Internal problem ID [2415]
Internal file name [OUTPUT/2415_Tuesday_February_27_2024_08_37_07_AM_35431443/index.tex]
Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath.
Boston. 1964
Section: Exercise 42, page 206
Problem number: 14.
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]]
\[ \boxed {x^{2} \left (x +1\right ) y^{\prime \prime }+x \left (x -4\right ) y^{\prime }+4 y=0} \] With the expansion point for the power series method at \(x = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ \left (x^{3}+x^{2}\right ) y^{\prime \prime }+\left (x^{2}-4 x \right ) y^{\prime }+4 y = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} y^{\prime \prime }+p(x) y^{\prime } + q(x) y &=0 \end {align*}
Where \begin {align*} p(x) &= \frac {x -4}{x \left (x +1\right )}\\ q(x) &= \frac {4}{x^{2} \left (x +1\right )}\\ \end {align*}
| \(p(x)=\frac {x -4}{x \left (x +1\right )}\) | |
| singularity | type |
| \(x = -1\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
| \(q(x)=\frac {4}{x^{2} \left (x +1\right )}\) | |
| singularity | type |
| \(x = -1\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([-1, 0, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} \left (x +1\right ) y^{\prime \prime }+\left (x^{2}-4 x \right ) y^{\prime }+4 y = 0 \] Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} \] Then \begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} x^{2} \left (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 (x^{2}-4 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+4 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 a_{n} x^{n +r}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r\) 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 }}x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}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 }}a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-4 x^{n +r} a_{n} \left (n +r \right )+4 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )-4 x^{r} a_{0} r +4 a_{0} x^{r} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )-4 x^{r} r +4 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r^{2}-5 r +4\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r^{2}-5 r +4 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 4\\ r_2 &= 1 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r^{2}-5 r +4\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([4, 1]\).
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 ) &= 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 )+x \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^{1+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 -1} \left (n +r -1\right ) \left (n +r -2\right )+a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n -1} \left (n +r -1\right )-4 a_{n} \left (n +r \right )+4 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -1} \left (n +r -1\right )}{n +r -4}\tag {4} \] Which for the root \(r = 4\) becomes \[ a_{n} = -\frac {a_{n -1} \left (n +3\right )}{n}\tag {5} \] 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 \[ a_{1}=-\frac {r}{-3+r} \] Which for the root \(r = 4\) becomes \[ a_{1}=-4 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r}{-3+r}\) | \(-4\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {\left (1+r \right ) r}{\left (-3+r \right ) \left (-2+r \right )} \] Which for the root \(r = 4\) becomes \[ a_{2}=10 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r}{-3+r}\) | \(-4\) |
| \(a_{2}\) | \(\frac {\left (1+r \right ) r}{\left (-3+r \right ) \left (-2+r \right )}\) | \(10\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {r \left (1+r \right ) \left (2+r \right )}{\left (-3+r \right ) \left (-2+r \right ) \left (-1+r \right )} \] Which for the root \(r = 4\) becomes \[ a_{3}=-20 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r}{-3+r}\) | \(-4\) |
| \(a_{2}\) | \(\frac {\left (1+r \right ) r}{\left (-3+r \right ) \left (-2+r \right )}\) | \(10\) |
| \(a_{3}\) | \(-\frac {r \left (1+r \right ) \left (2+r \right )}{\left (-3+r \right ) \left (-2+r \right ) \left (-1+r \right )}\) | \(-20\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {\left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{\left (-3+r \right ) \left (-2+r \right ) \left (-1+r \right )} \] Which for the root \(r = 4\) becomes \[ a_{4}=35 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r}{-3+r}\) | \(-4\) |
| \(a_{2}\) | \(\frac {\left (1+r \right ) r}{\left (-3+r \right ) \left (-2+r \right )}\) | \(10\) |
| \(a_{3}\) | \(-\frac {r \left (1+r \right ) \left (2+r \right )}{\left (-3+r \right ) \left (-2+r \right ) \left (-1+r \right )}\) | \(-20\) |
| \(a_{4}\) | \(\frac {\left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{\left (-3+r \right ) \left (-2+r \right ) \left (-1+r \right )}\) | \(35\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {\left (4+r \right ) \left (2+r \right ) \left (3+r \right )}{\left (-3+r \right ) \left (-2+r \right ) \left (-1+r \right )} \] Which for the root \(r = 4\) becomes \[ a_{5}=-56 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r}{-3+r}\) | \(-4\) |
| \(a_{2}\) | \(\frac {\left (1+r \right ) r}{\left (-3+r \right ) \left (-2+r \right )}\) | \(10\) |
| \(a_{3}\) | \(-\frac {r \left (1+r \right ) \left (2+r \right )}{\left (-3+r \right ) \left (-2+r \right ) \left (-1+r \right )}\) | \(-20\) |
| \(a_{4}\) | \(\frac {\left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{\left (-3+r \right ) \left (-2+r \right ) \left (-1+r \right )}\) | \(35\) |
| \(a_{5}\) | \(-\frac {\left (4+r \right ) \left (2+r \right ) \left (3+r \right )}{\left (-3+r \right ) \left (-2+r \right ) \left (-1+r \right )}\) | \(-56\) |
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}\dots \right ) \\ &= x^{4} \left (1-4 x +10 x^{2}-20 x^{3}+35 x^{4}-56 x^{5}+O\left (x^{6}\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 \(N=3\). 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 \left (1+r \right ) \left (2+r \right )}{\left (-3+r \right ) \left (-2+r \right ) \left (-1+r \right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}-\frac {r \left (1+r \right ) \left (2+r \right )}{\left (-3+r \right ) \left (-2+r \right ) \left (-1+r \right )}&= \lim _{r\rightarrow 1}-\frac {r \left (1+r \right ) \left (2+r \right )}{\left (-3+r \right ) \left (-2+r \right ) \left (-1+r \right )}\\ &= \textit {undefined} \end {align*}
Since the limit does not exist then the log term is needed. Therefore the second solution has the form \[ y_{2}\left (x \right ) = C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) \] Therefore \begin{align*} \frac {d}{d x}y_{2}\left (x \right ) &= C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right ) \\ &= C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right ) \\ \frac {d^{2}}{d x^{2}}y_{2}\left (x \right ) &= C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right ) \\ &= C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-2+n +r_{2}} b_{n} \left (n +r_{2}\right ) \left (-1+n +r_{2}\right )\right ) \\ \end{align*} Substituting these back into the given ode \(x^{2} \left (x +1\right ) y^{\prime \prime }+\left (x^{2}-4 x \right ) y^{\prime }+4 y = 0\) gives \[ x^{2} \left (x +1\right ) \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+\left (x^{2}-4 x \right ) \left (C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )\right )+4 C y_{1}\left (x \right ) \ln \left (x \right )+4 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (x^{2} \left (x +1\right ) y_{1}^{\prime \prime }\left (x \right )+\left (x^{2}-4 x \right ) y_{1}^{\prime }\left (x \right )+4 y_{1}\left (x \right )\right ) \ln \left (x \right )+x^{2} \left (x +1\right ) \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )+\frac {\left (x^{2}-4 x \right ) y_{1}\left (x \right )}{x}\right ) C +x^{2} \left (x +1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+\left (x^{2}-4 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )+4 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation} But since \(y_{1}\left (x \right )\) is a solution to the ode, then \[ x^{2} \left (x +1\right ) y_{1}^{\prime \prime }\left (x \right )+\left (x^{2}-4 x \right ) y_{1}^{\prime }\left (x \right )+4 y_{1}\left (x \right ) = 0 \] Eq (7) simplifes to \begin{equation} \tag{8} \left (x^{2} \left (x +1\right ) \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )+\frac {\left (x^{2}-4 x \right ) y_{1}\left (x \right )}{x}\right ) C +x^{2} \left (x +1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )+\left (x^{2}-4 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )+4 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation} Substituting \(y_{1} = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\) into the above gives \begin{equation} \tag{9} \left (2 x \left (x +1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{1}} a_{n} \left (n +r_{1}\right )\right )-5 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\right )\right ) C +x^{2} \left (x +1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-2+n +r_{2}} b_{n} \left (n +r_{2}\right ) \left (-1+n +r_{2}\right )\right )+\left (x^{2}-4 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right )+4 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation} Since \(r_{1} = 4\) and \(r_{2} = 1\) then the above becomes \begin{equation} \tag{10} \left (2 x \left (x +1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +3} a_{n} \left (n +4\right )\right )-5 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +4}\right )\right ) C +x^{2} \left (x +1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -1} b_{n} \left (1+n \right ) n \right )+\left (x^{2}-4 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n} b_{n} \left (1+n \right )\right )+4 \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{1+n}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n +5} a_{n} \left (n +4\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n +4} a_{n} \left (n +4\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-5 C a_{n} x^{n +4}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{n +2} b_{n} \left (1+n \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{1+n} b_{n} \left (1+n \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +2} b_{n} \left (1+n \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 x^{1+n} b_{n} \left (1+n \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 b_{n} x^{1+n}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(1+n\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{1+n}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n +5} a_{n} \left (n +4\right ) &= \moverset {\infty }{\munderset {n =4}{\sum }}2 C a_{-4+n} n \,x^{1+n} \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n +4} a_{n} \left (n +4\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}2 C a_{n -3} \left (1+n \right ) x^{1+n} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-5 C a_{n} x^{n +4}\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-5 C a_{n -3} x^{1+n}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{n +2} b_{n} \left (1+n \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (n -1\right ) b_{n -1} n \,x^{1+n} \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{n +2} b_{n} \left (1+n \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}b_{n -1} n \,x^{1+n} \\ \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 \(1+n\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =4}{\sum }}2 C a_{-4+n} n \,x^{1+n}\right )+\left (\moverset {\infty }{\munderset {n =3}{\sum }}2 C a_{n -3} \left (1+n \right ) x^{1+n}\right )+\moverset {\infty }{\munderset {n =3}{\sum }}\left (-5 C a_{n -3} x^{1+n}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}\left (n -1\right ) b_{n -1} n \,x^{1+n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{1+n} b_{n} \left (1+n \right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n -1} n \,x^{1+n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 x^{1+n} b_{n} \left (1+n \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 b_{n} x^{1+n}\right ) = 0 \end{equation} For \(n=0\) in Eq. (2B), we choose arbitray value for \(b_{0}\) as \(b_{0} = 1\). For \(n=1\), Eq (2B) gives \[ -2 b_{1}+b_{0} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -2 b_{1}+1 = 0 \] Solving the above for \(b_{1}\) gives \[ b_{1}={\frac {1}{2}} \] For \(n=2\), Eq (2B) gives \[ 4 b_{1}-2 b_{2} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 2-2 b_{2} = 0 \] Solving the above for \(b_{2}\) gives \[ b_{2}=1 \] For \(n=N\), where \(N=3\) which is the difference between the two roots, we are free to choose \(b_{3} = 0\). Hence for \(n=3\), Eq (2B) gives \[ 3 C +9 = 0 \] Which is solved for \(C\). Solving for \(C\) gives \[ C=-3 \] For \(n=4\), Eq (2B) gives \[ \left (8 a_{0}+5 a_{1}\right ) C +16 b_{3}+4 b_{4} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 36+4 b_{4} = 0 \] Solving the above for \(b_{4}\) gives \[ b_{4}=-9 \] For \(n=5\), Eq (2B) gives \[ \left (10 a_{1}+7 a_{2}\right ) C +25 b_{4}+10 b_{5} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -315+10 b_{5} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}={\frac {63}{2}} \] Now that we found all \(b_{n}\) and \(C\), we can calculate the second solution from \[ y_{2}\left (x \right ) = C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) \] Using the above value found for \(C=-3\) and all \(b_{n}\), then the second solution becomes \[ y_{2}\left (x \right )= \left (-3\right )\eslowast \left (x^{4} \left (1-4 x +10 x^{2}-20 x^{3}+35 x^{4}-56 x^{5}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+x \left (1+\frac {x}{2}+x^{2}-9 x^{4}+\frac {63 x^{5}}{2}+O\left (x^{6}\right )\right ) \] 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 (1-4 x +10 x^{2}-20 x^{3}+35 x^{4}-56 x^{5}+O\left (x^{6}\right )\right ) + c_{2} \left (\left (-3\right )\eslowast \left (x^{4} \left (1-4 x +10 x^{2}-20 x^{3}+35 x^{4}-56 x^{5}+O\left (x^{6}\right )\right )\right ) \ln \left (x \right )+x \left (1+\frac {x}{2}+x^{2}-9 x^{4}+\frac {63 x^{5}}{2}+O\left (x^{6}\right )\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{4} \left (1-4 x +10 x^{2}-20 x^{3}+35 x^{4}-56 x^{5}+O\left (x^{6}\right )\right )+c_{2} \left (-3 x^{4} \left (1-4 x +10 x^{2}-20 x^{3}+35 x^{4}-56 x^{5}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x \left (1+\frac {x}{2}+x^{2}-9 x^{4}+\frac {63 x^{5}}{2}+O\left (x^{6}\right )\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{4} \left (1-4 x +10 x^{2}-20 x^{3}+35 x^{4}-56 x^{5}+O\left (x^{6}\right )\right )+c_{2} \left (-3 x^{4} \left (1-4 x +10 x^{2}-20 x^{3}+35 x^{4}-56 x^{5}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x \left (1+\frac {x}{2}+x^{2}-9 x^{4}+\frac {63 x^{5}}{2}+O\left (x^{6}\right )\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{4} \left (1-4 x +10 x^{2}-20 x^{3}+35 x^{4}-56 x^{5}+O\left (x^{6}\right )\right )+c_{2} \left (-3 x^{4} \left (1-4 x +10 x^{2}-20 x^{3}+35 x^{4}-56 x^{5}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x \left (1+\frac {x}{2}+x^{2}-9 x^{4}+\frac {63 x^{5}}{2}+O\left (x^{6}\right )\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (x +1\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (x^{2}-4 x \right ) y^{\prime }+4 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}{x^{2} \left (x +1\right )}-\frac {\left (x -4\right ) y^{\prime }}{x \left (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} \\ {} & {} & \frac {d}{d x}y^{\prime }+\frac {\left (x -4\right ) y^{\prime }}{x \left (x +1\right )}+\frac {4 y}{x^{2} \left (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 {x -4}{x \left (x +1\right )}, P_{3}\left (x \right )=\frac {4}{x^{2} \left (x +1\right )}\right ] \\ {} & \circ & \left (x +1\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=5 \\ {} & \circ & \left (x +1\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (x +1\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=0 \\ {} & \circ & x =-1\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}=-1 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x^{2} \left (x +1\right ) \left (\frac {d}{d x}y^{\prime }\right )+x \left (x -4\right ) y^{\prime }+4 y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -1\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (u^{3}-2 u^{2}+u \right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (u^{2}-6 u +5\right ) \left (\frac {d}{d u}y \left (u \right )\right )+4 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^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & u^{m}\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+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +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 =1..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 +r -1\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} \\ {} & {} & a_{0} r \left (4+r \right ) u^{-1+r}+\left (a_{1} \left (1+r \right ) \left (5+r \right )-2 a_{0} \left (r^{2}+2 r -2\right )\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +5+r \right )-2 a_{k} \left (k^{2}+2 k r +r^{2}+2 k +2 r -2\right )+a_{k -1} \left (k +r -1\right )^{2}\right ) u^{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 \{-4, 0\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right ) \left (5+r \right )-2 a_{0} \left (r^{2}+2 r -2\right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k -1} \left (k +r -1\right )^{2}+a_{k +1} \left (k +1+r \right ) \left (k +5+r \right )-2 \left (k^{2}+\left (2 r +2\right ) k +r^{2}+2 r -2\right ) a_{k}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k} \left (k +r \right )^{2}+a_{k +2} \left (k +2+r \right ) \left (k +6+r \right )-2 \left (\left (k +1\right )^{2}+\left (2 r +2\right ) \left (k +1\right )+r^{2}+2 r -2\right ) a_{k +1}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-2 k^{2} a_{k +1}+2 k r a_{k}-4 k r a_{k +1}+r^{2} a_{k}-2 r^{2} a_{k +1}-8 k a_{k +1}-8 r a_{k +1}-2 a_{k +1}}{\left (k +2+r \right ) \left (k +6+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-4 \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-2 k^{2} a_{k +1}-8 k a_{k}+8 k a_{k +1}+16 a_{k}-2 a_{k +1}}{\left (k -2\right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Series not valid for}\hspace {3pt} r =-4\hspace {3pt}\textrm {, division by}\hspace {3pt} 0\hspace {3pt}\textrm {in the recursion relation at}\hspace {3pt} k =2 \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-2 k^{2} a_{k +1}-8 k a_{k}+8 k a_{k +1}+16 a_{k}-2 a_{k +1}}{\left (k -2\right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-2 k^{2} a_{k +1}-8 k a_{k +1}-2 a_{k +1}}{\left (k +2\right ) \left (k +6\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k}, a_{k +2}=-\frac {k^{2} a_{k}-2 k^{2} a_{k +1}-8 k a_{k +1}-2 a_{k +1}}{\left (k +2\right ) \left (k +6\right )}, 5 a_{1}+4 a_{0}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +1\right )^{k}, a_{k +2}=-\frac {k^{2} a_{k}-2 k^{2} a_{k +1}-8 k a_{k +1}-2 a_{k +1}}{\left (k +2\right ) \left (k +6\right )}, 5 a_{1}+4 a_{0}=0\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) Group is reducible, not completely reducible <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.032 (sec). Leaf size: 61
Order:=6; dsolve(x^2*(x+1)*diff(y(x),x$2)+x*(x-4)*diff(y(x),x)+4*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \left (c_{1} x^{3} \left (1-4 x +10 x^{2}-20 x^{3}+35 x^{4}-56 x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (\ln \left (x \right ) \left (\left (-36\right ) x^{3}+144 x^{4}-360 x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\left (12+6 x +12 x^{2}-240 x^{3}+852 x^{4}-2022 x^{5}+\operatorname {O}\left (x^{6}\right )\right )\right )\right ) x \]
✓ Solution by Mathematica
Time used: 0.023 (sec). Leaf size: 70
AsymptoticDSolveValue[x^2*(x+1)*y''[x]+x*(x-4)*y'[x]+4*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (3 (4 x-1) x^4 \log (x)+\frac {1}{2} \left (62 x^4-20 x^3+2 x^2+x+2\right ) x\right )+c_2 \left (35 x^8-20 x^7+10 x^6-4 x^5+x^4\right ) \]