Internal problem ID [6980]
Internal file name [OUTPUT/6223_Friday_August_12_2022_11_06_01_PM_93983477/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: 16.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference is integer"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x^{2} \left (4 x -1\right ) y^{\prime \prime }+x \left (5 x +1\right ) y^{\prime }+3 y=0} \] 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. \[ \left (4 x^{3}-x^{2}\right ) y^{\prime \prime }+\left (5 x^{2}+x \right ) y^{\prime }+3 y = 0 \] 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 {5 x +1}{x \left (4 x -1\right )}\\ q(x) &= \frac {3}{x^{2} \left (4 x -1\right )}\\ \end {align*}
| \(p(x)=\frac {5 x +1}{x \left (4 x -1\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = {\frac {1}{4}}\) | \(\text {``regular''}\) |
| \(q(x)=\frac {3}{x^{2} \left (4 x -1\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = {\frac {1}{4}}\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \(\left [0, {\frac {1}{4}}, \infty \right ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} \left (4 x -1\right ) y^{\prime \prime }+\left (5 x^{2}+x \right ) y^{\prime }+3 y = 0 \] 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^{2} \left (4 x -1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (5 x^{2}+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 }}4 x^{1+n +r} 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 ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}5 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\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 \(n +r\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}4 x^{1+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} \\ \moverset {\infty }{\munderset {n =0}{\sum }}5 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}5 a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}5 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ -x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+x^{n +r} a_{n} \left (n +r \right )+3 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ -x^{r} a_{0} r \left (-1+r \right )+x^{r} a_{0} r +3 a_{0} x^{r} = 0 \] Or \[ \left (-x^{r} r \left (-1+r \right )+x^{r} r +3 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (-r^{2}+2 r +3\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ -r^{2}+2 r +3 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 3\\ r_2 &= -1 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (-r^{2}+2 r +3\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([3, -1]\).
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^{3} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\frac {\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}}{x} \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +3}\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -1}\right ) \end {align*}
Where \(C\) above can be zero. We start by finding \(y_{1}\). Eq (2B) derived above is now used to find all \(a_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} 4 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )-a_{n} \left (n +r \right ) \left (n +r -1\right )+5 a_{n -1} \left (n +r -1\right )+a_{n} \left (n +r \right )+3 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {a_{n -1} \left (4 n^{2}+8 n r +4 r^{2}-7 n -7 r +3\right )}{n^{2}+2 n r +r^{2}-2 n -2 r -3}\tag {4} \] Which for the root \(r = 3\) becomes \[ a_{n} = \frac {a_{n -1} \left (4 n^{2}+17 n +18\right )}{n \left (n +4\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 3\) 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 \[ a_{1}=\frac {4 r^{2}+r}{r^{2}-4} \] Which for the root \(r = 3\) becomes \[ a_{1}={\frac {39}{5}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {4 r^{2}+r}{r^{2}-4}\) | \(\frac {39}{5}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {r \left (4 r +1\right ) \left (4 r^{2}+9 r +5\right )}{\left (r^{2}-4\right ) \left (r^{2}+2 r -3\right )} \] Which for the root \(r = 3\) becomes \[ a_{2}={\frac {221}{5}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {4 r^{2}+r}{r^{2}-4}\) | \(\frac {39}{5}\) |
| \(a_{2}\) | \(\frac {r \left (4 r +1\right ) \left (4 r^{2}+9 r +5\right )}{\left (r^{2}-4\right ) \left (r^{2}+2 r -3\right )}\) | \(\frac {221}{5}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {64 r^{4}+304 r^{3}+476 r^{2}+281 r +45}{\left (r +4\right ) \left (r +3\right ) \left (-1+r \right ) \left (r -2\right )} \] Which for the root \(r = 3\) becomes \[ a_{3}=221 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {4 r^{2}+r}{r^{2}-4}\) | \(\frac {39}{5}\) |
| \(a_{2}\) | \(\frac {r \left (4 r +1\right ) \left (4 r^{2}+9 r +5\right )}{\left (r^{2}-4\right ) \left (r^{2}+2 r -3\right )}\) | \(\frac {221}{5}\) |
| \(a_{3}\) | \(\frac {64 r^{4}+304 r^{3}+476 r^{2}+281 r +45}{\left (r +4\right ) \left (r +3\right ) \left (-1+r \right ) \left (r -2\right )}\) | \(221\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {256 r^{4}+1792 r^{3}+4064 r^{2}+3248 r +585}{\left (r +5\right ) \left (r -2\right ) \left (-1+r \right ) \left (r +4\right )} \] Which for the root \(r = 3\) becomes \[ a_{4}={\frac {16575}{16}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {4 r^{2}+r}{r^{2}-4}\) | \(\frac {39}{5}\) |
| \(a_{2}\) | \(\frac {r \left (4 r +1\right ) \left (4 r^{2}+9 r +5\right )}{\left (r^{2}-4\right ) \left (r^{2}+2 r -3\right )}\) | \(\frac {221}{5}\) |
| \(a_{3}\) | \(\frac {64 r^{4}+304 r^{3}+476 r^{2}+281 r +45}{\left (r +4\right ) \left (r +3\right ) \left (-1+r \right ) \left (r -2\right )}\) | \(221\) |
| \(a_{4}\) | \(\frac {256 r^{4}+1792 r^{3}+4064 r^{2}+3248 r +585}{\left (r +5\right ) \left (r -2\right ) \left (-1+r \right ) \left (r +4\right )}\) | \(\frac {16575}{16}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {1024 r^{5}+11520 r^{4}+46720 r^{3}+82080 r^{2}+57556 r +9945}{\left (r +6\right ) \left (r +2\right ) \left (-1+r \right ) \left (r -2\right ) \left (r +5\right )} \] Which for the root \(r = 3\) becomes \[ a_{5}={\frac {224315}{48}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {4 r^{2}+r}{r^{2}-4}\) | \(\frac {39}{5}\) |
| \(a_{2}\) | \(\frac {r \left (4 r +1\right ) \left (4 r^{2}+9 r +5\right )}{\left (r^{2}-4\right ) \left (r^{2}+2 r -3\right )}\) | \(\frac {221}{5}\) |
| \(a_{3}\) | \(\frac {64 r^{4}+304 r^{3}+476 r^{2}+281 r +45}{\left (r +4\right ) \left (r +3\right ) \left (-1+r \right ) \left (r -2\right )}\) | \(221\) |
| \(a_{4}\) | \(\frac {256 r^{4}+1792 r^{3}+4064 r^{2}+3248 r +585}{\left (r +5\right ) \left (r -2\right ) \left (-1+r \right ) \left (r +4\right )}\) | \(\frac {16575}{16}\) |
| \(a_{5}\) | \(\frac {1024 r^{5}+11520 r^{4}+46720 r^{3}+82080 r^{2}+57556 r +9945}{\left (r +6\right ) \left (r +2\right ) \left (-1+r \right ) \left (r -2\right ) \left (r +5\right )}\) | \(\frac {224315}{48}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {4096 r^{6}+67584 r^{5}+428800 r^{4}+1309440 r^{3}+1953904 r^{2}+1248456 r +208845}{\left (-1+r \right ) \left (r +6\right ) \left (r^{2}+10 r +21\right ) \left (r^{2}-4\right )} \] Which for the root \(r = 3\) becomes \[ a_{6}={\frac {493493}{24}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {4 r^{2}+r}{r^{2}-4}\) | \(\frac {39}{5}\) |
| \(a_{2}\) | \(\frac {r \left (4 r +1\right ) \left (4 r^{2}+9 r +5\right )}{\left (r^{2}-4\right ) \left (r^{2}+2 r -3\right )}\) | \(\frac {221}{5}\) |
| \(a_{3}\) | \(\frac {64 r^{4}+304 r^{3}+476 r^{2}+281 r +45}{\left (r +4\right ) \left (r +3\right ) \left (-1+r \right ) \left (r -2\right )}\) | \(221\) |
| \(a_{4}\) | \(\frac {256 r^{4}+1792 r^{3}+4064 r^{2}+3248 r +585}{\left (r +5\right ) \left (r -2\right ) \left (-1+r \right ) \left (r +4\right )}\) | \(\frac {16575}{16}\) |
| \(a_{5}\) | \(\frac {1024 r^{5}+11520 r^{4}+46720 r^{3}+82080 r^{2}+57556 r +9945}{\left (r +6\right ) \left (r +2\right ) \left (-1+r \right ) \left (r -2\right ) \left (r +5\right )}\) | \(\frac {224315}{48}\) |
| \(a_{6}\) | \(\frac {4096 r^{6}+67584 r^{5}+428800 r^{4}+1309440 r^{3}+1953904 r^{2}+1248456 r +208845}{\left (-1+r \right ) \left (r +6\right ) \left (r^{2}+10 r +21\right ) \left (r^{2}-4\right )}\) | \(\frac {493493}{24}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=\frac {16384 r^{7}+372736 r^{6}+3404800 r^{5}+15957760 r^{4}+40551616 r^{3}+53841424 r^{2}+32046780 r +5221125}{\left (r^{2}-4\right ) \left (r^{2}+10 r +21\right ) \left (-1+r \right ) \left (r^{2}+12 r +32\right )} \] Which for the root \(r = 3\) becomes \[ a_{7}={\frac {711399}{8}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {4 r^{2}+r}{r^{2}-4}\) | \(\frac {39}{5}\) |
| \(a_{2}\) | \(\frac {r \left (4 r +1\right ) \left (4 r^{2}+9 r +5\right )}{\left (r^{2}-4\right ) \left (r^{2}+2 r -3\right )}\) | \(\frac {221}{5}\) |
| \(a_{3}\) | \(\frac {64 r^{4}+304 r^{3}+476 r^{2}+281 r +45}{\left (r +4\right ) \left (r +3\right ) \left (-1+r \right ) \left (r -2\right )}\) | \(221\) |
| \(a_{4}\) | \(\frac {256 r^{4}+1792 r^{3}+4064 r^{2}+3248 r +585}{\left (r +5\right ) \left (r -2\right ) \left (-1+r \right ) \left (r +4\right )}\) | \(\frac {16575}{16}\) |
| \(a_{5}\) | \(\frac {1024 r^{5}+11520 r^{4}+46720 r^{3}+82080 r^{2}+57556 r +9945}{\left (r +6\right ) \left (r +2\right ) \left (-1+r \right ) \left (r -2\right ) \left (r +5\right )}\) | \(\frac {224315}{48}\) |
| \(a_{6}\) | \(\frac {4096 r^{6}+67584 r^{5}+428800 r^{4}+1309440 r^{3}+1953904 r^{2}+1248456 r +208845}{\left (-1+r \right ) \left (r +6\right ) \left (r^{2}+10 r +21\right ) \left (r^{2}-4\right )}\) | \(\frac {493493}{24}\) |
| \(a_{7}\) | \(\frac {16384 r^{7}+372736 r^{6}+3404800 r^{5}+15957760 r^{4}+40551616 r^{3}+53841424 r^{2}+32046780 r +5221125}{\left (r^{2}-4\right ) \left (r^{2}+10 r +21\right ) \left (-1+r \right ) \left (r^{2}+12 r +32\right )}\) | \(\frac {711399}{8}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{3} \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^{3} \left (1+\frac {39 x}{5}+\frac {221 x^{2}}{5}+221 x^{3}+\frac {16575 x^{4}}{16}+\frac {224315 x^{5}}{48}+\frac {493493 x^{6}}{24}+\frac {711399 x^{7}}{8}+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 {256 r^{4}+1792 r^{3}+4064 r^{2}+3248 r +585}{\left (r +5\right ) \left (r -2\right ) \left (-1+r \right ) \left (r +4\right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {256 r^{4}+1792 r^{3}+4064 r^{2}+3248 r +585}{\left (r +5\right ) \left (r -2\right ) \left (-1+r \right ) \left (r +4\right )}&= \lim _{r\rightarrow -1}\frac {256 r^{4}+1792 r^{3}+4064 r^{2}+3248 r +585}{\left (r +5\right ) \left (r -2\right ) \left (-1+r \right ) \left (r +4\right )}\\ &= -{\frac {15}{8}} \end {align*}
The limit is \(-{\frac {15}{8}}\). 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 -1} \end {align*}
Eq (3) derived above is used to find all \(b_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(b_{0}\) is arbitrary and taken as \(b_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{4} 4 b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )-b_{n} \left (n +r \right ) \left (n +r -1\right )+5 b_{n -1} \left (n +r -1\right )+b_{n} \left (n +r \right )+3 b_{n} = 0 \end{equation} Which for for the root \(r = -1\) becomes \begin{equation} \tag{4A} 4 b_{n -1} \left (n -2\right ) \left (n -3\right )-b_{n} \left (n -1\right ) \left (n -2\right )+5 b_{n -1} \left (n -2\right )+b_{n} \left (n -1\right )+3 b_{n} = 0 \end{equation} Solving for \(b_{n}\) from the recursive equation (4) gives \[ b_{n} = \frac {b_{n -1} \left (4 n^{2}+8 n r +4 r^{2}-7 n -7 r +3\right )}{n^{2}+2 n r +r^{2}-2 n -2 r -3}\tag {5} \] Which for the root \(r = -1\) becomes \[ b_{n} = \frac {b_{n -1} \left (4 n^{2}-15 n +14\right )}{n^{2}-4 n}\tag {6} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -1\) 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 \[ b_{1}=\frac {r \left (4 r +1\right )}{r^{2}-4} \] Which for the root \(r = -1\) becomes \[ b_{1}=-1 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {4 r^{2}+r}{r^{2}-4}\) | \(-1\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {r \left (4 r +1\right ) \left (4 r^{2}+9 r +5\right )}{\left (r^{2}-4\right ) \left (r^{2}+2 r -3\right )} \] Which for the root \(r = -1\) becomes \[ b_{2}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {4 r^{2}+r}{r^{2}-4}\) | \(-1\) |
| \(b_{2}\) | \(\frac {r \left (4 r +1\right ) \left (4 r^{2}+9 r +5\right )}{\left (r^{2}-4\right ) \left (r^{2}+2 r -3\right )}\) | \(0\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {\left (4 r +9\right ) \left (4 r +1\right ) \left (4 r^{2}+9 r +5\right )}{\left (r +4\right ) \left (r^{2}+2 r -3\right ) \left (r -2\right )} \] Which for the root \(r = -1\) becomes \[ b_{3}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {4 r^{2}+r}{r^{2}-4}\) | \(-1\) |
| \(b_{2}\) | \(\frac {r \left (4 r +1\right ) \left (4 r^{2}+9 r +5\right )}{\left (r^{2}-4\right ) \left (r^{2}+2 r -3\right )}\) | \(0\) |
| \(b_{3}\) | \(\frac {64 r^{4}+304 r^{3}+476 r^{2}+281 r +45}{\left (r +4\right ) \left (r +3\right ) \left (-1+r \right ) \left (r -2\right )}\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {\left (4 r +5\right ) \left (4 r +1\right ) \left (4 r +9\right ) \left (4 r +13\right )}{\left (r +5\right ) \left (r -2\right ) \left (-1+r \right ) \left (r +4\right )} \] Which for the root \(r = -1\) becomes \[ b_{4}=-{\frac {15}{8}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {4 r^{2}+r}{r^{2}-4}\) | \(-1\) |
| \(b_{2}\) | \(\frac {r \left (4 r +1\right ) \left (4 r^{2}+9 r +5\right )}{\left (r^{2}-4\right ) \left (r^{2}+2 r -3\right )}\) | \(0\) |
| \(b_{3}\) | \(\frac {64 r^{4}+304 r^{3}+476 r^{2}+281 r +45}{\left (r +4\right ) \left (r +3\right ) \left (-1+r \right ) \left (r -2\right )}\) | \(0\) |
| \(b_{4}\) | \(\frac {\left (4 r +5\right ) \left (4 r +1\right ) \left (4 r +9\right ) \left (4 r +13\right )}{\left (r +5\right ) \left (r -2\right ) \left (-1+r \right ) \left (r +4\right )}\) | \(-{\frac {15}{8}}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {\left (4 r +17\right ) \left (4 r +5\right ) \left (4 r +1\right ) \left (4 r +9\right ) \left (4 r +13\right )}{\left (-1+r \right ) \left (r -2\right ) \left (r +5\right ) \left (r^{2}+8 r +12\right )} \] Which for the root \(r = -1\) becomes \[ b_{5}=-{\frac {117}{8}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {4 r^{2}+r}{r^{2}-4}\) | \(-1\) |
| \(b_{2}\) | \(\frac {r \left (4 r +1\right ) \left (4 r^{2}+9 r +5\right )}{\left (r^{2}-4\right ) \left (r^{2}+2 r -3\right )}\) | \(0\) |
| \(b_{3}\) | \(\frac {64 r^{4}+304 r^{3}+476 r^{2}+281 r +45}{\left (r +4\right ) \left (r +3\right ) \left (-1+r \right ) \left (r -2\right )}\) | \(0\) |
| \(b_{4}\) | \(\frac {\left (4 r +5\right ) \left (4 r +1\right ) \left (4 r +9\right ) \left (4 r +13\right )}{\left (r +5\right ) \left (r -2\right ) \left (-1+r \right ) \left (r +4\right )}\) | \(-{\frac {15}{8}}\) |
| \(b_{5}\) | \(\frac {\left (4 r +17\right ) \left (4 r +5\right ) \left (4 r +1\right ) \left (4 r +9\right ) \left (4 r +13\right )}{\left (-1+r \right ) \left (r -2\right ) \left (r +5\right ) \left (r^{2}+8 r +12\right )}\) | \(-{\frac {117}{8}}\) |
For \(n = 6\), using the above recursive equation gives \[ b_{6}=\frac {\left (4 r +21\right ) \left (4 r +17\right ) \left (4 r +5\right ) \left (4 r +1\right ) \left (4 r +9\right ) \left (4 r +13\right )}{\left (r^{2}+8 r +12\right ) \left (-1+r \right ) \left (r -2\right ) \left (r^{2}+10 r +21\right )} \] Which for the root \(r = -1\) becomes \[ b_{6}=-{\frac {663}{8}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {4 r^{2}+r}{r^{2}-4}\) | \(-1\) |
| \(b_{2}\) | \(\frac {r \left (4 r +1\right ) \left (4 r^{2}+9 r +5\right )}{\left (r^{2}-4\right ) \left (r^{2}+2 r -3\right )}\) | \(0\) |
| \(b_{3}\) | \(\frac {64 r^{4}+304 r^{3}+476 r^{2}+281 r +45}{\left (r +4\right ) \left (r +3\right ) \left (-1+r \right ) \left (r -2\right )}\) | \(0\) |
| \(b_{4}\) | \(\frac {\left (4 r +5\right ) \left (4 r +1\right ) \left (4 r +9\right ) \left (4 r +13\right )}{\left (r +5\right ) \left (r -2\right ) \left (-1+r \right ) \left (r +4\right )}\) | \(-{\frac {15}{8}}\) |
| \(b_{5}\) | \(\frac {\left (4 r +17\right ) \left (4 r +5\right ) \left (4 r +1\right ) \left (4 r +9\right ) \left (4 r +13\right )}{\left (-1+r \right ) \left (r -2\right ) \left (r +5\right ) \left (r^{2}+8 r +12\right )}\) | \(-{\frac {117}{8}}\) |
| \(b_{6}\) | \(\frac {\left (4 r +21\right ) \left (4 r +17\right ) \left (4 r +5\right ) \left (4 r +1\right ) \left (4 r +9\right ) \left (4 r +13\right )}{\left (r^{2}+8 r +12\right ) \left (-1+r \right ) \left (r -2\right ) \left (r^{2}+10 r +21\right )}\) | \(-{\frac {663}{8}}\) |
For \(n = 7\), using the above recursive equation gives \[ b_{7}=\frac {\left (4 r +25\right ) \left (4 r +21\right ) \left (4 r +17\right ) \left (4 r +5\right ) \left (4 r +1\right ) \left (4 r +9\right ) \left (4 r +13\right )}{\left (r^{2}+10 r +21\right ) \left (r -2\right ) \left (-1+r \right ) \left (r +2\right ) \left (r^{2}+12 r +32\right )} \] Which for the root \(r = -1\) becomes \[ b_{7}=-{\frac {3315}{8}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {4 r^{2}+r}{r^{2}-4}\) | \(-1\) |
| \(b_{2}\) | \(\frac {r \left (4 r +1\right ) \left (4 r^{2}+9 r +5\right )}{\left (r^{2}-4\right ) \left (r^{2}+2 r -3\right )}\) | \(0\) |
| \(b_{3}\) | \(\frac {64 r^{4}+304 r^{3}+476 r^{2}+281 r +45}{\left (r +4\right ) \left (r +3\right ) \left (-1+r \right ) \left (r -2\right )}\) | \(0\) |
| \(b_{4}\) | \(\frac {\left (4 r +5\right ) \left (4 r +1\right ) \left (4 r +9\right ) \left (4 r +13\right )}{\left (r +5\right ) \left (r -2\right ) \left (-1+r \right ) \left (r +4\right )}\) | \(-{\frac {15}{8}}\) |
| \(b_{5}\) | \(\frac {\left (4 r +17\right ) \left (4 r +5\right ) \left (4 r +1\right ) \left (4 r +9\right ) \left (4 r +13\right )}{\left (-1+r \right ) \left (r -2\right ) \left (r +5\right ) \left (r^{2}+8 r +12\right )}\) | \(-{\frac {117}{8}}\) |
| \(b_{6}\) | \(\frac {\left (4 r +21\right ) \left (4 r +17\right ) \left (4 r +5\right ) \left (4 r +1\right ) \left (4 r +9\right ) \left (4 r +13\right )}{\left (r^{2}+8 r +12\right ) \left (-1+r \right ) \left (r -2\right ) \left (r^{2}+10 r +21\right )}\) | \(-{\frac {663}{8}}\) |
| \(b_{7}\) | \(\frac {\left (4 r +25\right ) \left (4 r +21\right ) \left (4 r +17\right ) \left (4 r +5\right ) \left (4 r +1\right ) \left (4 r +9\right ) \left (4 r +13\right )}{\left (r^{2}+10 r +21\right ) \left (-1+r \right ) \left (r^{2}+12 r +32\right ) \left (r^{2}-4\right )}\) | \(-{\frac {3315}{8}}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x^{3} \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-x -\frac {15 x^{4}}{8}-\frac {117 x^{5}}{8}-\frac {663 x^{6}}{8}-\frac {3315 x^{7}}{8}+O\left (x^{8}\right )}{x} \end {align*}
Therefore the homogeneous solution is \begin{align*} y_h(x) &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= c_{1} x^{3} \left (1+\frac {39 x}{5}+\frac {221 x^{2}}{5}+221 x^{3}+\frac {16575 x^{4}}{16}+\frac {224315 x^{5}}{48}+\frac {493493 x^{6}}{24}+\frac {711399 x^{7}}{8}+O\left (x^{8}\right )\right ) + \frac {c_{2} \left (1-x -\frac {15 x^{4}}{8}-\frac {117 x^{5}}{8}-\frac {663 x^{6}}{8}-\frac {3315 x^{7}}{8}+O\left (x^{8}\right )\right )}{x} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{3} \left (1+\frac {39 x}{5}+\frac {221 x^{2}}{5}+221 x^{3}+\frac {16575 x^{4}}{16}+\frac {224315 x^{5}}{48}+\frac {493493 x^{6}}{24}+\frac {711399 x^{7}}{8}+O\left (x^{8}\right )\right )+\frac {c_{2} \left (1-x -\frac {15 x^{4}}{8}-\frac {117 x^{5}}{8}-\frac {663 x^{6}}{8}-\frac {3315 x^{7}}{8}+O\left (x^{8}\right )\right )}{x} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{3} \left (1+\frac {39 x}{5}+\frac {221 x^{2}}{5}+221 x^{3}+\frac {16575 x^{4}}{16}+\frac {224315 x^{5}}{48}+\frac {493493 x^{6}}{24}+\frac {711399 x^{7}}{8}+O\left (x^{8}\right )\right )+\frac {c_{2} \left (1-x -\frac {15 x^{4}}{8}-\frac {117 x^{5}}{8}-\frac {663 x^{6}}{8}-\frac {3315 x^{7}}{8}+O\left (x^{8}\right )\right )}{x} \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{3} \left (1+\frac {39 x}{5}+\frac {221 x^{2}}{5}+221 x^{3}+\frac {16575 x^{4}}{16}+\frac {224315 x^{5}}{48}+\frac {493493 x^{6}}{24}+\frac {711399 x^{7}}{8}+O\left (x^{8}\right )\right )+\frac {c_{2} \left (1-x -\frac {15 x^{4}}{8}-\frac {117 x^{5}}{8}-\frac {663 x^{6}}{8}-\frac {3315 x^{7}}{8}+O\left (x^{8}\right )\right )}{x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (4 x -1\right ) y^{\prime \prime }+\left (5 x^{2}+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^{2} \left (4 x -1\right )}-\frac {\left (5 x +1\right ) y^{\prime }}{x \left (4 x -1\right )} \\ \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 (5 x +1\right ) y^{\prime }}{x \left (4 x -1\right )}+\frac {3 y}{x^{2} \left (4 x -1\right )}=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 {5 x +1}{x \left (4 x -1\right )}, P_{3}\left (x \right )=\frac {3}{x^{2} \left (4 x -1\right )}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=-1 \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=-3 \\ {} & \circ & x =0\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}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x^{2} \left (4 x -1\right ) y^{\prime \prime }+x \left (5 x +1\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 =1..2 \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..3 \\ {} & {} & x^{m}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & x^{m}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -a_{0} \left (1+r \right ) \left (-3+r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (-a_{k} \left (k +r +1\right ) \left (k +r -3\right )+a_{k -1} \left (k +r -1\right ) \left (4 k -3+4 r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -\left (1+r \right ) \left (-3+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-1, 3\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 4 \left (k +r -1\right ) \left (k -\frac {3}{4}+r \right ) a_{k -1}-a_{k} \left (k +r +1\right ) \left (k +r -3\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 4 \left (k +r \right ) \left (k +\frac {1}{4}+r \right ) a_{k}-a_{k +1} \left (k +2+r \right ) \left (k -2+r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {\left (k +r \right ) \left (4 k +4 r +1\right ) a_{k}}{\left (k +2+r \right ) \left (k -2+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =1 \\ {} & {} & a_{k +1}=\frac {\left (k -1\right ) \left (4 k -3\right ) a_{k}}{\left (k +1\right ) \left (k -3\right )} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =0 \\ {} & {} & a_{1}=-a_{0} \\ \bullet & {} & \textrm {Terminating series solution of the ODE for}\hspace {3pt} r =-1\hspace {3pt}\textrm {. Use reduction of order to find the second linearly independent solution}\hspace {3pt} \\ {} & {} & y=a_{0}\cdot \left (1-x \right ) \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =3 \\ {} & {} & a_{k +1}=\frac {\left (k +3\right ) \left (4 k +13\right ) a_{k}}{\left (k +5\right ) \left (k +1\right )} \\ \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 {\left (k +3\right ) \left (4 k +13\right ) a_{k}}{\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-x \right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +3}\right ), b_{k +1}=\frac {\left (k +3\right ) \left (4 k +13\right ) b_{k}}{\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: 51
Order:=8; dsolve(x^2*(4*x-1)*diff(y(x),x$2)+x*(5*x+1)*diff(y(x),x)+3*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{3} \left (1+\frac {39}{5} x +\frac {221}{5} x^{2}+221 x^{3}+\frac {16575}{16} x^{4}+\frac {224315}{48} x^{5}+\frac {493493}{24} x^{6}+\frac {711399}{8} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+\frac {c_{2} \left (-144+144 x +270 x^{4}+2106 x^{5}+11934 x^{6}+59670 x^{7}+\operatorname {O}\left (x^{8}\right )\right )}{x} \]
✓ Solution by Mathematica
Time used: 0.171 (sec). Leaf size: 80
AsymptoticDSolveValue[x^2*(4*x-1)*y''[x]+x*(5*x+1)*y'[x]+3*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_1 \left (-\frac {663 x^5}{8}-\frac {117 x^4}{8}-\frac {15 x^3}{8}+\frac {1}{x}-1\right )+c_2 \left (\frac {493493 x^9}{24}+\frac {224315 x^8}{48}+\frac {16575 x^7}{16}+221 x^6+\frac {221 x^5}{5}+\frac {39 x^4}{5}+x^3\right ) \]