Internal problem ID [1296]
Internal file name [OUTPUT/1297_Sunday_June_05_2022_02_08_51_AM_48638098/index.tex]
Book: Elementary differential equations with boundary value problems. William F. Trench.
Brooks/Cole 2001
Section: Chapter 7 Series Solutions of Linear Second Equations. 7.5 THE METHOD OF
FROBENIUS I. Exercises 7.5. Page 358
Problem number: 2.
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], [_2nd_order, _linear, `_with_symmetry_[0,F(x)]`]]
\[ \boxed {3 x^{2} y^{\prime \prime }+2 x \left (-2 x^{2}+x +1\right ) y^{\prime }+\left (-8 x^{2}+2 x \right ) y=0} \] 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. \[ 3 x^{2} y^{\prime \prime }+\left (-4 x^{3}+2 x^{2}+2 x \right ) y^{\prime }+\left (-8 x^{2}+2 x \right ) y = 0 \] 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) &= -\frac {2 \left (2 x^{2}-x -1\right )}{3 x}\\ q(x) &= -\frac {2 \left (4 x -1\right )}{3 x}\\ \end {align*}
| \(p(x)=-\frac {2 \left (2 x^{2}-x -1\right )}{3 x}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = \infty \) | \(\text {``regular''}\) |
| \(x = -\infty \) | \(\text {``regular''}\) |
| \(q(x)=-\frac {2 \left (4 x -1\right )}{3 x}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, \infty , -\infty ]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ 3 x^{2} y^{\prime \prime }+\left (-4 x^{3}+2 x^{2}+2 x \right ) y^{\prime }+\left (-8 x^{2}+2 x \right ) y = 0 \] 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} 3 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 )+\left (-4 x^{3}+2 x^{2}+2 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (-8 x^{2}+2 x \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 }}3 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 x^{n +r +2} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-8 x^{n +r +2} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{1+n +r} a_{n}\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^{n +r +2} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-4 a_{n -2} \left (n +r -2\right ) x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-8 x^{n +r +2} a_{n}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-8 a_{n -2} x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}2 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 }}3 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-4 a_{n -2} \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-8 a_{n -2} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 3 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+2 x^{n +r} a_{n} \left (n +r \right ) = 0 \] When \(n = 0\) the above becomes \[ 3 x^{r} a_{0} r \left (-1+r \right )+2 x^{r} a_{0} r = 0 \] Or \[ \left (3 x^{r} r \left (-1+r \right )+2 x^{r} r \right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ x^{r} r \left (-1+3 r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 3 r^{2}-r = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {1}{3}}\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ x^{r} r \left (-1+3 r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {1}{3}}, 0\right ]\).
Since \(r_1 - r_2 = {\frac {1}{3}}\) 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}{3}}\\ 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 \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = -\frac {2}{3 r +2} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} 3 a_{n} \left (n +r \right ) \left (n +r -1\right )-4 a_{n -2} \left (n +r -2\right )+2 a_{n -1} \left (n +r -1\right )+2 a_{n} \left (n +r \right )-8 a_{n -2}+2 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {4 a_{n -2}-2 a_{n -1}}{3 n -1+3 r}\tag {4} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ a_{n} = \frac {4 a_{n -2}-2 a_{n -1}}{3 n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {1}{3}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2}{3 r +2}\) | \(-{\frac {2}{3}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {12+12 r}{9 r^{2}+21 r +10} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ a_{2}={\frac {8}{9}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2}{3 r +2}\) | \(-{\frac {2}{3}}\) |
| \(a_{2}\) | \(\frac {12+12 r}{9 r^{2}+21 r +10}\) | \(\frac {8}{9}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-64-48 r}{27 r^{3}+135 r^{2}+198 r +80} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ a_{3}=-{\frac {40}{81}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2}{3 r +2}\) | \(-{\frac {2}{3}}\) |
| \(a_{2}\) | \(\frac {12+12 r}{9 r^{2}+21 r +10}\) | \(\frac {8}{9}\) |
| \(a_{3}\) | \(\frac {-64-48 r}{27 r^{3}+135 r^{2}+198 r +80}\) | \(-{\frac {40}{81}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {144 r^{2}+624 r +512}{81 r^{4}+702 r^{3}+2079 r^{2}+2418 r +880} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ a_{4}={\frac {92}{243}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2}{3 r +2}\) | \(-{\frac {2}{3}}\) |
| \(a_{2}\) | \(\frac {12+12 r}{9 r^{2}+21 r +10}\) | \(\frac {8}{9}\) |
| \(a_{3}\) | \(\frac {-64-48 r}{27 r^{3}+135 r^{2}+198 r +80}\) | \(-{\frac {40}{81}}\) |
| \(a_{4}\) | \(\frac {144 r^{2}+624 r +512}{81 r^{4}+702 r^{3}+2079 r^{2}+2418 r +880}\) | \(\frac {92}{243}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-864 r^{2}-4128 r -3840}{243 r^{5}+3240 r^{4}+16065 r^{3}+36360 r^{2}+36492 r +12320} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ a_{5}=-{\frac {664}{3645}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2}{3 r +2}\) | \(-{\frac {2}{3}}\) |
| \(a_{2}\) | \(\frac {12+12 r}{9 r^{2}+21 r +10}\) | \(\frac {8}{9}\) |
| \(a_{3}\) | \(\frac {-64-48 r}{27 r^{3}+135 r^{2}+198 r +80}\) | \(-{\frac {40}{81}}\) |
| \(a_{4}\) | \(\frac {144 r^{2}+624 r +512}{81 r^{4}+702 r^{3}+2079 r^{2}+2418 r +880}\) | \(\frac {92}{243}\) |
| \(a_{5}\) | \(\frac {-864 r^{2}-4128 r -3840}{243 r^{5}+3240 r^{4}+16065 r^{3}+36360 r^{2}+36492 r +12320}\) | \(-{\frac {664}{3645}}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{\frac {1}{3}} \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^{\frac {1}{3}} \left (1-\frac {2 x}{3}+\frac {8 x^{2}}{9}-\frac {40 x^{3}}{81}+\frac {92 x^{4}}{243}-\frac {664 x^{5}}{3645}+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 \(b_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ b_{1} = -\frac {2}{3 r +2} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} 3 b_{n} \left (n +r \right ) \left (n +r -1\right )-4 b_{n -2} \left (n +r -2\right )+2 b_{n -1} \left (n +r -1\right )+2 b_{n} \left (n +r \right )-8 b_{n -2}+2 b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = \frac {4 b_{n -2}-2 b_{n -1}}{3 n -1+3 r}\tag {4} \] Which for the root \(r = 0\) becomes \[ b_{n} = \frac {4 b_{n -2}-2 b_{n -1}}{3 n -1}\tag {5} \] 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.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {2}{3 r +2}\) | \(-1\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {12+12 r}{9 r^{2}+21 r +10} \] Which for the root \(r = 0\) becomes \[ b_{2}={\frac {6}{5}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {2}{3 r +2}\) | \(-1\) |
| \(b_{2}\) | \(\frac {12+12 r}{9 r^{2}+21 r +10}\) | \(\frac {6}{5}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {-64-48 r}{27 r^{3}+135 r^{2}+198 r +80} \] Which for the root \(r = 0\) becomes \[ b_{3}=-{\frac {4}{5}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {2}{3 r +2}\) | \(-1\) |
| \(b_{2}\) | \(\frac {12+12 r}{9 r^{2}+21 r +10}\) | \(\frac {6}{5}\) |
| \(b_{3}\) | \(\frac {-64-48 r}{27 r^{3}+135 r^{2}+198 r +80}\) | \(-{\frac {4}{5}}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {144 r^{2}+624 r +512}{81 r^{4}+702 r^{3}+2079 r^{2}+2418 r +880} \] Which for the root \(r = 0\) becomes \[ b_{4}={\frac {32}{55}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {2}{3 r +2}\) | \(-1\) |
| \(b_{2}\) | \(\frac {12+12 r}{9 r^{2}+21 r +10}\) | \(\frac {6}{5}\) |
| \(b_{3}\) | \(\frac {-64-48 r}{27 r^{3}+135 r^{2}+198 r +80}\) | \(-{\frac {4}{5}}\) |
| \(b_{4}\) | \(\frac {144 r^{2}+624 r +512}{81 r^{4}+702 r^{3}+2079 r^{2}+2418 r +880}\) | \(\frac {32}{55}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {-864 r^{2}-4128 r -3840}{243 r^{5}+3240 r^{4}+16065 r^{3}+36360 r^{2}+36492 r +12320} \] Which for the root \(r = 0\) becomes \[ b_{5}=-{\frac {24}{77}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {2}{3 r +2}\) | \(-1\) |
| \(b_{2}\) | \(\frac {12+12 r}{9 r^{2}+21 r +10}\) | \(\frac {6}{5}\) |
| \(b_{3}\) | \(\frac {-64-48 r}{27 r^{3}+135 r^{2}+198 r +80}\) | \(-{\frac {4}{5}}\) |
| \(b_{4}\) | \(\frac {144 r^{2}+624 r +512}{81 r^{4}+702 r^{3}+2079 r^{2}+2418 r +880}\) | \(\frac {32}{55}\) |
| \(b_{5}\) | \(\frac {-864 r^{2}-4128 r -3840}{243 r^{5}+3240 r^{4}+16065 r^{3}+36360 r^{2}+36492 r +12320}\) | \(-{\frac {24}{77}}\) |
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 +\frac {6 x^{2}}{5}-\frac {4 x^{3}}{5}+\frac {32 x^{4}}{55}-\frac {24 x^{5}}{77}+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} x^{\frac {1}{3}} \left (1-\frac {2 x}{3}+\frac {8 x^{2}}{9}-\frac {40 x^{3}}{81}+\frac {92 x^{4}}{243}-\frac {664 x^{5}}{3645}+O\left (x^{6}\right )\right ) + c_{2} \left (1-x +\frac {6 x^{2}}{5}-\frac {4 x^{3}}{5}+\frac {32 x^{4}}{55}-\frac {24 x^{5}}{77}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {1}{3}} \left (1-\frac {2 x}{3}+\frac {8 x^{2}}{9}-\frac {40 x^{3}}{81}+\frac {92 x^{4}}{243}-\frac {664 x^{5}}{3645}+O\left (x^{6}\right )\right )+c_{2} \left (1-x +\frac {6 x^{2}}{5}-\frac {4 x^{3}}{5}+\frac {32 x^{4}}{55}-\frac {24 x^{5}}{77}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {1}{3}} \left (1-\frac {2 x}{3}+\frac {8 x^{2}}{9}-\frac {40 x^{3}}{81}+\frac {92 x^{4}}{243}-\frac {664 x^{5}}{3645}+O\left (x^{6}\right )\right )+c_{2} \left (1-x +\frac {6 x^{2}}{5}-\frac {4 x^{3}}{5}+\frac {32 x^{4}}{55}-\frac {24 x^{5}}{77}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {1}{3}} \left (1-\frac {2 x}{3}+\frac {8 x^{2}}{9}-\frac {40 x^{3}}{81}+\frac {92 x^{4}}{243}-\frac {664 x^{5}}{3645}+O\left (x^{6}\right )\right )+c_{2} \left (1-x +\frac {6 x^{2}}{5}-\frac {4 x^{3}}{5}+\frac {32 x^{4}}{55}-\frac {24 x^{5}}{77}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 3 x^{2} y^{\prime \prime }+\left (-4 x^{3}+2 x^{2}+2 x \right ) y^{\prime }+\left (-8 x^{2}+2 x \right ) 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 {2 \left (4 x -1\right ) y}{3 x}+\frac {2 \left (2 x^{2}-x -1\right ) y^{\prime }}{3 x} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime }-\frac {2 \left (2 x^{2}-x -1\right ) y^{\prime }}{3 x}-\frac {2 \left (4 x -1\right ) y}{3 x}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=-\frac {2 \left (2 x^{2}-x -1\right )}{3 x}, P_{3}\left (x \right )=-\frac {2 \left (4 x -1\right )}{3 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 {2}{3} \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=0 \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 3 x y^{\prime \prime }+\left (-4 x^{2}+2 x +2\right ) y^{\prime }+\left (-8 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^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..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 \cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & x \cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (-1+3 r \right ) x^{-1+r}+\left (a_{1} \left (1+r \right ) \left (2+3 r \right )+2 a_{0} \left (1+r \right )\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (3 k +2+3 r \right )+2 a_{k} \left (k +1+r \right )-4 a_{k -1} \left (k +1+r \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+3 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {1}{3}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right ) \left (2+3 r \right )+2 a_{0} \left (1+r \right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k +1+r \right ) \left (3 k a_{k +1}+3 r a_{k +1}+2 a_{k}-4 a_{k -1}+2 a_{k +1}\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (k +r +2\right ) \left (3 \left (k +1\right ) a_{k +2}+3 r a_{k +2}+2 a_{k +1}-4 a_{k}+2 a_{k +2}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {2 \left (-a_{k +1}+2 a_{k}\right )}{3 k +5+3 r} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=\frac {2 \left (-a_{k +1}+2 a_{k}\right )}{3 k +5} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +2}=\frac {2 \left (-a_{k +1}+2 a_{k}\right )}{3 k +5}, 2 a_{1}+2 a_{0}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{3} \\ {} & {} & a_{k +2}=\frac {2 \left (-a_{k +1}+2 a_{k}\right )}{3 k +6} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{3} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{3}}, a_{k +2}=\frac {2 \left (-a_{k +1}+2 a_{k}\right )}{3 k +6}, 4 a_{1}+\frac {8 a_{0}}{3}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {1}{3}}\right ), a_{k +2}=\frac {2 \left (-a_{1+k}+2 a_{k}\right )}{3 k +5}, 2 a_{1}+2 a_{0}=0, b_{k +2}=\frac {2 \left (-b_{1+k}+2 b_{k}\right )}{3 k +6}, 4 b_{1}+\frac {8 b_{0}}{3}=0\right ] \end {array} \]
Maple trace Kovacic algorithm successful
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm A Liouvillian solution exists Reducible group (found an exponential solution) 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 -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius <- Heun successful: received ODE is equivalent to the HeunB ODE, case c = 0 <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.032 (sec). Leaf size: 44
Order:=6; dsolve(3*x^2*diff(y(x),x$2)+2*x*(1+x-2*x^2)*diff(y(x),x)+(2*x-8*x^2)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{\frac {1}{3}} \left (1-\frac {2}{3} x +\frac {8}{9} x^{2}-\frac {40}{81} x^{3}+\frac {92}{243} x^{4}-\frac {664}{3645} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (1-x +\frac {6}{5} x^{2}-\frac {4}{5} x^{3}+\frac {32}{55} x^{4}-\frac {24}{77} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 83
AsymptoticDSolveValue[3*x^2*y''[x]+2*x*(1+x-2*x^2)*y'[x]+(2*x-8*x^2)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \sqrt [3]{x} \left (-\frac {664 x^5}{3645}+\frac {92 x^4}{243}-\frac {40 x^3}{81}+\frac {8 x^2}{9}-\frac {2 x}{3}+1\right )+c_2 \left (-\frac {24 x^5}{77}+\frac {32 x^4}{55}-\frac {4 x^3}{5}+\frac {6 x^2}{5}-x+1\right ) \]