Internal problem ID [1348]
Internal file name [OUTPUT/1349_Sunday_June_05_2022_02_12_18_AM_35802698/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: 68.
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]]
\[ \boxed {4 x^{2} \left (x^{2}+2 x +3\right ) y^{\prime \prime }-x \left (-15 x^{2}-14 x +3\right ) y^{\prime }+\left (7 x^{2}+3\right ) y=0} \] 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. \[ \left (4 x^{4}+8 x^{3}+12 x^{2}\right ) y^{\prime \prime }+\left (15 x^{3}+14 x^{2}-3 x \right ) y^{\prime }+\left (7 x^{2}+3\right ) y = 0 \] 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 {15 x^{2}+14 x -3}{4 x \left (x^{2}+2 x +3\right )}\\ q(x) &= \frac {7 x^{2}+3}{4 x^{2} \left (x^{2}+2 x +3\right )}\\ \end {align*}
| \(p(x)=\frac {15 x^{2}+14 x -3}{4 x \left (x^{2}+2 x +3\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -1-i \sqrt {2}\) | \(\text {``regular''}\) |
| \(x = -1+i \sqrt {2}\) | \(\text {``regular''}\) |
| \(q(x)=\frac {7 x^{2}+3}{4 x^{2} \left (x^{2}+2 x +3\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -1-i \sqrt {2}\) | \(\text {``regular''}\) |
| \(x = -1+i \sqrt {2}\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \(\left [0, -1-i \sqrt {2}, -1+i \sqrt {2}, \infty \right ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ 4 x^{2} \left (x^{2}+2 x +3\right ) y^{\prime \prime }+\left (15 x^{3}+14 x^{2}-3 x \right ) y^{\prime }+\left (7 x^{2}+3\right ) y = 0 \] 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} 4 x^{2} \left (x^{2}+2 x +3\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (15 x^{3}+14 x^{2}-3 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (7 x^{2}+3\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 }}4 x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}8 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}12 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}15 x^{n +r +2} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}14 x^{1+n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}7 x^{n +r +2} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}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 }}4 x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}4 a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}8 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}8 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}15 x^{n +r +2} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}15 a_{n -2} \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}14 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}14 a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}7 x^{n +r +2} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}7 a_{n -2} 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 =2}{\sum }}4 a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}8 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}12 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}15 a_{n -2} \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}14 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}7 a_{n -2} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 12 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-3 x^{n +r} a_{n} \left (n +r \right )+3 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 12 x^{r} a_{0} r \left (-1+r \right )-3 x^{r} a_{0} r +3 a_{0} x^{r} = 0 \] Or \[ \left (12 x^{r} r \left (-1+r \right )-3 x^{r} r +3 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (12 r^{2}-15 r +3\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 12 r^{2}-15 r +3 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 1\\ r_2 &= {\frac {1}{4}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (12 r^{2}-15 r +3\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [1, {\frac {1}{4}}\right ]\).
Since \(r_1 - r_2 = {\frac {3}{4}}\) 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 {1}{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 \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = -{\frac {2}{3}} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} 4 a_{n -2} \left (n +r -2\right ) \left (n -3+r \right )+8 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+12 a_{n} \left (n +r \right ) \left (n +r -1\right )+15 a_{n -2} \left (n +r -2\right )+14 a_{n -1} \left (n +r -1\right )-3 a_{n} \left (n +r \right )+7 a_{n -2}+3 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -2}}{3}-\frac {2 a_{n -1}}{3}\tag {4} \] Which for the root \(r = 1\) becomes \[ a_{n} = -\frac {a_{n -2}}{3}-\frac {2 a_{n -1}}{3}\tag {5} \] 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.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-{\frac {2}{3}}\) | \(-{\frac {2}{3}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}={\frac {1}{9}} \] Which for the root \(r = 1\) becomes \[ a_{2}={\frac {1}{9}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-{\frac {2}{3}}\) | \(-{\frac {2}{3}}\) |
| \(a_{2}\) | \(\frac {1}{9}\) | \(\frac {1}{9}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}={\frac {4}{27}} \] Which for the root \(r = 1\) becomes \[ a_{3}={\frac {4}{27}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-{\frac {2}{3}}\) | \(-{\frac {2}{3}}\) |
| \(a_{2}\) | \(\frac {1}{9}\) | \(\frac {1}{9}\) |
| \(a_{3}\) | \(\frac {4}{27}\) | \(\frac {4}{27}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=-{\frac {11}{81}} \] Which for the root \(r = 1\) becomes \[ a_{4}=-{\frac {11}{81}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-{\frac {2}{3}}\) | \(-{\frac {2}{3}}\) |
| \(a_{2}\) | \(\frac {1}{9}\) | \(\frac {1}{9}\) |
| \(a_{3}\) | \(\frac {4}{27}\) | \(\frac {4}{27}\) |
| \(a_{4}\) | \(-{\frac {11}{81}}\) | \(-{\frac {11}{81}}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}={\frac {10}{243}} \] Which for the root \(r = 1\) becomes \[ a_{5}={\frac {10}{243}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-{\frac {2}{3}}\) | \(-{\frac {2}{3}}\) |
| \(a_{2}\) | \(\frac {1}{9}\) | \(\frac {1}{9}\) |
| \(a_{3}\) | \(\frac {4}{27}\) | \(\frac {4}{27}\) |
| \(a_{4}\) | \(-{\frac {11}{81}}\) | \(-{\frac {11}{81}}\) |
| \(a_{5}\) | \(\frac {10}{243}\) | \(\frac {10}{243}\) |
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 {2 x}{3}+\frac {x^{2}}{9}+\frac {4 x^{3}}{27}-\frac {11 x^{4}}{81}+\frac {10 x^{5}}{243}+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 \[ b_{1} = -{\frac {2}{3}} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} 4 b_{n -2} \left (n +r -2\right ) \left (n -3+r \right )+8 b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+12 b_{n} \left (n +r \right ) \left (n +r -1\right )+15 b_{n -2} \left (n +r -2\right )+14 b_{n -1} \left (n +r -1\right )-3 b_{n} \left (n +r \right )+7 b_{n -2}+3 b_{n} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {b_{n -2}}{3}-\frac {2 b_{n -1}}{3}\tag {4} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ b_{n} = -\frac {b_{n -2}}{3}-\frac {2 b_{n -1}}{3}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = {\frac {1}{4}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-{\frac {2}{3}}\) | \(-{\frac {2}{3}}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}={\frac {1}{9}} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ b_{2}={\frac {1}{9}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-{\frac {2}{3}}\) | \(-{\frac {2}{3}}\) |
| \(b_{2}\) | \(\frac {1}{9}\) | \(\frac {1}{9}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}={\frac {4}{27}} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ b_{3}={\frac {4}{27}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-{\frac {2}{3}}\) | \(-{\frac {2}{3}}\) |
| \(b_{2}\) | \(\frac {1}{9}\) | \(\frac {1}{9}\) |
| \(b_{3}\) | \(\frac {4}{27}\) | \(\frac {4}{27}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=-{\frac {11}{81}} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ b_{4}=-{\frac {11}{81}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-{\frac {2}{3}}\) | \(-{\frac {2}{3}}\) |
| \(b_{2}\) | \(\frac {1}{9}\) | \(\frac {1}{9}\) |
| \(b_{3}\) | \(\frac {4}{27}\) | \(\frac {4}{27}\) |
| \(b_{4}\) | \(-{\frac {11}{81}}\) | \(-{\frac {11}{81}}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}={\frac {10}{243}} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ b_{5}={\frac {10}{243}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-{\frac {2}{3}}\) | \(-{\frac {2}{3}}\) |
| \(b_{2}\) | \(\frac {1}{9}\) | \(\frac {1}{9}\) |
| \(b_{3}\) | \(\frac {4}{27}\) | \(\frac {4}{27}\) |
| \(b_{4}\) | \(-{\frac {11}{81}}\) | \(-{\frac {11}{81}}\) |
| \(b_{5}\) | \(\frac {10}{243}\) | \(\frac {10}{243}\) |
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 ) \\ &= x^{\frac {1}{4}} \left (1-\frac {2 x}{3}+\frac {x^{2}}{9}+\frac {4 x^{3}}{27}-\frac {11 x^{4}}{81}+\frac {10 x^{5}}{243}+O\left (x^{6}\right )\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 \left (1-\frac {2 x}{3}+\frac {x^{2}}{9}+\frac {4 x^{3}}{27}-\frac {11 x^{4}}{81}+\frac {10 x^{5}}{243}+O\left (x^{6}\right )\right ) + c_{2} x^{\frac {1}{4}} \left (1-\frac {2 x}{3}+\frac {x^{2}}{9}+\frac {4 x^{3}}{27}-\frac {11 x^{4}}{81}+\frac {10 x^{5}}{243}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x \left (1-\frac {2 x}{3}+\frac {x^{2}}{9}+\frac {4 x^{3}}{27}-\frac {11 x^{4}}{81}+\frac {10 x^{5}}{243}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {1}{4}} \left (1-\frac {2 x}{3}+\frac {x^{2}}{9}+\frac {4 x^{3}}{27}-\frac {11 x^{4}}{81}+\frac {10 x^{5}}{243}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x \left (1-\frac {2 x}{3}+\frac {x^{2}}{9}+\frac {4 x^{3}}{27}-\frac {11 x^{4}}{81}+\frac {10 x^{5}}{243}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {1}{4}} \left (1-\frac {2 x}{3}+\frac {x^{2}}{9}+\frac {4 x^{3}}{27}-\frac {11 x^{4}}{81}+\frac {10 x^{5}}{243}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x \left (1-\frac {2 x}{3}+\frac {x^{2}}{9}+\frac {4 x^{3}}{27}-\frac {11 x^{4}}{81}+\frac {10 x^{5}}{243}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {1}{4}} \left (1-\frac {2 x}{3}+\frac {x^{2}}{9}+\frac {4 x^{3}}{27}-\frac {11 x^{4}}{81}+\frac {10 x^{5}}{243}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 x^{2} \left (x^{2}+2 x +3\right ) y^{\prime \prime }+\left (15 x^{3}+14 x^{2}-3 x \right ) y^{\prime }+\left (7 x^{2}+3\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 {\left (7 x^{2}+3\right ) y}{4 x^{2} \left (x^{2}+2 x +3\right )}-\frac {\left (15 x^{2}+14 x -3\right ) y^{\prime }}{4 x \left (x^{2}+2 x +3\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 {\left (15 x^{2}+14 x -3\right ) y^{\prime }}{4 x \left (x^{2}+2 x +3\right )}+\frac {\left (7 x^{2}+3\right ) y}{4 x^{2} \left (x^{2}+2 x +3\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 {15 x^{2}+14 x -3}{4 x \left (x^{2}+2 x +3\right )}, P_{3}\left (x \right )=\frac {7 x^{2}+3}{4 x^{2} \left (x^{2}+2 x +3\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 {1}{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 {1}{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 x^{2} \left (x^{2}+2 x +3\right ) y^{\prime \prime }+x \left (15 x^{2}+14 x -3\right ) y^{\prime }+\left (7 x^{2}+3\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..2 \\ {} & {} & 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 =1..3 \\ {} & {} & 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..4 \\ {} & {} & 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} \\ {} & {} & 3 a_{0} \left (-1+4 r \right ) \left (-1+r \right ) x^{r}+\left (3 a_{1} \left (3+4 r \right ) r +2 a_{0} r \left (3+4 r \right )\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (3 a_{k} \left (4 k +4 r -1\right ) \left (k +r -1\right )+2 a_{k -1} \left (k +r -1\right ) \left (4 k +4 r -1\right )+a_{k -2} \left (4 k +4 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} \\ {} & {} & 3 \left (-1+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 {1}{4}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & 3 a_{1} \left (3+4 r \right ) r +2 a_{0} r \left (3+4 r \right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=-\frac {2 a_{0}}{3} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (4 k +4 r -1\right ) \left (k +r -1\right ) \left (3 a_{k}+2 a_{k -1}+a_{k -2}\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \left (4 k +4 r +7\right ) \left (k +r +1\right ) \left (3 a_{k +2}+2 a_{k +1}+a_{k}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {2 a_{k +1}}{3}-\frac {a_{k}}{3} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +2}=-\frac {2 a_{k +1}}{3}-\frac {a_{k}}{3} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1}, a_{k +2}=-\frac {2 a_{k +1}}{3}-\frac {a_{k}}{3}, a_{1}=-\frac {2 a_{0}}{3}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{4} \\ {} & {} & a_{k +2}=-\frac {2 a_{k +1}}{3}-\frac {a_{k}}{3} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{4} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{4}}, a_{k +2}=-\frac {2 a_{k +1}}{3}-\frac {a_{k}}{3}, a_{1}=-\frac {2 a_{0}}{3}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{1+k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {1}{4}}\right ), a_{k +2}=-\frac {2 a_{1+k}}{3}-\frac {a_{k}}{3}, a_{1}=-\frac {2 a_{0}}{3}, b_{k +2}=-\frac {2 b_{1+k}}{3}-\frac {b_{k}}{3}, b_{1}=-\frac {2 b_{0}}{3}\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: 45
Order:=6; dsolve(4*x^2*(3+2*x+x^2)*diff(y(x),x$2)-x*(3-14*x-15*x^2)*diff(y(x),x)+(3+7*x^2)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \left (1-\frac {2}{3} x +\frac {1}{9} x^{2}+\frac {4}{27} x^{3}-\frac {11}{81} x^{4}+\frac {10}{243} x^{5}\right ) \left (c_{1} x^{\frac {1}{4}}+c_{2} x \right )+O\left (x^{6}\right ) \]
✓ Solution by Mathematica
Time used: 0.007 (sec). Leaf size: 86
AsymptoticDSolveValue[4*x^2*(3+2*x+x^2)*y''[x]-x*(3-14*x-15*x^2)*y'[x]+(3+7*x^2)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 x \left (\frac {10 x^5}{243}-\frac {11 x^4}{81}+\frac {4 x^3}{27}+\frac {x^2}{9}-\frac {2 x}{3}+1\right )+c_2 \sqrt [4]{x} \left (\frac {10 x^5}{243}-\frac {11 x^4}{81}+\frac {4 x^3}{27}+\frac {x^2}{9}-\frac {2 x}{3}+1\right ) \]