Internal problem ID [5568]
Internal file name [OUTPUT/4816_Sunday_June_05_2022_03_06_40_PM_69573029/index.tex]
Book: A FIRST COURSE IN DIFFERENTIAL EQUATIONS with Modeling Applications.
Dennis G. Zill. 9th edition. Brooks/Cole. CA, USA.
Section: Chapter 6. SERIES SOLUTIONS OF LINEAR EQUATIONS. Exercises. 6.2 page
239
Problem number: 13.
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]]
\[ \boxed {x^{2} y^{\prime \prime }+\left (\frac {5}{3} x +x^{2}\right ) y^{\prime }-\frac {y}{3}=0} \] 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. \[ x^{2} y^{\prime \prime }+\left (\frac {5}{3} x +x^{2}\right ) y^{\prime }-\frac {y}{3} = 0 \] 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 x +5}{3 x}\\ q(x) &= -\frac {1}{3 x^{2}}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} y^{\prime \prime }+\left (\frac {5}{3} x +x^{2}\right ) y^{\prime }-\frac {y}{3} = 0 \] 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} 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 (\frac {5}{3} x +x^{2}\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )-\frac {\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right )}{3} = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {5 x^{n +r} a_{n} \left (n +r \right )}{3}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {a_{n} x^{n +r}}{3}\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 }}x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}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} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\frac {5 x^{n +r} a_{n} \left (n +r \right )}{3}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\frac {a_{n} x^{n +r}}{3}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+\frac {5 x^{n +r} a_{n} \left (n +r \right )}{3}-\frac {a_{n} x^{n +r}}{3} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )+\frac {5 x^{r} a_{0} r}{3}-\frac {a_{0} x^{r}}{3} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )+\frac {5 x^{r} r}{3}-\frac {x^{r}}{3}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \frac {\left (3 r^{2}+2 r -1\right ) x^{r}}{3} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r^{2}+\frac {2}{3} r -\frac {1}{3} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {1}{3}}\\ r_2 &= -1 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \frac {\left (3 r^{2}+2 r -1\right ) x^{r}}{3} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {1}{3}}, -1\right ]\).
Since \(r_1 - r_2 = {\frac {4}{3}}\) 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 -1} \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} \left (n +r \right ) \left (n +r -1\right )+a_{n -1} \left (n +r -1\right )+\frac {5 a_{n} \left (n +r \right )}{3}-\frac {a_{n}}{3} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {3 a_{n -1} \left (n +r -1\right )}{3 n^{2}+6 n r +3 r^{2}+2 n +2 r -1}\tag {4} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ a_{n} = \frac {a_{n -1} \left (2-3 n \right )}{3 n^{2}+4 n}\tag {5} \] 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.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ a_{1}=-\frac {3 r}{3 r^{2}+8 r +4} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ a_{1}=-{\frac {1}{7}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3 r}{3 r^{2}+8 r +4}\) | \(-{\frac {1}{7}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {9 r \left (1+r \right )}{\left (3 r^{2}+8 r +4\right ) \left (3 r^{2}+14 r +15\right )} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ a_{2}={\frac {1}{35}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3 r}{3 r^{2}+8 r +4}\) | \(-{\frac {1}{7}}\) |
| \(a_{2}\) | \(\frac {9 r \left (1+r \right )}{\left (3 r^{2}+8 r +4\right ) \left (3 r^{2}+14 r +15\right )}\) | \(\frac {1}{35}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {27 r \left (1+r \right )}{\left (3 r^{2}+20 r +32\right ) \left (3 r^{2}+14 r +15\right ) \left (3 r +2\right )} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ a_{3}=-{\frac {1}{195}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3 r}{3 r^{2}+8 r +4}\) | \(-{\frac {1}{7}}\) |
| \(a_{2}\) | \(\frac {9 r \left (1+r \right )}{\left (3 r^{2}+8 r +4\right ) \left (3 r^{2}+14 r +15\right )}\) | \(\frac {1}{35}\) |
| \(a_{3}\) | \(-\frac {27 r \left (1+r \right )}{\left (3 r^{2}+20 r +32\right ) \left (3 r^{2}+14 r +15\right ) \left (3 r +2\right )}\) | \(-{\frac {1}{195}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {81 r \left (1+r \right )}{\left (3 r^{2}+26 r +55\right ) \left (3 r +2\right ) \left (3 r +5\right ) \left (3 r^{2}+20 r +32\right )} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ a_{4}={\frac {1}{1248}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3 r}{3 r^{2}+8 r +4}\) | \(-{\frac {1}{7}}\) |
| \(a_{2}\) | \(\frac {9 r \left (1+r \right )}{\left (3 r^{2}+8 r +4\right ) \left (3 r^{2}+14 r +15\right )}\) | \(\frac {1}{35}\) |
| \(a_{3}\) | \(-\frac {27 r \left (1+r \right )}{\left (3 r^{2}+20 r +32\right ) \left (3 r^{2}+14 r +15\right ) \left (3 r +2\right )}\) | \(-{\frac {1}{195}}\) |
| \(a_{4}\) | \(\frac {81 r \left (1+r \right )}{\left (3 r^{2}+26 r +55\right ) \left (3 r +2\right ) \left (3 r +5\right ) \left (3 r^{2}+20 r +32\right )}\) | \(\frac {1}{1248}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {243 r \left (1+r \right )}{\left (3 r^{2}+32 r +84\right ) \left (3 r +8\right ) \left (3 r +5\right ) \left (3 r +2\right ) \left (3 r^{2}+26 r +55\right )} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ a_{5}=-{\frac {1}{9120}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3 r}{3 r^{2}+8 r +4}\) | \(-{\frac {1}{7}}\) |
| \(a_{2}\) | \(\frac {9 r \left (1+r \right )}{\left (3 r^{2}+8 r +4\right ) \left (3 r^{2}+14 r +15\right )}\) | \(\frac {1}{35}\) |
| \(a_{3}\) | \(-\frac {27 r \left (1+r \right )}{\left (3 r^{2}+20 r +32\right ) \left (3 r^{2}+14 r +15\right ) \left (3 r +2\right )}\) | \(-{\frac {1}{195}}\) |
| \(a_{4}\) | \(\frac {81 r \left (1+r \right )}{\left (3 r^{2}+26 r +55\right ) \left (3 r +2\right ) \left (3 r +5\right ) \left (3 r^{2}+20 r +32\right )}\) | \(\frac {1}{1248}\) |
| \(a_{5}\) | \(-\frac {243 r \left (1+r \right )}{\left (3 r^{2}+32 r +84\right ) \left (3 r +8\right ) \left (3 r +5\right ) \left (3 r +2\right ) \left (3 r^{2}+26 r +55\right )}\) | \(-{\frac {1}{9120}}\) |
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 {x}{7}+\frac {x^{2}}{35}-\frac {x^{3}}{195}+\frac {x^{4}}{1248}-\frac {x^{5}}{9120}+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} b_{n} \left (n +r \right ) \left (n +r -1\right )+b_{n -1} \left (n +r -1\right )+\frac {5 b_{n} \left (n +r \right )}{3}-\frac {b_{n}}{3} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {3 b_{n -1} \left (n +r -1\right )}{3 n^{2}+6 n r +3 r^{2}+2 n +2 r -1}\tag {4} \] Which for the root \(r = -1\) becomes \[ b_{n} = -\frac {3 b_{n -1} \left (n -2\right )}{n \left (3 n -4\right )}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -1\) 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 \[ b_{1}=-\frac {3 r}{3 r^{2}+8 r +4} \] Which for the root \(r = -1\) becomes \[ b_{1}=-3 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {3 r}{3 r^{2}+8 r +4}\) | \(-3\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {9 r \left (1+r \right )}{\left (3 r^{2}+8 r +4\right ) \left (3 r^{2}+14 r +15\right )} \] Which for the root \(r = -1\) becomes \[ b_{2}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {3 r}{3 r^{2}+8 r +4}\) | \(-3\) |
| \(b_{2}\) | \(\frac {9 r \left (1+r \right )}{\left (3 r^{2}+8 r +4\right ) \left (3 r^{2}+14 r +15\right )}\) | \(0\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=-\frac {27 r \left (1+r \right )}{\left (3 r^{2}+20 r +32\right ) \left (3 r^{2}+14 r +15\right ) \left (3 r +2\right )} \] Which for the root \(r = -1\) becomes \[ b_{3}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {3 r}{3 r^{2}+8 r +4}\) | \(-3\) |
| \(b_{2}\) | \(\frac {9 r \left (1+r \right )}{\left (3 r^{2}+8 r +4\right ) \left (3 r^{2}+14 r +15\right )}\) | \(0\) |
| \(b_{3}\) | \(-\frac {27 r \left (1+r \right )}{\left (3 r^{2}+20 r +32\right ) \left (3 r^{2}+14 r +15\right ) \left (3 r +2\right )}\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {81 r \left (1+r \right )}{\left (3 r^{2}+26 r +55\right ) \left (3 r +2\right ) \left (3 r +5\right ) \left (3 r^{2}+20 r +32\right )} \] Which for the root \(r = -1\) becomes \[ b_{4}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {3 r}{3 r^{2}+8 r +4}\) | \(-3\) |
| \(b_{2}\) | \(\frac {9 r \left (1+r \right )}{\left (3 r^{2}+8 r +4\right ) \left (3 r^{2}+14 r +15\right )}\) | \(0\) |
| \(b_{3}\) | \(-\frac {27 r \left (1+r \right )}{\left (3 r^{2}+20 r +32\right ) \left (3 r^{2}+14 r +15\right ) \left (3 r +2\right )}\) | \(0\) |
| \(b_{4}\) | \(\frac {81 r \left (1+r \right )}{\left (3 r^{2}+26 r +55\right ) \left (3 r +2\right ) \left (3 r +5\right ) \left (3 r^{2}+20 r +32\right )}\) | \(0\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-\frac {243 r \left (1+r \right )}{\left (3 r^{2}+32 r +84\right ) \left (3 r +8\right ) \left (3 r +5\right ) \left (3 r +2\right ) \left (3 r^{2}+26 r +55\right )} \] Which for the root \(r = -1\) becomes \[ b_{5}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {3 r}{3 r^{2}+8 r +4}\) | \(-3\) |
| \(b_{2}\) | \(\frac {9 r \left (1+r \right )}{\left (3 r^{2}+8 r +4\right ) \left (3 r^{2}+14 r +15\right )}\) | \(0\) |
| \(b_{3}\) | \(-\frac {27 r \left (1+r \right )}{\left (3 r^{2}+20 r +32\right ) \left (3 r^{2}+14 r +15\right ) \left (3 r +2\right )}\) | \(0\) |
| \(b_{4}\) | \(\frac {81 r \left (1+r \right )}{\left (3 r^{2}+26 r +55\right ) \left (3 r +2\right ) \left (3 r +5\right ) \left (3 r^{2}+20 r +32\right )}\) | \(0\) |
| \(b_{5}\) | \(-\frac {243 r \left (1+r \right )}{\left (3 r^{2}+32 r +84\right ) \left (3 r +8\right ) \left (3 r +5\right ) \left (3 r +2\right ) \left (3 r^{2}+26 r +55\right )}\) | \(0\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x^{\frac {1}{3}} \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 {1-3 x +O\left (x^{6}\right )}{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} x^{\frac {1}{3}} \left (1-\frac {x}{7}+\frac {x^{2}}{35}-\frac {x^{3}}{195}+\frac {x^{4}}{1248}-\frac {x^{5}}{9120}+O\left (x^{6}\right )\right ) + \frac {c_{2} \left (1-3 x +O\left (x^{6}\right )\right )}{x} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {1}{3}} \left (1-\frac {x}{7}+\frac {x^{2}}{35}-\frac {x^{3}}{195}+\frac {x^{4}}{1248}-\frac {x^{5}}{9120}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-3 x +O\left (x^{6}\right )\right )}{x} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {1}{3}} \left (1-\frac {x}{7}+\frac {x^{2}}{35}-\frac {x^{3}}{195}+\frac {x^{4}}{1248}-\frac {x^{5}}{9120}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-3 x +O\left (x^{6}\right )\right )}{x} \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {1}{3}} \left (1-\frac {x}{7}+\frac {x^{2}}{35}-\frac {x^{3}}{195}+\frac {x^{4}}{1248}-\frac {x^{5}}{9120}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-3 x +O\left (x^{6}\right )\right )}{x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime \prime }+\left (\frac {5}{3} x +x^{2}\right ) y^{\prime }-\frac {y}{3}=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 {y}{3 x^{2}}-\frac {\left (3 x +5\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 {\left (3 x +5\right ) y^{\prime }}{3 x}-\frac {y}{3 x^{2}}=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 {3 x +5}{3 x}, P_{3}\left (x \right )=-\frac {1}{3 x^{2}}\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 {5}{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}}}=-\frac {1}{3} \\ {} & \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^{2} y^{\prime \prime }+x \left (3 x +5\right ) y^{\prime }-y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =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^{2}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (1+r \right ) \left (-1+3 r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (k +r +1\right ) \left (3 k +3 r -1\right )+3 a_{k -1} \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} \\ {} & {} & \left (1+r \right ) \left (-1+3 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-1, \frac {1}{3}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 3 \left (k +r -\frac {1}{3}\right ) \left (k +r +1\right ) a_{k}+3 a_{k -1} \left (k +r -1\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 3 \left (k +\frac {2}{3}+r \right ) \left (k +2+r \right ) a_{k +1}+3 a_{k} \left (k +r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {3 a_{k} \left (k +r \right )}{\left (3 k +2+3 r \right ) \left (k +2+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =1 \\ {} & {} & a_{k +1}=-\frac {3 a_{k} \left (k -1\right )}{\left (3 k -1\right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =0 \\ {} & {} & a_{1}=-3 a_{0} \\ \bullet & {} & \textrm {Terminating series solution of the ODE for}\hspace {3pt} r =-1\hspace {3pt}\textrm {. Use reduction of order to find the second linearly independent solution}\hspace {3pt} \\ {} & {} & y=a_{0}\cdot \left (1-3 x \right ) \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{3} \\ {} & {} & a_{k +1}=-\frac {3 a_{k} \left (k +\frac {1}{3}\right )}{\left (3 k +3\right ) \left (k +\frac {7}{3}\right )} \\ \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 +1}=-\frac {3 a_{k} \left (k +\frac {1}{3}\right )}{\left (3 k +3\right ) \left (k +\frac {7}{3}\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=a_{0}\cdot \left (1-3 x \right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {1}{3}}\right ), b_{k +1}=-\frac {3 b_{k} \left (k +\frac {1}{3}\right )}{\left (3 k +3\right ) \left (k +\frac {7}{3}\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) 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 successful <- special function solution successful Solution using Kummer functions still has integrals. Trying a hypergeometric solution. -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 1F1 ODE -> 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: 39
Order:=6; dsolve(x^2*diff(y(x),x$2)+(5/3*x+x^2)*diff(y(x),x)-1/3*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{2} x^{\frac {4}{3}} \left (1-\frac {1}{7} x +\frac {1}{35} x^{2}-\frac {1}{195} x^{3}+\frac {1}{1248} x^{4}-\frac {1}{9120} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{1} \left (1-3 x +\operatorname {O}\left (x^{6}\right )\right )}{x} \]
✓ Solution by Mathematica
Time used: 0.003 (sec). Leaf size: 58
AsymptoticDSolveValue[x^2*y''[x]+(5/3*x+x^2)*y'[x]-1/3*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \sqrt [3]{x} \left (-\frac {x^5}{9120}+\frac {x^4}{1248}-\frac {x^3}{195}+\frac {x^2}{35}-\frac {x}{7}+1\right )+\frac {c_2 (1-3 x)}{x} \]