8.6 problem 8

8.6.1 Maple step by step solution

Internal problem ID [7000]
Internal file name [OUTPUT/6243_Thursday_August_18_2022_07_11_27_AM_31591721/index.tex]

Book: Elementary differential equations. Rainville, Bedient, Bedient. Prentice Hall. NJ. 8th edition. 1997.
Section: CHAPTER 18. Power series solutions. 18.11 Many-Term Recurrence Relations. Exercises page 391
Problem number: 8.
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]]

\[ \boxed {x \left (x -2\right )^{2} y^{\prime \prime }-2 \left (x -2\right ) y^{\prime }+2 y=0} \] 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. \[ \left (x^{3}-4 x^{2}+4 x \right ) y^{\prime \prime }+\left (-2 x +4\right ) y^{\prime }+2 y = 0 \] 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 \left (x -2\right )}\\ q(x) &= \frac {2}{x \left (x -2\right )^{2}}\\ \end {align*}

Table 85: Table \(p(x),q(x)\) singularites.
\(p(x)=-\frac {2}{x \left (x -2\right )}\)
singularity type
\(x = 0\) \(\text {``regular''}\)
\(x = 2\) \(\text {``regular''}\)
\(q(x)=\frac {2}{x \left (x -2\right )^{2}}\)
singularity type
\(x = 0\) \(\text {``regular''}\)
\(x = 2\) \(\text {``regular''}\)

Combining everything together gives the following summary of singularities for the ode as

Regular singular points : \([0, 2, \infty ]\)

Irregular singular points : \([]\)

Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x \left (x^{2}-4 x +4\right ) y^{\prime \prime }+\left (-2 x +4\right ) y^{\prime }+2 y = 0 \] 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}-4 x +4\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 +4\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^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}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^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 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 (-2 x^{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 -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 a_{n} x^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} 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 =2}{\sum }}a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 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 }}4 x^{n +r -1} 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 -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}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 \[ 4 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+4 \left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ 4 x^{-1+r} a_{0} r \left (-1+r \right )+4 r a_{0} x^{-1+r} = 0 \] Or \[ \left (4 x^{-1+r} r \left (-1+r \right )+4 r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ 4 x^{-1+r} r^{2} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 4 r^{2} = 0 \] 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 \[ 4 x^{-1+r} r^{2} = 0 \] 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 \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = \frac {2 r^{2}-r -1}{2 \left (1+r \right )^{2}} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} a_{n -2} \left (n +r -2\right ) \left (n -3+r \right )-4 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+4 a_{n} \left (n +r \right ) \left (n +r -1\right )-2 a_{n -1} \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 \[ a_{n} = -\frac {n^{2} a_{n -2}-4 n^{2} a_{n -1}+2 n r a_{n -2}-8 n r a_{n -1}+r^{2} a_{n -2}-4 r^{2} a_{n -1}-5 n a_{n -2}+10 n a_{n -1}-5 r a_{n -2}+10 r a_{n -1}+6 a_{n -2}-4 a_{n -1}}{4 \left (n^{2}+2 n r +r^{2}\right )}\tag {4} \] Which for the root \(r = 0\) becomes \[ a_{n} = -\frac {\left (\left (a_{n -2}-4 a_{n -1}\right ) n -3 a_{n -2}+2 a_{n -1}\right ) \left (n -2\right )}{4 n^{2}}\tag {5} \] 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.

\(n\) \(a_{n ,r}\) \(a_{n}\)
\(a_{0}\) \(1\) \(1\)
\(a_{1}\) \(\frac {2 r^{2}-r -1}{2 \left (1+r \right )^{2}}\) \(-{\frac {1}{2}}\)

For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {3 r^{4}+3 r^{3}-4 r^{2}-2 r}{4 \left (1+r \right )^{2} \left (r +2\right )^{2}} \] Which for the root \(r = 0\) becomes \[ a_{2}=0 \] And the table now becomes

\(n\) \(a_{n ,r}\) \(a_{n}\)
\(a_{0}\) \(1\) \(1\)
\(a_{1}\) \(\frac {2 r^{2}-r -1}{2 \left (1+r \right )^{2}}\) \(-{\frac {1}{2}}\)
\(a_{2}\) \(\frac {3 r^{4}+3 r^{3}-4 r^{2}-2 r}{4 \left (1+r \right )^{2} \left (r +2\right )^{2}}\) \(0\)

For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {2 r^{5}+7 r^{4}+2 r^{3}-8 r^{2}-3 r}{4 \left (r +3\right )^{2} \left (r +2\right )^{2} \left (1+r \right )} \] Which for the root \(r = 0\) becomes \[ a_{3}=0 \] And the table now becomes

\(n\) \(a_{n ,r}\) \(a_{n}\)
\(a_{0}\) \(1\) \(1\)
\(a_{1}\) \(\frac {2 r^{2}-r -1}{2 \left (1+r \right )^{2}}\) \(-{\frac {1}{2}}\)
\(a_{2}\) \(\frac {3 r^{4}+3 r^{3}-4 r^{2}-2 r}{4 \left (1+r \right )^{2} \left (r +2\right )^{2}}\) \(0\)
\(a_{3}\) \(\frac {2 r^{5}+7 r^{4}+2 r^{3}-8 r^{2}-3 r}{4 \left (r +3\right )^{2} \left (r +2\right )^{2} \left (1+r \right )}\) \(0\)

For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {5 r^{6}+35 r^{5}+65 r^{4}-5 r^{3}-76 r^{2}-24 r}{16 \left (r +4\right )^{2} \left (r +3\right )^{2} \left (r +2\right ) \left (1+r \right )} \] Which for the root \(r = 0\) becomes \[ a_{4}=0 \] And the table now becomes

\(n\) \(a_{n ,r}\) \(a_{n}\)
\(a_{0}\) \(1\) \(1\)
\(a_{1}\) \(\frac {2 r^{2}-r -1}{2 \left (1+r \right )^{2}}\) \(-{\frac {1}{2}}\)
\(a_{2}\) \(\frac {3 r^{4}+3 r^{3}-4 r^{2}-2 r}{4 \left (1+r \right )^{2} \left (r +2\right )^{2}}\) \(0\)
\(a_{3}\) \(\frac {2 r^{5}+7 r^{4}+2 r^{3}-8 r^{2}-3 r}{4 \left (r +3\right )^{2} \left (r +2\right )^{2} \left (1+r \right )}\) \(0\)
\(a_{4}\) \(\frac {5 r^{6}+35 r^{5}+65 r^{4}-5 r^{3}-76 r^{2}-24 r}{16 \left (r +4\right )^{2} \left (r +3\right )^{2} \left (r +2\right ) \left (1+r \right )}\) \(0\)

For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {6 r^{7}+69 r^{6}+265 r^{5}+335 r^{4}-127 r^{3}-428 r^{2}-120 r}{32 \left (r +5\right )^{2} \left (r +4\right )^{2} \left (1+r \right ) \left (r +2\right ) \left (r +3\right )} \] Which for the root \(r = 0\) becomes \[ a_{5}=0 \] And the table now becomes

\(n\) \(a_{n ,r}\) \(a_{n}\)
\(a_{0}\) \(1\) \(1\)
\(a_{1}\) \(\frac {2 r^{2}-r -1}{2 \left (1+r \right )^{2}}\) \(-{\frac {1}{2}}\)
\(a_{2}\) \(\frac {3 r^{4}+3 r^{3}-4 r^{2}-2 r}{4 \left (1+r \right )^{2} \left (r +2\right )^{2}}\) \(0\)
\(a_{3}\) \(\frac {2 r^{5}+7 r^{4}+2 r^{3}-8 r^{2}-3 r}{4 \left (r +3\right )^{2} \left (r +2\right )^{2} \left (1+r \right )}\) \(0\)
\(a_{4}\) \(\frac {5 r^{6}+35 r^{5}+65 r^{4}-5 r^{3}-76 r^{2}-24 r}{16 \left (r +4\right )^{2} \left (r +3\right )^{2} \left (r +2\right ) \left (1+r \right )}\) \(0\)
\(a_{5}\) \(\frac {6 r^{7}+69 r^{6}+265 r^{5}+335 r^{4}-127 r^{3}-428 r^{2}-120 r}{32 \left (r +5\right )^{2} \left (r +4\right )^{2} \left (1+r \right ) \left (r +2\right ) \left (r +3\right )}\) \(0\)

For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {7 r \left (r^{6}+18 r^{5}+125 r^{4}+420 r^{3}+696 r^{2}+504 r +\frac {720}{7}\right ) \left (-1+r \right )}{64 \left (r +6\right )^{2} \left (r +5\right )^{2} \left (1+r \right ) \left (r +2\right ) \left (r +3\right ) \left (r +4\right )} \] Which for the root \(r = 0\) becomes \[ a_{6}=0 \] And the table now becomes

\(n\) \(a_{n ,r}\) \(a_{n}\)
\(a_{0}\) \(1\) \(1\)
\(a_{1}\) \(\frac {2 r^{2}-r -1}{2 \left (1+r \right )^{2}}\) \(-{\frac {1}{2}}\)
\(a_{2}\) \(\frac {3 r^{4}+3 r^{3}-4 r^{2}-2 r}{4 \left (1+r \right )^{2} \left (r +2\right )^{2}}\) \(0\)
\(a_{3}\) \(\frac {2 r^{5}+7 r^{4}+2 r^{3}-8 r^{2}-3 r}{4 \left (r +3\right )^{2} \left (r +2\right )^{2} \left (1+r \right )}\) \(0\)
\(a_{4}\) \(\frac {5 r^{6}+35 r^{5}+65 r^{4}-5 r^{3}-76 r^{2}-24 r}{16 \left (r +4\right )^{2} \left (r +3\right )^{2} \left (r +2\right ) \left (1+r \right )}\) \(0\)
\(a_{5}\) \(\frac {6 r^{7}+69 r^{6}+265 r^{5}+335 r^{4}-127 r^{3}-428 r^{2}-120 r}{32 \left (r +5\right )^{2} \left (r +4\right )^{2} \left (1+r \right ) \left (r +2\right ) \left (r +3\right )}\) \(0\)
\(a_{6}\) \(\frac {7 r \left (r^{6}+18 r^{5}+125 r^{4}+420 r^{3}+696 r^{2}+504 r +\frac {720}{7}\right ) \left (-1+r \right )}{64 \left (r +6\right )^{2} \left (r +5\right )^{2} \left (1+r \right ) \left (r +2\right ) \left (r +3\right ) \left (r +4\right )}\) \(0\)

For \(n = 7\), using the above recursive equation gives \[ a_{7}=\frac {r \left (r^{6}+21 r^{5}+168 r^{4}+637 r^{3}+1155 r^{2}+882 r +180\right ) \left (\frac {7}{2}+r \right ) \left (-1+r \right )}{16 \left (r +7\right )^{2} \left (r +6\right )^{2} \left (r +3\right ) \left (r +2\right ) \left (1+r \right ) \left (r +4\right ) \left (r +5\right )} \] Which for the root \(r = 0\) becomes \[ a_{7}=0 \] And the table now becomes

\(n\) \(a_{n ,r}\) \(a_{n}\)
\(a_{0}\) \(1\) \(1\)
\(a_{1}\) \(\frac {2 r^{2}-r -1}{2 \left (1+r \right )^{2}}\) \(-{\frac {1}{2}}\)
\(a_{2}\) \(\frac {3 r^{4}+3 r^{3}-4 r^{2}-2 r}{4 \left (1+r \right )^{2} \left (r +2\right )^{2}}\) \(0\)
\(a_{3}\) \(\frac {2 r^{5}+7 r^{4}+2 r^{3}-8 r^{2}-3 r}{4 \left (r +3\right )^{2} \left (r +2\right )^{2} \left (1+r \right )}\) \(0\)
\(a_{4}\) \(\frac {5 r^{6}+35 r^{5}+65 r^{4}-5 r^{3}-76 r^{2}-24 r}{16 \left (r +4\right )^{2} \left (r +3\right )^{2} \left (r +2\right ) \left (1+r \right )}\) \(0\)
\(a_{5}\) \(\frac {6 r^{7}+69 r^{6}+265 r^{5}+335 r^{4}-127 r^{3}-428 r^{2}-120 r}{32 \left (r +5\right )^{2} \left (r +4\right )^{2} \left (1+r \right ) \left (r +2\right ) \left (r +3\right )}\) \(0\)
\(a_{6}\) \(\frac {7 r \left (r^{6}+18 r^{5}+125 r^{4}+420 r^{3}+696 r^{2}+504 r +\frac {720}{7}\right ) \left (-1+r \right )}{64 \left (r +6\right )^{2} \left (r +5\right )^{2} \left (1+r \right ) \left (r +2\right ) \left (r +3\right ) \left (r +4\right )}\) \(0\)
\(a_{7}\) \(\frac {r \left (r^{6}+21 r^{5}+168 r^{4}+637 r^{3}+1155 r^{2}+882 r +180\right ) \left (\frac {7}{2}+r \right ) \left (-1+r \right )}{16 \left (r +7\right )^{2} \left (r +6\right )^{2} \left (r +3\right ) \left (r +2\right ) \left (1+r \right ) \left (r +4\right ) \left (r +5\right )}\) \(0\)

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}+a_{7} x^{7}+a_{8} x^{8}\dots \\ &= 1-\frac {x}{2}+O\left (x^{8}\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
\(b_{1}\) \(\frac {2 r^{2}-r -1}{2 \left (1+r \right )^{2}}\) \(-{\frac {1}{2}}\) \(\frac {5 r +1}{2 \left (1+r \right )^{3}}\) \(\frac {1}{2}\)
\(b_{2}\) \(\frac {3 r^{4}+3 r^{3}-4 r^{2}-2 r}{4 \left (1+r \right )^{2} \left (r +2\right )^{2}}\) \(0\) \(\frac {15 r^{4}+41 r^{3}+24 r^{2}-10 r -4}{4 \left (1+r \right )^{3} \left (r +2\right )^{3}}\) \(-{\frac {1}{8}}\)
\(b_{3}\) \(\frac {2 r^{5}+7 r^{4}+2 r^{3}-8 r^{2}-3 r}{4 \left (r +3\right )^{2} \left (r +2\right )^{2} \left (1+r \right )}\) \(0\) \(\frac {15 r^{6}+109 r^{5}+278 r^{4}+270 r^{3}+27 r^{2}-81 r -18}{4 \left (r +3\right )^{3} \left (r +2\right )^{3} \left (1+r \right )^{2}}\) \(-{\frac {1}{48}}\)
\(b_{4}\) \(\frac {5 r^{6}+35 r^{5}+65 r^{4}-5 r^{3}-76 r^{2}-24 r}{16 \left (r +4\right )^{2} \left (r +3\right )^{2} \left (r +2\right ) \left (1+r \right )}\) \(0\) \(\frac {25 r^{8}+345 r^{7}+1900 r^{6}+5262 r^{5}+7429 r^{4}+4251 r^{3}-828 r^{2}-1656 r -288}{8 \left (r +4\right )^{3} \left (r +3\right )^{3} \left (r +2\right )^{2} \left (1+r \right )^{2}}\) \(-{\frac {1}{192}}\)
\(b_{5}\) \(\frac {6 r^{7}+69 r^{6}+265 r^{5}+335 r^{4}-127 r^{3}-428 r^{2}-120 r}{32 \left (r +5\right )^{2} \left (r +4\right )^{2} \left (1+r \right ) \left (r +2\right ) \left (r +3\right )}\) \(0\) \(\frac {75 r^{10}+1675 r^{9}+15840 r^{8}+82364 r^{7}+255031 r^{6}+471575 r^{5}+478630 r^{4}+175090 r^{3}-99560 r^{2}-96240 r -14400}{32 \left (r +5\right )^{3} \left (r +4\right )^{3} \left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\) \(-{\frac {1}{640}}\)
\(b_{6}\) \(\frac {7 r \left (r^{6}+18 r^{5}+125 r^{4}+420 r^{3}+696 r^{2}+504 r +\frac {720}{7}\right ) \left (-1+r \right )}{64 \left (r +6\right )^{2} \left (r +5\right )^{2} \left (1+r \right ) \left (r +2\right ) \left (r +3\right ) \left (r +4\right )}\) \(0\) \(\frac {105 r^{12}+3451 r^{11}+49700 r^{10}+411754 r^{9}+2163917 r^{8}+7487227 r^{7}+17063126 r^{6}+24603840 r^{5}+19702512 r^{4}+4264128 r^{3}-5515200 r^{2}-3853440 r -518400}{64 \left (r +6\right )^{3} \left (r +5\right )^{3} \left (1+r \right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) \(-{\frac {1}{1920}}\)
\(b_{7}\) \(\frac {r \left (r^{6}+21 r^{5}+168 r^{4}+637 r^{3}+1155 r^{2}+882 r +180\right ) \left (\frac {7}{2}+r \right ) \left (-1+r \right )}{16 \left (r +7\right )^{2} \left (r +6\right )^{2} \left (r +3\right ) \left (r +2\right ) \left (1+r \right ) \left (r +4\right ) \left (r +5\right )}\) \(0\) \(\frac {35 r^{14}+1589 r^{13}+32326 r^{12}+388906 r^{11}+3074148 r^{10}+16764690 r^{9}+64366780 r^{8}+173944834 r^{7}+323551445 r^{6}+390161093 r^{5}+257237778 r^{4}+22108968 r^{3}-89479152 r^{2}-51196320 r -6350400}{32 \left (r +7\right )^{3} \left (r +6\right )^{3} \left (r +3\right )^{2} \left (r +2\right )^{2} \left (1+r \right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\) \(-{\frac {1}{5376}}\)

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}+b_{7} x^{7}+b_{8} x^{8}\dots \\ &= \left (1-\frac {x}{2}+O\left (x^{8}\right )\right ) \ln \left (x \right )+\frac {x}{2}-\frac {x^{2}}{8}-\frac {x^{3}}{48}-\frac {x^{4}}{192}-\frac {x^{5}}{640}-\frac {x^{6}}{1920}-\frac {x^{7}}{5376}+O\left (x^{8}\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 {x}{2}+O\left (x^{8}\right )\right ) + c_{2} \left (\left (1-\frac {x}{2}+O\left (x^{8}\right )\right ) \ln \left (x \right )+\frac {x}{2}-\frac {x^{2}}{8}-\frac {x^{3}}{48}-\frac {x^{4}}{192}-\frac {x^{5}}{640}-\frac {x^{6}}{1920}-\frac {x^{7}}{5376}+O\left (x^{8}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \left (1-\frac {x}{2}+O\left (x^{8}\right )\right )+c_{2} \left (\left (1-\frac {x}{2}+O\left (x^{8}\right )\right ) \ln \left (x \right )+\frac {x}{2}-\frac {x^{2}}{8}-\frac {x^{3}}{48}-\frac {x^{4}}{192}-\frac {x^{5}}{640}-\frac {x^{6}}{1920}-\frac {x^{7}}{5376}+O\left (x^{8}\right )\right ) \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (1-\frac {x}{2}+O\left (x^{8}\right )\right )+c_{2} \left (\left (1-\frac {x}{2}+O\left (x^{8}\right )\right ) \ln \left (x \right )+\frac {x}{2}-\frac {x^{2}}{8}-\frac {x^{3}}{48}-\frac {x^{4}}{192}-\frac {x^{5}}{640}-\frac {x^{6}}{1920}-\frac {x^{7}}{5376}+O\left (x^{8}\right )\right ) \\ \end{align*}

Verification of solutions

\[ y = c_{1} \left (1-\frac {x}{2}+O\left (x^{8}\right )\right )+c_{2} \left (\left (1-\frac {x}{2}+O\left (x^{8}\right )\right ) \ln \left (x \right )+\frac {x}{2}-\frac {x^{2}}{8}-\frac {x^{3}}{48}-\frac {x^{4}}{192}-\frac {x^{5}}{640}-\frac {x^{6}}{1920}-\frac {x^{7}}{5376}+O\left (x^{8}\right )\right ) \] Verified OK.

8.6.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (x^{2}-4 x +4\right ) y^{\prime \prime }+\left (-2 x +4\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}-4 x +4\right )}+\frac {2 y^{\prime }}{\left (x -2\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 {2 y^{\prime }}{\left (x -2\right ) x}+\frac {2 y}{x \left (x^{2}-4 x +4\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 \left (x -2\right )}, P_{3}\left (x \right )=\frac {2}{x \left (x^{2}-4 x +4\right )}\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}}}=1 \\ {} & \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} \\ {} & {} & x \left (x -2\right ) y^{\prime \prime } \left (x^{2}-4 x +4\right )+\left (-2 x^{2}+8 x -8\right ) y^{\prime }+\left (2 x -4\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\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & 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^{m}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..4 \\ {} & {} & 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} \\ {} & {} & -8 a_{0} r^{2} x^{-1+r}+\left (-8 a_{1} \left (1+r \right )^{2}+4 a_{0} \left (3 r^{2}-r -1\right )\right ) x^{r}+\left (-8 a_{2} \left (2+r \right )^{2}+4 a_{1} \left (3 r^{2}+5 r +1\right )-2 a_{0} \left (1+3 r \right ) \left (-1+r \right )\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (-8 a_{k +1} \left (k +1+r \right )^{2}+4 a_{k} \left (3 k^{2}+6 k r +3 r^{2}-k -r -1\right )-2 a_{k -1} \left (3 k -2+3 r \right ) \left (k -2+r \right )+a_{k -2} \left (k -2+r \right ) \left (k -3+r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -8 r^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =0 \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} x \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [-8 a_{1} \left (1+r \right )^{2}+4 a_{0} \left (3 r^{2}-r -1\right )=0, -8 a_{2} \left (2+r \right )^{2}+4 a_{1} \left (3 r^{2}+5 r +1\right )-2 a_{0} \left (1+3 r \right ) \left (-1+r \right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=\frac {a_{0} \left (3 r^{2}-r -1\right )}{2 \left (r^{2}+2 r +1\right )}, a_{2}=\frac {a_{0} r \left (6 r^{3}+8 r^{2}-3 r -2\right )}{4 \left (r^{4}+6 r^{3}+13 r^{2}+12 r +4\right )}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k^{2}-5 k +6\right ) a_{k -2}+\left (-6 k^{2}+16 k -8\right ) a_{k -1}+\left (12 k^{2}-4 k -4\right ) a_{k}-8 a_{k +1} \left (k +1\right )^{2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \left (\left (k +2\right )^{2}-5 k -4\right ) a_{k}+\left (-6 \left (k +2\right )^{2}+16 k +24\right ) a_{k +1}+\left (12 \left (k +2\right )^{2}-4 k -12\right ) a_{k +2}-8 a_{k +3} \left (k +3\right )^{2}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=\frac {k^{2} a_{k}-6 k^{2} a_{k +1}+12 k^{2} a_{k +2}-k a_{k}-8 k a_{k +1}+44 k a_{k +2}+36 a_{k +2}}{8 \left (k +3\right )^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +3}=\frac {k^{2} a_{k}-6 k^{2} a_{k +1}+12 k^{2} a_{k +2}-k a_{k}-8 k a_{k +1}+44 k a_{k +2}+36 a_{k +2}}{8 \left (k +3\right )^{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +3}=\frac {k^{2} a_{k}-6 k^{2} a_{k +1}+12 k^{2} a_{k +2}-k a_{k}-8 k a_{k +1}+44 k a_{k +2}+36 a_{k +2}}{8 \left (k +3\right )^{2}}, a_{1}=-\frac {a_{0}}{2}, a_{2}=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: 0.047 (sec). Leaf size: 47

Order:=8; 
dsolve(x*(x-2)^2*diff(y(x),x$2)-2*(x-2)*diff(y(x),x)+2*y(x)=0,y(x),type='series',x=0);
 

\[ y \left (x \right ) = \left (c_{2} \ln \left (x \right )+c_{1} \right ) \left (1-\frac {1}{2} x +\operatorname {O}\left (x^{8}\right )\right )+\left (\frac {1}{2} x -\frac {1}{8} x^{2}-\frac {1}{48} x^{3}-\frac {1}{192} x^{4}-\frac {1}{640} x^{5}-\frac {1}{1920} x^{6}-\frac {1}{5376} x^{7}+\operatorname {O}\left (x^{8}\right )\right ) c_{2} \]

Solution by Mathematica

Time used: 0.016 (sec). Leaf size: 75

AsymptoticDSolveValue[x*(x-2)^2*y''[x]-2*(x-2)*y'[x]+2*y[x]==0,y[x],{x,0,7}]
 

\[ y(x)\to c_2 \left (-\frac {x^7}{5376}-\frac {x^6}{1920}-\frac {x^5}{640}-\frac {x^4}{192}-\frac {x^3}{48}-\frac {x^2}{8}+\frac {x}{2}+\left (1-\frac {x}{2}\right ) \log (x)\right )+c_1 \left (1-\frac {x}{2}\right ) \]