Internal problem ID [14012]
Internal file name [OUTPUT/13184_Friday_February_23_2024_06_58_30_AM_52144743/index.tex]
Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell.
second edition. CRC Press. FL, USA. 2020
Section: Chapter 36. The big theorem on the the Frobenius method. Additional Exercises. page
739
Problem number: 36.2 (i).
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]]
\[ \boxed {x y^{\prime \prime }+4 y^{\prime }+\frac {12 y}{\left (x +2\right )^{2}}=0} \] 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. \[ x y^{\prime \prime }+4 y^{\prime }+\frac {12 y}{\left (x +2\right )^{2}} = 0 \] 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}{x}\\ q(x) &= \frac {12}{x \left (x +2\right )^{2}}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, -2, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ y^{\prime \prime } x \left (x +2\right )^{2}+4 y^{\prime } \left (x +2\right )^{2}+12 y = 0 \] 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 \left (x +2\right )^{2}+4 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right ) \left (x +2\right )^{2}+12 \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^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}16 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}16 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}12 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 }}x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}4 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}4 a_{n -2} \left (n +r -2\right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}16 x^{n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}16 a_{n -1} \left (n +r -1\right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}12 a_{n} x^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}12 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 =2}{\sum }}a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}4 a_{n -2} \left (n +r -2\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}16 a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}16 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}12 a_{n -1} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 4 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+16 \left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ 4 x^{-1+r} a_{0} r \left (-1+r \right )+16 r a_{0} x^{-1+r} = 0 \] Or \[ \left (4 x^{-1+r} r \left (-1+r \right )+16 r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ 4 r \,x^{-1+r} \left (3+r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 4 r \left (3+r \right ) = 0 \] 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 \[ 4 r \,x^{-1+r} \left (3+r \right ) = 0 \] 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 \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = \frac {-r^{2}-3 r -3}{r^{2}+5 r +4} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} a_{n -2} \left (n +r -2\right ) \left (n -3+r \right )+4 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+4 a_{n} \left (n +r \right ) \left (n +r -1\right )+4 a_{n -2} \left (n +r -2\right )+16 a_{n -1} \left (n +r -1\right )+16 a_{n} \left (n +r \right )+12 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {n^{2} a_{n -2}+4 n^{2} a_{n -1}+2 n r a_{n -2}+8 n r a_{n -1}+r^{2} a_{n -2}+4 r^{2} a_{n -1}-n a_{n -2}+4 n a_{n -1}-r a_{n -2}+4 r a_{n -1}-2 a_{n -2}+4 a_{n -1}}{4 \left (n^{2}+2 n r +r^{2}+3 n +3 r \right )}\tag {4} \] Which for the root \(r = 0\) becomes \[ a_{n} = \frac {\left (-a_{n -2}-4 a_{n -1}\right ) n^{2}+\left (a_{n -2}-4 a_{n -1}\right ) n +2 a_{n -2}-4 a_{n -1}}{4 n \left (n +3\right )}\tag {5} \] 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.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}-3 r -3}{r^{2}+5 r +4}\) | \(-{\frac {3}{4}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {\frac {3}{4} r^{3}+\frac {9}{2} r^{2}+\frac {45}{4} r +\frac {21}{2}}{\left (r +5\right ) \left (r +4\right ) \left (1+r \right )} \] Which for the root \(r = 0\) becomes \[ a_{2}={\frac {21}{40}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}-3 r -3}{r^{2}+5 r +4}\) | \(-{\frac {3}{4}}\) |
| \(a_{2}\) | \(\frac {\frac {3}{4} r^{3}+\frac {9}{2} r^{2}+\frac {45}{4} r +\frac {21}{2}}{\left (r +5\right ) \left (r +4\right ) \left (1+r \right )}\) | \(\frac {21}{40}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-r^{4}-10 r^{3}-44 r^{2}-95 r -81}{2 \left (r +6\right ) \left (r +5\right ) \left (r +4\right ) \left (1+r \right )} \] Which for the root \(r = 0\) becomes \[ a_{3}=-{\frac {27}{80}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}-3 r -3}{r^{2}+5 r +4}\) | \(-{\frac {3}{4}}\) |
| \(a_{2}\) | \(\frac {\frac {3}{4} r^{3}+\frac {9}{2} r^{2}+\frac {45}{4} r +\frac {21}{2}}{\left (r +5\right ) \left (r +4\right ) \left (1+r \right )}\) | \(\frac {21}{40}\) |
| \(a_{3}\) | \(\frac {-r^{4}-10 r^{3}-44 r^{2}-95 r -81}{2 \left (r +6\right ) \left (r +5\right ) \left (r +4\right ) \left (1+r \right )}\) | \(-{\frac {27}{80}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {5 r^{5}+75 r^{4}+505 r^{3}+1845 r^{2}+3534 r +2772}{16 \left (r +7\right ) \left (r +4\right ) \left (1+r \right ) \left (r +5\right ) \left (r +6\right )} \] Which for the root \(r = 0\) becomes \[ a_{4}={\frac {33}{160}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}-3 r -3}{r^{2}+5 r +4}\) | \(-{\frac {3}{4}}\) |
| \(a_{2}\) | \(\frac {\frac {3}{4} r^{3}+\frac {9}{2} r^{2}+\frac {45}{4} r +\frac {21}{2}}{\left (r +5\right ) \left (r +4\right ) \left (1+r \right )}\) | \(\frac {21}{40}\) |
| \(a_{3}\) | \(\frac {-r^{4}-10 r^{3}-44 r^{2}-95 r -81}{2 \left (r +6\right ) \left (r +5\right ) \left (r +4\right ) \left (1+r \right )}\) | \(-{\frac {27}{80}}\) |
| \(a_{4}\) | \(\frac {5 r^{5}+75 r^{4}+505 r^{3}+1845 r^{2}+3534 r +2772}{16 \left (r +7\right ) \left (r +4\right ) \left (1+r \right ) \left (r +5\right ) \left (r +6\right )}\) | \(\frac {33}{160}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {3 \left (3+r \right ) \left (r^{2}+7 r +28\right ) \left (r^{2}+7 r +13\right )}{16 \left (r +8\right ) \left (r +5\right ) \left (r +6\right ) \left (1+r \right ) \left (r +7\right )} \] Which for the root \(r = 0\) becomes \[ a_{5}=-{\frac {39}{320}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}-3 r -3}{r^{2}+5 r +4}\) | \(-{\frac {3}{4}}\) |
| \(a_{2}\) | \(\frac {\frac {3}{4} r^{3}+\frac {9}{2} r^{2}+\frac {45}{4} r +\frac {21}{2}}{\left (r +5\right ) \left (r +4\right ) \left (1+r \right )}\) | \(\frac {21}{40}\) |
| \(a_{3}\) | \(\frac {-r^{4}-10 r^{3}-44 r^{2}-95 r -81}{2 \left (r +6\right ) \left (r +5\right ) \left (r +4\right ) \left (1+r \right )}\) | \(-{\frac {27}{80}}\) |
| \(a_{4}\) | \(\frac {5 r^{5}+75 r^{4}+505 r^{3}+1845 r^{2}+3534 r +2772}{16 \left (r +7\right ) \left (r +4\right ) \left (1+r \right ) \left (r +5\right ) \left (r +6\right )}\) | \(\frac {33}{160}\) |
| \(a_{5}\) | \(-\frac {3 \left (3+r \right ) \left (r^{2}+7 r +28\right ) \left (r^{2}+7 r +13\right )}{16 \left (r +8\right ) \left (r +5\right ) \left (r +6\right ) \left (1+r \right ) \left (r +7\right )}\) | \(-{\frac {39}{320}}\) |
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}\dots \\ &= 1-\frac {3 x}{4}+\frac {21 x^{2}}{40}-\frac {27 x^{3}}{80}+\frac {33 x^{4}}{160}-\frac {39 x^{5}}{320}+O\left (x^{6}\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 {-r^{4}-10 r^{3}-44 r^{2}-95 r -81}{2 \left (r +6\right ) \left (r +5\right ) \left (r +4\right ) \left (1+r \right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {-r^{4}-10 r^{3}-44 r^{2}-95 r -81}{2 \left (r +6\right ) \left (r +5\right ) \left (r +4\right ) \left (1+r \right )}&= \lim _{r\rightarrow -3}\frac {-r^{4}-10 r^{3}-44 r^{2}-95 r -81}{2 \left (r +6\right ) \left (r +5\right ) \left (r +4\right ) \left (1+r \right )}\\ &= {\frac {1}{8}} \end {align*}
The limit is \(\frac {1}{8}\). 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 \(b_{0} = 1\). Substituting \(n = 1\) in Eq(3) gives \[ b_{1} = -\frac {r^{2}+3 r +3}{r^{2}+5 r +4} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{4} b_{n -2} \left (n +r -2\right ) \left (n -3+r \right )+4 b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+4 b_{n} \left (n +r \right ) \left (n +r -1\right )+4 b_{n -2} \left (n +r -2\right )+16 b_{n -1} \left (n +r -1\right )+16 \left (n +r \right ) b_{n}+12 b_{n -1} = 0 \end{equation} Which for for the root \(r = -3\) becomes \begin{equation} \tag{4A} b_{n -2} \left (n -5\right ) \left (n -6\right )+4 b_{n -1} \left (n -4\right ) \left (n -5\right )+4 b_{n} \left (n -3\right ) \left (n -4\right )+4 b_{n -2} \left (n -5\right )+16 b_{n -1} \left (n -4\right )+16 \left (n -3\right ) b_{n}+12 b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from the recursive equation (4) gives \[ b_{n} = -\frac {n^{2} b_{n -2}+4 n^{2} b_{n -1}+2 n r b_{n -2}+8 n r b_{n -1}+r^{2} b_{n -2}+4 r^{2} b_{n -1}-n b_{n -2}+4 n b_{n -1}-r b_{n -2}+4 r b_{n -1}-2 b_{n -2}+4 b_{n -1}}{4 \left (n^{2}+2 n r +r^{2}+3 n +3 r \right )}\tag {5} \] Which for the root \(r = -3\) becomes \[ b_{n} = -\frac {n^{2} b_{n -2}+4 n^{2} b_{n -1}-7 n b_{n -2}-20 n b_{n -1}+10 b_{n -2}+28 b_{n -1}}{4 \left (n^{2}-3 n \right )}\tag {6} \] 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.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-r^{2}-3 r -3}{r^{2}+5 r +4}\) | \(\frac {3}{2}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {\frac {3}{4} r^{3}+\frac {9}{2} r^{2}+\frac {45}{4} r +\frac {21}{2}}{\left (r +5\right ) \left (r^{2}+5 r +4\right )} \] Which for the root \(r = -3\) becomes \[ b_{2}={\frac {3}{4}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-r^{2}-3 r -3}{r^{2}+5 r +4}\) | \(\frac {3}{2}\) |
| \(b_{2}\) | \(\frac {\frac {3}{4} r^{3}+\frac {9}{2} r^{2}+\frac {45}{4} r +\frac {21}{2}}{\left (r +5\right ) \left (r +4\right ) \left (1+r \right )}\) | \(\frac {3}{4}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=-\frac {r^{4}+10 r^{3}+44 r^{2}+95 r +81}{2 \left (r +6\right ) \left (r +5\right ) \left (r^{2}+5 r +4\right )} \] Which for the root \(r = -3\) becomes \[ b_{3}={\frac {1}{8}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-r^{2}-3 r -3}{r^{2}+5 r +4}\) | \(\frac {3}{2}\) |
| \(b_{2}\) | \(\frac {\frac {3}{4} r^{3}+\frac {9}{2} r^{2}+\frac {45}{4} r +\frac {21}{2}}{\left (r +5\right ) \left (r +4\right ) \left (1+r \right )}\) | \(\frac {3}{4}\) |
| \(b_{3}\) | \(\frac {-r^{4}-10 r^{3}-44 r^{2}-95 r -81}{2 \left (r +6\right ) \left (r +5\right ) \left (r +4\right ) \left (1+r \right )}\) | \(\frac {1}{8}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {5 r^{5}+75 r^{4}+505 r^{3}+1845 r^{2}+3534 r +2772}{16 \left (r^{2}+11 r +28\right ) \left (1+r \right ) \left (r +5\right ) \left (r +6\right )} \] Which for the root \(r = -3\) becomes \[ b_{4}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-r^{2}-3 r -3}{r^{2}+5 r +4}\) | \(\frac {3}{2}\) |
| \(b_{2}\) | \(\frac {\frac {3}{4} r^{3}+\frac {9}{2} r^{2}+\frac {45}{4} r +\frac {21}{2}}{\left (r +5\right ) \left (r +4\right ) \left (1+r \right )}\) | \(\frac {3}{4}\) |
| \(b_{3}\) | \(\frac {-r^{4}-10 r^{3}-44 r^{2}-95 r -81}{2 \left (r +6\right ) \left (r +5\right ) \left (r +4\right ) \left (1+r \right )}\) | \(\frac {1}{8}\) |
| \(b_{4}\) | \(\frac {5 r^{5}+75 r^{4}+505 r^{3}+1845 r^{2}+3534 r +2772}{16 \left (r +7\right ) \left (r +4\right ) \left (1+r \right ) \left (r +5\right ) \left (r +6\right )}\) | \(0\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-\frac {3 \left (r^{5}+17 r^{4}+132 r^{3}+557 r^{2}+1225 r +1092\right )}{16 \left (r^{2}+13 r +40\right ) \left (r +6\right ) \left (1+r \right ) \left (r +7\right )} \] Which for the root \(r = -3\) becomes \[ b_{5}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-r^{2}-3 r -3}{r^{2}+5 r +4}\) | \(\frac {3}{2}\) |
| \(b_{2}\) | \(\frac {\frac {3}{4} r^{3}+\frac {9}{2} r^{2}+\frac {45}{4} r +\frac {21}{2}}{\left (r +5\right ) \left (r +4\right ) \left (1+r \right )}\) | \(\frac {3}{4}\) |
| \(b_{3}\) | \(\frac {-r^{4}-10 r^{3}-44 r^{2}-95 r -81}{2 \left (r +6\right ) \left (r +5\right ) \left (r +4\right ) \left (1+r \right )}\) | \(\frac {1}{8}\) |
| \(b_{4}\) | \(\frac {5 r^{5}+75 r^{4}+505 r^{3}+1845 r^{2}+3534 r +2772}{16 \left (r +7\right ) \left (r +4\right ) \left (1+r \right ) \left (r +5\right ) \left (r +6\right )}\) | \(0\) |
| \(b_{5}\) | \(-\frac {3 \left (3+r \right ) \left (r^{2}+7 r +28\right ) \left (r^{2}+7 r +13\right )}{16 \left (r +8\right ) \left (r +5\right ) \left (r +6\right ) \left (1+r \right ) \left (r +7\right )}\) | \(0\) |
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}\dots \right ) \\ &= \frac {1+\frac {3 x}{2}+\frac {3 x^{2}}{4}+\frac {x^{3}}{8}+O\left (x^{6}\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 {21 x^{2}}{40}-\frac {27 x^{3}}{80}+\frac {33 x^{4}}{160}-\frac {39 x^{5}}{320}+O\left (x^{6}\right )\right ) + \frac {c_{2} \left (1+\frac {3 x}{2}+\frac {3 x^{2}}{4}+\frac {x^{3}}{8}+O\left (x^{6}\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 {21 x^{2}}{40}-\frac {27 x^{3}}{80}+\frac {33 x^{4}}{160}-\frac {39 x^{5}}{320}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+\frac {3 x}{2}+\frac {3 x^{2}}{4}+\frac {x^{3}}{8}+O\left (x^{6}\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 {21 x^{2}}{40}-\frac {27 x^{3}}{80}+\frac {33 x^{4}}{160}-\frac {39 x^{5}}{320}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+\frac {3 x}{2}+\frac {3 x^{2}}{4}+\frac {x^{3}}{8}+O\left (x^{6}\right )\right )}{x^{3}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (1-\frac {3 x}{4}+\frac {21 x^{2}}{40}-\frac {27 x^{3}}{80}+\frac {33 x^{4}}{160}-\frac {39 x^{5}}{320}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+\frac {3 x}{2}+\frac {3 x^{2}}{4}+\frac {x^{3}}{8}+O\left (x^{6}\right )\right )}{x^{3}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) x \left (x +2\right )^{2}+4 y^{\prime } \left (x +2\right )^{2}+12 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 {4 y^{\prime }}{x}-\frac {12 y}{\left (x +2\right )^{2} 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 {4 y^{\prime }}{x}+\frac {12 y}{\left (x +2\right )^{2} x}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {4}{x}, P_{3}\left (x \right )=\frac {12}{x \left (x +2\right )^{2}}\right ] \\ {} & \circ & \left (x +2\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-2 \\ {} & {} & \left (\left (x +2\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-2}}}=0 \\ {} & \circ & \left (x +2\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-2 \\ {} & {} & \left (\left (x +2\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-2}}}=-6 \\ {} & \circ & x =-2\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=-2 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) x \left (x +2\right )^{2}+4 y^{\prime } \left (x +2\right )^{2}+12 y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -2\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (u^{3}-2 u^{2}\right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+4 u^{2} \left (\frac {d}{d u}y \left (u \right )\right )+12 y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{2}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & u^{2}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r +1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -1 \\ {} & {} & u^{2}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =1}{\sum }}a_{k -1} \left (k -1+r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..3 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k -1+r \right ) u^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -2 a_{0} \left (2+r \right ) \left (-3+r \right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (-2 a_{k} \left (k +r +2\right ) \left (k +r -3\right )+a_{k -1} \left (k -1+r \right ) \left (k +r +2\right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -2 \left (2+r \right ) \left (-3+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-2, 3\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -2 \left (k +r +2\right ) \left (\frac {\left (-k -r +1\right ) a_{k -1}}{2}+a_{k} \left (k +r -3\right )\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & -2 \left (k +r +3\right ) \left (\frac {\left (-k -r \right ) a_{k}}{2}+a_{k +1} \left (k -2+r \right )\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {\left (k +r \right ) a_{k}}{2 \left (k -2+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-2\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =2 \\ {} & {} & a_{k +1}=\frac {\left (k -2\right ) a_{k}}{2 \left (k -4\right )} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =0 \\ {} & {} & a_{1}=\frac {a_{0}}{4} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =1 \\ {} & {} & a_{2}=\frac {a_{1}}{6} \\ \bullet & {} & \textrm {Express in terms of}\hspace {3pt} a_{0} \\ {} & {} & a_{2}=\frac {a_{0}}{24} \\ \bullet & {} & \textrm {Terminating series solution of the ODE for}\hspace {3pt} r =-2\hspace {3pt}\textrm {. Use reduction of order to find the second linearly independent solution}\hspace {3pt} \\ {} & {} & y \left (u \right )=a_{0}\cdot \left (1+\frac {1}{4} u +\frac {1}{24} u^{2}\right ) \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +2 \\ {} & {} & \left [y=a_{0} \left (\frac {5}{3}+\frac {5}{12} x +\frac {1}{24} x^{2}\right )\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =3 \\ {} & {} & a_{k +1}=\frac {\left (k +3\right ) a_{k}}{2 \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =3 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +3}, a_{k +1}=\frac {\left (k +3\right ) a_{k}}{2 \left (k +1\right )}\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +2\right )^{k +3}, a_{k +1}=\frac {\left (k +3\right ) a_{k}}{2 \left (k +1\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=a_{0} \left (\frac {5}{3}+\frac {5}{12} x +\frac {1}{24} x^{2}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (x +2\right )^{k +3}\right ), b_{k +1}=\frac {\left (k +3\right ) b_{k}}{2 \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) <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 42
Order:=6; dsolve(x*diff(y(x),x$2)+4*diff(y(x),x)+12/(x+2)^2*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} \left (1-\frac {3}{4} x +\frac {21}{40} x^{2}-\frac {27}{80} x^{3}+\frac {33}{160} x^{4}-\frac {39}{320} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\frac {c_{2} \left (12+18 x +9 x^{2}+\frac {9}{8} x^{4}-\frac {63}{80} x^{5}+\operatorname {O}\left (x^{6}\right )\right )}{x^{3}} \]
✓ Solution by Mathematica
Time used: 0.065 (sec). Leaf size: 59
AsymptoticDSolveValue[x*y''[x]+4*y'[x]+12/(x+2)^2*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (\frac {1}{x^3}+\frac {3}{2 x^2}+\frac {3}{4 x}+\frac {1}{8}\right )+c_2 \left (\frac {33 x^4}{160}-\frac {27 x^3}{80}+\frac {21 x^2}{40}-\frac {3 x}{4}+1\right ) \]