Internal problem ID [6974]
Internal file name [OUTPUT/6217_Friday_August_12_2022_11_05_47_PM_772028/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: 10.
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, _exact, _linear, _homogeneous]]
\[ \boxed {x y^{\prime \prime }+\left (4+3 x \right ) y^{\prime }+3 y=0} \] 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. \[ x y^{\prime \prime }+\left (4+3 x \right ) y^{\prime }+3 y = 0 \] 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 {4+3 x}{x}\\ q(x) &= \frac {3}{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 \[ x y^{\prime \prime }+\left (4+3 x \right ) y^{\prime }+3 y = 0 \] 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 (4+3 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+3 \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 }}3 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 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 }}3 x^{n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}3 a_{n -1} \left (n +r -1\right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}3 a_{n} x^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}3 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 }}3 a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}3 a_{n -1} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+4 \left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ x^{-1+r} a_{0} r \left (-1+r \right )+4 r a_{0} x^{-1+r} = 0 \] Or \[ \left (x^{-1+r} r \left (-1+r \right )+4 r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ r \,x^{-1+r} \left (3+r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r \left (3+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 0\\ r_2 &= -3 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ r \,x^{-1+r} \left (3+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([0, -3]\).
Since \(r_1 - r_2 = 3\) 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^{3}} \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 -3}\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 )+3 a_{n -1} \left (n +r -1\right )+4 a_{n} \left (n +r \right )+3 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {3 a_{n -1}}{n +3+r}\tag {4} \] Which for the root \(r = 0\) becomes \[ a_{n} = -\frac {3 a_{n -1}}{n +3}\tag {5} \] 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 \[ a_{1}=-\frac {3}{4+r} \] Which for the root \(r = 0\) becomes \[ a_{1}=-{\frac {3}{4}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3}{4+r}\) | \(-{\frac {3}{4}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {9}{\left (4+r \right ) \left (5+r \right )} \] Which for the root \(r = 0\) becomes \[ a_{2}={\frac {9}{20}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3}{4+r}\) | \(-{\frac {3}{4}}\) |
| \(a_{2}\) | \(\frac {9}{\left (4+r \right ) \left (5+r \right )}\) | \(\frac {9}{20}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {27}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right )} \] Which for the root \(r = 0\) becomes \[ a_{3}=-{\frac {9}{40}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3}{4+r}\) | \(-{\frac {3}{4}}\) |
| \(a_{2}\) | \(\frac {9}{\left (4+r \right ) \left (5+r \right )}\) | \(\frac {9}{20}\) |
| \(a_{3}\) | \(-\frac {27}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right )}\) | \(-{\frac {9}{40}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {81}{\left (5+r \right ) \left (6+r \right ) \left (7+r \right ) \left (4+r \right )} \] Which for the root \(r = 0\) becomes \[ a_{4}={\frac {27}{280}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3}{4+r}\) | \(-{\frac {3}{4}}\) |
| \(a_{2}\) | \(\frac {9}{\left (4+r \right ) \left (5+r \right )}\) | \(\frac {9}{20}\) |
| \(a_{3}\) | \(-\frac {27}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right )}\) | \(-{\frac {9}{40}}\) |
| \(a_{4}\) | \(\frac {81}{\left (5+r \right ) \left (6+r \right ) \left (7+r \right ) \left (4+r \right )}\) | \(\frac {27}{280}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {243}{\left (6+r \right ) \left (7+r \right ) \left (4+r \right ) \left (5+r \right ) \left (8+r \right )} \] Which for the root \(r = 0\) becomes \[ a_{5}=-{\frac {81}{2240}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3}{4+r}\) | \(-{\frac {3}{4}}\) |
| \(a_{2}\) | \(\frac {9}{\left (4+r \right ) \left (5+r \right )}\) | \(\frac {9}{20}\) |
| \(a_{3}\) | \(-\frac {27}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right )}\) | \(-{\frac {9}{40}}\) |
| \(a_{4}\) | \(\frac {81}{\left (5+r \right ) \left (6+r \right ) \left (7+r \right ) \left (4+r \right )}\) | \(\frac {27}{280}\) |
| \(a_{5}\) | \(-\frac {243}{\left (6+r \right ) \left (7+r \right ) \left (4+r \right ) \left (5+r \right ) \left (8+r \right )}\) | \(-{\frac {81}{2240}}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {729}{\left (7+r \right ) \left (4+r \right ) \left (5+r \right ) \left (8+r \right ) \left (9+r \right ) \left (6+r \right )} \] Which for the root \(r = 0\) becomes \[ a_{6}={\frac {27}{2240}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3}{4+r}\) | \(-{\frac {3}{4}}\) |
| \(a_{2}\) | \(\frac {9}{\left (4+r \right ) \left (5+r \right )}\) | \(\frac {9}{20}\) |
| \(a_{3}\) | \(-\frac {27}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right )}\) | \(-{\frac {9}{40}}\) |
| \(a_{4}\) | \(\frac {81}{\left (5+r \right ) \left (6+r \right ) \left (7+r \right ) \left (4+r \right )}\) | \(\frac {27}{280}\) |
| \(a_{5}\) | \(-\frac {243}{\left (6+r \right ) \left (7+r \right ) \left (4+r \right ) \left (5+r \right ) \left (8+r \right )}\) | \(-{\frac {81}{2240}}\) |
| \(a_{6}\) | \(\frac {729}{\left (7+r \right ) \left (4+r \right ) \left (5+r \right ) \left (8+r \right ) \left (9+r \right ) \left (6+r \right )}\) | \(\frac {27}{2240}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=-\frac {2187}{\left (4+r \right ) \left (5+r \right ) \left (8+r \right ) \left (9+r \right ) \left (6+r \right ) \left (10+r \right ) \left (7+r \right )} \] Which for the root \(r = 0\) becomes \[ a_{7}=-{\frac {81}{22400}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3}{4+r}\) | \(-{\frac {3}{4}}\) |
| \(a_{2}\) | \(\frac {9}{\left (4+r \right ) \left (5+r \right )}\) | \(\frac {9}{20}\) |
| \(a_{3}\) | \(-\frac {27}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right )}\) | \(-{\frac {9}{40}}\) |
| \(a_{4}\) | \(\frac {81}{\left (5+r \right ) \left (6+r \right ) \left (7+r \right ) \left (4+r \right )}\) | \(\frac {27}{280}\) |
| \(a_{5}\) | \(-\frac {243}{\left (6+r \right ) \left (7+r \right ) \left (4+r \right ) \left (5+r \right ) \left (8+r \right )}\) | \(-{\frac {81}{2240}}\) |
| \(a_{6}\) | \(\frac {729}{\left (7+r \right ) \left (4+r \right ) \left (5+r \right ) \left (8+r \right ) \left (9+r \right ) \left (6+r \right )}\) | \(\frac {27}{2240}\) |
| \(a_{7}\) | \(-\frac {2187}{\left (4+r \right ) \left (5+r \right ) \left (8+r \right ) \left (9+r \right ) \left (6+r \right ) \left (10+r \right ) \left (7+r \right )}\) | \(-{\frac {81}{22400}}\) |
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 {3 x}{4}+\frac {9 x^{2}}{20}-\frac {9 x^{3}}{40}+\frac {27 x^{4}}{280}-\frac {81 x^{5}}{2240}+\frac {27 x^{6}}{2240}-\frac {81 x^{7}}{22400}+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 \(N=3\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{3}\left (r \right )\). 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_{3} \\ &= -\frac {27}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}-\frac {27}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right )}&= \lim _{r\rightarrow -3}-\frac {27}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right )}\\ &= -{\frac {9}{2}} \end {align*}
The limit is \(-{\frac {9}{2}}\). 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 -3} \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 )+3 b_{n -1} \left (n +r -1\right )+4 \left (n +r \right ) b_{n}+3 b_{n -1} = 0 \end{equation} Which for for the root \(r = -3\) becomes \begin{equation} \tag{4A} b_{n} \left (n -3\right ) \left (n -4\right )+3 b_{n -1} \left (n -4\right )+4 \left (n -3\right ) b_{n}+3 b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from the recursive equation (4) gives \[ b_{n} = -\frac {3 b_{n -1}}{n +3+r}\tag {5} \] Which for the root \(r = -3\) becomes \[ b_{n} = -\frac {3 b_{n -1}}{n}\tag {6} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -3\) 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 \[ b_{1}=-\frac {3}{4+r} \] Which for the root \(r = -3\) becomes \[ b_{1}=-3 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {3}{4+r}\) | \(-3\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {9}{\left (4+r \right ) \left (5+r \right )} \] Which for the root \(r = -3\) becomes \[ b_{2}={\frac {9}{2}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {3}{4+r}\) | \(-3\) |
| \(b_{2}\) | \(\frac {9}{\left (4+r \right ) \left (5+r \right )}\) | \(\frac {9}{2}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=-\frac {27}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right )} \] Which for the root \(r = -3\) becomes \[ b_{3}=-{\frac {9}{2}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {3}{4+r}\) | \(-3\) |
| \(b_{2}\) | \(\frac {9}{\left (4+r \right ) \left (5+r \right )}\) | \(\frac {9}{2}\) |
| \(b_{3}\) | \(-\frac {27}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right )}\) | \(-{\frac {9}{2}}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {81}{\left (5+r \right ) \left (6+r \right ) \left (7+r \right ) \left (4+r \right )} \] Which for the root \(r = -3\) becomes \[ b_{4}={\frac {27}{8}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {3}{4+r}\) | \(-3\) |
| \(b_{2}\) | \(\frac {9}{\left (4+r \right ) \left (5+r \right )}\) | \(\frac {9}{2}\) |
| \(b_{3}\) | \(-\frac {27}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right )}\) | \(-{\frac {9}{2}}\) |
| \(b_{4}\) | \(\frac {81}{\left (5+r \right ) \left (6+r \right ) \left (7+r \right ) \left (4+r \right )}\) | \(\frac {27}{8}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-\frac {243}{\left (6+r \right ) \left (7+r \right ) \left (4+r \right ) \left (5+r \right ) \left (8+r \right )} \] Which for the root \(r = -3\) becomes \[ b_{5}=-{\frac {81}{40}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {3}{4+r}\) | \(-3\) |
| \(b_{2}\) | \(\frac {9}{\left (4+r \right ) \left (5+r \right )}\) | \(\frac {9}{2}\) |
| \(b_{3}\) | \(-\frac {27}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right )}\) | \(-{\frac {9}{2}}\) |
| \(b_{4}\) | \(\frac {81}{\left (5+r \right ) \left (6+r \right ) \left (7+r \right ) \left (4+r \right )}\) | \(\frac {27}{8}\) |
| \(b_{5}\) | \(-\frac {243}{\left (6+r \right ) \left (7+r \right ) \left (4+r \right ) \left (5+r \right ) \left (8+r \right )}\) | \(-{\frac {81}{40}}\) |
For \(n = 6\), using the above recursive equation gives \[ b_{6}=\frac {729}{\left (7+r \right ) \left (4+r \right ) \left (5+r \right ) \left (8+r \right ) \left (9+r \right ) \left (6+r \right )} \] Which for the root \(r = -3\) becomes \[ b_{6}={\frac {81}{80}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {3}{4+r}\) | \(-3\) |
| \(b_{2}\) | \(\frac {9}{\left (4+r \right ) \left (5+r \right )}\) | \(\frac {9}{2}\) |
| \(b_{3}\) | \(-\frac {27}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right )}\) | \(-{\frac {9}{2}}\) |
| \(b_{4}\) | \(\frac {81}{\left (5+r \right ) \left (6+r \right ) \left (7+r \right ) \left (4+r \right )}\) | \(\frac {27}{8}\) |
| \(b_{5}\) | \(-\frac {243}{\left (6+r \right ) \left (7+r \right ) \left (4+r \right ) \left (5+r \right ) \left (8+r \right )}\) | \(-{\frac {81}{40}}\) |
| \(b_{6}\) | \(\frac {729}{\left (7+r \right ) \left (4+r \right ) \left (5+r \right ) \left (8+r \right ) \left (9+r \right ) \left (6+r \right )}\) | \(\frac {81}{80}\) |
For \(n = 7\), using the above recursive equation gives \[ b_{7}=-\frac {2187}{\left (4+r \right ) \left (5+r \right ) \left (8+r \right ) \left (9+r \right ) \left (6+r \right ) \left (10+r \right ) \left (7+r \right )} \] Which for the root \(r = -3\) becomes \[ b_{7}=-{\frac {243}{560}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {3}{4+r}\) | \(-3\) |
| \(b_{2}\) | \(\frac {9}{\left (4+r \right ) \left (5+r \right )}\) | \(\frac {9}{2}\) |
| \(b_{3}\) | \(-\frac {27}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right )}\) | \(-{\frac {9}{2}}\) |
| \(b_{4}\) | \(\frac {81}{\left (5+r \right ) \left (6+r \right ) \left (7+r \right ) \left (4+r \right )}\) | \(\frac {27}{8}\) |
| \(b_{5}\) | \(-\frac {243}{\left (6+r \right ) \left (7+r \right ) \left (4+r \right ) \left (5+r \right ) \left (8+r \right )}\) | \(-{\frac {81}{40}}\) |
| \(b_{6}\) | \(\frac {729}{\left (7+r \right ) \left (4+r \right ) \left (5+r \right ) \left (8+r \right ) \left (9+r \right ) \left (6+r \right )}\) | \(\frac {81}{80}\) |
| \(b_{7}\) | \(-\frac {2187}{\left (4+r \right ) \left (5+r \right ) \left (8+r \right ) \left (9+r \right ) \left (6+r \right ) \left (10+r \right ) \left (7+r \right )}\) | \(-{\frac {243}{560}}\) |
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 {1-3 x +\frac {9 x^{2}}{2}-\frac {9 x^{3}}{2}+\frac {27 x^{4}}{8}-\frac {81 x^{5}}{40}+\frac {81 x^{6}}{80}-\frac {243 x^{7}}{560}+O\left (x^{8}\right )}{x^{3}} \end {align*}
Therefore the homogeneous solution is \begin{align*} y_h(x) &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= c_{1} \left (1-\frac {3 x}{4}+\frac {9 x^{2}}{20}-\frac {9 x^{3}}{40}+\frac {27 x^{4}}{280}-\frac {81 x^{5}}{2240}+\frac {27 x^{6}}{2240}-\frac {81 x^{7}}{22400}+O\left (x^{8}\right )\right ) + \frac {c_{2} \left (1-3 x +\frac {9 x^{2}}{2}-\frac {9 x^{3}}{2}+\frac {27 x^{4}}{8}-\frac {81 x^{5}}{40}+\frac {81 x^{6}}{80}-\frac {243 x^{7}}{560}+O\left (x^{8}\right )\right )}{x^{3}} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \left (1-\frac {3 x}{4}+\frac {9 x^{2}}{20}-\frac {9 x^{3}}{40}+\frac {27 x^{4}}{280}-\frac {81 x^{5}}{2240}+\frac {27 x^{6}}{2240}-\frac {81 x^{7}}{22400}+O\left (x^{8}\right )\right )+\frac {c_{2} \left (1-3 x +\frac {9 x^{2}}{2}-\frac {9 x^{3}}{2}+\frac {27 x^{4}}{8}-\frac {81 x^{5}}{40}+\frac {81 x^{6}}{80}-\frac {243 x^{7}}{560}+O\left (x^{8}\right )\right )}{x^{3}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (1-\frac {3 x}{4}+\frac {9 x^{2}}{20}-\frac {9 x^{3}}{40}+\frac {27 x^{4}}{280}-\frac {81 x^{5}}{2240}+\frac {27 x^{6}}{2240}-\frac {81 x^{7}}{22400}+O\left (x^{8}\right )\right )+\frac {c_{2} \left (1-3 x +\frac {9 x^{2}}{2}-\frac {9 x^{3}}{2}+\frac {27 x^{4}}{8}-\frac {81 x^{5}}{40}+\frac {81 x^{6}}{80}-\frac {243 x^{7}}{560}+O\left (x^{8}\right )\right )}{x^{3}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (1-\frac {3 x}{4}+\frac {9 x^{2}}{20}-\frac {9 x^{3}}{40}+\frac {27 x^{4}}{280}-\frac {81 x^{5}}{2240}+\frac {27 x^{6}}{2240}-\frac {81 x^{7}}{22400}+O\left (x^{8}\right )\right )+\frac {c_{2} \left (1-3 x +\frac {9 x^{2}}{2}-\frac {9 x^{3}}{2}+\frac {27 x^{4}}{8}-\frac {81 x^{5}}{40}+\frac {81 x^{6}}{80}-\frac {243 x^{7}}{560}+O\left (x^{8}\right )\right )}{x^{3}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x y^{\prime \prime }+\left (4+3 x \right ) y^{\prime }+3 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 {3 y}{x}-\frac {\left (4+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} \\ {} & {} & y^{\prime \prime }+\frac {\left (4+3 x \right ) y^{\prime }}{x}+\frac {3 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 {4+3 x}{x}, P_{3}\left (x \right )=\frac {3}{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}}}=4 \\ {} & \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 (4+3 x \right ) y^{\prime }+3 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 (3+r \right ) x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +4+r \right )+3 a_{k} \left (k +1+r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (3+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-3, 0\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k +1+r \right ) \left (a_{k +1} \left (k +4+r \right )+3 a_{k}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {3 a_{k}}{k +4+r} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-3 \\ {} & {} & a_{k +1}=-\frac {3 a_{k}}{k +1} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-3 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -3}, a_{k +1}=-\frac {3 a_{k}}{k +1}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=-\frac {3 a_{k}}{k +4} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +1}=-\frac {3 a_{k}}{k +4}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -3}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k}\right ), a_{k +1}=-\frac {3 a_{k}}{k +1}, b_{k +1}=-\frac {3 b_{k}}{k +4}\right ] \end {array} \]
Maple trace
`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)] <- linear_1 successful`
✓ Solution by Maple
Time used: 0.032 (sec). Leaf size: 52
Order:=8; dsolve(x*diff(y(x),x$2)+(4+3*x)*diff(y(x),x)+3*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} \left (1-\frac {3}{4} x +\frac {9}{20} x^{2}-\frac {9}{40} x^{3}+\frac {27}{280} x^{4}-\frac {81}{2240} x^{5}+\frac {27}{2240} x^{6}-\frac {81}{22400} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+\frac {c_{2} \left (12-36 x +54 x^{2}-54 x^{3}+\frac {81}{2} x^{4}-\frac {243}{10} x^{5}+\frac {243}{20} x^{6}-\frac {729}{140} x^{7}+\operatorname {O}\left (x^{8}\right )\right )}{x^{3}} \]
✓ Solution by Mathematica
Time used: 0.11 (sec). Leaf size: 90
AsymptoticDSolveValue[x*y''[x]+(4+3*x)*y'[x]+3*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_1 \left (\frac {81 x^3}{80}+\frac {1}{x^3}-\frac {81 x^2}{40}-\frac {3}{x^2}+\frac {27 x}{8}+\frac {9}{2 x}-\frac {9}{2}\right )+c_2 \left (\frac {27 x^6}{2240}-\frac {81 x^5}{2240}+\frac {27 x^4}{280}-\frac {9 x^3}{40}+\frac {9 x^2}{20}-\frac {3 x}{4}+1\right ) \]