Internal problem ID [5645]
Internal file name [OUTPUT/4893_Sunday_June_05_2022_03_09_30_PM_19903692/index.tex]
Book: ADVANCED ENGINEERING MATHEMATICS. ERWIN KREYSZIG, HERBERT
KREYSZIG, EDWARD J. NORMINTON. 10th edition. John Wiley USA. 2011
Section: Chapter 5. Series Solutions of ODEs. Special Functions. Problem set 5.3. Extended
Power Series Method: Frobenius Method page 186
Problem number: 11.
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]]
\[ \boxed {x y^{\prime \prime }+\left (2-2 x \right ) y^{\prime }+y \left (x -2\right )=0} \] With the expansion point for the power series method at \(x = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ x y^{\prime \prime }+\left (2-2 x \right ) y^{\prime }+y \left (x -2\right ) = 0 \] 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 {2 \left (x -1\right )}{x}\\ q(x) &= \frac {x -2}{x}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x y^{\prime \prime }+\left (2-2 x \right ) y^{\prime }+y \left (x -2\right ) = 0 \] Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} \] Then \begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} x \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (2-2 x \right ) \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 ) \left (x -2\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 \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 )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 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 }}\left (-2 x^{n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right ) \\ \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} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 a_{n} x^{n +r}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} x^{n +r -1}\right ) \\ \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 =1}{\sum }}\left (-2 a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 \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 )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+2 \left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ x^{-1+r} a_{0} r \left (-1+r \right )+2 r a_{0} x^{-1+r} = 0 \] Or \[ \left (x^{-1+r} r \left (-1+r \right )+2 r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ r \,x^{-1+r} \left (1+r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r \left (1+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 0\\ r_2 &= -1 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ r \,x^{-1+r} \left (1+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([0, -1]\).
Since \(r_1 - r_2 = 1\) 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} \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 -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 \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = \frac {2}{r +2} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} a_{n} \left (n +r \right ) \left (n +r -1\right )-2 a_{n -1} \left (n +r -1\right )+2 a_{n} \left (n +r \right )+a_{n -2}-2 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {2 n a_{n -1}+2 r a_{n -1}-a_{n -2}}{n^{2}+2 n r +r^{2}+n +r}\tag {4} \] Which for the root \(r = 0\) becomes \[ a_{n} = \frac {2 n a_{n -1}-a_{n -2}}{n \left (1+n \right )}\tag {5} \] 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.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2}{r +2}\) | \(1\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {3}{r^{2}+5 r +6} \] Which for the root \(r = 0\) becomes \[ a_{2}={\frac {1}{2}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2}{r +2}\) | \(1\) |
| \(a_{2}\) | \(\frac {3}{r^{2}+5 r +6}\) | \(\frac {1}{2}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {4}{\left (r +4\right ) \left (r +3\right ) \left (r +2\right )} \] Which for the root \(r = 0\) becomes \[ a_{3}={\frac {1}{6}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2}{r +2}\) | \(1\) |
| \(a_{2}\) | \(\frac {3}{r^{2}+5 r +6}\) | \(\frac {1}{2}\) |
| \(a_{3}\) | \(\frac {4}{\left (r +4\right ) \left (r +3\right ) \left (r +2\right )}\) | \(\frac {1}{6}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {5}{\left (r^{2}+9 r +20\right ) \left (r^{2}+5 r +6\right )} \] Which for the root \(r = 0\) becomes \[ a_{4}={\frac {1}{24}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2}{r +2}\) | \(1\) |
| \(a_{2}\) | \(\frac {3}{r^{2}+5 r +6}\) | \(\frac {1}{2}\) |
| \(a_{3}\) | \(\frac {4}{\left (r +4\right ) \left (r +3\right ) \left (r +2\right )}\) | \(\frac {1}{6}\) |
| \(a_{4}\) | \(\frac {5}{\left (r^{2}+9 r +20\right ) \left (r^{2}+5 r +6\right )}\) | \(\frac {1}{24}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {6}{\left (r +6\right ) \left (r +5\right ) \left (r +4\right ) \left (r +3\right ) \left (r +2\right )} \] Which for the root \(r = 0\) becomes \[ a_{5}={\frac {1}{120}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2}{r +2}\) | \(1\) |
| \(a_{2}\) | \(\frac {3}{r^{2}+5 r +6}\) | \(\frac {1}{2}\) |
| \(a_{3}\) | \(\frac {4}{\left (r +4\right ) \left (r +3\right ) \left (r +2\right )}\) | \(\frac {1}{6}\) |
| \(a_{4}\) | \(\frac {5}{\left (r^{2}+9 r +20\right ) \left (r^{2}+5 r +6\right )}\) | \(\frac {1}{24}\) |
| \(a_{5}\) | \(\frac {6}{\left (r +6\right ) \left (r +5\right ) \left (r +4\right ) \left (r +3\right ) \left (r +2\right )}\) | \(\frac {1}{120}\) |
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+x +\frac {x^{2}}{2}+\frac {x^{3}}{6}+\frac {x^{4}}{24}+\frac {x^{5}}{120}+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 \(N=1\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{1}\left (r \right )\). 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_{1} \\ &= \frac {2}{r +2} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {2}{r +2}&= \lim _{r\rightarrow -1}\frac {2}{r +2}\\ &= 2 \end {align*}
The limit is \(2\). Since the limit exists then the log term is not needed and we can set \(C = 0\). Therefore the second solution has the form \begin {align*} y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r}\\ &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -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 \(b_{0} = 1\). Substituting \(n = 1\) in Eq(3) gives \[ b_{1} = \frac {2}{r +2} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{4} b_{n} \left (n +r \right ) \left (n +r -1\right )-2 b_{n -1} \left (n +r -1\right )+2 \left (n +r \right ) b_{n}+b_{n -2}-2 b_{n -1} = 0 \end{equation} Which for for the root \(r = -1\) becomes \begin{equation} \tag{4A} b_{n} \left (n -1\right ) \left (n -2\right )-2 b_{n -1} \left (n -2\right )+2 \left (n -1\right ) b_{n}+b_{n -2}-2 b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from the recursive equation (4) gives \[ b_{n} = \frac {2 n b_{n -1}+2 r b_{n -1}-b_{n -2}}{n^{2}+2 n r +r^{2}+n +r}\tag {5} \] Which for the root \(r = -1\) becomes \[ b_{n} = \frac {2 n b_{n -1}-b_{n -2}-2 b_{n -1}}{n^{2}-n}\tag {6} \] 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.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2}{r +2}\) | \(2\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {3}{r^{2}+5 r +6} \] Which for the root \(r = -1\) becomes \[ b_{2}={\frac {3}{2}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2}{r +2}\) | \(2\) |
| \(b_{2}\) | \(\frac {3}{r^{2}+5 r +6}\) | \(\frac {3}{2}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {4}{\left (r^{2}+7 r +12\right ) \left (r +2\right )} \] Which for the root \(r = -1\) becomes \[ b_{3}={\frac {2}{3}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2}{r +2}\) | \(2\) |
| \(b_{2}\) | \(\frac {3}{r^{2}+5 r +6}\) | \(\frac {3}{2}\) |
| \(b_{3}\) | \(\frac {4}{\left (r +4\right ) \left (r +3\right ) \left (r +2\right )}\) | \(\frac {2}{3}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {5}{\left (r +3\right ) \left (r +2\right ) \left (r^{2}+9 r +20\right )} \] Which for the root \(r = -1\) becomes \[ b_{4}={\frac {5}{24}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2}{r +2}\) | \(2\) |
| \(b_{2}\) | \(\frac {3}{r^{2}+5 r +6}\) | \(\frac {3}{2}\) |
| \(b_{3}\) | \(\frac {4}{\left (r +4\right ) \left (r +3\right ) \left (r +2\right )}\) | \(\frac {2}{3}\) |
| \(b_{4}\) | \(\frac {5}{\left (r +3\right ) \left (r +2\right ) \left (r +5\right ) \left (r +4\right )}\) | \(\frac {5}{24}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {6}{\left (r^{2}+7 r +12\right ) \left (r +2\right ) \left (r^{2}+11 r +30\right )} \] Which for the root \(r = -1\) becomes \[ b_{5}={\frac {1}{20}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2}{r +2}\) | \(2\) |
| \(b_{2}\) | \(\frac {3}{r^{2}+5 r +6}\) | \(\frac {3}{2}\) |
| \(b_{3}\) | \(\frac {4}{\left (r +4\right ) \left (r +3\right ) \left (r +2\right )}\) | \(\frac {2}{3}\) |
| \(b_{4}\) | \(\frac {5}{\left (r +3\right ) \left (r +2\right ) \left (r +5\right ) \left (r +4\right )}\) | \(\frac {5}{24}\) |
| \(b_{5}\) | \(\frac {6}{\left (r +6\right ) \left (r +5\right ) \left (r +4\right ) \left (r +3\right ) \left (r +2\right )}\) | \(\frac {1}{20}\) |
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+2 x +\frac {3 x^{2}}{2}+\frac {2 x^{3}}{3}+\frac {5 x^{4}}{24}+\frac {x^{5}}{20}+O\left (x^{6}\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} \left (1+x +\frac {x^{2}}{2}+\frac {x^{3}}{6}+\frac {x^{4}}{24}+\frac {x^{5}}{120}+O\left (x^{6}\right )\right ) + \frac {c_{2} \left (1+2 x +\frac {3 x^{2}}{2}+\frac {2 x^{3}}{3}+\frac {5 x^{4}}{24}+\frac {x^{5}}{20}+O\left (x^{6}\right )\right )}{x} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \left (1+x +\frac {x^{2}}{2}+\frac {x^{3}}{6}+\frac {x^{4}}{24}+\frac {x^{5}}{120}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+2 x +\frac {3 x^{2}}{2}+\frac {2 x^{3}}{3}+\frac {5 x^{4}}{24}+\frac {x^{5}}{20}+O\left (x^{6}\right )\right )}{x} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (1+x +\frac {x^{2}}{2}+\frac {x^{3}}{6}+\frac {x^{4}}{24}+\frac {x^{5}}{120}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+2 x +\frac {3 x^{2}}{2}+\frac {2 x^{3}}{3}+\frac {5 x^{4}}{24}+\frac {x^{5}}{20}+O\left (x^{6}\right )\right )}{x} \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (1+x +\frac {x^{2}}{2}+\frac {x^{3}}{6}+\frac {x^{4}}{24}+\frac {x^{5}}{120}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+2 x +\frac {3 x^{2}}{2}+\frac {2 x^{3}}{3}+\frac {5 x^{4}}{24}+\frac {x^{5}}{20}+O\left (x^{6}\right )\right )}{x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x y^{\prime \prime }+\left (2-2 x \right ) y^{\prime }+y \left (x -2\right )=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 {\left (x -2\right ) y}{x}+\frac {2 \left (x -1\right ) y^{\prime }}{x} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime }-\frac {2 \left (x -1\right ) y^{\prime }}{x}+\frac {\left (x -2\right ) y}{x}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=-\frac {2 \left (x -1\right )}{x}, P_{3}\left (x \right )=\frac {x -2}{x}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=2 \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=0 \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x y^{\prime \prime }+\left (2-2 x \right ) y^{\prime }+y \left (x -2\right )=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\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x \cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & x \cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (1+r \right ) x^{-1+r}+\left (a_{1} \left (1+r \right ) \left (2+r \right )-2 a_{0} \left (1+r \right )\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +2+r \right )-2 a_{k} \left (k +1+r \right )+a_{k -1}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-1, 0\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right ) \left (2+r \right )-2 a_{0} \left (1+r \right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +1} \left (k +1+r \right ) \left (k +2+r \right )-2 a_{k} k -2 a_{k} r -2 a_{k}+a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +2} \left (k +2+r \right ) \left (k +3+r \right )-2 a_{k +1} \left (k +1\right )-2 r a_{k +1}-2 a_{k +1}+a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {2 k a_{k +1}+2 r a_{k +1}-a_{k}+4 a_{k +1}}{\left (k +2+r \right ) \left (k +3+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1 \\ {} & {} & a_{k +2}=\frac {2 k a_{k +1}-a_{k}+2 a_{k +1}}{\left (k +1\right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -1}, a_{k +2}=\frac {2 k a_{k +1}-a_{k}+2 a_{k +1}}{\left (k +1\right ) \left (k +2\right )}, 0=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=\frac {2 k a_{k +1}-a_{k}+4 a_{k +1}}{\left (k +2\right ) \left (k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +2}=\frac {2 k a_{k +1}-a_{k}+4 a_{k +1}}{\left (k +2\right ) \left (k +3\right )}, 2 a_{1}-2 a_{0}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k}\right ), a_{k +2}=\frac {2 k a_{k +1}-a_{k}+2 a_{k +1}}{\left (k +1\right ) \left (k +2\right )}, 0=0, b_{k +2}=\frac {2 k b_{k +1}-b_{k}+4 b_{k +1}}{\left (k +2\right ) \left (k +3\right )}, 2 b_{1}-2 b_{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) <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 44
Order:=6; dsolve(x*diff(y(x),x$2)+(2-2*x)*diff(y(x),x)+(x-2)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} \left (1+x +\frac {1}{2} x^{2}+\frac {1}{6} x^{3}+\frac {1}{24} x^{4}+\frac {1}{120} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\frac {c_{2} \left (1+2 x +\frac {3}{2} x^{2}+\frac {2}{3} x^{3}+\frac {5}{24} x^{4}+\frac {1}{20} x^{5}+\operatorname {O}\left (x^{6}\right )\right )}{x} \]
✓ Solution by Mathematica
Time used: 0.026 (sec). Leaf size: 58
AsymptoticDSolveValue[x*y''[x]+(2-2*x)*y'[x]+(x-2)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (\frac {5 x^3}{24}+\frac {2 x^2}{3}+\frac {3 x}{2}+\frac {1}{x}+2\right )+c_2 \left (\frac {x^4}{24}+\frac {x^3}{6}+\frac {x^2}{2}+x+1\right ) \]