Internal problem ID [5472]
Internal file name [OUTPUT/4720_Sunday_June_05_2022_03_04_01_PM_46599014/index.tex]
Book: A treatise on Differential Equations by A. R. Forsyth. 6th edition. 1929. Macmillan Co. ltd.
New York, reprinted 1956
Section: Chapter VI. Note I. Integration of linear equations in series by the method of Frobenius.
page 243
Problem number: Ex. 6(i), page 257.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Repeated root"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x^{2} \left (1+x \right ) y^{\prime \prime }-\left (2 x +1\right ) \left (y^{\prime } x -y\right )=0} \] 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. \[ \left (x^{3}+x^{2}\right ) y^{\prime \prime }+\left (-2 x^{2}-x \right ) y^{\prime }+\left (2 x +1\right ) y = 0 \] 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 {2 x +1}{x \left (1+x \right )}\\ q(x) &= \frac {2 x +1}{x^{2} \left (1+x \right )}\\ \end {align*}
| \(p(x)=-\frac {2 x +1}{x \left (1+x \right )}\) | |
| singularity | type |
| \(x = -1\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
| \(q(x)=\frac {2 x +1}{x^{2} \left (1+x \right )}\) | |
| singularity | type |
| \(x = -1\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([-1, 0, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} \left (1+x \right ) y^{\prime \prime }+\left (-2 x^{2}-x \right ) y^{\prime }+\left (2 x +1\right ) y = 0 \] 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} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right ) x^{2} \left (1+x \right )+\left (-2 x^{2}-x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (2 x +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^{1+n +r} 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 ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{1+n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{1+n +r} a_{n}\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 ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{1+n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} \left (n +r -1\right ) x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} 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 =1}{\sum }}a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} 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 \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-x^{n +r} a_{n} \left (n +r \right )+a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )-x^{r} a_{0} r +a_{0} x^{r} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )-x^{r} r +x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (-1+r \right )^{2} x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ \left (-1+r \right )^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 1\\ r_2 &= 1 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (-1+r \right )^{2} x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([1, 1]\).
Since the root of the indicial equation is repeated, then we can construct two linearly independent solutions. The first solution has the form \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\tag {1A} \end {align*}
Now the second solution \(y_{2}\) is found using \begin {align*} y_{2}\left (x \right ) &= y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +r}\right )\tag {1B} \end {align*}
Then the general solution will be \[ y = c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \] In Eq (1B) the sum starts from 1 and not zero. In Eq (1A), \(a_{0}\) is never zero, and is arbitrary and is typically taken as \(a_{0} = 1\), and \(\{c_{1}, c_{2}\}\) are two arbitray constants of integration which can be found from initial conditions. Using the value of the indicial root found earlier, \(r = 1\), Eqs (1A,1B) become \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{1+n}\\ y_{2}\left (x \right ) &= y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{1+n}\right ) \end {align*}
We start by finding the first solution \(y_{1}\left (x \right )\). 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 -1} \left (n +r -1\right ) \left (n +r -2\right )+a_{n} \left (n +r \right ) \left (n +r -1\right )-2 a_{n -1} \left (n +r -1\right )-a_{n} \left (n +r \right )+2 a_{n -1}+a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -1} \left (n^{2}+2 n r +r^{2}-5 n -5 r +6\right )}{n^{2}+2 n r +r^{2}-2 n -2 r +1}\tag {4} \] Which for the root \(r = 1\) becomes \[ a_{n} = -\frac {a_{n -1} \left (n^{2}-3 n +2\right )}{n^{2}}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 1\) 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 \[ a_{1}=\frac {-r^{2}+3 r -2}{r^{2}} \] Which for the root \(r = 1\) becomes \[ a_{1}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}+3 r -2}{r^{2}}\) | \(0\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {\left (-2+r \right ) \left (-1+r \right )^{2}}{r \left (r +1\right )^{2}} \] Which for the root \(r = 1\) becomes \[ a_{2}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}+3 r -2}{r^{2}}\) | \(0\) |
| \(a_{2}\) | \(\frac {\left (-2+r \right ) \left (-1+r \right )^{2}}{r \left (r +1\right )^{2}}\) | \(0\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {\left (-2+r \right ) \left (-1+r \right )^{2}}{\left (r +2\right )^{2} \left (r +1\right )} \] Which for the root \(r = 1\) becomes \[ a_{3}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}+3 r -2}{r^{2}}\) | \(0\) |
| \(a_{2}\) | \(\frac {\left (-2+r \right ) \left (-1+r \right )^{2}}{r \left (r +1\right )^{2}}\) | \(0\) |
| \(a_{3}\) | \(-\frac {\left (-2+r \right ) \left (-1+r \right )^{2}}{\left (r +2\right )^{2} \left (r +1\right )}\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {\left (-2+r \right ) \left (-1+r \right )^{2}}{\left (r +3\right )^{2} \left (r +2\right )} \] Which for the root \(r = 1\) becomes \[ a_{4}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}+3 r -2}{r^{2}}\) | \(0\) |
| \(a_{2}\) | \(\frac {\left (-2+r \right ) \left (-1+r \right )^{2}}{r \left (r +1\right )^{2}}\) | \(0\) |
| \(a_{3}\) | \(-\frac {\left (-2+r \right ) \left (-1+r \right )^{2}}{\left (r +2\right )^{2} \left (r +1\right )}\) | \(0\) |
| \(a_{4}\) | \(\frac {\left (-2+r \right ) \left (-1+r \right )^{2}}{\left (r +3\right )^{2} \left (r +2\right )}\) | \(0\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {\left (-2+r \right ) \left (-1+r \right )^{2}}{\left (r +4\right )^{2} \left (r +3\right )} \] Which for the root \(r = 1\) becomes \[ a_{5}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}+3 r -2}{r^{2}}\) | \(0\) |
| \(a_{2}\) | \(\frac {\left (-2+r \right ) \left (-1+r \right )^{2}}{r \left (r +1\right )^{2}}\) | \(0\) |
| \(a_{3}\) | \(-\frac {\left (-2+r \right ) \left (-1+r \right )^{2}}{\left (r +2\right )^{2} \left (r +1\right )}\) | \(0\) |
| \(a_{4}\) | \(\frac {\left (-2+r \right ) \left (-1+r \right )^{2}}{\left (r +3\right )^{2} \left (r +2\right )}\) | \(0\) |
| \(a_{5}\) | \(-\frac {\left (-2+r \right ) \left (-1+r \right )^{2}}{\left (r +4\right )^{2} \left (r +3\right )}\) | \(0\) |
Using the above table, then the first solution \(y_{1}\left (x \right )\) is \begin{align*} y_{1}\left (x \right )&= x \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 \left (1+O\left (x^{6}\right )\right ) \\ \end{align*} Now the second solution is found. The second solution is given by \[ y_{2}\left (x \right ) = y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +r}\right ) \] Where \(b_{n}\) is found using \[ b_{n} = \frac {d}{d r}a_{n ,r} \] And the above is then evaluated at \(r = 1\). The above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table
| \(n\) | \(b_{n ,r}\) | \(a_{n}\) | \(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) | \(b_{n}\left (r =1\right )\) |
| \(b_{0}\) | \(1\) | \(1\) | N/A since \(b_{n}\) starts from 1 | N/A |
| \(b_{1}\) | \(\frac {-r^{2}+3 r -2}{r^{2}}\) | \(0\) | \(\frac {-3 r +4}{r^{3}}\) | \(1\) |
| \(b_{2}\) | \(\frac {\left (-2+r \right ) \left (-1+r \right )^{2}}{r \left (r +1\right )^{2}}\) | \(0\) | \(\frac {6 r^{3}-14 r^{2}+6 r +2}{r^{2} \left (r +1\right )^{3}}\) | \(0\) |
| \(b_{3}\) | \(-\frac {\left (-2+r \right ) \left (-1+r \right )^{2}}{\left (r +2\right )^{2} \left (r +1\right )}\) | \(0\) | \(-\frac {9 \left (r^{2}-\frac {1}{3} r -2\right ) \left (-1+r \right )}{\left (r +2\right )^{3} \left (r +1\right )^{2}}\) | \(0\) |
| \(b_{4}\) | \(\frac {\left (-2+r \right ) \left (-1+r \right )^{2}}{\left (r +3\right )^{2} \left (r +2\right )}\) | \(0\) | \(\frac {12 r^{3}-4 r^{2}-52 r +44}{\left (r +3\right )^{3} \left (r +2\right )^{2}}\) | \(0\) |
| \(b_{5}\) | \(-\frac {\left (-2+r \right ) \left (-1+r \right )^{2}}{\left (r +4\right )^{2} \left (r +3\right )}\) | \(0\) | \(\frac {-15 r^{3}-10 r^{2}+105 r -80}{\left (r +4\right )^{3} \left (r +3\right )^{2}}\) | \(0\) |
The above table gives all values of \(b_{n}\) needed. Hence the second solution is \begin{align*} y_{2}\left (x \right )&=y_{1}\left (x \right ) \ln \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 \\ &= x \left (1+O\left (x^{6}\right )\right ) \ln \left (x \right )+x \left (x +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} x \left (1+O\left (x^{6}\right )\right ) + c_{2} \left (x \left (1+O\left (x^{6}\right )\right ) \ln \left (x \right )+x \left (x +O\left (x^{6}\right )\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x \left (1+O\left (x^{6}\right )\right )+c_{2} \left (x \left (1+O\left (x^{6}\right )\right ) \ln \left (x \right )+x \left (x +O\left (x^{6}\right )\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x \left (1+O\left (x^{6}\right )\right )+c_{2} \left (x \left (1+O\left (x^{6}\right )\right ) \ln \left (x \right )+x \left (x +O\left (x^{6}\right )\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x \left (1+O\left (x^{6}\right )\right )+c_{2} \left (x \left (1+O\left (x^{6}\right )\right ) \ln \left (x \right )+x \left (x +O\left (x^{6}\right )\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime } x^{2} \left (1+x \right )+\left (-2 x^{2}-x \right ) y^{\prime }+\left (2 x +1\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 (2 x +1\right ) y}{x^{2} \left (1+x \right )}+\frac {\left (2 x +1\right ) y^{\prime }}{x \left (1+x \right )} \\ \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 (2 x +1\right ) y^{\prime }}{x \left (1+x \right )}+\frac {\left (2 x +1\right ) y}{x^{2} \left (1+x \right )}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=-\frac {2 x +1}{x \left (1+x \right )}, P_{3}\left (x \right )=\frac {2 x +1}{x^{2} \left (1+x \right )}\right ] \\ {} & \circ & \left (1+x \right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (1+x \right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=-1 \\ {} & \circ & \left (1+x \right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (1+x \right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=0 \\ {} & \circ & x =-1\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=-1 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & y^{\prime \prime } x^{2} \left (1+x \right )-x \left (2 x +1\right ) y^{\prime }+\left (2 x +1\right ) y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -1\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (u^{3}-2 u^{2}+u \right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (-2 u^{2}+3 u -1\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (2 u -1\right ) y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot y \left (u \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (-2+r \right ) u^{-1+r}+\left (a_{1} \left (1+r \right ) \left (-1+r \right )-a_{0} \left (2 r^{2}-5 r +1\right )\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +r -1\right )-a_{k} \left (2 k^{2}+4 k r +2 r^{2}-5 k -5 r +1\right )+a_{k -1} \left (k +r -2\right ) \left (k -3+r \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (-2+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 2\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right ) \left (-1+r \right )-a_{0} \left (2 r^{2}-5 r +1\right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (-2 a_{k}+a_{k -1}+a_{k +1}\right ) k^{2}+\left (\left (-4 a_{k}+2 a_{k -1}+2 a_{k +1}\right ) r +5 a_{k}-5 a_{k -1}\right ) k +\left (-2 a_{k}+a_{k -1}+a_{k +1}\right ) r^{2}+\left (5 a_{k}-5 a_{k -1}\right ) r -a_{k}+6 a_{k -1}-a_{k +1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (-2 a_{k +1}+a_{k}+a_{k +2}\right ) \left (k +1\right )^{2}+\left (\left (-4 a_{k +1}+2 a_{k}+2 a_{k +2}\right ) r +5 a_{k +1}-5 a_{k}\right ) \left (k +1\right )+\left (-2 a_{k +1}+a_{k}+a_{k +2}\right ) r^{2}+\left (5 a_{k +1}-5 a_{k}\right ) r -a_{k +1}+6 a_{k}-a_{k +2}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-2 k^{2} a_{k +1}+2 k r a_{k}-4 k r a_{k +1}+r^{2} a_{k}-2 r^{2} a_{k +1}-3 k a_{k}+k a_{k +1}-3 r a_{k}+r a_{k +1}+2 a_{k}+2 a_{k +1}}{k^{2}+2 k r +r^{2}+2 k +2 r} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-2 k^{2} a_{k +1}-3 k a_{k}+k a_{k +1}+2 a_{k}+2 a_{k +1}}{k^{2}+2 k} \\ \bullet & {} & \textrm {Series not valid for}\hspace {3pt} r =0\hspace {3pt}\textrm {, division by}\hspace {3pt} 0\hspace {3pt}\textrm {in the recursion relation at}\hspace {3pt} k =0 \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-2 k^{2} a_{k +1}-3 k a_{k}+k a_{k +1}+2 a_{k}+2 a_{k +1}}{k^{2}+2 k} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-2 k^{2} a_{k +1}+k a_{k}-7 k a_{k +1}-4 a_{k +1}}{k^{2}+6 k +8} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +2}, a_{k +2}=-\frac {k^{2} a_{k}-2 k^{2} a_{k +1}+k a_{k}-7 k a_{k +1}-4 a_{k +1}}{k^{2}+6 k +8}, 3 a_{1}+a_{0}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =1+x \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (1+x \right )^{k +2}, a_{k +2}=-\frac {k^{2} a_{k}-2 k^{2} a_{k +1}+k a_{k}-7 k a_{k +1}-4 a_{k +1}}{k^{2}+6 k +8}, 3 a_{1}+a_{0}=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) Group is reducible, not completely reducible <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 35
Order:=6; dsolve(x^2*(1+x)*diff(y(x),x$2)-(1+2*x)*(x*diff(y(x),x)-y(x))=0,y(x),type='series',x=0);
\[ y \left (x \right ) = x \left (\left (c_{2} \ln \left (x \right )+c_{1} \right ) \left (1+\operatorname {O}\left (x^{6}\right )\right )+\left (x +\operatorname {O}\left (x^{6}\right )\right ) c_{2} \right ) \]
✓ Solution by Mathematica
Time used: 0.006 (sec). Leaf size: 2760
AsymptoticDSolveValue[x^2*(1+x)*y''[x]-(1+2*x)*(x*y'[x]+y[x])==0,y[x],{x,0,5}]
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