Internal problem ID [5587]
Internal file name [OUTPUT/4835_Sunday_June_05_2022_03_07_28_PM_18175668/index.tex]
Book: A FIRST COURSE IN DIFFERENTIAL EQUATIONS with Modeling Applications.
Dennis G. Zill. 9th edition. Brooks/Cole. CA, USA.
Section: Chapter 6. SERIES SOLUTIONS OF LINEAR EQUATIONS. Exercises. 6.2 page
239
Problem number: 32.
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]]
\[ \boxed {x \left (x -1\right ) y^{\prime \prime }+3 y^{\prime }-2 y=0} \] 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. \[ \left (x^{2}-x \right ) y^{\prime \prime }-2 y+3 y^{\prime } = 0 \] 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 {3}{\left (x -1\right ) x}\\ q(x) &= -\frac {2}{\left (x -1\right ) x}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, 1, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x \left (x -1\right ) y^{\prime \prime }+3 y^{\prime }-2 y = 0 \] 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} \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 -1\right ) x +3 \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 )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 \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 )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 \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 \[ -x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+3 \left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ -x^{-1+r} a_{0} r \left (-1+r \right )+3 r a_{0} x^{-1+r} = 0 \] Or \[ \left (-x^{-1+r} r \left (-1+r \right )+3 r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ r \,x^{-1+r} \left (4-r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ -r \left (-4+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 4\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ r \,x^{-1+r} \left (4-r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([4, 0]\).
Since \(r_1 - r_2 = 4\) 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^{4} \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 +4}\\ 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 )+3 a_{n} \left (n +r \right )-2 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {\left (n +r -3\right ) a_{n -1}}{n -4+r}\tag {4} \] Which for the root \(r = 4\) becomes \[ a_{n} = \frac {\left (n +1\right ) a_{n -1}}{n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 4\) 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 \[ a_{1}=\frac {-2+r}{r -3} \] Which for the root \(r = 4\) becomes \[ a_{1}=2 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2+r}{r -3}\) | \(2\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {-1+r}{r -3} \] Which for the root \(r = 4\) becomes \[ a_{2}=3 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2+r}{r -3}\) | \(2\) |
| \(a_{2}\) | \(\frac {-1+r}{r -3}\) | \(3\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {r}{r -3} \] Which for the root \(r = 4\) becomes \[ a_{3}=4 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2+r}{r -3}\) | \(2\) |
| \(a_{2}\) | \(\frac {-1+r}{r -3}\) | \(3\) |
| \(a_{3}\) | \(\frac {r}{r -3}\) | \(4\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {1+r}{r -3} \] Which for the root \(r = 4\) becomes \[ a_{4}=5 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2+r}{r -3}\) | \(2\) |
| \(a_{2}\) | \(\frac {-1+r}{r -3}\) | \(3\) |
| \(a_{3}\) | \(\frac {r}{r -3}\) | \(4\) |
| \(a_{4}\) | \(\frac {1+r}{r -3}\) | \(5\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {2+r}{r -3} \] Which for the root \(r = 4\) becomes \[ a_{5}=6 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2+r}{r -3}\) | \(2\) |
| \(a_{2}\) | \(\frac {-1+r}{r -3}\) | \(3\) |
| \(a_{3}\) | \(\frac {r}{r -3}\) | \(4\) |
| \(a_{4}\) | \(\frac {1+r}{r -3}\) | \(5\) |
| \(a_{5}\) | \(\frac {2+r}{r -3}\) | \(6\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{4} \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^{4} \left (1+2 x +3 x^{2}+4 x^{3}+5 x^{4}+6 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 \(N=4\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{4}\left (r \right )\). 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_{4} \\ &= \frac {1+r}{r -3} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {1+r}{r -3}&= \lim _{r\rightarrow 0}\frac {1+r}{r -3}\\ &= -{\frac {1}{3}} \end {align*}
The limit is \(-{\frac {1}{3}}\). 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 )+3 \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 )+3 n b_{n}-2 b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from the recursive equation (4) gives \[ b_{n} = \frac {\left (n +r -3\right ) b_{n -1}}{n -4+r}\tag {5} \] Which for the root \(r = 0\) becomes \[ b_{n} = \frac {\left (n -3\right ) b_{n -1}}{n -4}\tag {6} \] 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 \[ b_{1}=\frac {-2+r}{r -3} \] Which for the root \(r = 0\) becomes \[ b_{1}={\frac {2}{3}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-2+r}{r -3}\) | \(\frac {2}{3}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {-1+r}{r -3} \] Which for the root \(r = 0\) becomes \[ b_{2}={\frac {1}{3}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-2+r}{r -3}\) | \(\frac {2}{3}\) |
| \(b_{2}\) | \(\frac {-1+r}{r -3}\) | \(\frac {1}{3}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {r}{r -3} \] Which for the root \(r = 0\) becomes \[ b_{3}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-2+r}{r -3}\) | \(\frac {2}{3}\) |
| \(b_{2}\) | \(\frac {-1+r}{r -3}\) | \(\frac {1}{3}\) |
| \(b_{3}\) | \(\frac {r}{r -3}\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {1+r}{r -3} \] Which for the root \(r = 0\) becomes \[ b_{4}=-{\frac {1}{3}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-2+r}{r -3}\) | \(\frac {2}{3}\) |
| \(b_{2}\) | \(\frac {-1+r}{r -3}\) | \(\frac {1}{3}\) |
| \(b_{3}\) | \(\frac {r}{r -3}\) | \(0\) |
| \(b_{4}\) | \(\frac {1+r}{r -3}\) | \(-{\frac {1}{3}}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {2+r}{r -3} \] Which for the root \(r = 0\) becomes \[ b_{5}=-{\frac {2}{3}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-2+r}{r -3}\) | \(\frac {2}{3}\) |
| \(b_{2}\) | \(\frac {-1+r}{r -3}\) | \(\frac {1}{3}\) |
| \(b_{3}\) | \(\frac {r}{r -3}\) | \(0\) |
| \(b_{4}\) | \(\frac {1+r}{r -3}\) | \(-{\frac {1}{3}}\) |
| \(b_{5}\) | \(\frac {2+r}{r -3}\) | \(-{\frac {2}{3}}\) |
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 {2 x}{3}+\frac {x^{2}}{3}-\frac {x^{4}}{3}-\frac {2 x^{5}}{3}+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^{4} \left (1+2 x +3 x^{2}+4 x^{3}+5 x^{4}+6 x^{5}+O\left (x^{6}\right )\right ) + c_{2} \left (1+\frac {2 x}{3}+\frac {x^{2}}{3}-\frac {x^{4}}{3}-\frac {2 x^{5}}{3}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{4} \left (1+2 x +3 x^{2}+4 x^{3}+5 x^{4}+6 x^{5}+O\left (x^{6}\right )\right )+c_{2} \left (1+\frac {2 x}{3}+\frac {x^{2}}{3}-\frac {x^{4}}{3}-\frac {2 x^{5}}{3}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{4} \left (1+2 x +3 x^{2}+4 x^{3}+5 x^{4}+6 x^{5}+O\left (x^{6}\right )\right )+c_{2} \left (1+\frac {2 x}{3}+\frac {x^{2}}{3}-\frac {x^{4}}{3}-\frac {2 x^{5}}{3}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{4} \left (1+2 x +3 x^{2}+4 x^{3}+5 x^{4}+6 x^{5}+O\left (x^{6}\right )\right )+c_{2} \left (1+\frac {2 x}{3}+\frac {x^{2}}{3}-\frac {x^{4}}{3}-\frac {2 x^{5}}{3}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime } \left (x -1\right ) x +3 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}{\left (x -1\right ) x}-\frac {3 y^{\prime }}{\left (x -1\right ) 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} \\ {} & {} & y^{\prime \prime }+\frac {3 y^{\prime }}{\left (x -1\right ) x}-\frac {2 y}{\left (x -1\right ) x}=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 {3}{\left (x -1\right ) x}, P_{3}\left (x \right )=-\frac {2}{\left (x -1\right ) 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}}}=-3 \\ {} & \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}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & y^{\prime \prime } \left (x -1\right ) x +3 y^{\prime }-2 y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & y^{\prime }=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & x^{m}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & x^{m}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -a_{0} r \left (-4+r \right ) x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (-a_{k +1} \left (k +1+r \right ) \left (k -3+r \right )+a_{k} \left (k +1+r \right ) \left (k +r -2\right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -r \left (-4+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 4\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (\left (-k -r +3\right ) a_{k +1}+a_{k} \left (k +r -2\right )\right ) \left (k +1+r \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 -3+r} \\ \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 -3} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =0 \\ {} & {} & a_{1}=\frac {2 a_{0}}{3} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =1 \\ {} & {} & a_{2}=\frac {a_{1}}{2} \\ \bullet & {} & \textrm {Express in terms of}\hspace {3pt} a_{0} \\ {} & {} & a_{2}=\frac {a_{0}}{3} \\ \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=a_{0}\cdot \left (1+\frac {2}{3} x +\frac {1}{3} x^{2}\right ) \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =4 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k +2\right )}{k +1} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =4 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +4}, a_{k +1}=\frac {a_{k} \left (k +2\right )}{k +1}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=a_{0}\cdot \left (1+\frac {2}{3} x +\frac {1}{3} x^{2}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +4}\right ), b_{k +1}=\frac {b_{k} \left (k +2\right )}{k +1}\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: 0.016 (sec). Leaf size: 42
Order:=6; dsolve(x*(x-1)*diff(y(x),x$2)+3*diff(y(x),x)-2*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{4} \left (1+2 x +3 x^{2}+4 x^{3}+5 x^{4}+6 x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (-144-96 x -48 x^{2}+48 x^{4}+96 x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.041 (sec). Leaf size: 55
AsymptoticDSolveValue[x*(x-1)*y''[x]+3*y'[x]-2*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (-\frac {x^4}{3}+\frac {x^2}{3}+\frac {2 x}{3}+1\right )+c_2 \left (5 x^8+4 x^7+3 x^6+2 x^5+x^4\right ) \]