Internal problem ID [6470]
Internal file name [OUTPUT/5718_Sunday_June_05_2022_03_48_39_PM_66849501/index.tex]
Book: Differential Equations: Theory, Technique, and Practice by George Simmons, Steven
Krantz. McGraw-Hill NY. 2007. 1st Edition.
Section: Chapter 4. Power Series Solutions and Special Functions. Section 4.6. Gauss’s
Hypergeometric Equation. Page 187
Problem number: 2(a).
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]
\[ \boxed {x \left (1-x \right ) y^{\prime \prime }+\left (\frac {3}{2}-2 x \right ) y^{\prime }+2 y=0} \] With the expansion point for the power series method at \(x = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ \left (-x^{2}+x \right ) y^{\prime \prime }+\left (\frac {3}{2}-2 x \right ) y^{\prime }+2 y = 0 \] 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 {4 x -3}{2 \left (x -1\right ) x}\\ q(x) &= -\frac {2}{\left (x -1\right ) x}\\ \end {align*}
| \(p(x)=\frac {4 x -3}{2 \left (x -1\right ) x}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = 1\) | \(\text {``regular''}\) |
| \(q(x)=-\frac {2}{\left (x -1\right ) x}\) | |
| 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 \[ -\left (x -1\right ) x y^{\prime \prime }+\left (\frac {3}{2}-2 x \right ) y^{\prime }+2 y = 0 \] 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} -\left (x -1\right ) 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 (\frac {3}{2}-2 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\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 }}\frac {3 \left (n +r \right ) a_{n} x^{n +r -1}}{2}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}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 (-x^{n +r} 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 ) \left (n +r -2\right ) x^{n +r -1}\right ) \\ \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 }}2 a_{n} x^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}2 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} \moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r -1}\right )+\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 }}\frac {3 \left (n +r \right ) a_{n} x^{n +r -1}}{2}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}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 \[ x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+\frac {3 \left (n +r \right ) a_{n} x^{n +r -1}}{2} = 0 \] When \(n = 0\) the above becomes \[ x^{-1+r} a_{0} r \left (-1+r \right )+\frac {3 r a_{0} x^{-1+r}}{2} = 0 \] Or \[ \left (x^{-1+r} r \left (-1+r \right )+\frac {3 r \,x^{-1+r}}{2}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ r \,x^{-1+r} \left (\frac {1}{2}+r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r^{2}+\frac {1}{2} r = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 0\\ r_2 &= -{\frac {1}{2}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ r \,x^{-1+r} \left (\frac {1}{2}+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [0, -{\frac {1}{2}}\right ]\).
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}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -\frac {1}{2}} \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} -a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+a_{n} \left (n +r \right ) \left (n +r -1\right )-2 a_{n -1} \left (n +r -1\right )+\frac {3 a_{n} \left (n +r \right )}{2}+2 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {2 a_{n -1} \left (n^{2}+2 n r +r^{2}-n -r -2\right )}{2 n^{2}+4 n r +2 r^{2}+n +r}\tag {4} \] Which for the root \(r = 0\) becomes \[ a_{n} = \frac {2 a_{n -1} \left (n^{2}-n -2\right )}{2 n^{2}+n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 0\) 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 \[ a_{1}=\frac {2 r^{2}+2 r -4}{2 r^{2}+5 r +3} \] Which for the root \(r = 0\) becomes \[ a_{1}=-{\frac {4}{3}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2 r^{2}+2 r -4}{2 r^{2}+5 r +3}\) | \(-{\frac {4}{3}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {4 \left (-1+r \right ) r \left (r +3\right )}{4 r^{3}+20 r^{2}+31 r +15} \] Which for the root \(r = 0\) becomes \[ a_{2}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2 r^{2}+2 r -4}{2 r^{2}+5 r +3}\) | \(-{\frac {4}{3}}\) |
| \(a_{2}\) | \(\frac {4 \left (-1+r \right ) r \left (r +3\right )}{4 r^{3}+20 r^{2}+31 r +15}\) | \(0\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {8 \left (-1+r \right ) r \left (r +4\right )}{8 r^{3}+60 r^{2}+142 r +105} \] Which for the root \(r = 0\) becomes \[ a_{3}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2 r^{2}+2 r -4}{2 r^{2}+5 r +3}\) | \(-{\frac {4}{3}}\) |
| \(a_{2}\) | \(\frac {4 \left (-1+r \right ) r \left (r +3\right )}{4 r^{3}+20 r^{2}+31 r +15}\) | \(0\) |
| \(a_{3}\) | \(\frac {8 \left (-1+r \right ) r \left (r +4\right )}{8 r^{3}+60 r^{2}+142 r +105}\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {16 \left (r +5\right ) \left (r +2\right ) r \left (-1+r \right )}{16 r^{4}+192 r^{3}+824 r^{2}+1488 r +945} \] Which for the root \(r = 0\) becomes \[ a_{4}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2 r^{2}+2 r -4}{2 r^{2}+5 r +3}\) | \(-{\frac {4}{3}}\) |
| \(a_{2}\) | \(\frac {4 \left (-1+r \right ) r \left (r +3\right )}{4 r^{3}+20 r^{2}+31 r +15}\) | \(0\) |
| \(a_{3}\) | \(\frac {8 \left (-1+r \right ) r \left (r +4\right )}{8 r^{3}+60 r^{2}+142 r +105}\) | \(0\) |
| \(a_{4}\) | \(\frac {16 \left (r +5\right ) \left (r +2\right ) r \left (-1+r \right )}{16 r^{4}+192 r^{3}+824 r^{2}+1488 r +945}\) | \(0\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {32 \left (r +6\right ) \left (r +3\right ) \left (-1+r \right ) r \left (r +2\right )}{32 r^{5}+560 r^{4}+3760 r^{3}+12040 r^{2}+18258 r +10395} \] Which for the root \(r = 0\) becomes \[ a_{5}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2 r^{2}+2 r -4}{2 r^{2}+5 r +3}\) | \(-{\frac {4}{3}}\) |
| \(a_{2}\) | \(\frac {4 \left (-1+r \right ) r \left (r +3\right )}{4 r^{3}+20 r^{2}+31 r +15}\) | \(0\) |
| \(a_{3}\) | \(\frac {8 \left (-1+r \right ) r \left (r +4\right )}{8 r^{3}+60 r^{2}+142 r +105}\) | \(0\) |
| \(a_{4}\) | \(\frac {16 \left (r +5\right ) \left (r +2\right ) r \left (-1+r \right )}{16 r^{4}+192 r^{3}+824 r^{2}+1488 r +945}\) | \(0\) |
| \(a_{5}\) | \(\frac {32 \left (r +6\right ) \left (r +3\right ) \left (-1+r \right ) r \left (r +2\right )}{32 r^{5}+560 r^{4}+3760 r^{3}+12040 r^{2}+18258 r +10395}\) | \(0\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {64 \left (7+r \right ) \left (r +4\right ) \left (r +2\right ) \left (-1+r \right ) r \left (r +3\right )}{64 r^{6}+1536 r^{5}+14800 r^{4}+72960 r^{3}+193036 r^{2}+258144 r +135135} \] Which for the root \(r = 0\) becomes \[ a_{6}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2 r^{2}+2 r -4}{2 r^{2}+5 r +3}\) | \(-{\frac {4}{3}}\) |
| \(a_{2}\) | \(\frac {4 \left (-1+r \right ) r \left (r +3\right )}{4 r^{3}+20 r^{2}+31 r +15}\) | \(0\) |
| \(a_{3}\) | \(\frac {8 \left (-1+r \right ) r \left (r +4\right )}{8 r^{3}+60 r^{2}+142 r +105}\) | \(0\) |
| \(a_{4}\) | \(\frac {16 \left (r +5\right ) \left (r +2\right ) r \left (-1+r \right )}{16 r^{4}+192 r^{3}+824 r^{2}+1488 r +945}\) | \(0\) |
| \(a_{5}\) | \(\frac {32 \left (r +6\right ) \left (r +3\right ) \left (-1+r \right ) r \left (r +2\right )}{32 r^{5}+560 r^{4}+3760 r^{3}+12040 r^{2}+18258 r +10395}\) | \(0\) |
| \(a_{6}\) | \(\frac {64 \left (7+r \right ) \left (r +4\right ) \left (r +2\right ) \left (-1+r \right ) r \left (r +3\right )}{64 r^{6}+1536 r^{5}+14800 r^{4}+72960 r^{3}+193036 r^{2}+258144 r +135135}\) | \(0\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=\frac {128 \left (r +8\right ) \left (r +5\right ) \left (-1+r \right ) r \left (r +3\right ) \left (r +2\right ) \left (r +4\right )}{128 r^{7}+4032 r^{6}+52640 r^{5}+367920 r^{4}+1480472 r^{3}+3411828 r^{2}+4142430 r +2027025} \] Which for the root \(r = 0\) becomes \[ a_{7}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2 r^{2}+2 r -4}{2 r^{2}+5 r +3}\) | \(-{\frac {4}{3}}\) |
| \(a_{2}\) | \(\frac {4 \left (-1+r \right ) r \left (r +3\right )}{4 r^{3}+20 r^{2}+31 r +15}\) | \(0\) |
| \(a_{3}\) | \(\frac {8 \left (-1+r \right ) r \left (r +4\right )}{8 r^{3}+60 r^{2}+142 r +105}\) | \(0\) |
| \(a_{4}\) | \(\frac {16 \left (r +5\right ) \left (r +2\right ) r \left (-1+r \right )}{16 r^{4}+192 r^{3}+824 r^{2}+1488 r +945}\) | \(0\) |
| \(a_{5}\) | \(\frac {32 \left (r +6\right ) \left (r +3\right ) \left (-1+r \right ) r \left (r +2\right )}{32 r^{5}+560 r^{4}+3760 r^{3}+12040 r^{2}+18258 r +10395}\) | \(0\) |
| \(a_{6}\) | \(\frac {64 \left (7+r \right ) \left (r +4\right ) \left (r +2\right ) \left (-1+r \right ) r \left (r +3\right )}{64 r^{6}+1536 r^{5}+14800 r^{4}+72960 r^{3}+193036 r^{2}+258144 r +135135}\) | \(0\) |
| \(a_{7}\) | \(\frac {128 \left (r +8\right ) \left (r +5\right ) \left (-1+r \right ) r \left (r +3\right ) \left (r +2\right ) \left (r +4\right )}{128 r^{7}+4032 r^{6}+52640 r^{5}+367920 r^{4}+1480472 r^{3}+3411828 r^{2}+4142430 r +2027025}\) | \(0\) |
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}+a_{7} x^{7}+a_{8} x^{8}\dots \\ &= 1-\frac {4 x}{3}+O\left (x^{8}\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} -b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+b_{n} \left (n +r \right ) \left (n +r -1\right )-2 b_{n -1} \left (n +r -1\right )+\frac {3 \left (n +r \right ) b_{n}}{2}+2 b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = \frac {2 b_{n -1} \left (n^{2}+2 n r +r^{2}-n -r -2\right )}{2 n^{2}+4 n r +2 r^{2}+n +r}\tag {4} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ b_{n} = \frac {b_{n -1} \left (4 n^{2}-8 n -5\right )}{4 n^{2}-2 n}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -{\frac {1}{2}}\) 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 \[ b_{1}=\frac {2 r^{2}+2 r -4}{2 r^{2}+5 r +3} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ b_{1}=-{\frac {9}{2}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2 r^{2}+2 r -4}{2 r^{2}+5 r +3}\) | \(-{\frac {9}{2}}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {4 \left (-1+r \right ) r \left (r +3\right )}{4 r^{3}+20 r^{2}+31 r +15} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ b_{2}={\frac {15}{8}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2 r^{2}+2 r -4}{2 r^{2}+5 r +3}\) | \(-{\frac {9}{2}}\) |
| \(b_{2}\) | \(\frac {4 \left (-1+r \right ) r \left (r +3\right )}{4 r^{3}+20 r^{2}+31 r +15}\) | \(\frac {15}{8}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {8 \left (-1+r \right ) r \left (r +4\right )}{8 r^{3}+60 r^{2}+142 r +105} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ b_{3}={\frac {7}{16}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2 r^{2}+2 r -4}{2 r^{2}+5 r +3}\) | \(-{\frac {9}{2}}\) |
| \(b_{2}\) | \(\frac {4 \left (-1+r \right ) r \left (r +3\right )}{4 r^{3}+20 r^{2}+31 r +15}\) | \(\frac {15}{8}\) |
| \(b_{3}\) | \(\frac {8 \left (-1+r \right ) r \left (r +4\right )}{8 r^{3}+60 r^{2}+142 r +105}\) | \(\frac {7}{16}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {16 \left (r +5\right ) \left (r +2\right ) r \left (-1+r \right )}{16 r^{4}+192 r^{3}+824 r^{2}+1488 r +945} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ b_{4}={\frac {27}{128}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2 r^{2}+2 r -4}{2 r^{2}+5 r +3}\) | \(-{\frac {9}{2}}\) |
| \(b_{2}\) | \(\frac {4 \left (-1+r \right ) r \left (r +3\right )}{4 r^{3}+20 r^{2}+31 r +15}\) | \(\frac {15}{8}\) |
| \(b_{3}\) | \(\frac {8 \left (-1+r \right ) r \left (r +4\right )}{8 r^{3}+60 r^{2}+142 r +105}\) | \(\frac {7}{16}\) |
| \(b_{4}\) | \(\frac {16 \left (r +5\right ) \left (r +2\right ) r \left (-1+r \right )}{16 r^{4}+192 r^{3}+824 r^{2}+1488 r +945}\) | \(\frac {27}{128}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {32 \left (r +6\right ) \left (r +3\right ) \left (-1+r \right ) r \left (r +2\right )}{32 r^{5}+560 r^{4}+3760 r^{3}+12040 r^{2}+18258 r +10395} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ b_{5}={\frac {33}{256}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2 r^{2}+2 r -4}{2 r^{2}+5 r +3}\) | \(-{\frac {9}{2}}\) |
| \(b_{2}\) | \(\frac {4 \left (-1+r \right ) r \left (r +3\right )}{4 r^{3}+20 r^{2}+31 r +15}\) | \(\frac {15}{8}\) |
| \(b_{3}\) | \(\frac {8 \left (-1+r \right ) r \left (r +4\right )}{8 r^{3}+60 r^{2}+142 r +105}\) | \(\frac {7}{16}\) |
| \(b_{4}\) | \(\frac {16 \left (r +5\right ) \left (r +2\right ) r \left (-1+r \right )}{16 r^{4}+192 r^{3}+824 r^{2}+1488 r +945}\) | \(\frac {27}{128}\) |
| \(b_{5}\) | \(\frac {32 \left (r +6\right ) \left (r +3\right ) \left (-1+r \right ) r \left (r +2\right )}{32 r^{5}+560 r^{4}+3760 r^{3}+12040 r^{2}+18258 r +10395}\) | \(\frac {33}{256}\) |
For \(n = 6\), using the above recursive equation gives \[ b_{6}=\frac {64 \left (7+r \right ) \left (r +4\right ) \left (r +2\right ) \left (-1+r \right ) r \left (r +3\right )}{64 r^{6}+1536 r^{5}+14800 r^{4}+72960 r^{3}+193036 r^{2}+258144 r +135135} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ b_{6}={\frac {91}{1024}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2 r^{2}+2 r -4}{2 r^{2}+5 r +3}\) | \(-{\frac {9}{2}}\) |
| \(b_{2}\) | \(\frac {4 \left (-1+r \right ) r \left (r +3\right )}{4 r^{3}+20 r^{2}+31 r +15}\) | \(\frac {15}{8}\) |
| \(b_{3}\) | \(\frac {8 \left (-1+r \right ) r \left (r +4\right )}{8 r^{3}+60 r^{2}+142 r +105}\) | \(\frac {7}{16}\) |
| \(b_{4}\) | \(\frac {16 \left (r +5\right ) \left (r +2\right ) r \left (-1+r \right )}{16 r^{4}+192 r^{3}+824 r^{2}+1488 r +945}\) | \(\frac {27}{128}\) |
| \(b_{5}\) | \(\frac {32 \left (r +6\right ) \left (r +3\right ) \left (-1+r \right ) r \left (r +2\right )}{32 r^{5}+560 r^{4}+3760 r^{3}+12040 r^{2}+18258 r +10395}\) | \(\frac {33}{256}\) |
| \(b_{6}\) | \(\frac {64 \left (7+r \right ) \left (r +4\right ) \left (r +2\right ) \left (-1+r \right ) r \left (r +3\right )}{64 r^{6}+1536 r^{5}+14800 r^{4}+72960 r^{3}+193036 r^{2}+258144 r +135135}\) | \(\frac {91}{1024}\) |
For \(n = 7\), using the above recursive equation gives \[ b_{7}=\frac {128 \left (r +8\right ) \left (r +5\right ) \left (-1+r \right ) r \left (r +3\right ) \left (r +2\right ) \left (r +4\right )}{128 r^{7}+4032 r^{6}+52640 r^{5}+367920 r^{4}+1480472 r^{3}+3411828 r^{2}+4142430 r +2027025} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ b_{7}={\frac {135}{2048}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2 r^{2}+2 r -4}{2 r^{2}+5 r +3}\) | \(-{\frac {9}{2}}\) |
| \(b_{2}\) | \(\frac {4 \left (-1+r \right ) r \left (r +3\right )}{4 r^{3}+20 r^{2}+31 r +15}\) | \(\frac {15}{8}\) |
| \(b_{3}\) | \(\frac {8 \left (-1+r \right ) r \left (r +4\right )}{8 r^{3}+60 r^{2}+142 r +105}\) | \(\frac {7}{16}\) |
| \(b_{4}\) | \(\frac {16 \left (r +5\right ) \left (r +2\right ) r \left (-1+r \right )}{16 r^{4}+192 r^{3}+824 r^{2}+1488 r +945}\) | \(\frac {27}{128}\) |
| \(b_{5}\) | \(\frac {32 \left (r +6\right ) \left (r +3\right ) \left (-1+r \right ) r \left (r +2\right )}{32 r^{5}+560 r^{4}+3760 r^{3}+12040 r^{2}+18258 r +10395}\) | \(\frac {33}{256}\) |
| \(b_{6}\) | \(\frac {64 \left (7+r \right ) \left (r +4\right ) \left (r +2\right ) \left (-1+r \right ) r \left (r +3\right )}{64 r^{6}+1536 r^{5}+14800 r^{4}+72960 r^{3}+193036 r^{2}+258144 r +135135}\) | \(\frac {91}{1024}\) |
| \(b_{7}\) | \(\frac {128 \left (r +8\right ) \left (r +5\right ) \left (-1+r \right ) r \left (r +3\right ) \left (r +2\right ) \left (r +4\right )}{128 r^{7}+4032 r^{6}+52640 r^{5}+367920 r^{4}+1480472 r^{3}+3411828 r^{2}+4142430 r +2027025}\) | \(\frac {135}{2048}\) |
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}+b_{7} x^{7}+b_{8} x^{8}\dots \right ) \\ &= \frac {1-\frac {9 x}{2}+\frac {15 x^{2}}{8}+\frac {7 x^{3}}{16}+\frac {27 x^{4}}{128}+\frac {33 x^{5}}{256}+\frac {91 x^{6}}{1024}+\frac {135 x^{7}}{2048}+O\left (x^{8}\right )}{\sqrt {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-\frac {4 x}{3}+O\left (x^{8}\right )\right ) + \frac {c_{2} \left (1-\frac {9 x}{2}+\frac {15 x^{2}}{8}+\frac {7 x^{3}}{16}+\frac {27 x^{4}}{128}+\frac {33 x^{5}}{256}+\frac {91 x^{6}}{1024}+\frac {135 x^{7}}{2048}+O\left (x^{8}\right )\right )}{\sqrt {x}} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \left (1-\frac {4 x}{3}+O\left (x^{8}\right )\right )+\frac {c_{2} \left (1-\frac {9 x}{2}+\frac {15 x^{2}}{8}+\frac {7 x^{3}}{16}+\frac {27 x^{4}}{128}+\frac {33 x^{5}}{256}+\frac {91 x^{6}}{1024}+\frac {135 x^{7}}{2048}+O\left (x^{8}\right )\right )}{\sqrt {x}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (1-\frac {4 x}{3}+O\left (x^{8}\right )\right )+\frac {c_{2} \left (1-\frac {9 x}{2}+\frac {15 x^{2}}{8}+\frac {7 x^{3}}{16}+\frac {27 x^{4}}{128}+\frac {33 x^{5}}{256}+\frac {91 x^{6}}{1024}+\frac {135 x^{7}}{2048}+O\left (x^{8}\right )\right )}{\sqrt {x}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (1-\frac {4 x}{3}+O\left (x^{8}\right )\right )+\frac {c_{2} \left (1-\frac {9 x}{2}+\frac {15 x^{2}}{8}+\frac {7 x^{3}}{16}+\frac {27 x^{4}}{128}+\frac {33 x^{5}}{256}+\frac {91 x^{6}}{1024}+\frac {135 x^{7}}{2048}+O\left (x^{8}\right )\right )}{\sqrt {x}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -\left (x -1\right ) x y^{\prime \prime }+\left (\frac {3}{2}-2 x \right ) y^{\prime }+2 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 (4 x -3\right ) y^{\prime }}{2 x \left (x -1\right )}+\frac {2 y}{\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 (4 x -3\right ) y^{\prime }}{2 x \left (x -1\right )}-\frac {2 y}{\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 {4 x -3}{2 \left (x -1\right ) x}, P_{3}\left (x \right )=-\frac {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 {3}{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 \left (x -1\right ) x y^{\prime \prime }+\left (4 x -3\right ) y^{\prime }-4 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 (2 r +1\right ) x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-a_{k +1} \left (k +1+r \right ) \left (2 k +3+2 r \right )+2 a_{k} \left (k +r +2\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 (2 r +1\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 +r +\frac {3}{2}\right ) \left (k +1+r \right ) a_{k +1}+2 a_{k} \left (k +r +2\right ) \left (k +r -1\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 +r -1\right )}{\left (2 k +3+2 r \right ) \left (k +1+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 {2 a_{k} \left (k +2\right ) \left (k -1\right )}{\left (2 k +3\right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =0 \\ {} & {} & a_{1}=-\frac {4 a_{0}}{3} \\ \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-\frac {4 x}{3}\right ) \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {1}{2} \\ {} & {} & a_{k +1}=\frac {2 a_{k} \left (k +\frac {3}{2}\right ) \left (k -\frac {3}{2}\right )}{\left (2 k +2\right ) \left (k +\frac {1}{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 a_{k} \left (k +\frac {3}{2}\right ) \left (k -\frac {3}{2}\right )}{\left (2 k +2\right ) \left (k +\frac {1}{2}\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=a_{0}\cdot \left (1-\frac {4 x}{3}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k -\frac {1}{2}}\right ), b_{k +1}=\frac {2 b_{k} \left (k +\frac {3}{2}\right ) \left (k -\frac {3}{2}\right )}{\left (2 k +2\right ) \left (k +\frac {1}{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: 0.047 (sec). Leaf size: 40
Order:=8; dsolve(x*(1-x)*diff(y(x),x$2)+(3/2-2*x)*diff(y(x),x)+2*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{1} \left (1-\frac {9}{2} x +\frac {15}{8} x^{2}+\frac {7}{16} x^{3}+\frac {27}{128} x^{4}+\frac {33}{256} x^{5}+\frac {91}{1024} x^{6}+\frac {135}{2048} x^{7}+\operatorname {O}\left (x^{8}\right )\right )}{\sqrt {x}}+c_{2} \left (1-\frac {4}{3} x +\operatorname {O}\left (x^{8}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.005 (sec). Leaf size: 71
AsymptoticDSolveValue[x*(1-x)*y''[x]+(3/2-2*x)*y'[x]+2*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to \frac {c_2 \left (\frac {135 x^7}{2048}+\frac {91 x^6}{1024}+\frac {33 x^5}{256}+\frac {27 x^4}{128}+\frac {7 x^3}{16}+\frac {15 x^2}{8}-\frac {9 x}{2}+1\right )}{\sqrt {x}}+c_1 \left (1-\frac {4 x}{3}\right ) \]