Internal problem ID [6968]
Internal file name [OUTPUT/6211_Friday_August_12_2022_11_05_32_PM_10666093/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: 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
[_Laguerre]
\[ \boxed {x y^{\prime \prime }-\left (3+x \right ) y^{\prime }+2 y=0} \] 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. \[ x y^{\prime \prime }+\left (-3-x \right ) y^{\prime }+2 y = 0 \] 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+x}{x}\\ q(x) &= \frac {2}{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 \[ x y^{\prime \prime }+\left (-3-x \right ) y^{\prime }+2 y = 0 \] 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-x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+2 \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 )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 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 }}\left (-x^{n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 a_{n} x^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}2 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 )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 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 \[ r \,x^{-1+r} \left (-4+r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r \left (-4+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 4\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ r \,x^{-1+r} \left (-4+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([4, 0]\).
Since \(r_1 - r_2 = 4\) 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^{4} \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 )+\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 +4}\\ 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}\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 )-a_{n -1} \left (n +r -1\right )-3 a_{n} \left (n +r \right )+2 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {a_{n -1} \left (n +r -3\right )}{n^{2}+2 n r +r^{2}-4 n -4 r}\tag {4} \] Which for the root \(r = 4\) becomes \[ a_{n} = \frac {a_{n -1} \left (n +1\right )}{n \left (n +4\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 4\) 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 \[ a_{1}=\frac {-2+r}{r^{2}-2 r -3} \] Which for the root \(r = 4\) becomes \[ a_{1}={\frac {2}{5}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2+r}{r^{2}-2 r -3}\) | \(\frac {2}{5}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {-1+r}{r^{3}-7 r -6} \] Which for the root \(r = 4\) becomes \[ a_{2}={\frac {1}{10}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2+r}{r^{2}-2 r -3}\) | \(\frac {2}{5}\) |
| \(a_{2}\) | \(\frac {-1+r}{r^{3}-7 r -6}\) | \(\frac {1}{10}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {r}{\left (r +3\right ) \left (r^{3}-7 r -6\right )} \] Which for the root \(r = 4\) becomes \[ a_{3}={\frac {2}{105}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2+r}{r^{2}-2 r -3}\) | \(\frac {2}{5}\) |
| \(a_{2}\) | \(\frac {-1+r}{r^{3}-7 r -6}\) | \(\frac {1}{10}\) |
| \(a_{3}\) | \(\frac {r}{\left (r +3\right ) \left (r^{3}-7 r -6\right )}\) | \(\frac {2}{105}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {1}{\left (r +4\right ) \left (r +2\right ) \left (r -3\right ) \left (r +3\right )} \] Which for the root \(r = 4\) becomes \[ a_{4}={\frac {1}{336}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2+r}{r^{2}-2 r -3}\) | \(\frac {2}{5}\) |
| \(a_{2}\) | \(\frac {-1+r}{r^{3}-7 r -6}\) | \(\frac {1}{10}\) |
| \(a_{3}\) | \(\frac {r}{\left (r +3\right ) \left (r^{3}-7 r -6\right )}\) | \(\frac {2}{105}\) |
| \(a_{4}\) | \(\frac {1}{\left (r +4\right ) \left (r +2\right ) \left (r -3\right ) \left (r +3\right )}\) | \(\frac {1}{336}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {1}{\left (r +4\right ) \left (r^{2}+6 r +5\right ) \left (r^{2}-9\right )} \] Which for the root \(r = 4\) becomes \[ a_{5}={\frac {1}{2520}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2+r}{r^{2}-2 r -3}\) | \(\frac {2}{5}\) |
| \(a_{2}\) | \(\frac {-1+r}{r^{3}-7 r -6}\) | \(\frac {1}{10}\) |
| \(a_{3}\) | \(\frac {r}{\left (r +3\right ) \left (r^{3}-7 r -6\right )}\) | \(\frac {2}{105}\) |
| \(a_{4}\) | \(\frac {1}{\left (r +4\right ) \left (r +2\right ) \left (r -3\right ) \left (r +3\right )}\) | \(\frac {1}{336}\) |
| \(a_{5}\) | \(\frac {1}{\left (r +4\right ) \left (r^{2}+6 r +5\right ) \left (r^{2}-9\right )}\) | \(\frac {1}{2520}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {1}{\left (6+r \right ) \left (r +2\right ) \left (r -3\right ) \left (5+r \right ) \left (1+r \right ) \left (r +4\right )} \] Which for the root \(r = 4\) becomes \[ a_{6}={\frac {1}{21600}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2+r}{r^{2}-2 r -3}\) | \(\frac {2}{5}\) |
| \(a_{2}\) | \(\frac {-1+r}{r^{3}-7 r -6}\) | \(\frac {1}{10}\) |
| \(a_{3}\) | \(\frac {r}{\left (r +3\right ) \left (r^{3}-7 r -6\right )}\) | \(\frac {2}{105}\) |
| \(a_{4}\) | \(\frac {1}{\left (r +4\right ) \left (r +2\right ) \left (r -3\right ) \left (r +3\right )}\) | \(\frac {1}{336}\) |
| \(a_{5}\) | \(\frac {1}{\left (r +4\right ) \left (r^{2}+6 r +5\right ) \left (r^{2}-9\right )}\) | \(\frac {1}{2520}\) |
| \(a_{6}\) | \(\frac {1}{\left (6+r \right ) \left (r +2\right ) \left (r -3\right ) \left (5+r \right ) \left (1+r \right ) \left (r +4\right )}\) | \(\frac {1}{21600}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=\frac {1}{\left (6+r \right ) \left (r +2\right ) \left (r -3\right ) \left (5+r \right ) \left (1+r \right ) \left (7+r \right ) \left (r +3\right )} \] Which for the root \(r = 4\) becomes \[ a_{7}={\frac {1}{207900}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2+r}{r^{2}-2 r -3}\) | \(\frac {2}{5}\) |
| \(a_{2}\) | \(\frac {-1+r}{r^{3}-7 r -6}\) | \(\frac {1}{10}\) |
| \(a_{3}\) | \(\frac {r}{\left (r +3\right ) \left (r^{3}-7 r -6\right )}\) | \(\frac {2}{105}\) |
| \(a_{4}\) | \(\frac {1}{\left (r +4\right ) \left (r +2\right ) \left (r -3\right ) \left (r +3\right )}\) | \(\frac {1}{336}\) |
| \(a_{5}\) | \(\frac {1}{\left (r +4\right ) \left (r^{2}+6 r +5\right ) \left (r^{2}-9\right )}\) | \(\frac {1}{2520}\) |
| \(a_{6}\) | \(\frac {1}{\left (6+r \right ) \left (r +2\right ) \left (r -3\right ) \left (5+r \right ) \left (1+r \right ) \left (r +4\right )}\) | \(\frac {1}{21600}\) |
| \(a_{7}\) | \(\frac {1}{\left (6+r \right ) \left (r +2\right ) \left (r -3\right ) \left (5+r \right ) \left (1+r \right ) \left (7+r \right ) \left (r +3\right )}\) | \(\frac {1}{207900}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{4} \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^{4} \left (1+\frac {2 x}{5}+\frac {x^{2}}{10}+\frac {2 x^{3}}{105}+\frac {x^{4}}{336}+\frac {x^{5}}{2520}+\frac {x^{6}}{21600}+\frac {x^{7}}{207900}+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 \(N=4\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{4}\left (r \right )\). 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_{4} \\ &= \frac {1}{\left (r +4\right ) \left (r +2\right ) \left (r -3\right ) \left (r +3\right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {1}{\left (r +4\right ) \left (r +2\right ) \left (r -3\right ) \left (r +3\right )}&= \lim _{r\rightarrow 0}\frac {1}{\left (r +4\right ) \left (r +2\right ) \left (r -3\right ) \left (r +3\right )}\\ &= -{\frac {1}{72}} \end {align*}
The limit is \(-{\frac {1}{72}}\). 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} \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 )-b_{n -1} \left (n +r -1\right )-3 \left (n +r \right ) b_{n}+2 b_{n -1} = 0 \end{equation} Which for for the root \(r = 0\) becomes \begin{equation} \tag{4A} b_{n} n \left (n -1\right )-b_{n -1} \left (n -1\right )-3 n b_{n}+2 b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from the recursive equation (4) gives \[ b_{n} = \frac {b_{n -1} \left (n +r -3\right )}{n^{2}+2 n r +r^{2}-4 n -4 r}\tag {5} \] Which for the root \(r = 0\) becomes \[ b_{n} = \frac {b_{n -1} \left (n -3\right )}{n^{2}-4 n}\tag {6} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = 0\) 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 \[ b_{1}=\frac {-2+r}{r^{2}-2 r -3} \] Which for the root \(r = 0\) becomes \[ b_{1}={\frac {2}{3}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-2+r}{r^{2}-2 r -3}\) | \(\frac {2}{3}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {-1+r}{\left (r +2\right ) \left (r^{2}-2 r -3\right )} \] Which for the root \(r = 0\) becomes \[ b_{2}={\frac {1}{6}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-2+r}{r^{2}-2 r -3}\) | \(\frac {2}{3}\) |
| \(b_{2}\) | \(\frac {-1+r}{r^{3}-7 r -6}\) | \(\frac {1}{6}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {r}{\left (r +3\right ) \left (r +2\right ) \left (r^{2}-2 r -3\right )} \] Which for the root \(r = 0\) becomes \[ b_{3}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-2+r}{r^{2}-2 r -3}\) | \(\frac {2}{3}\) |
| \(b_{2}\) | \(\frac {-1+r}{r^{3}-7 r -6}\) | \(\frac {1}{6}\) |
| \(b_{3}\) | \(\frac {r}{\left (r +3\right ) \left (r +2\right ) \left (1+r \right ) \left (r -3\right )}\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {1}{\left (r +4\right ) \left (r +2\right ) \left (r -3\right ) \left (r +3\right )} \] Which for the root \(r = 0\) becomes \[ b_{4}=-{\frac {1}{72}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-2+r}{r^{2}-2 r -3}\) | \(\frac {2}{3}\) |
| \(b_{2}\) | \(\frac {-1+r}{r^{3}-7 r -6}\) | \(\frac {1}{6}\) |
| \(b_{3}\) | \(\frac {r}{\left (r +3\right ) \left (r +2\right ) \left (1+r \right ) \left (r -3\right )}\) | \(0\) |
| \(b_{4}\) | \(\frac {1}{\left (r +4\right ) \left (r +2\right ) \left (r^{2}-9\right )}\) | \(-{\frac {1}{72}}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {1}{\left (r +4\right ) \left (r -3\right ) \left (r +3\right ) \left (r^{2}+6 r +5\right )} \] Which for the root \(r = 0\) becomes \[ b_{5}=-{\frac {1}{180}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-2+r}{r^{2}-2 r -3}\) | \(\frac {2}{3}\) |
| \(b_{2}\) | \(\frac {-1+r}{r^{3}-7 r -6}\) | \(\frac {1}{6}\) |
| \(b_{3}\) | \(\frac {r}{\left (r +3\right ) \left (r +2\right ) \left (1+r \right ) \left (r -3\right )}\) | \(0\) |
| \(b_{4}\) | \(\frac {1}{\left (r +4\right ) \left (r +2\right ) \left (r^{2}-9\right )}\) | \(-{\frac {1}{72}}\) |
| \(b_{5}\) | \(\frac {1}{\left (r +4\right ) \left (r^{2}+6 r +5\right ) \left (r^{2}-9\right )}\) | \(-{\frac {1}{180}}\) |
For \(n = 6\), using the above recursive equation gives \[ b_{6}=\frac {1}{\left (r^{2}+8 r +12\right ) \left (r -3\right ) \left (r^{2}+6 r +5\right ) \left (r +4\right )} \] Which for the root \(r = 0\) becomes \[ b_{6}=-{\frac {1}{720}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-2+r}{r^{2}-2 r -3}\) | \(\frac {2}{3}\) |
| \(b_{2}\) | \(\frac {-1+r}{r^{3}-7 r -6}\) | \(\frac {1}{6}\) |
| \(b_{3}\) | \(\frac {r}{\left (r +3\right ) \left (r +2\right ) \left (1+r \right ) \left (r -3\right )}\) | \(0\) |
| \(b_{4}\) | \(\frac {1}{\left (r +4\right ) \left (r +2\right ) \left (r^{2}-9\right )}\) | \(-{\frac {1}{72}}\) |
| \(b_{5}\) | \(\frac {1}{\left (r +4\right ) \left (r^{2}+6 r +5\right ) \left (r^{2}-9\right )}\) | \(-{\frac {1}{180}}\) |
| \(b_{6}\) | \(\frac {1}{\left (6+r \right ) \left (r +2\right ) \left (r -3\right ) \left (5+r \right ) \left (1+r \right ) \left (r +4\right )}\) | \(-{\frac {1}{720}}\) |
For \(n = 7\), using the above recursive equation gives \[ b_{7}=\frac {1}{\left (r^{2}+8 r +12\right ) \left (r -3\right ) \left (r^{2}+6 r +5\right ) \left (r^{2}+10 r +21\right )} \] Which for the root \(r = 0\) becomes \[ b_{7}=-{\frac {1}{3780}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-2+r}{r^{2}-2 r -3}\) | \(\frac {2}{3}\) |
| \(b_{2}\) | \(\frac {-1+r}{r^{3}-7 r -6}\) | \(\frac {1}{6}\) |
| \(b_{3}\) | \(\frac {r}{\left (r +3\right ) \left (r +2\right ) \left (1+r \right ) \left (r -3\right )}\) | \(0\) |
| \(b_{4}\) | \(\frac {1}{\left (r +4\right ) \left (r +2\right ) \left (r^{2}-9\right )}\) | \(-{\frac {1}{72}}\) |
| \(b_{5}\) | \(\frac {1}{\left (r +4\right ) \left (r^{2}+6 r +5\right ) \left (r^{2}-9\right )}\) | \(-{\frac {1}{180}}\) |
| \(b_{6}\) | \(\frac {1}{\left (6+r \right ) \left (r +2\right ) \left (r -3\right ) \left (5+r \right ) \left (1+r \right ) \left (r +4\right )}\) | \(-{\frac {1}{720}}\) |
| \(b_{7}\) | \(\frac {1}{\left (r^{2}+8 r +12\right ) \left (r -3\right ) \left (r^{2}+6 r +5\right ) \left (r^{2}+10 r +21\right )}\) | \(-{\frac {1}{3780}}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\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 \\ &= 1+\frac {2 x}{3}+\frac {x^{2}}{6}-\frac {x^{4}}{72}-\frac {x^{5}}{180}-\frac {x^{6}}{720}-\frac {x^{7}}{3780}+O\left (x^{8}\right ) \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^{4} \left (1+\frac {2 x}{5}+\frac {x^{2}}{10}+\frac {2 x^{3}}{105}+\frac {x^{4}}{336}+\frac {x^{5}}{2520}+\frac {x^{6}}{21600}+\frac {x^{7}}{207900}+O\left (x^{8}\right )\right ) + c_{2} \left (1+\frac {2 x}{3}+\frac {x^{2}}{6}-\frac {x^{4}}{72}-\frac {x^{5}}{180}-\frac {x^{6}}{720}-\frac {x^{7}}{3780}+O\left (x^{8}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{4} \left (1+\frac {2 x}{5}+\frac {x^{2}}{10}+\frac {2 x^{3}}{105}+\frac {x^{4}}{336}+\frac {x^{5}}{2520}+\frac {x^{6}}{21600}+\frac {x^{7}}{207900}+O\left (x^{8}\right )\right )+c_{2} \left (1+\frac {2 x}{3}+\frac {x^{2}}{6}-\frac {x^{4}}{72}-\frac {x^{5}}{180}-\frac {x^{6}}{720}-\frac {x^{7}}{3780}+O\left (x^{8}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{4} \left (1+\frac {2 x}{5}+\frac {x^{2}}{10}+\frac {2 x^{3}}{105}+\frac {x^{4}}{336}+\frac {x^{5}}{2520}+\frac {x^{6}}{21600}+\frac {x^{7}}{207900}+O\left (x^{8}\right )\right )+c_{2} \left (1+\frac {2 x}{3}+\frac {x^{2}}{6}-\frac {x^{4}}{72}-\frac {x^{5}}{180}-\frac {x^{6}}{720}-\frac {x^{7}}{3780}+O\left (x^{8}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{4} \left (1+\frac {2 x}{5}+\frac {x^{2}}{10}+\frac {2 x^{3}}{105}+\frac {x^{4}}{336}+\frac {x^{5}}{2520}+\frac {x^{6}}{21600}+\frac {x^{7}}{207900}+O\left (x^{8}\right )\right )+c_{2} \left (1+\frac {2 x}{3}+\frac {x^{2}}{6}-\frac {x^{4}}{72}-\frac {x^{5}}{180}-\frac {x^{6}}{720}-\frac {x^{7}}{3780}+O\left (x^{8}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}y^{\prime }\right )+\left (-3-x \right ) y^{\prime }+2 y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=-\frac {2 y}{x}+\frac {\left (3+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} \\ {} & {} & \frac {d}{d x}y^{\prime }-\frac {\left (3+x \right ) y^{\prime }}{x}+\frac {2 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+x}{x}, P_{3}\left (x \right )=\frac {2}{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 \left (\frac {d}{d x}y^{\prime }\right )+\left (-3-x \right ) y^{\prime }+2 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 \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot \left (\frac {d}{d x}y^{\prime }\right )=\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 \left (\frac {d}{d x}y^{\prime }\right )=\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 (-4+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 )-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 (-4+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 4\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 )-a_{k} \left (k +r -2\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {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 =0\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =2 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k -2\right )}{\left (k +1\right ) \left (k -3\right )} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =0 \\ {} & {} & a_{1}=\frac {2 a_{0}}{3} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =1 \\ {} & {} & a_{2}=\frac {a_{1}}{4} \\ \bullet & {} & \textrm {Express in terms of}\hspace {3pt} a_{0} \\ {} & {} & a_{2}=\frac {a_{0}}{6} \\ \bullet & {} & \textrm {Terminating series solution of the ODE for}\hspace {3pt} r =0\hspace {3pt}\textrm {. Use reduction of order to find the second linearly independent solution}\hspace {3pt} \\ {} & {} & y=a_{0}\cdot \left (1+\frac {2}{3} x +\frac {1}{6} x^{2}\right ) \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =4 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +2\right )}{\left (k +5\right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =4 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +4}, a_{k +1}=\frac {a_{k} \left (k +2\right )}{\left (k +5\right ) \left (k +1\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=a_{0}\cdot \left (1+\frac {2}{3} x +\frac {1}{6} x^{2}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +4}\right ), b_{k +1}=\frac {b_{k} \left (k +2\right )}{\left (k +5\right ) \left (k +1\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: 0.031 (sec). Leaf size: 50
Order:=8; dsolve(x*diff(y(x),x$2)-(3+x)*diff(y(x),x)+2*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{4} \left (1+\frac {2}{5} x +\frac {1}{10} x^{2}+\frac {2}{105} x^{3}+\frac {1}{336} x^{4}+\frac {1}{2520} x^{5}+\frac {1}{21600} x^{6}+\frac {1}{207900} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+c_{2} \left (-144-96 x -24 x^{2}+2 x^{4}+\frac {4}{5} x^{5}+\frac {1}{5} x^{6}+\frac {4}{105} x^{7}+\operatorname {O}\left (x^{8}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.101 (sec). Leaf size: 91
AsymptoticDSolveValue[x*y''[x]-(3+x)*y'[x]+2*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_1 \left (-\frac {x^6}{720}-\frac {x^5}{180}-\frac {x^4}{72}+\frac {x^2}{6}+\frac {2 x}{3}+1\right )+c_2 \left (\frac {x^{10}}{21600}+\frac {x^9}{2520}+\frac {x^8}{336}+\frac {2 x^7}{105}+\frac {x^6}{10}+\frac {2 x^5}{5}+x^4\right ) \]