Internal problem ID [5485]
Internal file name [OUTPUT/4733_Sunday_June_05_2022_03_04_21_PM_58288000/index.tex]
Book: Advanced Mathematical Methods for Scientists and Engineers, Bender and Orszag.
Springer October 29, 1999
Section: Chapter 3. APPROXIMATE SOLUTION OF LINEAR DIFFERENTIAL EQUATIONS.
page 136
Problem number: 3.24 (a).
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]]
\[ \boxed {x \left (x +2\right ) y^{\prime \prime }+2 \left (1+x \right ) y^{\prime }-2 y=0} \] 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. \[ \left (x^{2}+2 x \right ) y^{\prime \prime }+\left (2 x +2\right ) y^{\prime }-2 y = 0 \] 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) &= \frac {2 x +2}{x \left (x +2\right )}\\ q(x) &= -\frac {2}{x \left (x +2\right )}\\ \end {align*}
| \(p(x)=\frac {2 x +2}{x \left (x +2\right )}\) | |
| singularity | type |
| \(x = -2\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
| \(q(x)=-\frac {2}{x \left (x +2\right )}\) | |
| singularity | type |
| \(x = -2\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([-2, 0, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x \left (x +2\right ) y^{\prime \prime }+\left (2 x +2\right ) y^{\prime }-2 y = 0 \] 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 \left (x +2\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (2 x +2\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )-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 }}2 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 a_{n} x^{n +r}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r -1\) 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 ) \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 -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 a_{n} x^{n +r}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} x^{n +r -1}\right ) \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r -1\). \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 -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 2 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+2 \left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ 2 x^{-1+r} a_{0} r \left (-1+r \right )+2 r a_{0} x^{-1+r} = 0 \] Or \[ \left (2 x^{-1+r} r \left (-1+r \right )+2 r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ 2 x^{-1+r} r^{2} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 2 r^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 0\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ 2 x^{-1+r} r^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([0, 0]\).
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. 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 )+2 a_{n} \left (n +r \right ) \left (n +r -1\right )+2 a_{n -1} \left (n +r -1\right )+2 a_{n} \left (n +r \right )-2 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -1} \left (n^{2}+2 n r +r^{2}-n -r -2\right )}{2 \left (n^{2}+2 n r +r^{2}\right )}\tag {4} \] Which for the root \(r = 0\) becomes \[ a_{n} = -\frac {a_{n -1} \left (n^{2}-n -2\right )}{2 n^{2}}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 0\) 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 \[ a_{1}=\frac {-r^{2}-r +2}{2 \left (r +1\right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{1}=1 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}-r +2}{2 \left (r +1\right )^{2}}\) | \(1\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {\left (r +3\right ) r \left (-1+r \right )}{4 \left (r +2\right ) \left (r +1\right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{2}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}-r +2}{2 \left (r +1\right )^{2}}\) | \(1\) |
| \(a_{2}\) | \(\frac {\left (r +3\right ) r \left (-1+r \right )}{4 \left (r +2\right ) \left (r +1\right )^{2}}\) | \(0\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {\left (-1+r \right ) r \left (r +4\right )}{8 \left (r +3\right ) \left (r +1\right ) \left (r +2\right )} \] Which for the root \(r = 0\) becomes \[ a_{3}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}-r +2}{2 \left (r +1\right )^{2}}\) | \(1\) |
| \(a_{2}\) | \(\frac {\left (r +3\right ) r \left (-1+r \right )}{4 \left (r +2\right ) \left (r +1\right )^{2}}\) | \(0\) |
| \(a_{3}\) | \(-\frac {\left (-1+r \right ) r \left (r +4\right )}{8 \left (r +3\right ) \left (r +1\right ) \left (r +2\right )}\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {r \left (-1+r \right ) \left (r +5\right )}{16 \left (r +4\right ) \left (r +1\right ) \left (r +3\right )} \] Which for the root \(r = 0\) becomes \[ a_{4}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}-r +2}{2 \left (r +1\right )^{2}}\) | \(1\) |
| \(a_{2}\) | \(\frac {\left (r +3\right ) r \left (-1+r \right )}{4 \left (r +2\right ) \left (r +1\right )^{2}}\) | \(0\) |
| \(a_{3}\) | \(-\frac {\left (-1+r \right ) r \left (r +4\right )}{8 \left (r +3\right ) \left (r +1\right ) \left (r +2\right )}\) | \(0\) |
| \(a_{4}\) | \(\frac {r \left (-1+r \right ) \left (r +5\right )}{16 \left (r +4\right ) \left (r +1\right ) \left (r +3\right )}\) | \(0\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {\left (-1+r \right ) r \left (r +6\right )}{32 \left (r +5\right ) \left (r +1\right ) \left (r +4\right )} \] Which for the root \(r = 0\) becomes \[ a_{5}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r^{2}-r +2}{2 \left (r +1\right )^{2}}\) | \(1\) |
| \(a_{2}\) | \(\frac {\left (r +3\right ) r \left (-1+r \right )}{4 \left (r +2\right ) \left (r +1\right )^{2}}\) | \(0\) |
| \(a_{3}\) | \(-\frac {\left (-1+r \right ) r \left (r +4\right )}{8 \left (r +3\right ) \left (r +1\right ) \left (r +2\right )}\) | \(0\) |
| \(a_{4}\) | \(\frac {r \left (-1+r \right ) \left (r +5\right )}{16 \left (r +4\right ) \left (r +1\right ) \left (r +3\right )}\) | \(0\) |
| \(a_{5}\) | \(-\frac {\left (-1+r \right ) r \left (r +6\right )}{32 \left (r +5\right ) \left (r +1\right ) \left (r +4\right )}\) | \(0\) |
Using the above table, then the first solution \(y_{1}\left (x \right )\) becomes \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+x +O\left (x^{6}\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 \(r = 0\). The above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table
| \(n\) | \(b_{n ,r}\) | \(a_{n}\) | \(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) | \(b_{n}\left (r =0\right )\) |
| \(b_{0}\) | \(1\) | \(1\) | N/A since \(b_{n}\) starts from 1 | N/A |
| \(b_{1}\) | \(\frac {-r^{2}-r +2}{2 \left (r +1\right )^{2}}\) | \(1\) | \(\frac {-r -5}{2 \left (r +1\right )^{3}}\) | \(-{\frac {5}{2}}\) |
| \(b_{2}\) | \(\frac {\left (r +3\right ) r \left (-1+r \right )}{4 \left (r +2\right ) \left (r +1\right )^{2}}\) | \(0\) | \(\frac {r^{3}+7 r^{2}+7 r -3}{2 \left (r +2\right )^{2} \left (r +1\right )^{3}}\) | \(-{\frac {3}{8}}\) |
| \(b_{3}\) | \(-\frac {\left (-1+r \right ) r \left (r +4\right )}{8 \left (r +3\right ) \left (r +1\right ) \left (r +2\right )}\) | \(0\) | \(\frac {3-\frac {75}{8} r^{2}-\frac {3}{8} r^{4}-\frac {15}{4} r^{3}-\frac {9}{2} r}{\left (r +3\right )^{2} \left (r +1\right )^{2} \left (r +2\right )^{2}}\) | \(\frac {1}{12}\) |
| \(b_{4}\) | \(\frac {r \left (-1+r \right ) \left (r +5\right )}{16 \left (r +4\right ) \left (r +1\right ) \left (r +3\right )}\) | \(0\) | \(\frac {r^{4}+12 r^{3}+38 r^{2}+24 r -15}{4 \left (r +4\right )^{2} \left (r +1\right )^{2} \left (r +3\right )^{2}}\) | \(-{\frac {5}{192}}\) |
| \(b_{5}\) | \(-\frac {\left (-1+r \right ) r \left (r +6\right )}{32 \left (r +5\right ) \left (r +1\right ) \left (r +4\right )}\) | \(0\) | \(\frac {\frac {15}{4}-\frac {265}{32} r^{2}-\frac {5}{32} r^{4}-\frac {35}{16} r^{3}-\frac {25}{4} r}{\left (r +5\right )^{2} \left (r +1\right )^{2} \left (r +4\right )^{2}}\) | \(\frac {3}{320}\) |
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 \\ &= \left (1+x +O\left (x^{6}\right )\right ) \ln \left (x \right )-\frac {5 x}{2}-\frac {3 x^{2}}{8}+\frac {x^{3}}{12}-\frac {5 x^{4}}{192}+\frac {3 x^{5}}{320}+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} \left (1+x +O\left (x^{6}\right )\right ) + c_{2} \left (\left (1+x +O\left (x^{6}\right )\right ) \ln \left (x \right )-\frac {5 x}{2}-\frac {3 x^{2}}{8}+\frac {x^{3}}{12}-\frac {5 x^{4}}{192}+\frac {3 x^{5}}{320}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \left (1+x +O\left (x^{6}\right )\right )+c_{2} \left (\left (1+x +O\left (x^{6}\right )\right ) \ln \left (x \right )-\frac {5 x}{2}-\frac {3 x^{2}}{8}+\frac {x^{3}}{12}-\frac {5 x^{4}}{192}+\frac {3 x^{5}}{320}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (1+x +O\left (x^{6}\right )\right )+c_{2} \left (\left (1+x +O\left (x^{6}\right )\right ) \ln \left (x \right )-\frac {5 x}{2}-\frac {3 x^{2}}{8}+\frac {x^{3}}{12}-\frac {5 x^{4}}{192}+\frac {3 x^{5}}{320}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (1+x +O\left (x^{6}\right )\right )+c_{2} \left (\left (1+x +O\left (x^{6}\right )\right ) \ln \left (x \right )-\frac {5 x}{2}-\frac {3 x^{2}}{8}+\frac {x^{3}}{12}-\frac {5 x^{4}}{192}+\frac {3 x^{5}}{320}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (x +2\right ) y^{\prime \prime }+\left (2 x +2\right ) y^{\prime }-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 }=\frac {2 y}{x \left (x +2\right )}-\frac {2 \left (1+x \right ) y^{\prime }}{x \left (x +2\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 {2 \left (1+x \right ) y^{\prime }}{x \left (x +2\right )}-\frac {2 y}{x \left (x +2\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 \left (1+x \right )}{x \left (x +2\right )}, P_{3}\left (x \right )=-\frac {2}{x \left (x +2\right )}\right ] \\ {} & \circ & \left (x +2\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-2 \\ {} & {} & \left (\left (x +2\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-2}}}=1 \\ {} & \circ & \left (x +2\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-2 \\ {} & {} & \left (\left (x +2\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-2}}}=0 \\ {} & \circ & x =-2\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}=-2 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x \left (x +2\right ) y^{\prime \prime }+\left (2 x +2\right ) y^{\prime }-2 y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -2\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (u^{2}-2 u \right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (2 u -2\right ) \left (\frac {d}{d u}y \left (u \right )\right )-2 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 \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & 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..2 \\ {} & {} & 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} \\ {} & {} & -2 a_{0} r^{2} u^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-2 a_{k +1} \left (k +1+r \right )^{2}+a_{k} \left (k +r +2\right ) \left (k +r -1\right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -2 r^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -2 a_{k +1} \left (k +1\right )^{2}+a_{k} \left (k +2\right ) \left (k -1\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +2\right ) \left (k -1\right )}{2 \left (k +1\right )^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =1 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +2\right ) \left (k -1\right )}{2 \left (k +1\right )^{2}} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =0 \\ {} & {} & a_{1}=-a_{0} \\ \bullet & {} & \textrm {Terminating series solution of the ODE for}\hspace {3pt} r =0\hspace {3pt}\textrm {. Use reduction of order to find the second linearly independent solution}\hspace {3pt} \\ {} & {} & y \left (u \right )=a_{0}\cdot \left (-u +1\right ) \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +2 \\ {} & {} & \left [y=a_{0} \left (-1-x \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 <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 43
Order:=6; dsolve(x*(x+2)*diff(y(x),x$2)+2*(x+1)*diff(y(x),x)-2*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \left (c_{2} \ln \left (x \right )+c_{1} \right ) \left (1+x +\operatorname {O}\left (x^{6}\right )\right )+\left (-\frac {5}{2} x -\frac {3}{8} x^{2}+\frac {1}{12} x^{3}-\frac {5}{192} x^{4}+\frac {3}{320} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) c_{2} \]
✓ Solution by Mathematica
Time used: 0.006 (sec). Leaf size: 53
AsymptoticDSolveValue[x*(x+2)*y''[x]+2*(x+1)*y'[x]-2*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_2 \left (\frac {3 x^5}{320}-\frac {5 x^4}{192}+\frac {x^3}{12}-\frac {3 x^2}{8}-\frac {5 x}{2}+(x+1) \log (x)\right )+c_1 (x+1) \]