Internal problem ID [5643]
Internal file name [OUTPUT/4891_Sunday_June_05_2022_03_09_25_PM_35616607/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: 9.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference not integer"
Maple gives the following as the ode type
[_Jacobi]
\[ \boxed {2 x \left (x -1\right ) y^{\prime \prime }-\left (1+x \right ) y^{\prime }+y=0} \] 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. \[ \left (2 x^{2}-2 x \right ) y^{\prime \prime }+\left (-1-x \right ) y^{\prime }+y = 0 \] 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 {1+x}{2 \left (x -1\right ) x}\\ q(x) &= \frac {1}{2 \left (x -1\right ) x}\\ \end {align*}
| \(p(x)=-\frac {1+x}{2 \left (x -1\right ) x}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = 1\) | \(\text {``regular''}\) |
| \(q(x)=\frac {1}{2 \left (x -1\right ) x}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = 1\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, 1, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ 2 x \left (x -1\right ) y^{\prime \prime }+\left (-1-x \right ) y^{\prime }+y = 0 \] 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} 2 \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 -1\right ) x +\left (-1-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 ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}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 }}2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} 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 =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ -2 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )-\left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ -2 x^{-1+r} a_{0} r \left (-1+r \right )-r a_{0} x^{-1+r} = 0 \] Or \[ \left (-2 x^{-1+r} r \left (-1+r \right )-r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (-2 r^{2}+r \right ) x^{-1+r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ -2 r^{2}+r = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {1}{2}}\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (-2 r^{2}+r \right ) x^{-1+r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {1}{2}}, 0\right ]\).
Since \(r_1 - r_2 = {\frac {1}{2}}\) is not 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 ) &= 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 +\frac {1}{2}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n} \end {align*}
We start by finding \(y_{1}\left (x \right )\). 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} 2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )-2 a_{n} \left (n +r \right ) \left (n +r -1\right )-a_{n -1} \left (n +r -1\right )-a_{n} \left (n +r \right )+a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {a_{n -1} \left (2 n^{2}+4 n r +2 r^{2}-7 n -7 r +6\right )}{2 n^{2}+4 n r +2 r^{2}-n -r}\tag {4} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{n} = \frac {a_{n -1} \left (2 n^{2}-5 n +3\right )}{2 n^{2}+n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {1}{2}}\) 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 \[ a_{1}=\frac {2 r^{2}-3 r +1}{2 r^{2}+3 r +1} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{1}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2 r^{2}-3 r +1}{2 r^{2}+3 r +1}\) | \(0\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {2 r^{3}-3 r^{2}+r}{2 r^{3}+9 r^{2}+13 r +6} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{2}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2 r^{2}-3 r +1}{2 r^{2}+3 r +1}\) | \(0\) |
| \(a_{2}\) | \(\frac {2 r^{3}-3 r^{2}+r}{2 r^{3}+9 r^{2}+13 r +6}\) | \(0\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {2 r^{3}-3 r^{2}+r}{2 r^{3}+15 r^{2}+37 r +30} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{3}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2 r^{2}-3 r +1}{2 r^{2}+3 r +1}\) | \(0\) |
| \(a_{2}\) | \(\frac {2 r^{3}-3 r^{2}+r}{2 r^{3}+9 r^{2}+13 r +6}\) | \(0\) |
| \(a_{3}\) | \(\frac {2 r^{3}-3 r^{2}+r}{2 r^{3}+15 r^{2}+37 r +30}\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {2 r^{3}-3 r^{2}+r}{2 r^{3}+21 r^{2}+73 r +84} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{4}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2 r^{2}-3 r +1}{2 r^{2}+3 r +1}\) | \(0\) |
| \(a_{2}\) | \(\frac {2 r^{3}-3 r^{2}+r}{2 r^{3}+9 r^{2}+13 r +6}\) | \(0\) |
| \(a_{3}\) | \(\frac {2 r^{3}-3 r^{2}+r}{2 r^{3}+15 r^{2}+37 r +30}\) | \(0\) |
| \(a_{4}\) | \(\frac {2 r^{3}-3 r^{2}+r}{2 r^{3}+21 r^{2}+73 r +84}\) | \(0\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {2 r^{3}-3 r^{2}+r}{2 r^{3}+27 r^{2}+121 r +180} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{5}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2 r^{2}-3 r +1}{2 r^{2}+3 r +1}\) | \(0\) |
| \(a_{2}\) | \(\frac {2 r^{3}-3 r^{2}+r}{2 r^{3}+9 r^{2}+13 r +6}\) | \(0\) |
| \(a_{3}\) | \(\frac {2 r^{3}-3 r^{2}+r}{2 r^{3}+15 r^{2}+37 r +30}\) | \(0\) |
| \(a_{4}\) | \(\frac {2 r^{3}-3 r^{2}+r}{2 r^{3}+21 r^{2}+73 r +84}\) | \(0\) |
| \(a_{5}\) | \(\frac {2 r^{3}-3 r^{2}+r}{2 r^{3}+27 r^{2}+121 r +180}\) | \(0\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= \sqrt {x} \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 ) \\ &= \sqrt {x}\, \left (1+O\left (x^{6}\right )\right ) \end {align*}
Now the second solution \(y_{2}\left (x \right )\) is found. Eq (2B) derived above is now used to find all \(b_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(b_{0}\) is arbitrary and taken as \(b_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} 2 b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )-2 b_{n} \left (n +r \right ) \left (n +r -1\right )-b_{n -1} \left (n +r -1\right )-\left (n +r \right ) b_{n}+b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = \frac {b_{n -1} \left (2 n^{2}+4 n r +2 r^{2}-7 n -7 r +6\right )}{2 n^{2}+4 n r +2 r^{2}-n -r}\tag {4} \] Which for the root \(r = 0\) becomes \[ b_{n} = \frac {b_{n -1} \left (2 n^{2}-7 n +6\right )}{2 n^{2}-n}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = 0\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ b_{1}=\frac {2 r^{2}-3 r +1}{2 r^{2}+3 r +1} \] Which for the root \(r = 0\) becomes \[ b_{1}=1 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2 r^{2}-3 r +1}{2 r^{2}+3 r +1}\) | \(1\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {2 r^{3}-3 r^{2}+r}{2 r^{3}+9 r^{2}+13 r +6} \] Which for the root \(r = 0\) becomes \[ b_{2}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2 r^{2}-3 r +1}{2 r^{2}+3 r +1}\) | \(1\) |
| \(b_{2}\) | \(\frac {2 r^{3}-3 r^{2}+r}{2 r^{3}+9 r^{2}+13 r +6}\) | \(0\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {2 r^{3}-3 r^{2}+r}{2 r^{3}+15 r^{2}+37 r +30} \] Which for the root \(r = 0\) becomes \[ b_{3}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2 r^{2}-3 r +1}{2 r^{2}+3 r +1}\) | \(1\) |
| \(b_{2}\) | \(\frac {2 r^{3}-3 r^{2}+r}{2 r^{3}+9 r^{2}+13 r +6}\) | \(0\) |
| \(b_{3}\) | \(\frac {2 r^{3}-3 r^{2}+r}{2 r^{3}+15 r^{2}+37 r +30}\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {2 r^{3}-3 r^{2}+r}{2 r^{3}+21 r^{2}+73 r +84} \] Which for the root \(r = 0\) becomes \[ b_{4}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2 r^{2}-3 r +1}{2 r^{2}+3 r +1}\) | \(1\) |
| \(b_{2}\) | \(\frac {2 r^{3}-3 r^{2}+r}{2 r^{3}+9 r^{2}+13 r +6}\) | \(0\) |
| \(b_{3}\) | \(\frac {2 r^{3}-3 r^{2}+r}{2 r^{3}+15 r^{2}+37 r +30}\) | \(0\) |
| \(b_{4}\) | \(\frac {2 r^{3}-3 r^{2}+r}{2 r^{3}+21 r^{2}+73 r +84}\) | \(0\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {2 r^{3}-3 r^{2}+r}{2 r^{3}+27 r^{2}+121 r +180} \] Which for the root \(r = 0\) becomes \[ b_{5}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2 r^{2}-3 r +1}{2 r^{2}+3 r +1}\) | \(1\) |
| \(b_{2}\) | \(\frac {2 r^{3}-3 r^{2}+r}{2 r^{3}+9 r^{2}+13 r +6}\) | \(0\) |
| \(b_{3}\) | \(\frac {2 r^{3}-3 r^{2}+r}{2 r^{3}+15 r^{2}+37 r +30}\) | \(0\) |
| \(b_{4}\) | \(\frac {2 r^{3}-3 r^{2}+r}{2 r^{3}+21 r^{2}+73 r +84}\) | \(0\) |
| \(b_{5}\) | \(\frac {2 r^{3}-3 r^{2}+r}{2 r^{3}+27 r^{2}+121 r +180}\) | \(0\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= 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 \\ &= 1+x +O\left (x^{6}\right ) \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} \sqrt {x}\, \left (1+O\left (x^{6}\right )\right ) + c_{2} \left (1+x +O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \sqrt {x}\, \left (1+O\left (x^{6}\right )\right )+c_{2} \left (1+x +O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \sqrt {x}\, \left (1+O\left (x^{6}\right )\right )+c_{2} \left (1+x +O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \sqrt {x}\, \left (1+O\left (x^{6}\right )\right )+c_{2} \left (1+x +O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 y^{\prime \prime } \left (x -1\right ) x +\left (-1-x \right ) y^{\prime }+y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {\left (1+x \right ) y^{\prime }}{2 x \left (x -1\right )}-\frac {y}{2 \left (x -1\right ) 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 {\left (1+x \right ) y^{\prime }}{2 x \left (x -1\right )}+\frac {y}{2 \left (x -1\right ) x}=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 {1+x}{2 \left (x -1\right ) x}, P_{3}\left (x \right )=\frac {1}{2 \left (x -1\right ) 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}}}=\frac {1}{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}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 2 y^{\prime \prime } \left (x -1\right ) x +\left (-1-x \right ) y^{\prime }+y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =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^{m}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & x^{m}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & x^{m}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -a_{0} r \left (-1+2 r \right ) x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-a_{k +1} \left (k +1+r \right ) \left (2 k +1+2 r \right )+a_{k} \left (2 k +2 r -1\right ) \left (k +r -1\right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -r \left (-1+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {1}{2}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -2 \left (k +1+r \right ) \left (k +\frac {1}{2}+r \right ) a_{k +1}+2 \left (k +r -1\right ) a_{k} \left (k +r -\frac {1}{2}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {\left (k +r -1\right ) a_{k} \left (2 k +2 r -1\right )}{\left (k +1+r \right ) \left (2 k +1+2 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =1 \\ {} & {} & a_{k +1}=\frac {\left (k -1\right ) a_{k} \left (2 k -1\right )}{\left (k +1\right ) \left (2 k +1\right )} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =0 \\ {} & {} & a_{1}=a_{0} \\ \bullet & {} & \textrm {Terminating series solution of the ODE for}\hspace {3pt} r =0\hspace {3pt}\textrm {. Use reduction of order to find the second linearly independent solution}\hspace {3pt} \\ {} & {} & y=a_{0}\cdot \left (1+x \right ) \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & a_{k +1}=\frac {2 \left (k -\frac {1}{2}\right ) a_{k} k}{\left (k +\frac {3}{2}\right ) \left (2 k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}}, a_{k +1}=\frac {2 \left (k -\frac {1}{2}\right ) a_{k} k}{\left (k +\frac {3}{2}\right ) \left (2 k +2\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=a_{0}\cdot \left (1+x \right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {1}{2}}\right ), b_{k +1}=\frac {2 \left (k -\frac {1}{2}\right ) b_{k} k}{\left (k +\frac {3}{2}\right ) \left (2 k +2\right )}\right ] \end {array} \]
Maple trace Kovacic algorithm successful
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm A Liouvillian solution exists Reducible group (found an exponential solution) Reducible group (found another exponential solution) <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 26
Order:=6; dsolve(2*x*(x-1)*diff(y(x),x$2)-(x+1)*diff(y(x),x)+y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} \sqrt {x}\, \left (1+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (1+x +\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 18
AsymptoticDSolveValue[2*x*(x-1)*y''[x]-(x+1)*y'[x]+y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \sqrt {x}+c_2 (x+1) \]