Internal problem ID [2401]
Internal file name [OUTPUT/2401_Tuesday_February_27_2024_08_36_56_AM_15179823/index.tex]
Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath.
Boston. 1964
Section: Exercise 41, page 195
Problem number: 26.
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]]
\[ \boxed {2 x y^{\prime \prime }-\left (x^{3}+1\right ) y^{\prime }+y=0} \] 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. \[ 2 x y^{\prime \prime }+\left (-x^{3}-1\right ) y^{\prime }+y = 0 \] 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 {x^{3}+1}{2 x}\\ q(x) &= \frac {1}{2 x}\\ \end {align*}
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 \[ 2 x y^{\prime \prime }+\left (-x^{3}-1\right ) y^{\prime }+y = 0 \] 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} 2 x \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (-x^{3}-1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\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 }}2 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r +2} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r -1\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r -1}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r +2} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-a_{n -3} \left (n +r -3\right ) x^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r -1} \\ \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 \(n +r -1\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =3}{\sum }}\left (-a_{n -3} \left (n +r -3\right ) x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 2 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )-\left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ 2 x^{-1+r} a_{0} r \left (-1+r \right )-r a_{0} x^{-1+r} = 0 \] Or \[ \left (2 x^{-1+r} r \left (-1+r \right )-r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ r \,x^{-1+r} \left (-3+2 r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 2 r^{2}-3 r = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {3}{2}}\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ r \,x^{-1+r} \left (-3+2 r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {3}{2}}, 0\right ]\).
Since \(r_1 - r_2 = {\frac {3}{2}}\) 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 {3}{2}}\\ 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 \[ a_{1} = -\frac {1}{2 r^{2}+r -1} \] Substituting \(n = 2\) in Eq. (2B) gives \[ a_{2} = \frac {1}{4 r^{4}+12 r^{3}+7 r^{2}-3 r -2} \] For \(3\le n\) the recursive equation is \begin{equation} \tag{3} 2 a_{n} \left (n +r \right ) \left (n +r -1\right )-a_{n -3} \left (n +r -3\right )-a_{n} \left (n +r \right )+a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {n a_{n -3}+r a_{n -3}-3 a_{n -3}-a_{n -1}}{2 n^{2}+4 n r +2 r^{2}-3 n -3 r}\tag {4} \] Which for the root \(r = {\frac {3}{2}}\) becomes \[ a_{n} = \frac {2 n a_{n -3}-3 a_{n -3}-2 a_{n -1}}{4 n^{2}+6 n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {3}{2}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{2 r^{2}+r -1}\) | \(-{\frac {1}{5}}\) |
| \(a_{2}\) | \(\frac {1}{4 r^{4}+12 r^{3}+7 r^{2}-3 r -2}\) | \(\frac {1}{70}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {4 r^{5}+12 r^{4}+7 r^{3}-3 r^{2}-2 r -1}{8 r^{6}+60 r^{5}+158 r^{4}+165 r^{3}+32 r^{2}-45 r -18} \] Which for the root \(r = {\frac {3}{2}}\) becomes \[ a_{3}={\frac {52}{945}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{2 r^{2}+r -1}\) | \(-{\frac {1}{5}}\) |
| \(a_{2}\) | \(\frac {1}{4 r^{4}+12 r^{3}+7 r^{2}-3 r -2}\) | \(\frac {1}{70}\) |
| \(a_{3}\) | \(\frac {4 r^{5}+12 r^{4}+7 r^{3}-3 r^{2}-2 r -1}{8 r^{6}+60 r^{5}+158 r^{4}+165 r^{3}+32 r^{2}-45 r -18}\) | \(\frac {52}{945}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {-8 r^{5}-44 r^{4}-102 r^{3}-127 r^{2}-79 r -17}{16 r^{8}+224 r^{7}+1256 r^{6}+3584 r^{5}+5369 r^{4}+3626 r^{3}+19 r^{2}-1134 r -360} \] Which for the root \(r = {\frac {3}{2}}\) becomes \[ a_{4}=-{\frac {1049}{83160}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{2 r^{2}+r -1}\) | \(-{\frac {1}{5}}\) |
| \(a_{2}\) | \(\frac {1}{4 r^{4}+12 r^{3}+7 r^{2}-3 r -2}\) | \(\frac {1}{70}\) |
| \(a_{3}\) | \(\frac {4 r^{5}+12 r^{4}+7 r^{3}-3 r^{2}-2 r -1}{8 r^{6}+60 r^{5}+158 r^{4}+165 r^{3}+32 r^{2}-45 r -18}\) | \(\frac {52}{945}\) |
| \(a_{4}\) | \(\frac {-8 r^{5}-44 r^{4}-102 r^{3}-127 r^{2}-79 r -17}{16 r^{8}+224 r^{7}+1256 r^{6}+3584 r^{5}+5369 r^{4}+3626 r^{3}+19 r^{2}-1134 r -360}\) | \(-{\frac {1049}{83160}}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {12 r^{5}+96 r^{4}+365 r^{3}+774 r^{2}+853 r +377}{32 r^{10}+720 r^{9}+6880 r^{8}+36360 r^{7}+115626 r^{6}+223965 r^{5}+249595 r^{4}+124965 r^{3}-19333 r^{2}-45810 r -12600} \] Which for the root \(r = {\frac {3}{2}}\) becomes \[ a_{5}={\frac {5207}{5405400}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{2 r^{2}+r -1}\) | \(-{\frac {1}{5}}\) |
| \(a_{2}\) | \(\frac {1}{4 r^{4}+12 r^{3}+7 r^{2}-3 r -2}\) | \(\frac {1}{70}\) |
| \(a_{3}\) | \(\frac {4 r^{5}+12 r^{4}+7 r^{3}-3 r^{2}-2 r -1}{8 r^{6}+60 r^{5}+158 r^{4}+165 r^{3}+32 r^{2}-45 r -18}\) | \(\frac {52}{945}\) |
| \(a_{4}\) | \(\frac {-8 r^{5}-44 r^{4}-102 r^{3}-127 r^{2}-79 r -17}{16 r^{8}+224 r^{7}+1256 r^{6}+3584 r^{5}+5369 r^{4}+3626 r^{3}+19 r^{2}-1134 r -360}\) | \(-{\frac {1049}{83160}}\) |
| \(a_{5}\) | \(\frac {12 r^{5}+96 r^{4}+365 r^{3}+774 r^{2}+853 r +377}{32 r^{10}+720 r^{9}+6880 r^{8}+36360 r^{7}+115626 r^{6}+223965 r^{5}+249595 r^{4}+124965 r^{3}-19333 r^{2}-45810 r -12600}\) | \(\frac {5207}{5405400}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{\frac {3}{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^{\frac {3}{2}} \left (1-\frac {x}{5}+\frac {x^{2}}{70}+\frac {52 x^{3}}{945}-\frac {1049 x^{4}}{83160}+\frac {5207 x^{5}}{5405400}+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 \[ b_{1} = -\frac {1}{2 r^{2}+r -1} \] Substituting \(n = 2\) in Eq. (2B) gives \[ b_{2} = \frac {1}{4 r^{4}+12 r^{3}+7 r^{2}-3 r -2} \] For \(3\le n\) the recursive equation is \begin{equation} \tag{3} 2 b_{n} \left (n +r \right ) \left (n +r -1\right )-b_{n -3} \left (n +r -3\right )-\left (n +r \right ) b_{n}+b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = \frac {n b_{n -3}+r b_{n -3}-3 b_{n -3}-b_{n -1}}{2 n^{2}+4 n r +2 r^{2}-3 n -3 r}\tag {4} \] Which for the root \(r = 0\) becomes \[ b_{n} = \frac {\left (n -3\right ) b_{n -3}-b_{n -1}}{n \left (2 n -3\right )}\tag {5} \] 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.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {1}{2 r^{2}+r -1}\) | \(1\) |
| \(b_{2}\) | \(\frac {1}{4 r^{4}+12 r^{3}+7 r^{2}-3 r -2}\) | \(-{\frac {1}{2}}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {4 r^{5}+12 r^{4}+7 r^{3}-3 r^{2}-2 r -1}{8 r^{6}+60 r^{5}+158 r^{4}+165 r^{3}+32 r^{2}-45 r -18} \] Which for the root \(r = 0\) becomes \[ b_{3}={\frac {1}{18}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {1}{2 r^{2}+r -1}\) | \(1\) |
| \(b_{2}\) | \(\frac {1}{4 r^{4}+12 r^{3}+7 r^{2}-3 r -2}\) | \(-{\frac {1}{2}}\) |
| \(b_{3}\) | \(\frac {4 r^{5}+12 r^{4}+7 r^{3}-3 r^{2}-2 r -1}{8 r^{6}+60 r^{5}+158 r^{4}+165 r^{3}+32 r^{2}-45 r -18}\) | \(\frac {1}{18}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {-8 r^{5}-44 r^{4}-102 r^{3}-127 r^{2}-79 r -17}{16 r^{8}+224 r^{7}+1256 r^{6}+3584 r^{5}+5369 r^{4}+3626 r^{3}+19 r^{2}-1134 r -360} \] Which for the root \(r = 0\) becomes \[ b_{4}={\frac {17}{360}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {1}{2 r^{2}+r -1}\) | \(1\) |
| \(b_{2}\) | \(\frac {1}{4 r^{4}+12 r^{3}+7 r^{2}-3 r -2}\) | \(-{\frac {1}{2}}\) |
| \(b_{3}\) | \(\frac {4 r^{5}+12 r^{4}+7 r^{3}-3 r^{2}-2 r -1}{8 r^{6}+60 r^{5}+158 r^{4}+165 r^{3}+32 r^{2}-45 r -18}\) | \(\frac {1}{18}\) |
| \(b_{4}\) | \(\frac {-8 r^{5}-44 r^{4}-102 r^{3}-127 r^{2}-79 r -17}{16 r^{8}+224 r^{7}+1256 r^{6}+3584 r^{5}+5369 r^{4}+3626 r^{3}+19 r^{2}-1134 r -360}\) | \(\frac {17}{360}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {12 r^{5}+96 r^{4}+365 r^{3}+774 r^{2}+853 r +377}{32 r^{10}+720 r^{9}+6880 r^{8}+36360 r^{7}+115626 r^{6}+223965 r^{5}+249595 r^{4}+124965 r^{3}-19333 r^{2}-45810 r -12600} \] Which for the root \(r = 0\) becomes \[ b_{5}=-{\frac {377}{12600}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {1}{2 r^{2}+r -1}\) | \(1\) |
| \(b_{2}\) | \(\frac {1}{4 r^{4}+12 r^{3}+7 r^{2}-3 r -2}\) | \(-{\frac {1}{2}}\) |
| \(b_{3}\) | \(\frac {4 r^{5}+12 r^{4}+7 r^{3}-3 r^{2}-2 r -1}{8 r^{6}+60 r^{5}+158 r^{4}+165 r^{3}+32 r^{2}-45 r -18}\) | \(\frac {1}{18}\) |
| \(b_{4}\) | \(\frac {-8 r^{5}-44 r^{4}-102 r^{3}-127 r^{2}-79 r -17}{16 r^{8}+224 r^{7}+1256 r^{6}+3584 r^{5}+5369 r^{4}+3626 r^{3}+19 r^{2}-1134 r -360}\) | \(\frac {17}{360}\) |
| \(b_{5}\) | \(\frac {12 r^{5}+96 r^{4}+365 r^{3}+774 r^{2}+853 r +377}{32 r^{10}+720 r^{9}+6880 r^{8}+36360 r^{7}+115626 r^{6}+223965 r^{5}+249595 r^{4}+124965 r^{3}-19333 r^{2}-45810 r -12600}\) | \(-{\frac {377}{12600}}\) |
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 {x^{2}}{2}+\frac {x^{3}}{18}+\frac {17 x^{4}}{360}-\frac {377 x^{5}}{12600}+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 {3}{2}} \left (1-\frac {x}{5}+\frac {x^{2}}{70}+\frac {52 x^{3}}{945}-\frac {1049 x^{4}}{83160}+\frac {5207 x^{5}}{5405400}+O\left (x^{6}\right )\right ) + c_{2} \left (1+x -\frac {x^{2}}{2}+\frac {x^{3}}{18}+\frac {17 x^{4}}{360}-\frac {377 x^{5}}{12600}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {3}{2}} \left (1-\frac {x}{5}+\frac {x^{2}}{70}+\frac {52 x^{3}}{945}-\frac {1049 x^{4}}{83160}+\frac {5207 x^{5}}{5405400}+O\left (x^{6}\right )\right )+c_{2} \left (1+x -\frac {x^{2}}{2}+\frac {x^{3}}{18}+\frac {17 x^{4}}{360}-\frac {377 x^{5}}{12600}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {3}{2}} \left (1-\frac {x}{5}+\frac {x^{2}}{70}+\frac {52 x^{3}}{945}-\frac {1049 x^{4}}{83160}+\frac {5207 x^{5}}{5405400}+O\left (x^{6}\right )\right )+c_{2} \left (1+x -\frac {x^{2}}{2}+\frac {x^{3}}{18}+\frac {17 x^{4}}{360}-\frac {377 x^{5}}{12600}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {3}{2}} \left (1-\frac {x}{5}+\frac {x^{2}}{70}+\frac {52 x^{3}}{945}-\frac {1049 x^{4}}{83160}+\frac {5207 x^{5}}{5405400}+O\left (x^{6}\right )\right )+c_{2} \left (1+x -\frac {x^{2}}{2}+\frac {x^{3}}{18}+\frac {17 x^{4}}{360}-\frac {377 x^{5}}{12600}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 x y^{\prime \prime }+\left (-x^{3}-1\right ) y^{\prime }+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 {y}{2 x}+\frac {\left (x^{3}+1\right ) y^{\prime }}{2 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 (x^{3}+1\right ) y^{\prime }}{2 x}+\frac {y}{2 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 {x^{3}+1}{2 x}, P_{3}\left (x \right )=\frac {1}{2 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 {1}{2} \\ {} & \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} \\ {} & {} & 2 x y^{\prime \prime }+\left (-x^{3}-1\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 =0..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 \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 (-3+2 r \right ) x^{-1+r}+\left (a_{1} \left (1+r \right ) \left (-1+2 r \right )+a_{0}\right ) x^{r}+\left (a_{2} \left (2+r \right ) \left (1+2 r \right )+a_{1}\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (2 k -1+2 r \right )+a_{k}-a_{k -2} \left (k -2+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 (-3+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {3}{2}\right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} x \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [a_{1} \left (1+r \right ) \left (-1+2 r \right )+a_{0}=0, a_{2} \left (2+r \right ) \left (1+2 r \right )+a_{1}=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=-\frac {a_{0}}{2 r^{2}+r -1}, a_{2}=\frac {a_{0}}{4 r^{4}+12 r^{3}+7 r^{2}-3 r -2}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +1} \left (k +1+r \right ) \left (2 k -1+2 r \right )+a_{k}-a_{k -2} \left (k -2+r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +3} \left (k +3+r \right ) \left (2 k +3+2 r \right )+a_{k +2}-a_{k} \left (k +r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=\frac {k a_{k}+r a_{k}-a_{k +2}}{\left (k +3+r \right ) \left (2 k +3+2 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +3}=\frac {k a_{k}-a_{k +2}}{\left (k +3\right ) \left (2 k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +3}=\frac {k a_{k}-a_{k +2}}{\left (k +3\right ) \left (2 k +3\right )}, a_{1}=a_{0}, a_{2}=-\frac {a_{0}}{2}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {3}{2} \\ {} & {} & a_{k +3}=\frac {k a_{k}+\frac {3}{2} a_{k}-a_{k +2}}{\left (k +\frac {9}{2}\right ) \left (2 k +6\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {3}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {3}{2}}, a_{k +3}=\frac {k a_{k}+\frac {3}{2} a_{k}-a_{k +2}}{\left (k +\frac {9}{2}\right ) \left (2 k +6\right )}, a_{1}=-\frac {a_{0}}{5}, a_{2}=\frac {a_{0}}{70}\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 {3}{2}}\right ), a_{k +3}=\frac {k a_{k}-a_{k +2}}{\left (k +3\right ) \left (2 k +3\right )}, a_{1}=a_{0}, a_{2}=-\frac {a_{0}}{2}, b_{k +3}=\frac {k b_{k}+\frac {3}{2} b_{k}-b_{k +2}}{\left (k +\frac {9}{2}\right ) \left (2 k +6\right )}, b_{1}=-\frac {b_{0}}{5}, b_{2}=\frac {b_{0}}{70}\right ] \end {array} \]
Maple trace
`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 <- No Liouvillian solutions exists -> 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 trying a solution in terms of MeijerG functions -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) trying a symmetry of the form [xi=0, eta=F(x)] trying differential order: 2; exact nonlinear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients trying 2nd order, integrating factor of the form mu(x,y) -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Whittaker -> 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 trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients trying to convert to an ODE of Bessel type trying to convert to an ODE of Bessel type -> trying reduction of order to Riccati trying Riccati sub-methods: -> trying a symmetry pattern of the form [F(x)*G(y), 0] -> trying a symmetry pattern of the form [0, F(x)*G(y)] -> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] --- Trying Lie symmetry methods, 2nd order --- `, `-> Computing symmetries using: way = 3`[0, y]
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 44
Order:=6; dsolve(2*x*diff(y(x),x$2)-(1+x^3)*diff(y(x),x)+y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{\frac {3}{2}} \left (1-\frac {1}{5} x +\frac {1}{70} x^{2}+\frac {52}{945} x^{3}-\frac {1049}{83160} x^{4}+\frac {5207}{5405400} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (1+x -\frac {1}{2} x^{2}+\frac {1}{18} x^{3}+\frac {17}{360} x^{4}-\frac {377}{12600} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.003 (sec). Leaf size: 81
AsymptoticDSolveValue[2*x*y''[x]-(1+x^3)*y'[x]+y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_2 \left (-\frac {377 x^5}{12600}+\frac {17 x^4}{360}+\frac {x^3}{18}-\frac {x^2}{2}+x+1\right )+c_1 \left (\frac {5207 x^5}{5405400}-\frac {1049 x^4}{83160}+\frac {52 x^3}{945}+\frac {x^2}{70}-\frac {x}{5}+1\right ) x^{3/2} \]