Internal problem ID [2411]
Internal file name [OUTPUT/2411_Tuesday_February_27_2024_08_37_03_AM_40584997/index.tex]
Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath.
Boston. 1964
Section: Exercise 42, page 206
Problem number: 10.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference is integer"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x^{2} y^{\prime \prime }+2 y^{\prime } x^{2}-\left (3 x^{2}+2\right ) y=0} \] 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. \[ x^{2} y^{\prime \prime }+2 y^{\prime } x^{2}+\left (-3 x^{2}-2\right ) y = 0 \] 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) &= 2\\ q(x) &= -\frac {3 x^{2}+2}{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 \[ x^{2} y^{\prime \prime }+2 y^{\prime } x^{2}+\left (-3 x^{2}-2\right ) y = 0 \] 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} 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 )+2 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right ) x^{2}+\left (-3 x^{2}-2\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 }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{1+n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 x^{n +r +2} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 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 }}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 (-3 x^{n +r +2} a_{n}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-3 a_{n -2} x^{n +r}\right ) \\ \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 }}2 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-3 a_{n -2} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-2 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )-2 a_{0} x^{r} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )-2 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r^{2}-r -2\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r^{2}-r -2 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 2\\ r_2 &= -1 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r^{2}-r -2\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([2, -1]\).
Since \(r_1 - r_2 = 3\) is 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 ) &= C y_{1}\left (x \right ) \ln \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 ) &= x^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\frac {\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}}{x} \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +2}\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -1}\right ) \end {align*}
Where \(C\) above can be zero. We start by finding \(y_{1}\). 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 \[ a_{1} = -\frac {2 r}{r^{2}+r -2} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} a_{n} \left (n +r \right ) \left (n +r -1\right )+2 a_{n -1} \left (n +r -1\right )-3 a_{n -2}-2 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {2 n a_{n -1}+2 r a_{n -1}-3 a_{n -2}-2 a_{n -1}}{n^{2}+2 n r +r^{2}-n -r -2}\tag {4} \] Which for the root \(r = 2\) becomes \[ a_{n} = \frac {-2 n a_{n -1}+3 a_{n -2}-2 a_{n -1}}{n \left (n +3\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 2\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2 r}{r^{2}+r -2}\) | \(-1\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {7 r^{2}+7 r -6}{\left (r^{2}+r -2\right ) r \left (r +3\right )} \] Which for the root \(r = 2\) becomes \[ a_{2}={\frac {9}{10}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2 r}{r^{2}+r -2}\) | \(-1\) |
| \(a_{2}\) | \(\frac {7 r^{2}+7 r -6}{\left (r^{2}+r -2\right ) r \left (r +3\right )}\) | \(\frac {9}{10}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-20 r^{2}-40 r +24}{\left (r +4\right ) \left (r^{2}+r -2\right ) r \left (r +3\right )} \] Which for the root \(r = 2\) becomes \[ a_{3}=-{\frac {17}{30}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2 r}{r^{2}+r -2}\) | \(-1\) |
| \(a_{2}\) | \(\frac {7 r^{2}+7 r -6}{\left (r^{2}+r -2\right ) r \left (r +3\right )}\) | \(\frac {9}{10}\) |
| \(a_{3}\) | \(\frac {-20 r^{2}-40 r +24}{\left (r +4\right ) \left (r^{2}+r -2\right ) r \left (r +3\right )}\) | \(-{\frac {17}{30}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {61 r^{2}+183 r -108}{\left (r +5\right ) \left (r +2\right ) \left (r +3\right ) r \left (-1+r \right ) \left (r +4\right )} \] Which for the root \(r = 2\) becomes \[ a_{4}={\frac {251}{840}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2 r}{r^{2}+r -2}\) | \(-1\) |
| \(a_{2}\) | \(\frac {7 r^{2}+7 r -6}{\left (r^{2}+r -2\right ) r \left (r +3\right )}\) | \(\frac {9}{10}\) |
| \(a_{3}\) | \(\frac {-20 r^{2}-40 r +24}{\left (r +4\right ) \left (r^{2}+r -2\right ) r \left (r +3\right )}\) | \(-{\frac {17}{30}}\) |
| \(a_{4}\) | \(\frac {61 r^{2}+183 r -108}{\left (r +5\right ) \left (r +2\right ) \left (r +3\right ) r \left (-1+r \right ) \left (r +4\right )}\) | \(\frac {251}{840}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-182 r^{2}-728 r +408}{\left (r +6\right ) \left (r +3\right ) r \left (r +2\right ) \left (-1+r \right ) \left (r +4\right ) \left (r +5\right )} \] Which for the root \(r = 2\) becomes \[ a_{5}=-{\frac {37}{280}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2 r}{r^{2}+r -2}\) | \(-1\) |
| \(a_{2}\) | \(\frac {7 r^{2}+7 r -6}{\left (r^{2}+r -2\right ) r \left (r +3\right )}\) | \(\frac {9}{10}\) |
| \(a_{3}\) | \(\frac {-20 r^{2}-40 r +24}{\left (r +4\right ) \left (r^{2}+r -2\right ) r \left (r +3\right )}\) | \(-{\frac {17}{30}}\) |
| \(a_{4}\) | \(\frac {61 r^{2}+183 r -108}{\left (r +5\right ) \left (r +2\right ) \left (r +3\right ) r \left (-1+r \right ) \left (r +4\right )}\) | \(\frac {251}{840}\) |
| \(a_{5}\) | \(\frac {-182 r^{2}-728 r +408}{\left (r +6\right ) \left (r +3\right ) r \left (r +2\right ) \left (-1+r \right ) \left (r +4\right ) \left (r +5\right )}\) | \(-{\frac {37}{280}}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{2} \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^{2} \left (1-x +\frac {9 x^{2}}{10}-\frac {17 x^{3}}{30}+\frac {251 x^{4}}{840}-\frac {37 x^{5}}{280}+O\left (x^{6}\right )\right ) \end {align*}
Now the second solution \(y_{2}\left (x \right )\) is found. Let \[ r_{1}-r_{2} = N \] Where \(N\) is positive integer which is the difference between the two roots. \(r_{1}\) is taken as the larger root. Hence for this problem we have \(N=3\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{3}\left (r \right )\). If this limit exists, then \(C = 0\), else we need to keep the log term and \(C \neq 0\). The above table shows that \begin {align*} a_N &= a_{3} \\ &= \frac {-20 r^{2}-40 r +24}{\left (r +4\right ) \left (r^{2}+r -2\right ) r \left (r +3\right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {-20 r^{2}-40 r +24}{\left (r +4\right ) \left (r^{2}+r -2\right ) r \left (r +3\right )}&= \lim _{r\rightarrow -1}\frac {-20 r^{2}-40 r +24}{\left (r +4\right ) \left (r^{2}+r -2\right ) r \left (r +3\right )}\\ &= {\frac {11}{3}} \end {align*}
The limit is \(\frac {11}{3}\). Since the limit exists then the log term is not needed and we can set \(C = 0\). Therefore the second solution has the form \begin {align*} y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r}\\ &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -1} \end {align*}
Eq (3) derived above is 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(3) gives \[ b_{1} = -\frac {2 r}{r^{2}+r -2} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{4} b_{n} \left (n +r \right ) \left (n +r -1\right )+2 b_{n -1} \left (n +r -1\right )-3 b_{n -2}-2 b_{n} = 0 \end{equation} Which for for the root \(r = -1\) becomes \begin{equation} \tag{4A} b_{n} \left (n -1\right ) \left (n -2\right )+2 b_{n -1} \left (n -2\right )-3 b_{n -2}-2 b_{n} = 0 \end{equation} Solving for \(b_{n}\) from the recursive equation (4) gives \[ b_{n} = -\frac {2 n b_{n -1}+2 r b_{n -1}-3 b_{n -2}-2 b_{n -1}}{n^{2}+2 n r +r^{2}-n -r -2}\tag {5} \] Which for the root \(r = -1\) becomes \[ b_{n} = -\frac {2 n b_{n -1}-3 b_{n -2}-4 b_{n -1}}{n^{2}-3 n}\tag {6} \] 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.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {2 r}{r^{2}+r -2}\) | \(-1\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {7 r^{2}+7 r -6}{\left (r^{2}+r -2\right ) r \left (r +3\right )} \] Which for the root \(r = -1\) becomes \[ b_{2}=-{\frac {3}{2}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {2 r}{r^{2}+r -2}\) | \(-1\) |
| \(b_{2}\) | \(\frac {7 r^{2}+7 r -6}{\left (r^{2}+r -2\right ) r \left (r +3\right )}\) | \(-{\frac {3}{2}}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=-\frac {4 \left (5 r^{2}+10 r -6\right )}{\left (r +4\right ) \left (r^{2}+r -2\right ) r \left (r +3\right )} \] Which for the root \(r = -1\) becomes \[ b_{3}={\frac {11}{3}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {2 r}{r^{2}+r -2}\) | \(-1\) |
| \(b_{2}\) | \(\frac {7 r^{2}+7 r -6}{\left (r^{2}+r -2\right ) r \left (r +3\right )}\) | \(-{\frac {3}{2}}\) |
| \(b_{3}\) | \(\frac {-20 r^{2}-40 r +24}{\left (r +4\right ) \left (r^{2}+r -2\right ) r \left (r +3\right )}\) | \(\frac {11}{3}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {61 r^{2}+183 r -108}{\left (r^{2}+7 r +10\right ) \left (r +3\right ) r \left (-1+r \right ) \left (r +4\right )} \] Which for the root \(r = -1\) becomes \[ b_{4}=-{\frac {115}{24}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {2 r}{r^{2}+r -2}\) | \(-1\) |
| \(b_{2}\) | \(\frac {7 r^{2}+7 r -6}{\left (r^{2}+r -2\right ) r \left (r +3\right )}\) | \(-{\frac {3}{2}}\) |
| \(b_{3}\) | \(\frac {-20 r^{2}-40 r +24}{\left (r +4\right ) \left (r^{2}+r -2\right ) r \left (r +3\right )}\) | \(\frac {11}{3}\) |
| \(b_{4}\) | \(\frac {61 r^{2}+183 r -108}{\left (r +5\right ) \left (r +2\right ) \left (r +3\right ) r \left (-1+r \right ) \left (r +4\right )}\) | \(-{\frac {115}{24}}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-\frac {2 \left (91 r^{2}+364 r -204\right )}{\left (r^{2}+9 r +18\right ) r \left (r^{2}+r -2\right ) \left (r +4\right ) \left (r +5\right )} \] Which for the root \(r = -1\) becomes \[ b_{5}={\frac {159}{40}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {2 r}{r^{2}+r -2}\) | \(-1\) |
| \(b_{2}\) | \(\frac {7 r^{2}+7 r -6}{\left (r^{2}+r -2\right ) r \left (r +3\right )}\) | \(-{\frac {3}{2}}\) |
| \(b_{3}\) | \(\frac {-20 r^{2}-40 r +24}{\left (r +4\right ) \left (r^{2}+r -2\right ) r \left (r +3\right )}\) | \(\frac {11}{3}\) |
| \(b_{4}\) | \(\frac {61 r^{2}+183 r -108}{\left (r +5\right ) \left (r +2\right ) \left (r +3\right ) r \left (-1+r \right ) \left (r +4\right )}\) | \(-{\frac {115}{24}}\) |
| \(b_{5}\) | \(\frac {-182 r^{2}-728 r +408}{\left (r +6\right ) \left (r +3\right ) r \left (r +2\right ) \left (-1+r \right ) \left (r +4\right ) \left (r +5\right )}\) | \(\frac {159}{40}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x^{2} \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-x -\frac {3 x^{2}}{2}+\frac {11 x^{3}}{3}-\frac {115 x^{4}}{24}+\frac {159 x^{5}}{40}+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^{2} \left (1-x +\frac {9 x^{2}}{10}-\frac {17 x^{3}}{30}+\frac {251 x^{4}}{840}-\frac {37 x^{5}}{280}+O\left (x^{6}\right )\right ) + \frac {c_{2} \left (1-x -\frac {3 x^{2}}{2}+\frac {11 x^{3}}{3}-\frac {115 x^{4}}{24}+\frac {159 x^{5}}{40}+O\left (x^{6}\right )\right )}{x} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{2} \left (1-x +\frac {9 x^{2}}{10}-\frac {17 x^{3}}{30}+\frac {251 x^{4}}{840}-\frac {37 x^{5}}{280}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-x -\frac {3 x^{2}}{2}+\frac {11 x^{3}}{3}-\frac {115 x^{4}}{24}+\frac {159 x^{5}}{40}+O\left (x^{6}\right )\right )}{x} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{2} \left (1-x +\frac {9 x^{2}}{10}-\frac {17 x^{3}}{30}+\frac {251 x^{4}}{840}-\frac {37 x^{5}}{280}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-x -\frac {3 x^{2}}{2}+\frac {11 x^{3}}{3}-\frac {115 x^{4}}{24}+\frac {159 x^{5}}{40}+O\left (x^{6}\right )\right )}{x} \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{2} \left (1-x +\frac {9 x^{2}}{10}-\frac {17 x^{3}}{30}+\frac {251 x^{4}}{840}-\frac {37 x^{5}}{280}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-x -\frac {3 x^{2}}{2}+\frac {11 x^{3}}{3}-\frac {115 x^{4}}{24}+\frac {159 x^{5}}{40}+O\left (x^{6}\right )\right )}{x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime \prime }+2 y^{\prime } x^{2}+\left (-3 x^{2}-2\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 (3 x^{2}+2\right ) y}{x^{2}}-2 y^{\prime } \\ \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 }+2 y^{\prime }-\frac {\left (3 x^{2}+2\right ) y}{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 )=2, P_{3}\left (x \right )=-\frac {3 x^{2}+2}{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}}}=0 \\ {} & \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}}}=-2 \\ {} & \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} \\ {} & {} & x^{2} y^{\prime \prime }+2 y^{\prime } x^{2}+\left (-3 x^{2}-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..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^{2}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r +1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -1 \\ {} & {} & x^{2}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =1}{\sum }}a_{k -1} \left (k -1+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 -1+r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (1+r \right ) \left (-2+r \right ) x^{r}+\left (a_{1} \left (2+r \right ) \left (-1+r \right )+2 a_{0} r \right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (k +r +1\right ) \left (k +r -2\right )+2 a_{k -1} \left (k -1+r \right )-3 a_{k -2}\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 (-2+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-1, 2\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (2+r \right ) \left (-1+r \right )+2 a_{0} r =0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=-\frac {2 a_{0} r}{r^{2}+r -2} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (k +r +1\right ) \left (k +r -2\right )+2 a_{k -1} k +2 a_{k -1} r -3 a_{k -2}-2 a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +2} \left (k +3+r \right ) \left (k +r \right )+2 a_{k +1} \left (k +2\right )+2 a_{k +1} r -3 a_{k}-2 a_{k +1}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {2 k a_{k +1}+2 a_{k +1} r -3 a_{k}+2 a_{k +1}}{\left (k +3+r \right ) \left (k +r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1 \\ {} & {} & a_{k +2}=-\frac {2 k a_{k +1}-3 a_{k}}{\left (k +2\right ) \left (k -1\right )} \\ \bullet & {} & \textrm {Series not valid for}\hspace {3pt} r =-1\hspace {3pt}\textrm {, division by}\hspace {3pt} 0\hspace {3pt}\textrm {in the recursion relation at}\hspace {3pt} k =1 \\ {} & {} & a_{k +2}=-\frac {2 k a_{k +1}-3 a_{k}}{\left (k +2\right ) \left (k -1\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +2}=-\frac {2 k a_{k +1}-3 a_{k}+6 a_{k +1}}{\left (k +5\right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +2}, a_{k +2}=-\frac {2 k a_{k +1}-3 a_{k}+6 a_{k +1}}{\left (k +5\right ) \left (k +2\right )}, a_{1}=-a_{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) Reducible group (found another exponential solution) <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 47
Order:=6; dsolve(x^2*diff(y(x),x$2)+2*x^2*diff(y(x),x)-(3*x^2+2)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{2} \left (1-x +\frac {9}{10} x^{2}-\frac {17}{30} x^{3}+\frac {251}{840} x^{4}-\frac {37}{280} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\frac {c_{2} \left (12-12 x -18 x^{2}+44 x^{3}-\frac {115}{2} x^{4}+\frac {477}{10} x^{5}+\operatorname {O}\left (x^{6}\right )\right )}{x} \]
✓ Solution by Mathematica
Time used: 0.027 (sec). Leaf size: 64
AsymptoticDSolveValue[x^2*y''[x]+2*x^2*y'[x]-(3*x^2+2)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (-\frac {115 x^3}{24}+\frac {11 x^2}{3}-\frac {3 x}{2}+\frac {1}{x}-1\right )+c_2 \left (\frac {251 x^6}{840}-\frac {17 x^5}{30}+\frac {9 x^4}{10}-x^3+x^2\right ) \]