right )\right )}{x^{2}} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \left (1-\frac {4 x}{3}+x^{2}-\frac {8 x^{3}}{15}+\frac {2 x^{4}}{9}-\frac {8 x^{5}}{105}+\frac {x^{6}}{45}-\frac {16 x^{7}}{2835}+O\left (x^{8}\right )\right )+\frac {c_{2} \left (1-2 x^{2}+\frac {8 x^{3}}{3}-2 x^{4}+\frac {16 x^{5}}{15}-\frac {4 x^{6}}{9}+\frac {16 x^{7}}{105}+O\left (x^{8}\right )\right )}{x^{2}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (1-\frac {4 x}{3}+x^{2}-\frac {8 x^{3}}{15}+\frac {2 x^{4}}{9}-\frac {8 x^{5}}{105}+\frac {x^{6}}{45}-\frac {16 x^{7}}{2835}+O\left (x^{8}\right )\right )+\frac {c_{2} \left (1-2 x^{2}+\frac {8 x^{3}}{3}-2 x^{4}+\frac {16 x^{5}}{15}-\frac {4 x^{6}}{9}+\frac {16 x^{7}}{105}+O\left (x^{8}\right )\right )}{x^{2}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (1-\frac {4 x}{3}+x^{2}-\frac {8 x^{3}}{15}+\frac {2 x^{4}}{9}-\frac {8 x^{5}}{105}+\frac {x^{6}}{45}-\frac {16 x^{7}}{2835}+O\left (x^{8}\right )\right )+\frac {c_{2} \left (1-2 x^{2}+\frac {8 x^{3}}{3}-2 x^{4}+\frac {16 x^{5}}{15}-\frac {4 x^{6}}{9}+\frac {16 x^{7}}{105}+O\left (x^{8}\right )\right )}{x^{2}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x y^{\prime \prime }+\left (3+2 x \right ) y^{\prime }+4 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 {4 y}{x}-\frac {\left (3+2 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 (3+2 x \right ) y^{\prime }}{x}+\frac {4 y}{x}=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 {3+2 x}{x}, P_{3}\left (x \right )=\frac {4}{x}\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}}}=3 \\ {} & \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}}}=0 \\ {} & \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} \\ {} & {} & x y^{\prime \prime }+\left (3+2 x \right ) y^{\prime }+4 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^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & 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 \cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & x \cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (2+r \right ) x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +3+r \right )+2 a_{k} \left (k +r +2\right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (2+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-2, 0\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +1} \left (k +1+r \right ) \left (k +3+r \right )+2 a_{k} \left (k +r +2\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {2 a_{k} \left (k +r +2\right )}{\left (k +1+r \right ) \left (k +3+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-2 \\ {} & {} & a_{k +1}=-\frac {2 a_{k} k}{\left (k -1\right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Series not valid for}\hspace {3pt} r =-2\hspace {3pt}\textrm {, division by}\hspace {3pt} 0\hspace {3pt}\textrm {in the recursion relation at}\hspace {3pt} k =1 \\ {} & {} & a_{k +1}=-\frac {2 a_{k} k}{\left (k -1\right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=-\frac {2 a_{k} \left (k +2\right )}{\left (k +1\right ) \left (k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +1}=-\frac {2 a_{k} \left (k +2\right )}{\left (k +1\right ) \left (k +3\right )}\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: 0.015 (sec). Leaf size: 50
Order:=8; dsolve(x*diff(y(x),x$2)+(3+2*x)*diff(y(x),x)+4*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} \left (1-\frac {4}{3} x +x^{2}-\frac {8}{15} x^{3}+\frac {2}{9} x^{4}-\frac {8}{105} x^{5}+\frac {1}{45} x^{6}-\frac {16}{2835} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+\frac {c_{2} \left (-2+4 x^{2}-\frac {16}{3} x^{3}+4 x^{4}-\frac {32}{15} x^{5}+\frac {8}{9} x^{6}-\frac {32}{105} x^{7}+\operatorname {O}\left (x^{8}\right )\right )}{x^{2}} \]
✓ Solution by Mathematica
Time used: 0.106 (sec). Leaf size: 77
AsymptoticDSolveValue[x*y''[x]+(3+2*x)*y'[x]+4*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_1 \left (-\frac {4 x^4}{9}+\frac {16 x^3}{15}-2 x^2+\frac {1}{x^2}+\frac {8 x}{3}-2\right )+c_2 \left (\frac {x^6}{45}-\frac {8 x^5}{105}+\frac {2 x^4}{9}-\frac {8 x^3}{15}+x^2-\frac {4 x}{3}+1\right ) \]