Internal problem ID [2392]
Internal file name [OUTPUT/2392_Tuesday_February_27_2024_08_36_47_AM_67396664/index.tex]
Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath.
Boston. 1964
Section: Exercise 41, page 195
Problem number: 17.
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
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {4 x^{2} \left (1-x \right ) y^{\prime \prime }+3 x \left (1+2 x \right ) y^{\prime }-3 y=0} \] 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. \[ \left (-4 x^{3}+4 x^{2}\right ) y^{\prime \prime }+\left (6 x^{2}+3 x \right ) y^{\prime }-3 y = 0 \] 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 {3 \left (1+2 x \right )}{4 x \left (x -1\right )}\\ q(x) &= \frac {3}{4 x^{2} \left (x -1\right )}\\ \end {align*}
| \(p(x)=-\frac {3 \left (1+2 x \right )}{4 x \left (x -1\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = 1\) | \(\text {``regular''}\) |
| \(q(x)=\frac {3}{4 x^{2} \left (x -1\right )}\) | |
| 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 \[ -4 y^{\prime \prime } x^{2} \left (x -1\right )+\left (6 x^{2}+3 x \right ) y^{\prime }-3 y = 0 \] 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} -4 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right ) x^{2} \left (x -1\right )+\left (6 x^{2}+3 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )-3 \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 (-4 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}6 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 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 (-4 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}6 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}6 a_{n -1} \left (n +r -1\right ) 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} \moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}6 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+3 x^{n +r} a_{n} \left (n +r \right )-3 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 4 x^{r} a_{0} r \left (-1+r \right )+3 x^{r} a_{0} r -3 a_{0} x^{r} = 0 \] Or \[ \left (4 x^{r} r \left (-1+r \right )+3 x^{r} r -3 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (4 r^{2}-r -3\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 4 r^{2}-r -3 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 1\\ r_2 &= -{\frac {3}{4}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (4 r^{2}-r -3\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [1, -{\frac {3}{4}}\right ]\).
Since \(r_1 - r_2 = {\frac {7}{4}}\) 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^{1+n}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -\frac {3}{4}} \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} -4 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+4 a_{n} \left (n +r \right ) \left (n +r -1\right )+6 a_{n -1} \left (n +r -1\right )+3 a_{n} \left (n +r \right )-3 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {2 \left (2 n +2 r -7\right ) a_{n -1}}{4 n +4 r +3}\tag {4} \] Which for the root \(r = 1\) becomes \[ a_{n} = \frac {\left (4 n -10\right ) a_{n -1}}{4 n +7}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 1\) 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 \[ a_{1}=\frac {4 r -10}{7+4 r} \] Which for the root \(r = 1\) becomes \[ a_{1}=-{\frac {6}{11}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {4 r -10}{7+4 r}\) | \(-{\frac {6}{11}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {16 r^{2}-64 r +60}{16 r^{2}+72 r +77} \] Which for the root \(r = 1\) becomes \[ a_{2}={\frac {4}{55}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {4 r -10}{7+4 r}\) | \(-{\frac {6}{11}}\) |
| \(a_{2}\) | \(\frac {16 r^{2}-64 r +60}{16 r^{2}+72 r +77}\) | \(\frac {4}{55}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {64 r^{3}-288 r^{2}+368 r -120}{64 r^{3}+528 r^{2}+1388 r +1155} \] Which for the root \(r = 1\) becomes \[ a_{3}={\frac {8}{1045}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {4 r -10}{7+4 r}\) | \(-{\frac {6}{11}}\) |
| \(a_{2}\) | \(\frac {16 r^{2}-64 r +60}{16 r^{2}+72 r +77}\) | \(\frac {4}{55}\) |
| \(a_{3}\) | \(\frac {64 r^{3}-288 r^{2}+368 r -120}{64 r^{3}+528 r^{2}+1388 r +1155}\) | \(\frac {8}{1045}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {256 r^{4}-1024 r^{3}+896 r^{2}+256 r -240}{256 r^{4}+3328 r^{3}+15584 r^{2}+30992 r +21945} \] Which for the root \(r = 1\) becomes \[ a_{4}={\frac {48}{24035}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {4 r -10}{7+4 r}\) | \(-{\frac {6}{11}}\) |
| \(a_{2}\) | \(\frac {16 r^{2}-64 r +60}{16 r^{2}+72 r +77}\) | \(\frac {4}{55}\) |
| \(a_{3}\) | \(\frac {64 r^{3}-288 r^{2}+368 r -120}{64 r^{3}+528 r^{2}+1388 r +1155}\) | \(\frac {8}{1045}\) |
| \(a_{4}\) | \(\frac {256 r^{4}-1024 r^{3}+896 r^{2}+256 r -240}{256 r^{4}+3328 r^{3}+15584 r^{2}+30992 r +21945}\) | \(\frac {48}{24035}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {1024 r^{5}-2560 r^{4}-2560 r^{3}+6400 r^{2}+576 r -1440}{1024 r^{5}+19200 r^{4}+138880 r^{3}+482400 r^{2}+800596 r +504735} \] Which for the root \(r = 1\) becomes \[ a_{5}={\frac {32}{43263}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {4 r -10}{7+4 r}\) | \(-{\frac {6}{11}}\) |
| \(a_{2}\) | \(\frac {16 r^{2}-64 r +60}{16 r^{2}+72 r +77}\) | \(\frac {4}{55}\) |
| \(a_{3}\) | \(\frac {64 r^{3}-288 r^{2}+368 r -120}{64 r^{3}+528 r^{2}+1388 r +1155}\) | \(\frac {8}{1045}\) |
| \(a_{4}\) | \(\frac {256 r^{4}-1024 r^{3}+896 r^{2}+256 r -240}{256 r^{4}+3328 r^{3}+15584 r^{2}+30992 r +21945}\) | \(\frac {48}{24035}\) |
| \(a_{5}\) | \(\frac {1024 r^{5}-2560 r^{4}-2560 r^{3}+6400 r^{2}+576 r -1440}{1024 r^{5}+19200 r^{4}+138880 r^{3}+482400 r^{2}+800596 r +504735}\) | \(\frac {32}{43263}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= 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 ) \\ &= x \left (1-\frac {6 x}{11}+\frac {4 x^{2}}{55}+\frac {8 x^{3}}{1045}+\frac {48 x^{4}}{24035}+\frac {32 x^{5}}{43263}+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} -4 b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+4 b_{n} \left (n +r \right ) \left (n +r -1\right )+6 b_{n -1} \left (n +r -1\right )+3 b_{n} \left (n +r \right )-3 b_{n} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = \frac {2 \left (2 n +2 r -7\right ) b_{n -1}}{4 n +4 r +3}\tag {4} \] Which for the root \(r = -{\frac {3}{4}}\) becomes \[ b_{n} = \frac {\left (4 n -17\right ) b_{n -1}}{4 n}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -{\frac {3}{4}}\) 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 \[ b_{1}=\frac {4 r -10}{7+4 r} \] Which for the root \(r = -{\frac {3}{4}}\) becomes \[ b_{1}=-{\frac {13}{4}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {4 r -10}{7+4 r}\) | \(-{\frac {13}{4}}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {16 r^{2}-64 r +60}{16 r^{2}+72 r +77} \] Which for the root \(r = -{\frac {3}{4}}\) becomes \[ b_{2}={\frac {117}{32}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {4 r -10}{7+4 r}\) | \(-{\frac {13}{4}}\) |
| \(b_{2}\) | \(\frac {16 r^{2}-64 r +60}{16 r^{2}+72 r +77}\) | \(\frac {117}{32}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {64 r^{3}-288 r^{2}+368 r -120}{64 r^{3}+528 r^{2}+1388 r +1155} \] Which for the root \(r = -{\frac {3}{4}}\) becomes \[ b_{3}=-{\frac {195}{128}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {4 r -10}{7+4 r}\) | \(-{\frac {13}{4}}\) |
| \(b_{2}\) | \(\frac {16 r^{2}-64 r +60}{16 r^{2}+72 r +77}\) | \(\frac {117}{32}\) |
| \(b_{3}\) | \(\frac {64 r^{3}-288 r^{2}+368 r -120}{64 r^{3}+528 r^{2}+1388 r +1155}\) | \(-{\frac {195}{128}}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {256 r^{4}-1024 r^{3}+896 r^{2}+256 r -240}{256 r^{4}+3328 r^{3}+15584 r^{2}+30992 r +21945} \] Which for the root \(r = -{\frac {3}{4}}\) becomes \[ b_{4}={\frac {195}{2048}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {4 r -10}{7+4 r}\) | \(-{\frac {13}{4}}\) |
| \(b_{2}\) | \(\frac {16 r^{2}-64 r +60}{16 r^{2}+72 r +77}\) | \(\frac {117}{32}\) |
| \(b_{3}\) | \(\frac {64 r^{3}-288 r^{2}+368 r -120}{64 r^{3}+528 r^{2}+1388 r +1155}\) | \(-{\frac {195}{128}}\) |
| \(b_{4}\) | \(\frac {256 r^{4}-1024 r^{3}+896 r^{2}+256 r -240}{256 r^{4}+3328 r^{3}+15584 r^{2}+30992 r +21945}\) | \(\frac {195}{2048}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {1024 r^{5}-2560 r^{4}-2560 r^{3}+6400 r^{2}+576 r -1440}{1024 r^{5}+19200 r^{4}+138880 r^{3}+482400 r^{2}+800596 r +504735} \] Which for the root \(r = -{\frac {3}{4}}\) becomes \[ b_{5}={\frac {117}{8192}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {4 r -10}{7+4 r}\) | \(-{\frac {13}{4}}\) |
| \(b_{2}\) | \(\frac {16 r^{2}-64 r +60}{16 r^{2}+72 r +77}\) | \(\frac {117}{32}\) |
| \(b_{3}\) | \(\frac {64 r^{3}-288 r^{2}+368 r -120}{64 r^{3}+528 r^{2}+1388 r +1155}\) | \(-{\frac {195}{128}}\) |
| \(b_{4}\) | \(\frac {256 r^{4}-1024 r^{3}+896 r^{2}+256 r -240}{256 r^{4}+3328 r^{3}+15584 r^{2}+30992 r +21945}\) | \(\frac {195}{2048}\) |
| \(b_{5}\) | \(\frac {1024 r^{5}-2560 r^{4}-2560 r^{3}+6400 r^{2}+576 r -1440}{1024 r^{5}+19200 r^{4}+138880 r^{3}+482400 r^{2}+800596 r +504735}\) | \(\frac {117}{8192}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x \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-\frac {13 x}{4}+\frac {117 x^{2}}{32}-\frac {195 x^{3}}{128}+\frac {195 x^{4}}{2048}+\frac {117 x^{5}}{8192}+O\left (x^{6}\right )}{x^{\frac {3}{4}}} \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} x \left (1-\frac {6 x}{11}+\frac {4 x^{2}}{55}+\frac {8 x^{3}}{1045}+\frac {48 x^{4}}{24035}+\frac {32 x^{5}}{43263}+O\left (x^{6}\right )\right ) + \frac {c_{2} \left (1-\frac {13 x}{4}+\frac {117 x^{2}}{32}-\frac {195 x^{3}}{128}+\frac {195 x^{4}}{2048}+\frac {117 x^{5}}{8192}+O\left (x^{6}\right )\right )}{x^{\frac {3}{4}}} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x \left (1-\frac {6 x}{11}+\frac {4 x^{2}}{55}+\frac {8 x^{3}}{1045}+\frac {48 x^{4}}{24035}+\frac {32 x^{5}}{43263}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-\frac {13 x}{4}+\frac {117 x^{2}}{32}-\frac {195 x^{3}}{128}+\frac {195 x^{4}}{2048}+\frac {117 x^{5}}{8192}+O\left (x^{6}\right )\right )}{x^{\frac {3}{4}}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x \left (1-\frac {6 x}{11}+\frac {4 x^{2}}{55}+\frac {8 x^{3}}{1045}+\frac {48 x^{4}}{24035}+\frac {32 x^{5}}{43263}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-\frac {13 x}{4}+\frac {117 x^{2}}{32}-\frac {195 x^{3}}{128}+\frac {195 x^{4}}{2048}+\frac {117 x^{5}}{8192}+O\left (x^{6}\right )\right )}{x^{\frac {3}{4}}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} x \left (1-\frac {6 x}{11}+\frac {4 x^{2}}{55}+\frac {8 x^{3}}{1045}+\frac {48 x^{4}}{24035}+\frac {32 x^{5}}{43263}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-\frac {13 x}{4}+\frac {117 x^{2}}{32}-\frac {195 x^{3}}{128}+\frac {195 x^{4}}{2048}+\frac {117 x^{5}}{8192}+O\left (x^{6}\right )\right )}{x^{\frac {3}{4}}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -4 y^{\prime \prime } x^{2} \left (x -1\right )+\left (6 x^{2}+3 x \right ) y^{\prime }-3 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 {3 y}{4 x^{2} \left (x -1\right )}+\frac {3 \left (1+2 x \right ) y^{\prime }}{4 x \left (x -1\right )} \\ \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 {3 \left (1+2 x \right ) y^{\prime }}{4 x \left (x -1\right )}+\frac {3 y}{4 x^{2} \left (x -1\right )}=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 {3 \left (1+2 x \right )}{4 x \left (x -1\right )}, P_{3}\left (x \right )=\frac {3}{4 x^{2} \left (x -1\right )}\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}{4} \\ {} & \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}}}=-\frac {3}{4} \\ {} & \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} \\ {} & {} & 4 y^{\prime \prime } x^{2} \left (x -1\right )-3 x \left (1+2 x \right ) y^{\prime }+3 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 =1..2 \\ {} & {} & 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 =2..3 \\ {} & {} & 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} \left (3+4 r \right ) \left (-1+r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (-a_{k} \left (4 k +4 r +3\right ) \left (k +r -1\right )+2 a_{k -1} \left (k +r -1\right ) \left (2 k -7+2 r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -\left (3+4 r \right ) \left (-1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{1, -\frac {3}{4}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -4 \left (\left (-k -r +\frac {7}{2}\right ) a_{k -1}+a_{k} \left (k +r +\frac {3}{4}\right )\right ) \left (k +r -1\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & -4 \left (\left (-k +\frac {5}{2}-r \right ) a_{k}+a_{k +1} \left (k +\frac {7}{4}+r \right )\right ) \left (k +r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {2 \left (2 k +2 r -5\right ) a_{k}}{4 k +7+4 r} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +1}=\frac {2 \left (2 k -3\right ) a_{k}}{4 k +11} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1}, a_{k +1}=\frac {2 \left (2 k -3\right ) a_{k}}{4 k +11}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {3}{4} \\ {} & {} & a_{k +1}=\frac {2 \left (2 k -\frac {13}{2}\right ) a_{k}}{4 k +4} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {3}{4} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {3}{4}}, a_{k +1}=\frac {2 \left (2 k -\frac {13}{2}\right ) a_{k}}{4 k +4}\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 -\frac {3}{4}}\right ), a_{k +1}=\frac {2 \left (2 k -3\right ) a_{k}}{4 k +11}, b_{k +1}=\frac {2 \left (2 k -\frac {13}{2}\right ) b_{k}}{4 k +4}\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) Group is reducible, not completely reducible Solution has integrals. Trying a special function solution free of integrals... -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach <- heuristic approach successful <- hypergeometric successful <- special function solution successful -> Trying to convert hypergeometric functions to elementary form... <- elementary form is not straightforward to achieve - returning special function solution free of uncomputed integrals <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 45
Order:=6; dsolve(4*x^2*(1-x)*diff(y(x),x$2)+3*x*(1+2*x)*diff(y(x),x)-3*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{1} \left (1-\frac {13}{4} x +\frac {117}{32} x^{2}-\frac {195}{128} x^{3}+\frac {195}{2048} x^{4}+\frac {117}{8192} x^{5}+\operatorname {O}\left (x^{6}\right )\right )}{x^{\frac {3}{4}}}+c_{2} x \left (1-\frac {6}{11} x +\frac {4}{55} x^{2}+\frac {8}{1045} x^{3}+\frac {48}{24035} x^{4}+\frac {32}{43263} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.006 (sec). Leaf size: 86
AsymptoticDSolveValue[4*x^2*(1-x)*y''[x]+3*x*(1+2*x)*y'[x]-3*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 x \left (\frac {32 x^5}{43263}+\frac {48 x^4}{24035}+\frac {8 x^3}{1045}+\frac {4 x^2}{55}-\frac {6 x}{11}+1\right )+\frac {c_2 \left (\frac {117 x^5}{8192}+\frac {195 x^4}{2048}-\frac {195 x^3}{128}+\frac {117 x^2}{32}-\frac {13 x}{4}+1\right )}{x^{3/4}} \]