Internal problem ID [6061]
Internal file name [OUTPUT/5309_Sunday_June_05_2022_03_33_46_PM_14042674/index.tex]
Book: An introduction to Ordinary Differential Equations. Earl A. Coddington. Dover. NY
1961
Section: Chapter 4. Linear equations with Regular Singular Points. Page 166
Problem number: 3(f).
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], [_2nd_order, _linear, `_with_symmetry_[0,F(x)]`]]
\[ \boxed {x^{2} y^{\prime \prime }-2 x^{2} y^{\prime }+\left (4 x -2\right ) y=0} \] 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. \[ x^{2} y^{\prime \prime }-2 x^{2} y^{\prime }+\left (4 x -2\right ) y = 0 \] 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) &= -2\\ q(x) &= \frac {4 x -2}{x^{2}}\\ \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 \[ x^{2} y^{\prime \prime }-2 x^{2} y^{\prime }+\left (4 x -2\right ) y = 0 \] 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 (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )-2 x^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (4 x -2\right ) \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^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 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 \(n +r\) 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 }}\left (-2 x^{1+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}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}4 x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} 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 =0}{\sum }}x^{n +r} 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}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-2 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )-2 a_{0} x^{r} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )-2 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r^{2}-r -2\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r^{2}-r -2 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 2\\ r_2 &= -1 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r^{2}-r -2\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([2, -1]\).
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^{2} \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 +2}\\ 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} a_{n} \left (n +r \right ) \left (n +r -1\right )-2 a_{n -1} \left (n +r -1\right )+4 a_{n -1}-2 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {2 a_{n -1} \left (n +r -3\right )}{n^{2}+2 n r +r^{2}-n -r -2}\tag {4} \] Which for the root \(r = 2\) becomes \[ a_{n} = \frac {2 a_{n -1} \left (n -1\right )}{n \left (n +3\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 2\) 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 \[ a_{1}=\frac {-4+2 r}{r^{2}+r -2} \] Which for the root \(r = 2\) becomes \[ a_{1}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-4+2 r}{r^{2}+r -2}\) | \(0\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {-8+4 r}{\left (r +3\right ) r \left (r +2\right )} \] Which for the root \(r = 2\) becomes \[ a_{2}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-4+2 r}{r^{2}+r -2}\) | \(0\) |
| \(a_{2}\) | \(\frac {-8+4 r}{\left (r +3\right ) r \left (r +2\right )}\) | \(0\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-16+8 r}{\left (r +3\right ) \left (r +2\right ) \left (r +4\right ) \left (r +1\right )} \] Which for the root \(r = 2\) becomes \[ a_{3}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-4+2 r}{r^{2}+r -2}\) | \(0\) |
| \(a_{2}\) | \(\frac {-8+4 r}{\left (r +3\right ) r \left (r +2\right )}\) | \(0\) |
| \(a_{3}\) | \(\frac {-16+8 r}{\left (r +3\right ) \left (r +2\right ) \left (r +4\right ) \left (r +1\right )}\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {-32+16 r}{\left (r +3\right ) \left (r +2\right )^{2} \left (r +4\right ) \left (r +5\right )} \] Which for the root \(r = 2\) becomes \[ a_{4}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-4+2 r}{r^{2}+r -2}\) | \(0\) |
| \(a_{2}\) | \(\frac {-8+4 r}{\left (r +3\right ) r \left (r +2\right )}\) | \(0\) |
| \(a_{3}\) | \(\frac {-16+8 r}{\left (r +3\right ) \left (r +2\right ) \left (r +4\right ) \left (r +1\right )}\) | \(0\) |
| \(a_{4}\) | \(\frac {-32+16 r}{\left (r +3\right ) \left (r +2\right )^{2} \left (r +4\right ) \left (r +5\right )}\) | \(0\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-64+32 r}{\left (r +3\right )^{2} \left (r +2\right ) \left (r +4\right ) \left (r +5\right ) \left (r +6\right )} \] Which for the root \(r = 2\) becomes \[ a_{5}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-4+2 r}{r^{2}+r -2}\) | \(0\) |
| \(a_{2}\) | \(\frac {-8+4 r}{\left (r +3\right ) r \left (r +2\right )}\) | \(0\) |
| \(a_{3}\) | \(\frac {-16+8 r}{\left (r +3\right ) \left (r +2\right ) \left (r +4\right ) \left (r +1\right )}\) | \(0\) |
| \(a_{4}\) | \(\frac {-32+16 r}{\left (r +3\right ) \left (r +2\right )^{2} \left (r +4\right ) \left (r +5\right )}\) | \(0\) |
| \(a_{5}\) | \(\frac {-64+32 r}{\left (r +3\right )^{2} \left (r +2\right ) \left (r +4\right ) \left (r +5\right ) \left (r +6\right )}\) | \(0\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {-128+64 r}{\left (r +3\right ) \left (r +2\right ) \left (r +4\right )^{2} \left (r +5\right ) \left (r +6\right ) \left (r +7\right )} \] Which for the root \(r = 2\) becomes \[ a_{6}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-4+2 r}{r^{2}+r -2}\) | \(0\) |
| \(a_{2}\) | \(\frac {-8+4 r}{\left (r +3\right ) r \left (r +2\right )}\) | \(0\) |
| \(a_{3}\) | \(\frac {-16+8 r}{\left (r +3\right ) \left (r +2\right ) \left (r +4\right ) \left (r +1\right )}\) | \(0\) |
| \(a_{4}\) | \(\frac {-32+16 r}{\left (r +3\right ) \left (r +2\right )^{2} \left (r +4\right ) \left (r +5\right )}\) | \(0\) |
| \(a_{5}\) | \(\frac {-64+32 r}{\left (r +3\right )^{2} \left (r +2\right ) \left (r +4\right ) \left (r +5\right ) \left (r +6\right )}\) | \(0\) |
| \(a_{6}\) | \(\frac {-128+64 r}{\left (r +3\right ) \left (r +2\right ) \left (r +4\right )^{2} \left (r +5\right ) \left (r +6\right ) \left (r +7\right )}\) | \(0\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=\frac {-256+128 r}{\left (r +3\right ) \left (r +2\right ) \left (r +4\right ) \left (r +5\right )^{2} \left (r +6\right ) \left (r +7\right ) \left (r +8\right )} \] Which for the root \(r = 2\) becomes \[ a_{7}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-4+2 r}{r^{2}+r -2}\) | \(0\) |
| \(a_{2}\) | \(\frac {-8+4 r}{\left (r +3\right ) r \left (r +2\right )}\) | \(0\) |
| \(a_{3}\) | \(\frac {-16+8 r}{\left (r +3\right ) \left (r +2\right ) \left (r +4\right ) \left (r +1\right )}\) | \(0\) |
| \(a_{4}\) | \(\frac {-32+16 r}{\left (r +3\right ) \left (r +2\right )^{2} \left (r +4\right ) \left (r +5\right )}\) | \(0\) |
| \(a_{5}\) | \(\frac {-64+32 r}{\left (r +3\right )^{2} \left (r +2\right ) \left (r +4\right ) \left (r +5\right ) \left (r +6\right )}\) | \(0\) |
| \(a_{6}\) | \(\frac {-128+64 r}{\left (r +3\right ) \left (r +2\right ) \left (r +4\right )^{2} \left (r +5\right ) \left (r +6\right ) \left (r +7\right )}\) | \(0\) |
| \(a_{7}\) | \(\frac {-256+128 r}{\left (r +3\right ) \left (r +2\right ) \left (r +4\right ) \left (r +5\right )^{2} \left (r +6\right ) \left (r +7\right ) \left (r +8\right )}\) | \(0\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{2} \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^{2} \left (1+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 \(N=3\). 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 {-16+8 r}{\left (r +3\right ) \left (r +2\right ) \left (r +4\right ) \left (r +1\right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {-16+8 r}{\left (r +3\right ) \left (r +2\right ) \left (r +4\right ) \left (r +1\right )}&= \lim _{r\rightarrow -1}\frac {-16+8 r}{\left (r +3\right ) \left (r +2\right ) \left (r +4\right ) \left (r +1\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 \(x^{2} y^{\prime \prime }-2 x^{2} y^{\prime }+\left (4 x -2\right ) y = 0\) gives \[ x^{2} \left (C y_{1}^{\prime \prime }\left (x \right ) \ln \left (x \right )+\frac {2 C y_{1}^{\prime }\left (x \right )}{x}-\frac {C y_{1}\left (x \right )}{x^{2}}+\moverset {\infty }{\munderset {n =0}{\sum }}\left (\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )^{2}}{x^{2}}-\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x^{2}}\right )\right )-2 x^{2} \left (C y_{1}^{\prime }\left (x \right ) \ln \left (x \right )+\frac {C y_{1}\left (x \right )}{x}+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )\right )+\left (4 x -2\right ) \left (C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right )\right ) = 0 \] Which can be written as \begin{equation} \tag{7} \left (\left (y_{1}^{\prime \prime }\left (x \right ) x^{2}-2 x^{2} y_{1}^{\prime }\left (x \right )+\left (4 x -2\right ) y_{1}\left (x \right )\right ) \ln \left (x \right )+x^{2} \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )-2 x y_{1}\left (x \right )\right ) C +x^{2} \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 )-2 x^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )+\left (4 x -2\right ) \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 \[ y_{1}^{\prime \prime }\left (x \right ) x^{2}-2 x^{2} y_{1}^{\prime }\left (x \right )+\left (4 x -2\right ) y_{1}\left (x \right ) = 0 \] Eq (7) simplifes to \begin{equation} \tag{8} \left (x^{2} \left (\frac {2 y_{1}^{\prime }\left (x \right )}{x}-\frac {y_{1}\left (x \right )}{x^{2}}\right )-2 x y_{1}\left (x \right )\right ) C +x^{2} \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 )-2 x^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {b_{n} x^{n +r_{2}} \left (n +r_{2}\right )}{x}\right )+\left (4 x -2\right ) \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 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{1}} a_{n} \left (n +r_{1}\right )\right ) x +\left (-1-2 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r_{1}}\right )\right ) C +\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 ) x^{2}-2 x^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-1+n +r_{2}} b_{n} \left (n +r_{2}\right )\right )+2 \left (2 x -1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r_{2}}\right ) = 0 \end{equation} Since \(r_{1} = 2\) and \(r_{2} = -1\) then the above becomes \begin{equation} \tag{10} \left (2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n} a_{n} \left (n +2\right )\right ) x +\left (-1-2 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +2}\right )\right ) C +\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-3+n} b_{n} \left (n -1\right ) \left (-2+n \right )\right ) x^{2}-2 x^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{-2+n} b_{n} \left (n -1\right )\right )+2 \left (2 x -1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -1}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n +2} a_{n} \left (n +2\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-C \,x^{n +2} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 C \,x^{n +3} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -1} b_{n} \left (n^{2}-3 n +2\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n} b_{n} \left (n -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 b_{n} x^{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 b_{n} x^{n -1}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n -1\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n -1}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}2 C \,x^{n +2} a_{n} \left (n +2\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}2 C a_{-3+n} \left (n -1\right ) x^{n -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-C \,x^{n +2} a_{n}\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-C a_{-3+n} x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 C \,x^{n +3} a_{n}\right ) &= \moverset {\infty }{\munderset {n =4}{\sum }}\left (-2 C a_{n -4} x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n} b_{n} \left (n -1\right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 b_{n -1} \left (-2+n \right ) x^{n -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}4 b_{n} x^{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}4 b_{n -1} x^{n -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 \(n -1\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =3}{\sum }}2 C a_{-3+n} \left (n -1\right ) x^{n -1}\right )+\moverset {\infty }{\munderset {n =3}{\sum }}\left (-C a_{-3+n} x^{n -1}\right )+\moverset {\infty }{\munderset {n =4}{\sum }}\left (-2 C a_{n -4} x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n -1} b_{n} \left (n^{2}-3 n +2\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 b_{n -1} \left (-2+n \right ) x^{n -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}4 b_{n -1} x^{n -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 b_{n} x^{n -1}\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 \[ 6 b_{0}-2 b_{1} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 6-2 b_{1} = 0 \] Solving the above for \(b_{1}\) gives \[ b_{1}=3 \] 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 \[ 12-2 b_{2} = 0 \] Solving the above for \(b_{2}\) gives \[ b_{2}=6 \] 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 \[ 3 C +12 = 0 \] Which is solved for \(C\). Solving for \(C\) gives \[ C=-4 \] For \(n=4\), Eq (2B) gives \[ \left (-2 a_{0}+5 a_{1}\right ) C +4 b_{4} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 8+4 b_{4} = 0 \] Solving the above for \(b_{4}\) gives \[ b_{4}=-2 \] For \(n=5\), Eq (2B) gives \[ \left (-2 a_{1}+7 a_{2}\right ) C -2 b_{4}+10 b_{5} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ 4+10 b_{5} = 0 \] Solving the above for \(b_{5}\) gives \[ b_{5}=-{\frac {2}{5}} \] For \(n=6\), Eq (2B) gives \[ \left (-2 a_{2}+9 a_{3}\right ) C -4 b_{5}+18 b_{6} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {8}{5}+18 b_{6} = 0 \] Solving the above for \(b_{6}\) gives \[ b_{6}=-{\frac {4}{45}} \] For \(n=7\), Eq (2B) gives \[ \left (-2 a_{3}+11 a_{4}\right ) C -6 b_{6}+28 b_{7} = 0 \] Which when replacing the above values found already for \(b_{n}\) and the values found earlier for \(a_{n}\) and for \(C\), gives \[ \frac {8}{15}+28 b_{7} = 0 \] Solving the above for \(b_{7}\) gives \[ b_{7}=-{\frac {2}{105}} \] 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 \(C=-4\) and all \(b_{n}\), then the second solution becomes \[ y_{2}\left (x \right )= \left (-4\right )\eslowast \left (x^{2} \left (1+O\left (x^{8}\right )\right )\right ) \ln \left (x \right )+\frac {1+3 x +6 x^{2}-2 x^{4}-\frac {2 x^{5}}{5}-\frac {4 x^{6}}{45}-\frac {2 x^{7}}{105}+O\left (x^{8}\right )}{x} \] 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^{2} \left (1+O\left (x^{8}\right )\right ) + c_{2} \left (\left (-4\right )\eslowast \left (x^{2} \left (1+O\left (x^{8}\right )\right )\right ) \ln \left (x \right )+\frac {1+3 x +6 x^{2}-2 x^{4}-\frac {2 x^{5}}{5}-\frac {4 x^{6}}{45}-\frac {2 x^{7}}{105}+O\left (x^{8}\right )}{x}\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{2} \left (1+O\left (x^{8}\right )\right )+c_{2} \left (-4 x^{2} \left (1+O\left (x^{8}\right )\right ) \ln \left (x \right )+\frac {1+3 x +6 x^{2}-2 x^{4}-\frac {2 x^{5}}{5}-\frac {4 x^{6}}{45}-\frac {2 x^{7}}{105}+O\left (x^{8}\right )}{x}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{2} \left (1+O\left (x^{8}\right )\right )+c_{2} \left (-4 x^{2} \left (1+O\left (x^{8}\right )\right ) \ln \left (x \right )+\frac {1+3 x +6 x^{2}-2 x^{4}-\frac {2 x^{5}}{5}-\frac {4 x^{6}}{45}-\frac {2 x^{7}}{105}+O\left (x^{8}\right )}{x}\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{2} \left (1+O\left (x^{8}\right )\right )+c_{2} \left (-4 x^{2} \left (1+O\left (x^{8}\right )\right ) \ln \left (x \right )+\frac {1+3 x +6 x^{2}-2 x^{4}-\frac {2 x^{5}}{5}-\frac {4 x^{6}}{45}-\frac {2 x^{7}}{105}+O\left (x^{8}\right )}{x}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d}{d x}y^{\prime }\right )-2 x^{2} y^{\prime }+\left (4 x -2\right ) 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 {2 \left (2 x -1\right ) y}{x^{2}}+2 y^{\prime } \\ \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 }-2 y^{\prime }+\frac {2 \left (2 x -1\right ) y}{x^{2}}=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 )=-2, P_{3}\left (x \right )=\frac {2 \left (2 x -1\right )}{x^{2}}\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}}}=0 \\ {} & \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}}}=-2 \\ {} & \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^{2} \left (\frac {d}{d x}y^{\prime }\right )-2 x^{2} y^{\prime }+\left (4 x -2\right ) 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\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^{2}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r +1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -1 \\ {} & {} & x^{2}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =1}{\sum }}a_{k -1} \left (k -1+r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{2}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k -1+r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (1+r \right ) \left (-2+r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (k +r +1\right ) \left (k +r -2\right )-2 a_{k -1} \left (k -3+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 (-2+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-1, 2\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (k +r +1\right ) \left (k +r -2\right )-2 a_{k -1} \left (k -3+r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +1} \left (k +2+r \right ) \left (k -1+r \right )-2 a_{k} \left (k +r -2\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {2 a_{k} \left (k +r -2\right )}{\left (k +2+r \right ) \left (k -1+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =3 \\ {} & {} & a_{k +1}=\frac {2 a_{k} \left (k -3\right )}{\left (k +1\right ) \left (k -2\right )} \\ \bullet & {} & \textrm {Series not valid for}\hspace {3pt} r =-1\hspace {3pt}\textrm {, division by}\hspace {3pt} 0\hspace {3pt}\textrm {in the recursion relation at}\hspace {3pt} k =2 \\ {} & {} & a_{k +1}=\frac {2 a_{k} \left (k -3\right )}{\left (k +1\right ) \left (k -2\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +1}=\frac {2 a_{k} k}{\left (k +4\right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +2}, a_{k +1}=\frac {2 a_{k} k}{\left (k +4\right ) \left (k +1\right )}\right ] \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)] <- linear_1 successful`
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 55
Order:=8; dsolve(x^2*diff(y(x),x$2)-2*x^2*diff(y(x),x)+(4*x-2)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{2} \left (1+\operatorname {O}\left (x^{8}\right )\right )+\frac {c_{2} \left (\ln \left (x \right ) \left (\left (-48\right ) x^{3}+\operatorname {O}\left (x^{8}\right )\right )+\left (12+36 x +72 x^{2}+88 x^{3}-24 x^{4}-\frac {24}{5} x^{5}-\frac {16}{15} x^{6}-\frac {8}{35} x^{7}+\operatorname {O}\left (x^{8}\right )\right )\right )}{x} \]
✓ Solution by Mathematica
Time used: 0.1 (sec). Leaf size: 58
AsymptoticDSolveValue[x^2*y''[x]-2*x^2*y'[x]+(4*x-2)*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_2 x^2+c_1 \left (-4 x^2 \log (x)-\frac {4 x^6+18 x^5+90 x^4-390 x^3-270 x^2-135 x-45}{45 x}\right ) \]