Internal problem ID [1441]
Internal file name [OUTPUT/1442_Sunday_June_05_2022_02_17_41_AM_46253651/index.tex]
Book: Elementary differential equations with boundary value problems. William F. Trench.
Brooks/Cole 2001
Section: Chapter 7 Series Solutions of Linear Second Equations. 7.6 THE METHOD OF
FROBENIUS III. Exercises 7.7. Page 389
Problem number: 25.
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
[_Lienard]
\[ \boxed {x y^{\prime \prime }-5 y^{\prime }+y x=0} \] 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. \[ x y^{\prime \prime }-5 y^{\prime }+y x = 0 \] 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}\\ q(x) &= 1\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x y^{\prime \prime }-5 y^{\prime }+y x = 0 \] Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} \] Then \begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} x \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )-5 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) x = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-5 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n}\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} &= \moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} x^{n +r -1} \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r -1\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-5 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )-5 \left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ x^{-1+r} a_{0} r \left (-1+r \right )-5 r a_{0} x^{-1+r} = 0 \] Or \[ \left (x^{-1+r} r \left (-1+r \right )-5 r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ r \,x^{-1+r} \left (-6+r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r \left (-6+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 6\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ r \,x^{-1+r} \left (-6+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([6, 0]\).
Since \(r_1 - r_2 = 6\) 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^{6} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +6}\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end {align*}
Where \(C\) above can be zero. We start by finding \(y_{1}\). Eq (2B) derived above is now used to find all \(a_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = 0 \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} a_{n} \left (n +r \right ) \left (n +r -1\right )-5 a_{n} \left (n +r \right )+a_{n -2} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -2}}{n^{2}+2 n r +r^{2}-6 n -6 r}\tag {4} \] Which for the root \(r = 6\) becomes \[ a_{n} = -\frac {a_{n -2}}{n \left (n +6\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 6\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=-\frac {1}{r^{2}-2 r -8} \] Which for the root \(r = 6\) becomes \[ a_{2}=-{\frac {1}{16}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {1}{r^{2}-2 r -8}\) | \(-{\frac {1}{16}}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {1}{r^{2}-2 r -8}\) | \(-{\frac {1}{16}}\) |
| \(a_{3}\) | \(0\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {1}{r^{4}-20 r^{2}+64} \] Which for the root \(r = 6\) becomes \[ a_{4}={\frac {1}{640}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {1}{r^{2}-2 r -8}\) | \(-{\frac {1}{16}}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {1}{r^{4}-20 r^{2}+64}\) | \(\frac {1}{640}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {1}{r^{2}-2 r -8}\) | \(-{\frac {1}{16}}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {1}{r^{4}-20 r^{2}+64}\) | \(\frac {1}{640}\) |
| \(a_{5}\) | \(0\) | \(0\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=-\frac {1}{\left (r^{4}-20 r^{2}+64\right ) r \left (r +6\right )} \] Which for the root \(r = 6\) becomes \[ a_{6}=-{\frac {1}{46080}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(0\) | \(0\) |
| \(a_{2}\) | \(-\frac {1}{r^{2}-2 r -8}\) | \(-{\frac {1}{16}}\) |
| \(a_{3}\) | \(0\) | \(0\) |
| \(a_{4}\) | \(\frac {1}{r^{4}-20 r^{2}+64}\) | \(\frac {1}{640}\) |
| \(a_{5}\) | \(0\) | \(0\) |
| \(a_{6}\) | \(-\frac {1}{\left (r^{4}-20 r^{2}+64\right ) r \left (r +6\right )}\) | \(-{\frac {1}{46080}}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{6} \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}\dots \right ) \\ &= x^{6} \left (1-\frac {x^{2}}{16}+\frac {x^{4}}{640}-\frac {x^{6}}{46080}+O\left (x^{7}\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 \(N=6\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{6}\left (r \right )\). 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_{6} \\ &= -\frac {1}{\left (r^{4}-20 r^{2}+64\right ) r \left (r +6\right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}-\frac {1}{\left (r^{4}-20 r^{2}+64\right ) r \left (r +6\right )}&= \lim _{r\rightarrow 0}-\frac {1}{\left (r^{4}-20 r^{2}+64\right ) r \left (r +6\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 \(x y^{\prime \prime }-5 y^{\prime }+y x = 0\) gives \[ x \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )-5 C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )-\frac {5 C y_{1}\left (x \right )}{x}-5 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )+\left (C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right )\right ) x = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (y_{1}^{\prime \prime }\left (x \right ) x +y_{1}\left (x \right ) x -5 y_{1}^{\prime }\left (x \right )\right ) \ln \left (x \right )+x \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )-\frac {5 y_{1}\left (x \right )}{x}\right ) C +x \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 (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) x -5 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right ) = 0 \end{equation} But since \(y_{1}\left (x \right )\) is a solution to the ode, then \[ y_{1}^{\prime \prime }\left (x \right ) x +y_{1}\left (x \right ) x -5 y_{1}^{\prime }\left (x \right ) = 0 \] Eq (7) simplifes to \begin{equation} \tag{8} \left (x \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )-\frac {5 y_{1}\left (x \right )}{x}\right ) C +x \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 (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) x -5 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\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} \frac {\left (2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{1}} a_{n} \left (n +r_{1}\right )\right ) x -6 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\right )\right ) C}{x}+\frac {\left (\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\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 )\right ) x^{2}-5 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right ) x}{x} = 0 \end{equation} Since \(r_{1} = 6\) and \(r_{2} = 0\) then the above becomes \begin{equation} \tag{10} \frac {\left (2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{5+n} a_{n} \left (n +6\right )\right ) x -6 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +6}\right )\right ) C}{x}+\frac {\left (\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -2} b_{n} n \left (-1+n \right )\right )\right ) x^{2}-5 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n} b_{n} n \right ) x}{x} = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{5+n} a_{n} \left (n +6\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 C \,x^{5+n} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n} b_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{-1+n} b_{n} \left (-1+n \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-5 x^{-1+n} b_{n} 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^{5+n} a_{n} \left (n +6\right ) &= \moverset {\infty }{\munderset {n =6}{\sum }}2 C a_{-6+n} n \,x^{-1+n} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-6 C \,x^{5+n} a_{n}\right ) &= \moverset {\infty }{\munderset {n =6}{\sum }}\left (-6 C a_{-6+n} x^{-1+n}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n} b_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}b_{n -2} 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 =6}{\sum }}2 C a_{-6+n} n \,x^{-1+n}\right )+\moverset {\infty }{\munderset {n =6}{\sum }}\left (-6 C a_{-6+n} x^{-1+n}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}b_{n -2} x^{-1+n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{-1+n} b_{n} \left (-1+n \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-5 x^{-1+n} b_{n} 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 \[ -5 b_{1} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -5 b_{1} = 0 \] Solving the above for \(b_{1}\) gives \[ b_{1}=0 \] For \(n=2\), Eq (2B) gives \[ b_{0}-8 b_{2} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 1-8 b_{2} = 0 \] Solving the above for \(b_{2}\) gives \[ b_{2}={\frac {1}{8}} \] For \(n=3\), Eq (2B) gives \[ b_{1}-9 b_{3} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -9 b_{3} = 0 \] Solving the above for \(b_{3}\) gives \[ b_{3}=0 \] For \(n=4\), Eq (2B) gives \[ b_{2}-8 b_{4} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {1}{8}-8 b_{4} = 0 \] Solving the above for \(b_{4}\) gives \[ b_{4}={\frac {1}{64}} \] For \(n=5\), Eq (2B) gives \[ b_{3}-5 b_{5} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ -5 b_{5} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}=0 \] For \(n=N\), where \(N=6\) which is the difference between the two roots, we are free to choose \(b_{6} = 0\). Hence for \(n=6\), Eq (2B) gives \[ 6 C +\frac {1}{64} = 0 \] Which is solved for \(C\). Solving for \(C\) gives \[ C=-{\frac {1}{384}} \] 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 \(C=-{\frac {1}{384}}\) and all \(b_{n}\), then the second solution becomes \[ y_{2}\left (x \right )= -\frac {1}{384}\eslowast \left (x^{6} \left (1-\frac {x^{2}}{16}+\frac {x^{4}}{640}-\frac {x^{6}}{46080}+O\left (x^{7}\right )\right )\right ) \ln \left (x \right )+1+\frac {x^{2}}{8}+\frac {x^{4}}{64}+O\left (x^{7}\right ) \] 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^{6} \left (1-\frac {x^{2}}{16}+\frac {x^{4}}{640}-\frac {x^{6}}{46080}+O\left (x^{7}\right )\right ) + c_{2} \left (-\frac {1}{384}\eslowast \left (x^{6} \left (1-\frac {x^{2}}{16}+\frac {x^{4}}{640}-\frac {x^{6}}{46080}+O\left (x^{7}\right )\right )\right ) \ln \left (x \right )+1+\frac {x^{2}}{8}+\frac {x^{4}}{64}+O\left (x^{7}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{6} \left (1-\frac {x^{2}}{16}+\frac {x^{4}}{640}-\frac {x^{6}}{46080}+O\left (x^{7}\right )\right )+c_{2} \left (-\frac {x^{6} \left (1-\frac {x^{2}}{16}+\frac {x^{4}}{640}-\frac {x^{6}}{46080}+O\left (x^{7}\right )\right ) \ln \left (x \right )}{384}+1+\frac {x^{2}}{8}+\frac {x^{4}}{64}+O\left (x^{7}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{6} \left (1-\frac {x^{2}}{16}+\frac {x^{4}}{640}-\frac {x^{6}}{46080}+O\left (x^{7}\right )\right )+c_{2} \left (-\frac {x^{6} \left (1-\frac {x^{2}}{16}+\frac {x^{4}}{640}-\frac {x^{6}}{46080}+O\left (x^{7}\right )\right ) \ln \left (x \right )}{384}+1+\frac {x^{2}}{8}+\frac {x^{4}}{64}+O\left (x^{7}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{6} \left (1-\frac {x^{2}}{16}+\frac {x^{4}}{640}-\frac {x^{6}}{46080}+O\left (x^{7}\right )\right )+c_{2} \left (-\frac {x^{6} \left (1-\frac {x^{2}}{16}+\frac {x^{4}}{640}-\frac {x^{6}}{46080}+O\left (x^{7}\right )\right ) \ln \left (x \right )}{384}+1+\frac {x^{2}}{8}+\frac {x^{4}}{64}+O\left (x^{7}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}y^{\prime }\right )-5 y^{\prime }+y x =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 {5 y^{\prime }}{x}-y \\ \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 {5 y^{\prime }}{x}+y=0 \\ \bullet & {} & \textrm {Simplify ODE}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d}{d x}y^{\prime }\right )-5 x y^{\prime }+y x^{2}=0 \\ \bullet & {} & \textrm {Make a change of variables}\hspace {3pt} \\ {} & {} & y=x^{3} u \left (x \right ) \\ \bullet & {} & \textrm {Compute}\hspace {3pt} y^{\prime } \\ {} & {} & y^{\prime }=3 x^{2} u \left (x \right )+x^{3} u^{\prime }\left (x \right ) \\ \bullet & {} & \textrm {Compute}\hspace {3pt} \frac {d}{d x}y^{\prime } \\ {} & {} & \frac {d}{d x}y^{\prime }=6 x u \left (x \right )+6 x^{2} u^{\prime }\left (x \right )+x^{3} \left (\frac {d}{d x}u^{\prime }\left (x \right )\right ) \\ \bullet & {} & \textrm {Apply change of variables to the ODE}\hspace {3pt} \\ {} & {} & x^{2} u \left (x \right )+\left (\frac {d}{d x}u^{\prime }\left (x \right )\right ) x^{2}+u^{\prime }\left (x \right ) x -9 u \left (x \right )=0 \\ \bullet & {} & \textrm {ODE is now of the Bessel form}\hspace {3pt} \\ \bullet & {} & \textrm {Solution to Bessel ODE}\hspace {3pt} \\ {} & {} & u \left (x \right )=c_{1} \mathit {BesselJ}\left (3, x\right )+c_{2} \mathit {BesselY}\left (3, x\right ) \\ \bullet & {} & \textrm {Make the change from}\hspace {3pt} y\hspace {3pt}\textrm {back to}\hspace {3pt} y \\ {} & {} & y=\left (c_{1} \mathit {BesselJ}\left (3, x\right )+c_{2} \mathit {BesselY}\left (3, x\right )\right ) x^{3} \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel <- Bessel successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 32
Order:=6; dsolve(x*diff(y(x),x$2)-5*diff(y(x),x)+x*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{6} \left (1-\frac {1}{16} x^{2}+\frac {1}{640} x^{4}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (-86400-10800 x^{2}-1350 x^{4}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.008 (sec). Leaf size: 44
AsymptoticDSolveValue[x*y''[x]-5*y'[x]+x*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (\frac {x^4}{64}+\frac {x^2}{8}+1\right )+c_2 \left (\frac {x^{10}}{640}-\frac {x^8}{16}+x^6\right ) \]