Internal problem ID [1429]
Internal file name [OUTPUT/1430_Sunday_June_05_2022_02_16_56_AM_12855953/index.tex]
Book: Elementary differential equations with boundary value problems. William F. Trench.
Brooks/Cole 2001
Section: Chapter 7 Series Solutions of Linear Second Equations. 7.6 THE METHOD OF
FROBENIUS III. Exercises 7.7. Page 389
Problem number: 13.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference is integer"
Maple gives the following as the ode type
[[_2nd_order, _exact, _linear, _homogeneous]]
\[ \boxed {x \left (1+x \right ) y^{\prime \prime }-4 y^{\prime }-2 y=0} \] 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. \[ \left (x^{2}+x \right ) y^{\prime \prime }-2 y-4 y^{\prime } = 0 \] 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 {4}{x \left (1+x \right )}\\ q(x) &= -\frac {2}{x \left (1+x \right )}\\ \end {align*}
| \(p(x)=-\frac {4}{x \left (1+x \right )}\) | |
| singularity | type |
| \(x = -1\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
| \(q(x)=-\frac {2}{x \left (1+x \right )}\) | |
| 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 \[ x \left (1+x \right ) y^{\prime \prime }-4 y^{\prime }-2 y = 0 \] 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 (1+x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )-4 \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 }}x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 \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 }}\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 }}x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 \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 \[ x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )-4 \left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ x^{-1+r} a_{0} r \left (-1+r \right )-4 r a_{0} x^{-1+r} = 0 \] Or \[ \left (x^{-1+r} r \left (-1+r \right )-4 r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ r \,x^{-1+r} \left (-5+r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r \left (-5+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 5\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ r \,x^{-1+r} \left (-5+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([5, 0]\).
Since \(r_1 - r_2 = 5\) is 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 ) &= C y_{1}\left (x \right ) \ln \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 ) &= x^{5} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\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 +5}\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end {align*}
Where \(C\) above can be zero. We start by finding \(y_{1}\). 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 )-4 a_{n} \left (n +r \right )-2 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {\left (n +r -3\right ) a_{n -1}}{n -5+r}\tag {4} \] Which for the root \(r = 5\) becomes \[ a_{n} = -\frac {\left (n +2\right ) a_{n -1}}{n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 5\) 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 \[ a_{1}=\frac {2-r}{-4+r} \] Which for the root \(r = 5\) becomes \[ a_{1}=-3 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2-r}{-4+r}\) | \(-3\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {r^{2}-3 r +2}{\left (-4+r \right ) \left (r -3\right )} \] Which for the root \(r = 5\) becomes \[ a_{2}=6 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2-r}{-4+r}\) | \(-3\) |
| \(a_{2}\) | \(\frac {r^{2}-3 r +2}{\left (-4+r \right ) \left (r -3\right )}\) | \(6\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {r \left (-1+r \right )}{\left (-4+r \right ) \left (r -3\right )} \] Which for the root \(r = 5\) becomes \[ a_{3}=-10 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2-r}{-4+r}\) | \(-3\) |
| \(a_{2}\) | \(\frac {r^{2}-3 r +2}{\left (-4+r \right ) \left (r -3\right )}\) | \(6\) |
| \(a_{3}\) | \(-\frac {r \left (-1+r \right )}{\left (-4+r \right ) \left (r -3\right )}\) | \(-10\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {r \left (1+r \right )}{\left (-4+r \right ) \left (r -3\right )} \] Which for the root \(r = 5\) becomes \[ a_{4}=15 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2-r}{-4+r}\) | \(-3\) |
| \(a_{2}\) | \(\frac {r^{2}-3 r +2}{\left (-4+r \right ) \left (r -3\right )}\) | \(6\) |
| \(a_{3}\) | \(-\frac {r \left (-1+r \right )}{\left (-4+r \right ) \left (r -3\right )}\) | \(-10\) |
| \(a_{4}\) | \(\frac {r \left (1+r \right )}{\left (-4+r \right ) \left (r -3\right )}\) | \(15\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {\left (1+r \right ) \left (2+r \right )}{\left (-4+r \right ) \left (r -3\right )} \] Which for the root \(r = 5\) becomes \[ a_{5}=-21 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {2-r}{-4+r}\) | \(-3\) |
| \(a_{2}\) | \(\frac {r^{2}-3 r +2}{\left (-4+r \right ) \left (r -3\right )}\) | \(6\) |
| \(a_{3}\) | \(-\frac {r \left (-1+r \right )}{\left (-4+r \right ) \left (r -3\right )}\) | \(-10\) |
| \(a_{4}\) | \(\frac {r \left (1+r \right )}{\left (-4+r \right ) \left (r -3\right )}\) | \(15\) |
| \(a_{5}\) | \(-\frac {\left (1+r \right ) \left (2+r \right )}{\left (-4+r \right ) \left (r -3\right )}\) | \(-21\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{5} \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^{5} \left (1-3 x +6 x^{2}-10 x^{3}+15 x^{4}-21 x^{5}+O\left (x^{6}\right )\right ) \end {align*}
Now the second solution \(y_{2}\left (x \right )\) is found. Let \[ r_{1}-r_{2} = N \] Where \(N\) is positive integer which is the difference between the two roots. \(r_{1}\) is taken as the larger root. Hence for this problem we have \(N=5\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{5}\left (r \right )\). If this limit exists, then \(C = 0\), else we need to keep the log term and \(C \neq 0\). The above table shows that \begin {align*} a_N &= a_{5} \\ &= -\frac {\left (1+r \right ) \left (2+r \right )}{\left (-4+r \right ) \left (r -3\right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}-\frac {\left (1+r \right ) \left (2+r \right )}{\left (-4+r \right ) \left (r -3\right )}&= \lim _{r\rightarrow 0}-\frac {\left (1+r \right ) \left (2+r \right )}{\left (-4+r \right ) \left (r -3\right )}\\ &= -{\frac {1}{6}} \end {align*}
The limit is \(-{\frac {1}{6}}\). Since the limit exists then the log term is not needed and we can set \(C = 0\). Therefore the second solution has the form \begin {align*} y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r}\\ &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n} \end {align*}
Eq (3) derived above is 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{4} b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+b_{n} \left (n +r \right ) \left (n +r -1\right )-4 \left (n +r \right ) b_{n}-2 b_{n -1} = 0 \end{equation} Which for for the root \(r = 0\) becomes \begin{equation} \tag{4A} b_{n -1} \left (n -1\right ) \left (n -2\right )+b_{n} n \left (n -1\right )-4 n b_{n}-2 b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from the recursive equation (4) gives \[ b_{n} = -\frac {\left (n +r -3\right ) b_{n -1}}{n -5+r}\tag {5} \] Which for the root \(r = 0\) becomes \[ b_{n} = -\frac {\left (n -3\right ) b_{n -1}}{n -5}\tag {6} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = 0\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ b_{1}=-\frac {-2+r}{-4+r} \] Which for the root \(r = 0\) becomes \[ b_{1}=-{\frac {1}{2}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2-r}{-4+r}\) | \(-{\frac {1}{2}}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {r^{2}-3 r +2}{\left (-4+r \right ) \left (r -3\right )} \] Which for the root \(r = 0\) becomes \[ b_{2}={\frac {1}{6}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2-r}{-4+r}\) | \(-{\frac {1}{2}}\) |
| \(b_{2}\) | \(\frac {r^{2}-3 r +2}{\left (-4+r \right ) \left (r -3\right )}\) | \(\frac {1}{6}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=-\frac {r \left (-1+r \right )}{\left (-4+r \right ) \left (r -3\right )} \] Which for the root \(r = 0\) becomes \[ b_{3}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2-r}{-4+r}\) | \(-{\frac {1}{2}}\) |
| \(b_{2}\) | \(\frac {r^{2}-3 r +2}{\left (-4+r \right ) \left (r -3\right )}\) | \(\frac {1}{6}\) |
| \(b_{3}\) | \(-\frac {r \left (-1+r \right )}{\left (-4+r \right ) \left (r -3\right )}\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {r \left (1+r \right )}{\left (-4+r \right ) \left (r -3\right )} \] Which for the root \(r = 0\) becomes \[ b_{4}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2-r}{-4+r}\) | \(-{\frac {1}{2}}\) |
| \(b_{2}\) | \(\frac {r^{2}-3 r +2}{\left (-4+r \right ) \left (r -3\right )}\) | \(\frac {1}{6}\) |
| \(b_{3}\) | \(-\frac {r \left (-1+r \right )}{\left (-4+r \right ) \left (r -3\right )}\) | \(0\) |
| \(b_{4}\) | \(\frac {r \left (1+r \right )}{\left (-4+r \right ) \left (r -3\right )}\) | \(0\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-\frac {\left (1+r \right ) \left (2+r \right )}{\left (-4+r \right ) \left (r -3\right )} \] Which for the root \(r = 0\) becomes \[ b_{5}=-{\frac {1}{6}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {2-r}{-4+r}\) | \(-{\frac {1}{2}}\) |
| \(b_{2}\) | \(\frac {r^{2}-3 r +2}{\left (-4+r \right ) \left (r -3\right )}\) | \(\frac {1}{6}\) |
| \(b_{3}\) | \(-\frac {r \left (-1+r \right )}{\left (-4+r \right ) \left (r -3\right )}\) | \(0\) |
| \(b_{4}\) | \(\frac {r \left (1+r \right )}{\left (-4+r \right ) \left (r -3\right )}\) | \(0\) |
| \(b_{5}\) | \(-\frac {\left (1+r \right ) \left (2+r \right )}{\left (-4+r \right ) \left (r -3\right )}\) | \(-{\frac {1}{6}}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= b_{0}+b_{1} x +b_{2} x^{2}+b_{3} x^{3}+b_{4} x^{4}+b_{5} x^{5}+b_{6} x^{6}\dots \\ &= 1-\frac {x}{2}+\frac {x^{2}}{6}-\frac {x^{5}}{6}+O\left (x^{6}\right ) \end {align*}
Therefore the homogeneous solution is \begin{align*} y_h(x) &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= c_{1} x^{5} \left (1-3 x +6 x^{2}-10 x^{3}+15 x^{4}-21 x^{5}+O\left (x^{6}\right )\right ) + c_{2} \left (1-\frac {x}{2}+\frac {x^{2}}{6}-\frac {x^{5}}{6}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{5} \left (1-3 x +6 x^{2}-10 x^{3}+15 x^{4}-21 x^{5}+O\left (x^{6}\right )\right )+c_{2} \left (1-\frac {x}{2}+\frac {x^{2}}{6}-\frac {x^{5}}{6}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{5} \left (1-3 x +6 x^{2}-10 x^{3}+15 x^{4}-21 x^{5}+O\left (x^{6}\right )\right )+c_{2} \left (1-\frac {x}{2}+\frac {x^{2}}{6}-\frac {x^{5}}{6}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{5} \left (1-3 x +6 x^{2}-10 x^{3}+15 x^{4}-21 x^{5}+O\left (x^{6}\right )\right )+c_{2} \left (1-\frac {x}{2}+\frac {x^{2}}{6}-\frac {x^{5}}{6}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (1+x \right ) \left (\frac {d}{d x}y^{\prime }\right )-4 y^{\prime }-2 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 {2 y}{x \left (1+x \right )}+\frac {4 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} \\ {} & {} & \frac {d}{d x}y^{\prime }-\frac {4 y^{\prime }}{x \left (1+x \right )}-\frac {2 y}{x \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 {4}{x \left (1+x \right )}, P_{3}\left (x \right )=-\frac {2}{x \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}}}=4 \\ {} & \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} \\ {} & {} & x \left (1+x \right ) \left (\frac {d}{d x}y^{\prime }\right )-4 y^{\prime }-2 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^{2}-u \right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )-4 \frac {d}{d u}y \left (u \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} \frac {d}{d u}y \left (u \right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & \frac {d}{d u}y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \frac {d}{d u}y \left (u \right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}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}{d u}\frac {d}{d u}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 (3+r \right ) u^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-a_{k +1} \left (k +1+r \right ) \left (k +4+r \right )+a_{k} \left (k +1+r \right ) \left (k +r -2\right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -r \left (3+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-3, 0\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k +1+r \right ) \left (\left (-k -r -4\right ) a_{k +1}+a_{k} \left (k +r -2\right )\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +r -2\right )}{k +4+r} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-3\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =5 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k -5\right )}{k +1} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =0 \\ {} & {} & a_{1}=-5 a_{0} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =1 \\ {} & {} & a_{2}=-2 a_{1} \\ \bullet & {} & \textrm {Express in terms of}\hspace {3pt} a_{0} \\ {} & {} & a_{2}=10 a_{0} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =2 \\ {} & {} & a_{3}=-a_{2} \\ \bullet & {} & \textrm {Express in terms of}\hspace {3pt} a_{0} \\ {} & {} & a_{3}=-10 a_{0} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =3 \\ {} & {} & a_{4}=-\frac {a_{3}}{2} \\ \bullet & {} & \textrm {Express in terms of}\hspace {3pt} a_{0} \\ {} & {} & a_{4}=5 a_{0} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =4 \\ {} & {} & a_{5}=-\frac {a_{4}}{5} \\ \bullet & {} & \textrm {Express in terms of}\hspace {3pt} a_{0} \\ {} & {} & a_{5}=-a_{0} \\ \bullet & {} & \textrm {Terminating series solution of the ODE for}\hspace {3pt} r =-3\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^{5}+5 u^{4}-10 u^{3}+10 u^{2}-5 u +1\right ) \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =1+x \\ {} & {} & \left [y=-a_{0} x^{5}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =2 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k -2\right )}{k +4} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =0 \\ {} & {} & a_{1}=-\frac {a_{0}}{2} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =1 \\ {} & {} & a_{2}=-\frac {a_{1}}{5} \\ \bullet & {} & \textrm {Express in terms of}\hspace {3pt} a_{0} \\ {} & {} & a_{2}=\frac {a_{0}}{10} \\ \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 (1-\frac {1}{2} u +\frac {1}{10} u^{2}\right ) \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =1+x \\ {} & {} & \left [y=a_{0} \left (\frac {3}{5}-\frac {3}{10} x +\frac {1}{10} x^{2}\right )\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=-a_{0} x^{5}+b_{0} \left (\frac {3}{5}-\frac {3}{10} x +\frac {1}{10} x^{2}\right )\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)] <- linear_1 successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 40
Order:=6; dsolve(x*(1+x)*diff(y(x),x$2)-4*diff(y(x),x)-2*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{5} \left (1-3 x +6 x^{2}-10 x^{3}+15 x^{4}-21 x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (2880-1440 x +480 x^{2}-480 x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.041 (sec). Leaf size: 48
AsymptoticDSolveValue[x*(1+x)*y''[x]-4*y'[x]-2*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (\frac {x^2}{6}-\frac {x}{2}+1\right )+c_2 \left (15 x^9-10 x^8+6 x^7-3 x^6+x^5\right ) \]