Internal problem ID [4696]
Internal file name [OUTPUT/4189_Sunday_June_05_2022_12_39_07_PM_53153814/index.tex]
Book: A treatise on ordinary and partial differential equations by William Woolsey Johnson.
1913
Section: Chapter VII, Solutions in series. Examples XIV. page 177
Problem number: 1.
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]]
\[ \boxed {x y^{\prime \prime }+\left (x +n \right ) y^{\prime }+\left (n +1\right ) y=0} \] 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. \[ x y^{\prime \prime }+\left (x +n \right ) y^{\prime }+\left (n +1\right ) y = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} y^{\prime \prime }+p(x) y^{\prime } + q(x) y &=0 \end {align*}
Where \begin {align*} p(x) &= \frac {x +n}{x}\\ q(x) &= \frac {n +1}{x}\\ \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 \[ x y^{\prime \prime }+\left (x +n \right ) y^{\prime }+\left (n +1\right ) y = 0 \] 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 \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 +n \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (n +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 }}x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{n +r -1} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +1\right ) 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 }}x^{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 -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +1\right ) a_{n} x^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +1\right ) 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 }}x^{n +r -1} 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 -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}n \,x^{n +r -1} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +1\right ) x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+n \,x^{n +r -1} a_{n} \left (n +r \right ) = 0 \] When \(n = 0\) the above becomes \[ x^{-1+r} a_{0} r \left (-1+r \right )+n \,x^{-1+r} a_{0} r = 0 \] Or \[ \left (x^{-1+r} r \left (-1+r \right )+n \,x^{-1+r} r \right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ r \,x^{-1+r} \left (-1+r +n \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r \left (-1+r +n \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 0\\ r_2 &= -n +1 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ r \,x^{-1+r} \left (-1+r +n \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([0, -n +1]\).
Assuming the roots differ by non-integer Since \(r_1 - r_2 = n -1\) 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}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -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 )+n a_{n} \left (n +r \right )+a_{n -1} \left (n +1\right ) = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -1} \left (n +n +r \right )}{n n +n r +n^{2}+2 n r +r^{2}-n -r}\tag {4} \] Which for the root \(r = 0\) becomes \[ a_{n} = -\frac {a_{n -1} \left (n +n \right )}{n \left (n +n -1\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 0\) 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 \[ a_{1}=\frac {-1-n -r}{\left (r +1\right ) \left (r +n \right )} \] Which for the root \(r = 0\) becomes \[ a_{1}=\frac {-n -1}{n} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-1-n -r}{\left (r +1\right ) \left (r +n \right )}\) | \(\frac {-n -1}{n}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {n +2+r}{\left (r +1\right ) \left (r +n \right ) \left (r +2\right )} \] Which for the root \(r = 0\) becomes \[ a_{2}=\frac {n +2}{2 n} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-1-n -r}{\left (r +1\right ) \left (r +n \right )}\) | \(\frac {-n -1}{n}\) |
| \(a_{2}\) | \(\frac {n +2+r}{\left (r +1\right ) \left (r +n \right ) \left (r +2\right )}\) | \(\frac {n +2}{2 n}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-n -3-r}{\left (r +3\right ) \left (r +2\right ) \left (r +1\right ) \left (r +n \right )} \] Which for the root \(r = 0\) becomes \[ a_{3}=\frac {-n -3}{6 n} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-1-n -r}{\left (r +1\right ) \left (r +n \right )}\) | \(\frac {-n -1}{n}\) |
| \(a_{2}\) | \(\frac {n +2+r}{\left (r +1\right ) \left (r +n \right ) \left (r +2\right )}\) | \(\frac {n +2}{2 n}\) |
| \(a_{3}\) | \(\frac {-n -3-r}{\left (r +3\right ) \left (r +2\right ) \left (r +1\right ) \left (r +n \right )}\) | \(\frac {-n -3}{6 n}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {n +4+r}{\left (r +3\right ) \left (r +2\right ) \left (r +1\right ) \left (r +n \right ) \left (r +4\right )} \] Which for the root \(r = 0\) becomes \[ a_{4}=\frac {n +4}{24 n} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-1-n -r}{\left (r +1\right ) \left (r +n \right )}\) | \(\frac {-n -1}{n}\) |
| \(a_{2}\) | \(\frac {n +2+r}{\left (r +1\right ) \left (r +n \right ) \left (r +2\right )}\) | \(\frac {n +2}{2 n}\) |
| \(a_{3}\) | \(\frac {-n -3-r}{\left (r +3\right ) \left (r +2\right ) \left (r +1\right ) \left (r +n \right )}\) | \(\frac {-n -3}{6 n}\) |
| \(a_{4}\) | \(\frac {n +4+r}{\left (r +3\right ) \left (r +2\right ) \left (r +1\right ) \left (r +n \right ) \left (r +4\right )}\) | \(\frac {n +4}{24 n}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-n -5-r}{\left (r +5\right ) \left (r +4\right ) \left (r +3\right ) \left (r +2\right ) \left (r +1\right ) \left (r +n \right )} \] Which for the root \(r = 0\) becomes \[ a_{5}=\frac {-n -5}{120 n} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-1-n -r}{\left (r +1\right ) \left (r +n \right )}\) | \(\frac {-n -1}{n}\) |
| \(a_{2}\) | \(\frac {n +2+r}{\left (r +1\right ) \left (r +n \right ) \left (r +2\right )}\) | \(\frac {n +2}{2 n}\) |
| \(a_{3}\) | \(\frac {-n -3-r}{\left (r +3\right ) \left (r +2\right ) \left (r +1\right ) \left (r +n \right )}\) | \(\frac {-n -3}{6 n}\) |
| \(a_{4}\) | \(\frac {n +4+r}{\left (r +3\right ) \left (r +2\right ) \left (r +1\right ) \left (r +n \right ) \left (r +4\right )}\) | \(\frac {n +4}{24 n}\) |
| \(a_{5}\) | \(\frac {-n -5-r}{\left (r +5\right ) \left (r +4\right ) \left (r +3\right ) \left (r +2\right ) \left (r +1\right ) \left (r +n \right )}\) | \(\frac {-n -5}{120 n}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= 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 \\ &= 1+\frac {\left (-n -1\right ) x}{n}+\frac {\left (n +2\right ) x^{2}}{2 n}+\frac {\left (-n -3\right ) x^{3}}{6 n}+\frac {\left (n +4\right ) x^{4}}{24 n}+\frac {\left (-n -5\right ) x^{5}}{120 n}+O\left (x^{6}\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 )+n b_{n} \left (n +r \right )+b_{n -1} \left (n +1\right ) = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {b_{n -1} \left (n +n +r \right )}{n n +n r +n^{2}+2 n r +r^{2}-n -r}\tag {4} \] Which for the root \(r = -n +1\) becomes \[ b_{n} = \frac {b_{n -1} \left (n +1\right )}{n \left (n -n -1\right )}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -n +1\) 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 \[ b_{1}=\frac {-1-n -r}{\left (r +1\right ) \left (r +n \right )} \] Which for the root \(r = -n +1\) becomes \[ b_{1}=\frac {2}{-2+n} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-1-n -r}{\left (r +1\right ) \left (r +n \right )}\) | \(\frac {2}{-2+n}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {n +2+r}{\left (r +1\right ) \left (r +n \right ) \left (r +2\right )} \] Which for the root \(r = -n +1\) becomes \[ b_{2}=\frac {3}{\left (-2+n \right ) \left (n -3\right )} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-1-n -r}{\left (r +1\right ) \left (r +n \right )}\) | \(\frac {2}{-2+n}\) |
| \(b_{2}\) | \(\frac {n +2+r}{\left (r +1\right ) \left (r +n \right ) \left (r +2\right )}\) | \(\frac {3}{\left (-2+n \right ) \left (n -3\right )}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {-n -3-r}{\left (r +3\right ) \left (r +2\right ) \left (r +1\right ) \left (r +n \right )} \] Which for the root \(r = -n +1\) becomes \[ b_{3}=\frac {4}{\left (-2+n \right ) \left (n -3\right ) \left (n -4\right )} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-1-n -r}{\left (r +1\right ) \left (r +n \right )}\) | \(\frac {2}{-2+n}\) |
| \(b_{2}\) | \(\frac {n +2+r}{\left (r +1\right ) \left (r +n \right ) \left (r +2\right )}\) | \(\frac {3}{\left (-2+n \right ) \left (n -3\right )}\) |
| \(b_{3}\) | \(\frac {-n -3-r}{\left (r +3\right ) \left (r +2\right ) \left (r +1\right ) \left (r +n \right )}\) | \(\frac {4}{\left (-2+n \right ) \left (n -3\right ) \left (n -4\right )}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {n +4+r}{\left (r +3\right ) \left (r +2\right ) \left (r +1\right ) \left (r +n \right ) \left (r +4\right )} \] Which for the root \(r = -n +1\) becomes \[ b_{4}=\frac {5}{\left (n -4\right ) \left (n -3\right ) \left (-2+n \right ) \left (n -5\right )} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-1-n -r}{\left (r +1\right ) \left (r +n \right )}\) | \(\frac {2}{-2+n}\) |
| \(b_{2}\) | \(\frac {n +2+r}{\left (r +1\right ) \left (r +n \right ) \left (r +2\right )}\) | \(\frac {3}{\left (-2+n \right ) \left (n -3\right )}\) |
| \(b_{3}\) | \(\frac {-n -3-r}{\left (r +3\right ) \left (r +2\right ) \left (r +1\right ) \left (r +n \right )}\) | \(\frac {4}{\left (-2+n \right ) \left (n -3\right ) \left (n -4\right )}\) |
| \(b_{4}\) | \(\frac {n +4+r}{\left (r +3\right ) \left (r +2\right ) \left (r +1\right ) \left (r +n \right ) \left (r +4\right )}\) | \(\frac {5}{\left (n -4\right ) \left (n -3\right ) \left (-2+n \right ) \left (n -5\right )}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {-n -5-r}{\left (r +5\right ) \left (r +4\right ) \left (r +3\right ) \left (r +2\right ) \left (r +1\right ) \left (r +n \right )} \] Which for the root \(r = -n +1\) becomes \[ b_{5}=\frac {6}{\left (n -4\right ) \left (n -3\right ) \left (-2+n \right ) \left (n -5\right ) \left (n -6\right )} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-1-n -r}{\left (r +1\right ) \left (r +n \right )}\) | \(\frac {2}{-2+n}\) |
| \(b_{2}\) | \(\frac {n +2+r}{\left (r +1\right ) \left (r +n \right ) \left (r +2\right )}\) | \(\frac {3}{\left (-2+n \right ) \left (n -3\right )}\) |
| \(b_{3}\) | \(\frac {-n -3-r}{\left (r +3\right ) \left (r +2\right ) \left (r +1\right ) \left (r +n \right )}\) | \(\frac {4}{\left (-2+n \right ) \left (n -3\right ) \left (n -4\right )}\) |
| \(b_{4}\) | \(\frac {n +4+r}{\left (r +3\right ) \left (r +2\right ) \left (r +1\right ) \left (r +n \right ) \left (r +4\right )}\) | \(\frac {5}{\left (n -4\right ) \left (n -3\right ) \left (-2+n \right ) \left (n -5\right )}\) |
| \(b_{5}\) | \(\frac {-n -5-r}{\left (r +5\right ) \left (r +4\right ) \left (r +3\right ) \left (r +2\right ) \left (r +1\right ) \left (r +n \right )}\) | \(\frac {6}{\left (n -4\right ) \left (n -3\right ) \left (-2+n \right ) \left (n -5\right ) \left (n -6\right )}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= 1 \left (b_{0}+b_{1} x +b_{2} x^{2}+b_{3} x^{3}+b_{4} x^{4}+b_{5} x^{5}+b_{6} x^{6}\dots \right ) \\ &= x^{-n +1} \left (1+\frac {2 x}{-2+n}+\frac {3 x^{2}}{\left (-2+n \right ) \left (n -3\right )}+\frac {4 x^{3}}{\left (-2+n \right ) \left (n -3\right ) \left (n -4\right )}+\frac {5 x^{4}}{\left (n -4\right ) \left (n -3\right ) \left (-2+n \right ) \left (n -5\right )}+\frac {6 x^{5}}{\left (n -4\right ) \left (n -3\right ) \left (-2+n \right ) \left (n -5\right ) \left (n -6\right )}+O\left (x^{6}\right )\right ) \end {align*}
Therefore the homogeneous solution is \begin{align*} y_h(x) &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= c_{1} \left (1+\frac {\left (-n -1\right ) x}{n}+\frac {\left (n +2\right ) x^{2}}{2 n}+\frac {\left (-n -3\right ) x^{3}}{6 n}+\frac {\left (n +4\right ) x^{4}}{24 n}+\frac {\left (-n -5\right ) x^{5}}{120 n}+O\left (x^{6}\right )\right ) + c_{2} x^{-n +1} \left (1+\frac {2 x}{-2+n}+\frac {3 x^{2}}{\left (-2+n \right ) \left (n -3\right )}+\frac {4 x^{3}}{\left (-2+n \right ) \left (n -3\right ) \left (n -4\right )}+\frac {5 x^{4}}{\left (n -4\right ) \left (n -3\right ) \left (-2+n \right ) \left (n -5\right )}+\frac {6 x^{5}}{\left (n -4\right ) \left (n -3\right ) \left (-2+n \right ) \left (n -5\right ) \left (n -6\right )}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \left (1+\frac {\left (-n -1\right ) x}{n}+\frac {\left (n +2\right ) x^{2}}{2 n}+\frac {\left (-n -3\right ) x^{3}}{6 n}+\frac {\left (n +4\right ) x^{4}}{24 n}+\frac {\left (-n -5\right ) x^{5}}{120 n}+O\left (x^{6}\right )\right )+c_{2} x^{-n +1} \left (1+\frac {2 x}{-2+n}+\frac {3 x^{2}}{\left (-2+n \right ) \left (n -3\right )}+\frac {4 x^{3}}{\left (-2+n \right ) \left (n -3\right ) \left (n -4\right )}+\frac {5 x^{4}}{\left (n -4\right ) \left (n -3\right ) \left (-2+n \right ) \left (n -5\right )}+\frac {6 x^{5}}{\left (n -4\right ) \left (n -3\right ) \left (-2+n \right ) \left (n -5\right ) \left (n -6\right )}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (1+\frac {\left (-n -1\right ) x}{n}+\frac {\left (n +2\right ) x^{2}}{2 n}+\frac {\left (-n -3\right ) x^{3}}{6 n}+\frac {\left (n +4\right ) x^{4}}{24 n}+\frac {\left (-n -5\right ) x^{5}}{120 n}+O\left (x^{6}\right )\right )+c_{2} x^{-n +1} \left (1+\frac {2 x}{-2+n}+\frac {3 x^{2}}{\left (-2+n \right ) \left (n -3\right )}+\frac {4 x^{3}}{\left (-2+n \right ) \left (n -3\right ) \left (n -4\right )}+\frac {5 x^{4}}{\left (n -4\right ) \left (n -3\right ) \left (-2+n \right ) \left (n -5\right )}+\frac {6 x^{5}}{\left (n -4\right ) \left (n -3\right ) \left (-2+n \right ) \left (n -5\right ) \left (n -6\right )}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (1+\frac {\left (-n -1\right ) x}{n}+\frac {\left (n +2\right ) x^{2}}{2 n}+\frac {\left (-n -3\right ) x^{3}}{6 n}+\frac {\left (n +4\right ) x^{4}}{24 n}+\frac {\left (-n -5\right ) x^{5}}{120 n}+O\left (x^{6}\right )\right )+c_{2} x^{-n +1} \left (1+\frac {2 x}{-2+n}+\frac {3 x^{2}}{\left (-2+n \right ) \left (n -3\right )}+\frac {4 x^{3}}{\left (-2+n \right ) \left (n -3\right ) \left (n -4\right )}+\frac {5 x^{4}}{\left (n -4\right ) \left (n -3\right ) \left (-2+n \right ) \left (n -5\right )}+\frac {6 x^{5}}{\left (n -4\right ) \left (n -3\right ) \left (-2+n \right ) \left (n -5\right ) \left (n -6\right )}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}y^{\prime }\right )+\left (x +n \right ) y^{\prime }+\left (n +1\right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=-\frac {\left (n +1\right ) y}{x}-\frac {\left (x +n \right ) y^{\prime }}{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} \\ {} & {} & \frac {d}{d x}y^{\prime }+\frac {\left (x +n \right ) y^{\prime }}{x}+\frac {\left (n +1\right ) y}{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 +n}{x}, P_{3}\left (x \right )=\frac {n +1}{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}}}=n \\ {} & \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} \\ {} & {} & x \left (\frac {d}{d x}y^{\prime }\right )+\left (x +n \right ) y^{\prime }+\left (n +1\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^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & 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 \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot \left (\frac {d}{d x}y^{\prime }\right )=\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 \left (\frac {d}{d x}y^{\prime }\right )=\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 (-1+r +n \right ) x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +r +n \right )+a_{k} \left (k +r +n +1\right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (-1+r +n \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -n +1\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 (k +r +n \right )+a_{k} \left (k +r +n +1\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (k +r +n +1\right )}{\left (k +1+r \right ) \left (k +r +n \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (k +1+n \right )}{\left (k +1\right ) \left (k +n \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +1}=-\frac {a_{k} \left (k +1+n \right )}{\left (k +1\right ) \left (k +n \right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-n +1 \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (k +2\right )}{\left (k +2-n \right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-n +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -n +1}, a_{k +1}=-\frac {a_{k} \left (k +2\right )}{\left (k +2-n \right ) \left (k +1\right )}\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 -n +1}\right ), a_{k +1}=-\frac {a_{k} \left (k +1+n \right )}{\left (k +1\right ) \left (k +n \right )}, b_{k +1}=-\frac {b_{k} \left (k +2\right )}{\left (k +2-n \right ) \left (k +1\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 could result into a too large expression - returning special function form of solution, free of uncomputed <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 248
Order:=6; dsolve(x*diff(y(x),x$2)+(x+n)*diff(y(x),x)+(n+1)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{1-n} \left (1+2 \frac {1}{n -2} x +3 \frac {1}{\left (-3+n \right ) \left (n -2\right )} x^{2}+4 \frac {1}{\left (-4+n \right ) \left (-3+n \right ) \left (n -2\right )} x^{3}+5 \frac {1}{\left (-5+n \right ) \left (-4+n \right ) \left (-3+n \right ) \left (n -2\right )} x^{4}+6 \frac {1}{\left (-6+n \right ) \left (-5+n \right ) \left (-4+n \right ) \left (-3+n \right ) \left (n -2\right )} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\left (1+\frac {-1-n}{n} x +\frac {1}{2} \frac {n +2}{n} x^{2}-\frac {1}{6} \frac {n +3}{n} x^{3}+\frac {1}{24} \frac {n +4}{n} x^{4}-\frac {1}{120} \frac {n +5}{n} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) c_{2} \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 519
AsymptoticDSolveValue[x*y''[x]+(x+n)*y'[x]+(n+1)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_2 \left (\frac {(-n-1) (n+2) (n+3) (n+4) (n+5) x^5}{n (2 n+2) (3 n+6) (4 n+12) (5 n+20)}-\frac {(-n-1) (n+2) (n+3) (n+4) x^4}{n (2 n+2) (3 n+6) (4 n+12)}+\frac {(-n-1) (n+2) (n+3) x^3}{n (2 n+2) (3 n+6)}-\frac {(-n-1) (n+2) x^2}{n (2 n+2)}+\frac {(-n-1) x}{n}+1\right )+c_1 \left (-\frac {720 x^5}{((1-n) (2-n)+n (2-n)) ((2-n) (3-n)+n (3-n)) ((3-n) (4-n)+n (4-n)) ((4-n) (5-n)+n (5-n)) ((5-n) (6-n)+n (6-n))}+\frac {120 x^4}{((1-n) (2-n)+n (2-n)) ((2-n) (3-n)+n (3-n)) ((3-n) (4-n)+n (4-n)) ((4-n) (5-n)+n (5-n))}-\frac {24 x^3}{((1-n) (2-n)+n (2-n)) ((2-n) (3-n)+n (3-n)) ((3-n) (4-n)+n (4-n))}+\frac {6 x^2}{((1-n) (2-n)+n (2-n)) ((2-n) (3-n)+n (3-n))}-\frac {2 x}{(1-n) (2-n)+n (2-n)}+1\right ) x^{1-n} \]