Internal problem ID [414]
Internal file name [OUTPUT/414_Sunday_June_05_2022_01_40_52_AM_2426009/index.tex]
Book: Differential equations and linear algebra, 4th ed., Edwards and Penney
Section: Chapter 11 Power series methods. Section 11.1 Introduction and Review of power series.
Page 615
Problem number: problem 23.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Complex roots"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x^{2} y^{\prime \prime }+y^{\prime } x^{2}+y=0} \] 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. \[ x^{2} y^{\prime \prime }+y^{\prime } x^{2}+y = 0 \] 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) &= 1\\ q(x) &= \frac {1}{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 \[ x^{2} y^{\prime \prime }+y^{\prime } x^{2}+y = 0 \] 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 )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right ) x^{2}+\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 }}x^{1+n +r} a_{n} \left (n +r \right )\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 \(n +r\) 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 }}a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )+a_{0} x^{r} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )+x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r^{2}-r +1\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r^{2}-r +1 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= \frac {1}{2}+\frac {i \sqrt {3}}{2}\\ r_2 &= \frac {1}{2}-\frac {i \sqrt {3}}{2} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r^{2}-r +1\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [\frac {1}{2}+\frac {i \sqrt {3}}{2}, \frac {1}{2}-\frac {i \sqrt {3}}{2}\right ]\).
Since the roots are complex conjugates, then two linearly independent solutions can be constructed using \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}{2}+\frac {i \sqrt {3}}{2}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +\frac {1}{2}-\frac {i \sqrt {3}}{2}} \end {align*}
\(y_{1}\left (x \right )\) is found first. 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 )+a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -1} \left (n +r -1\right )}{n^{2}+2 n r +r^{2}-n -r +1}\tag {4} \] Which for the root \(r = \frac {1}{2}+\frac {i \sqrt {3}}{2}\) becomes \[ a_{n} = -\frac {a_{n -1} \left (i \sqrt {3}+2 n -1\right )}{2 n \left (i \sqrt {3}+n \right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = \frac {1}{2}+\frac {i \sqrt {3}}{2}\) 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 \[ a_{1}=-\frac {r}{r^{2}+r +1} \] Which for the root \(r = \frac {1}{2}+\frac {i \sqrt {3}}{2}\) becomes \[ a_{1}=-{\frac {1}{2}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r}{r^{2}+r +1}\) | \(-{\frac {1}{2}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {r \left (1+r \right )}{\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right )} \] Which for the root \(r = \frac {1}{2}+\frac {i \sqrt {3}}{2}\) becomes \[ a_{2}=\frac {i \sqrt {3}+3}{16+8 i \sqrt {3}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r}{r^{2}+r +1}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {r \left (1+r \right )}{\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right )}\) | \(\frac {i \sqrt {3}+3}{16+8 i \sqrt {3}}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {r \left (1+r \right ) \left (2+r \right )}{\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right )} \] Which for the root \(r = \frac {1}{2}+\frac {i \sqrt {3}}{2}\) becomes \[ a_{3}=\frac {-i \sqrt {3}-5}{48 i \sqrt {3}+96} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r}{r^{2}+r +1}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {r \left (1+r \right )}{\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right )}\) | \(\frac {i \sqrt {3}+3}{16+8 i \sqrt {3}}\) |
| \(a_{3}\) | \(-\frac {r \left (1+r \right ) \left (2+r \right )}{\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right )}\) | \(\frac {-i \sqrt {3}-5}{48 i \sqrt {3}+96}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right ) \left (r^{2}+7 r +13\right )} \] Which for the root \(r = \frac {1}{2}+\frac {i \sqrt {3}}{2}\) becomes \[ a_{4}=\frac {\left (i \sqrt {3}+5\right ) \left (i \sqrt {3}+7\right )}{384 \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r}{r^{2}+r +1}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {r \left (1+r \right )}{\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right )}\) | \(\frac {i \sqrt {3}+3}{16+8 i \sqrt {3}}\) |
| \(a_{3}\) | \(-\frac {r \left (1+r \right ) \left (2+r \right )}{\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right )}\) | \(\frac {-i \sqrt {3}-5}{48 i \sqrt {3}+96}\) |
| \(a_{4}\) | \(\frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right ) \left (r^{2}+7 r +13\right )}\) | \(\frac {\left (i \sqrt {3}+5\right ) \left (i \sqrt {3}+7\right )}{384 \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}{\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right ) \left (r^{2}+7 r +13\right ) \left (r^{2}+9 r +21\right )} \] Which for the root \(r = \frac {1}{2}+\frac {i \sqrt {3}}{2}\) becomes \[ a_{5}=-\frac {\left (i \sqrt {3}+7\right ) \left (i \sqrt {3}+9\right )}{3840 \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r}{r^{2}+r +1}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {r \left (1+r \right )}{\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right )}\) | \(\frac {i \sqrt {3}+3}{16+8 i \sqrt {3}}\) |
| \(a_{3}\) | \(-\frac {r \left (1+r \right ) \left (2+r \right )}{\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right )}\) | \(\frac {-i \sqrt {3}-5}{48 i \sqrt {3}+96}\) |
| \(a_{4}\) | \(\frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right ) \left (r^{2}+7 r +13\right )}\) | \(\frac {\left (i \sqrt {3}+5\right ) \left (i \sqrt {3}+7\right )}{384 \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )}\) |
| \(a_{5}\) | \(-\frac {r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}{\left (r^{2}+r +1\right ) \left (r^{2}+3 r +3\right ) \left (r^{2}+5 r +7\right ) \left (r^{2}+7 r +13\right ) \left (r^{2}+9 r +21\right )}\) | \(-\frac {\left (i \sqrt {3}+7\right ) \left (i \sqrt {3}+9\right )}{3840 \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin{align*} y_{1}\left (x \right )&= x^{\frac {1}{2}+\frac {i \sqrt {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 {1}{2}+\frac {i \sqrt {3}}{2}} \left (1-\frac {x}{2}+\frac {\left (i \sqrt {3}+3\right ) x^{2}}{16+8 i \sqrt {3}}+\frac {\left (-i \sqrt {3}-5\right ) x^{3}}{48 i \sqrt {3}+96}+\frac {\left (i \sqrt {3}+5\right ) \left (i \sqrt {3}+7\right ) x^{4}}{384 \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )}-\frac {\left (i \sqrt {3}+7\right ) \left (i \sqrt {3}+9\right ) x^{5}}{3840 \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )}+O\left (x^{6}\right )\right ) \\ \end{align*} The second solution \(y_{2}\left (x \right )\) is found by taking the complex conjugate of \(y_{1}\left (x \right )\) which gives \[ y_{2}\left (x \right )= x^{\frac {1}{2}-\frac {i \sqrt {3}}{2}} \left (1-\frac {x}{2}+\frac {\left (-i \sqrt {3}+3\right ) x^{2}}{16-8 i \sqrt {3}}+\frac {\left (i \sqrt {3}-5\right ) x^{3}}{-48 i \sqrt {3}+96}+\frac {\left (-i \sqrt {3}+5\right ) \left (-i \sqrt {3}+7\right ) x^{4}}{384 \left (-i \sqrt {3}+4\right ) \left (2-i \sqrt {3}\right )}-\frac {\left (-i \sqrt {3}+7\right ) \left (-i \sqrt {3}+9\right ) x^{5}}{3840 \left (-i \sqrt {3}+4\right ) \left (2-i \sqrt {3}\right )}+O\left (x^{6}\right )\right ) \] 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}{2}+\frac {i \sqrt {3}}{2}} \left (1-\frac {x}{2}+\frac {\left (i \sqrt {3}+3\right ) x^{2}}{16+8 i \sqrt {3}}+\frac {\left (-i \sqrt {3}-5\right ) x^{3}}{48 i \sqrt {3}+96}+\frac {\left (i \sqrt {3}+5\right ) \left (i \sqrt {3}+7\right ) x^{4}}{384 \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )}-\frac {\left (i \sqrt {3}+7\right ) \left (i \sqrt {3}+9\right ) x^{5}}{3840 \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )}+O\left (x^{6}\right )\right ) + c_{2} x^{\frac {1}{2}-\frac {i \sqrt {3}}{2}} \left (1-\frac {x}{2}+\frac {\left (-i \sqrt {3}+3\right ) x^{2}}{16-8 i \sqrt {3}}+\frac {\left (i \sqrt {3}-5\right ) x^{3}}{-48 i \sqrt {3}+96}+\frac {\left (-i \sqrt {3}+5\right ) \left (-i \sqrt {3}+7\right ) x^{4}}{384 \left (-i \sqrt {3}+4\right ) \left (2-i \sqrt {3}\right )}-\frac {\left (-i \sqrt {3}+7\right ) \left (-i \sqrt {3}+9\right ) x^{5}}{3840 \left (-i \sqrt {3}+4\right ) \left (2-i \sqrt {3}\right )}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {1}{2}+\frac {i \sqrt {3}}{2}} \left (1-\frac {x}{2}+\frac {\left (i \sqrt {3}+3\right ) x^{2}}{16+8 i \sqrt {3}}+\frac {\left (-i \sqrt {3}-5\right ) x^{3}}{48 i \sqrt {3}+96}+\frac {\left (i \sqrt {3}+5\right ) \left (i \sqrt {3}+7\right ) x^{4}}{384 \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )}-\frac {\left (i \sqrt {3}+7\right ) \left (i \sqrt {3}+9\right ) x^{5}}{3840 \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {1}{2}-\frac {i \sqrt {3}}{2}} \left (1-\frac {x}{2}+\frac {\left (-i \sqrt {3}+3\right ) x^{2}}{16-8 i \sqrt {3}}+\frac {\left (i \sqrt {3}-5\right ) x^{3}}{-48 i \sqrt {3}+96}+\frac {\left (-i \sqrt {3}+5\right ) \left (-i \sqrt {3}+7\right ) x^{4}}{384 \left (-i \sqrt {3}+4\right ) \left (2-i \sqrt {3}\right )}-\frac {\left (-i \sqrt {3}+7\right ) \left (-i \sqrt {3}+9\right ) x^{5}}{3840 \left (-i \sqrt {3}+4\right ) \left (2-i \sqrt {3}\right )}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {1}{2}+\frac {i \sqrt {3}}{2}} \left (1-\frac {x}{2}+\frac {\left (i \sqrt {3}+3\right ) x^{2}}{16+8 i \sqrt {3}}+\frac {\left (-i \sqrt {3}-5\right ) x^{3}}{48 i \sqrt {3}+96}+\frac {\left (i \sqrt {3}+5\right ) \left (i \sqrt {3}+7\right ) x^{4}}{384 \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )}-\frac {\left (i \sqrt {3}+7\right ) \left (i \sqrt {3}+9\right ) x^{5}}{3840 \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {1}{2}-\frac {i \sqrt {3}}{2}} \left (1-\frac {x}{2}+\frac {\left (-i \sqrt {3}+3\right ) x^{2}}{16-8 i \sqrt {3}}+\frac {\left (i \sqrt {3}-5\right ) x^{3}}{-48 i \sqrt {3}+96}+\frac {\left (-i \sqrt {3}+5\right ) \left (-i \sqrt {3}+7\right ) x^{4}}{384 \left (-i \sqrt {3}+4\right ) \left (2-i \sqrt {3}\right )}-\frac {\left (-i \sqrt {3}+7\right ) \left (-i \sqrt {3}+9\right ) x^{5}}{3840 \left (-i \sqrt {3}+4\right ) \left (2-i \sqrt {3}\right )}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {1}{2}+\frac {i \sqrt {3}}{2}} \left (1-\frac {x}{2}+\frac {\left (i \sqrt {3}+3\right ) x^{2}}{16+8 i \sqrt {3}}+\frac {\left (-i \sqrt {3}-5\right ) x^{3}}{48 i \sqrt {3}+96}+\frac {\left (i \sqrt {3}+5\right ) \left (i \sqrt {3}+7\right ) x^{4}}{384 \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )}-\frac {\left (i \sqrt {3}+7\right ) \left (i \sqrt {3}+9\right ) x^{5}}{3840 \left (i \sqrt {3}+4\right ) \left (2+i \sqrt {3}\right )}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {1}{2}-\frac {i \sqrt {3}}{2}} \left (1-\frac {x}{2}+\frac {\left (-i \sqrt {3}+3\right ) x^{2}}{16-8 i \sqrt {3}}+\frac {\left (i \sqrt {3}-5\right ) x^{3}}{-48 i \sqrt {3}+96}+\frac {\left (-i \sqrt {3}+5\right ) \left (-i \sqrt {3}+7\right ) x^{4}}{384 \left (-i \sqrt {3}+4\right ) \left (2-i \sqrt {3}\right )}-\frac {\left (-i \sqrt {3}+7\right ) \left (-i \sqrt {3}+9\right ) x^{5}}{3840 \left (-i \sqrt {3}+4\right ) \left (2-i \sqrt {3}\right )}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime \prime }+y^{\prime } x^{2}+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 }=-y^{\prime }-\frac {y}{x^{2}} \\ \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 }+y^{\prime }+\frac {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 )=1, P_{3}\left (x \right )=\frac {1}{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}}}=1 \\ {} & \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 }+y^{\prime } x^{2}+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^{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 (r^{2}-r +1\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (k^{2}+2 k r +r^{2}-k -r +1\right )+a_{k -1} \left (k -1+r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r^{2}-r +1=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}, \frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k^{2}+\left (2 r -1\right ) k +r^{2}-r +1\right ) a_{k}+a_{k -1} \left (k -1+r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (\left (k +1\right )^{2}+\left (2 r -1\right ) \left (k +1\right )+r^{2}-r +1\right ) a_{k +1}+a_{k} \left (k +r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (k +r \right )}{k^{2}+2 k r +r^{2}+k +r +1} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2} \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (k +\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )}{k^{2}+2 k \left (\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )+\left (\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}+k +\frac {3}{2}-\frac {\mathrm {I} \sqrt {3}}{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}}, a_{k +1}=-\frac {a_{k} \left (k +\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )}{k^{2}+2 k \left (\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )+\left (\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}+k +\frac {3}{2}-\frac {\mathrm {I} \sqrt {3}}{2}}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2} \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (k +\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )}{k^{2}+2 k \left (\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )+\left (\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}+k +\frac {3}{2}+\frac {\mathrm {I} \sqrt {3}}{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}}, a_{k +1}=-\frac {a_{k} \left (k +\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )}{k^{2}+2 k \left (\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )+\left (\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}+k +\frac {3}{2}+\frac {\mathrm {I} \sqrt {3}}{2}}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}}\right ), a_{k +1}=-\frac {a_{k} \left (k +\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )}{k^{2}+2 k \left (\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )+\left (\frac {1}{2}-\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}+k +\frac {3}{2}-\frac {\mathrm {I} \sqrt {3}}{2}}, b_{k +1}=-\frac {b_{k} \left (k +\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )}{k^{2}+2 k \left (\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )+\left (\frac {1}{2}+\frac {\mathrm {I} \sqrt {3}}{2}\right )^{2}+k +\frac {3}{2}+\frac {\mathrm {I} \sqrt {3}}{2}}\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 <- Bessel successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 907
Order:=6; dsolve(x^2*diff(y(x),x$2)+x^2*diff(y(x),x)+y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \sqrt {x}\, \left (c_{2} x^{\frac {i \sqrt {3}}{2}} \left (1-\frac {1}{2} x +\frac {i \sqrt {3}+3}{8 i \sqrt {3}+16} x^{2}+\frac {-i \sqrt {3}-5}{48 i \sqrt {3}+96} x^{3}+\frac {1}{384} \frac {\left (i \sqrt {3}+5\right ) \left (i \sqrt {3}+7\right )}{\left (i \sqrt {3}+4\right ) \left (i \sqrt {3}+2\right )} x^{4}-\frac {1}{3840} \frac {\left (i \sqrt {3}+7\right ) \left (i \sqrt {3}+9\right )}{\left (i \sqrt {3}+4\right ) \left (i \sqrt {3}+2\right )} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{1} x^{-\frac {i \sqrt {3}}{2}} \left (1-\frac {1}{2} x +\frac {\sqrt {3}+3 i}{8 \sqrt {3}+16 i} x^{2}+\frac {-\sqrt {3}-5 i}{48 \sqrt {3}+96 i} x^{3}+\frac {3 i \sqrt {3}-8}{576 i \sqrt {3}-480} x^{4}-\frac {1}{3840} \frac {\left (\sqrt {3}+7 i\right ) \left (\sqrt {3}+9 i\right )}{\left (\sqrt {3}+4 i\right ) \left (\sqrt {3}+2 i\right )} x^{5}+\operatorname {O}\left (x^{6}\right )\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 886
AsymptoticDSolveValue[x^2*y''[x]+x^2*y'[x]+y[x]==0,y[x],{x,0,5}]
\[ y(x)\to \left (\frac {(-1)^{2/3} \left (1-(-1)^{2/3}\right ) \left (2-(-1)^{2/3}\right ) \left (3-(-1)^{2/3}\right ) \left (4-(-1)^{2/3}\right ) x^5}{\left (1-(-1)^{2/3} \left (1-(-1)^{2/3}\right )\right ) \left (1+\left (1-(-1)^{2/3}\right ) \left (2-(-1)^{2/3}\right )\right ) \left (1+\left (2-(-1)^{2/3}\right ) \left (3-(-1)^{2/3}\right )\right ) \left (1+\left (3-(-1)^{2/3}\right ) \left (4-(-1)^{2/3}\right )\right ) \left (1+\left (4-(-1)^{2/3}\right ) \left (5-(-1)^{2/3}\right )\right )}-\frac {(-1)^{2/3} \left (1-(-1)^{2/3}\right ) \left (2-(-1)^{2/3}\right ) \left (3-(-1)^{2/3}\right ) x^4}{\left (1-(-1)^{2/3} \left (1-(-1)^{2/3}\right )\right ) \left (1+\left (1-(-1)^{2/3}\right ) \left (2-(-1)^{2/3}\right )\right ) \left (1+\left (2-(-1)^{2/3}\right ) \left (3-(-1)^{2/3}\right )\right ) \left (1+\left (3-(-1)^{2/3}\right ) \left (4-(-1)^{2/3}\right )\right )}+\frac {(-1)^{2/3} \left (1-(-1)^{2/3}\right ) \left (2-(-1)^{2/3}\right ) x^3}{\left (1-(-1)^{2/3} \left (1-(-1)^{2/3}\right )\right ) \left (1+\left (1-(-1)^{2/3}\right ) \left (2-(-1)^{2/3}\right )\right ) \left (1+\left (2-(-1)^{2/3}\right ) \left (3-(-1)^{2/3}\right )\right )}-\frac {(-1)^{2/3} \left (1-(-1)^{2/3}\right ) x^2}{\left (1-(-1)^{2/3} \left (1-(-1)^{2/3}\right )\right ) \left (1+\left (1-(-1)^{2/3}\right ) \left (2-(-1)^{2/3}\right )\right )}+\frac {(-1)^{2/3} x}{1-(-1)^{2/3} \left (1-(-1)^{2/3}\right )}+1\right ) c_1 x^{-(-1)^{2/3}}+\left (-\frac {\sqrt [3]{-1} \left (1+\sqrt [3]{-1}\right ) \left (2+\sqrt [3]{-1}\right ) \left (3+\sqrt [3]{-1}\right ) \left (4+\sqrt [3]{-1}\right ) x^5}{\left (1+\sqrt [3]{-1} \left (1+\sqrt [3]{-1}\right )\right ) \left (1+\left (1+\sqrt [3]{-1}\right ) \left (2+\sqrt [3]{-1}\right )\right ) \left (1+\left (2+\sqrt [3]{-1}\right ) \left (3+\sqrt [3]{-1}\right )\right ) \left (1+\left (3+\sqrt [3]{-1}\right ) \left (4+\sqrt [3]{-1}\right )\right ) \left (1+\left (4+\sqrt [3]{-1}\right ) \left (5+\sqrt [3]{-1}\right )\right )}+\frac {\sqrt [3]{-1} \left (1+\sqrt [3]{-1}\right ) \left (2+\sqrt [3]{-1}\right ) \left (3+\sqrt [3]{-1}\right ) x^4}{\left (1+\sqrt [3]{-1} \left (1+\sqrt [3]{-1}\right )\right ) \left (1+\left (1+\sqrt [3]{-1}\right ) \left (2+\sqrt [3]{-1}\right )\right ) \left (1+\left (2+\sqrt [3]{-1}\right ) \left (3+\sqrt [3]{-1}\right )\right ) \left (1+\left (3+\sqrt [3]{-1}\right ) \left (4+\sqrt [3]{-1}\right )\right )}-\frac {\sqrt [3]{-1} \left (1+\sqrt [3]{-1}\right ) \left (2+\sqrt [3]{-1}\right ) x^3}{\left (1+\sqrt [3]{-1} \left (1+\sqrt [3]{-1}\right )\right ) \left (1+\left (1+\sqrt [3]{-1}\right ) \left (2+\sqrt [3]{-1}\right )\right ) \left (1+\left (2+\sqrt [3]{-1}\right ) \left (3+\sqrt [3]{-1}\right )\right )}+\frac {\sqrt [3]{-1} \left (1+\sqrt [3]{-1}\right ) x^2}{\left (1+\sqrt [3]{-1} \left (1+\sqrt [3]{-1}\right )\right ) \left (1+\left (1+\sqrt [3]{-1}\right ) \left (2+\sqrt [3]{-1}\right )\right )}-\frac {\sqrt [3]{-1} x}{1+\sqrt [3]{-1} \left (1+\sqrt [3]{-1}\right )}+1\right ) c_2 x^{\sqrt [3]{-1}} \]