15.14 problem 10

15.14.1 Maple step by step solution

Internal problem ID [1362]
Internal file name [OUTPUT/1363_Sunday_June_05_2022_02_13_13_AM_74302315/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 II. Exercises 7.6. Page 374
Problem number: 10.
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]]

\[ \boxed {16 x^{2} y^{\prime \prime }+4 x \left (2 x^{2}+x +6\right ) y^{\prime }+\left (18 x^{2}+5 x +1\right ) y=0} \] 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. \[ 16 x^{2} y^{\prime \prime }+\left (8 x^{3}+4 x^{2}+24 x \right ) y^{\prime }+\left (18 x^{2}+5 x +1\right ) y = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} y^{\prime \prime }+p(x) y^{\prime } + q(x) y &=0 \end {align*}

Where \begin {align*} p(x) &= \frac {2 x^{2}+x +6}{4 x}\\ q(x) &= \frac {18 x^{2}+5 x +1}{16 x^{2}}\\ \end {align*}

Table 347: Table \(p(x),q(x)\) singularites.
\(p(x)=\frac {2 x^{2}+x +6}{4 x}\)
singularity type
\(x = 0\) \(\text {``regular''}\)
\(x = \infty \) \(\text {``regular''}\)
\(x = -\infty \) \(\text {``regular''}\)
\(q(x)=\frac {18 x^{2}+5 x +1}{16 x^{2}}\)
singularity type
\(x = 0\) \(\text {``regular''}\)

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

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

Irregular singular points : \([\infty ]\)

Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ 16 x^{2} y^{\prime \prime }+\left (8 x^{3}+4 x^{2}+24 x \right ) y^{\prime }+\left (18 x^{2}+5 x +1\right ) y = 0 \] Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} \] Then \begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} 16 x^{2} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (8 x^{3}+4 x^{2}+24 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (18 x^{2}+5 x +1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}16 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}8 x^{n +r +2} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}24 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}18 x^{n +r +2} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}5 x^{1+n +r} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}8 x^{n +r +2} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}8 a_{n -2} \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}4 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}18 x^{n +r +2} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}18 a_{n -2} x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}5 x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}5 a_{n -1} x^{n +r} \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}16 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}8 a_{n -2} \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}24 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}18 a_{n -2} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}5 a_{n -1} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 16 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+24 x^{n +r} a_{n} \left (n +r \right )+a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 16 x^{r} a_{0} r \left (-1+r \right )+24 x^{r} a_{0} r +a_{0} x^{r} = 0 \] Or \[ \left (16 x^{r} r \left (-1+r \right )+24 x^{r} r +x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ x^{r} \left (4 r +1\right )^{2} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ \left (4 r +1\right )^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= -{\frac {1}{4}}\\ r_2 &= -{\frac {1}{4}} \end {align*}

Since \(a_{0}\neq 0\) then the indicial equation becomes \[ x^{r} \left (4 r +1\right )^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [-{\frac {1}{4}}, -{\frac {1}{4}}\right ]\).

Since the root of the indicial equation is repeated, then we can construct two linearly independent solutions. The first solution has the form \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\tag {1A} \end {align*}

Now the second solution \(y_{2}\) is found using \begin {align*} y_{2}\left (x \right ) &= y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +r}\right )\tag {1B} \end {align*}

Then the general solution will be \[ y = c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \] In Eq (1B) the sum starts from 1 and not zero. In Eq (1A), \(a_{0}\) is never zero, and is arbitrary and is typically taken as \(a_{0} = 1\), and \(\{c_{1}, c_{2}\}\) are two arbitray constants of integration which can be found from initial conditions. Using the value of the indicial root found earlier, \(r = -{\frac {1}{4}}\), Eqs (1A,1B) become \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n -\frac {1}{4}}\\ 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 -\frac {1}{4}}\right ) \end {align*}

We start by finding the first solution \(y_{1}\left (x \right )\). Eq (2B) derived above is now used to find all \(a_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = -\frac {1}{4 r +5} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} 16 a_{n} \left (n +r \right ) \left (n +r -1\right )+8 a_{n -2} \left (n +r -2\right )+4 a_{n -1} \left (n +r -1\right )+24 a_{n} \left (n +r \right )+18 a_{n -2}+5 a_{n -1}+a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {2 a_{n -2}+a_{n -1}}{4 n +4 r +1}\tag {4} \] Which for the root \(r = -{\frac {1}{4}}\) becomes \[ a_{n} = \frac {-2 a_{n -2}-a_{n -1}}{4 n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = -{\frac {1}{4}}\) and after as more terms are found using the above recursive equation.

\(n\) \(a_{n ,r}\) \(a_{n}\)
\(a_{0}\) \(1\) \(1\)
\(a_{1}\) \(-\frac {1}{4 r +5}\) \(-{\frac {1}{4}}\)

For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {-9-8 r}{16 r^{2}+56 r +45} \] Which for the root \(r = -{\frac {1}{4}}\) becomes \[ a_{2}=-{\frac {7}{32}} \] And the table now becomes

\(n\) \(a_{n ,r}\) \(a_{n}\)
\(a_{0}\) \(1\) \(1\)
\(a_{1}\) \(-\frac {1}{4 r +5}\) \(-{\frac {1}{4}}\)
\(a_{2}\) \(\frac {-9-8 r}{16 r^{2}+56 r +45}\) \(-{\frac {7}{32}}\)

For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {16 r +27}{64 r^{3}+432 r^{2}+908 r +585} \] Which for the root \(r = -{\frac {1}{4}}\) becomes \[ a_{3}={\frac {23}{384}} \] And the table now becomes

\(n\) \(a_{n ,r}\) \(a_{n}\)
\(a_{0}\) \(1\) \(1\)
\(a_{1}\) \(-\frac {1}{4 r +5}\) \(-{\frac {1}{4}}\)
\(a_{2}\) \(\frac {-9-8 r}{16 r^{2}+56 r +45}\) \(-{\frac {7}{32}}\)
\(a_{3}\) \(\frac {16 r +27}{64 r^{3}+432 r^{2}+908 r +585}\) \(\frac {23}{384}\)

For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {64 r^{2}+264 r +207}{256 r^{4}+2816 r^{3}+10976 r^{2}+17776 r +9945} \] Which for the root \(r = -{\frac {1}{4}}\) becomes \[ a_{4}={\frac {145}{6144}} \] And the table now becomes

\(n\) \(a_{n ,r}\) \(a_{n}\)
\(a_{0}\) \(1\) \(1\)
\(a_{1}\) \(-\frac {1}{4 r +5}\) \(-{\frac {1}{4}}\)
\(a_{2}\) \(\frac {-9-8 r}{16 r^{2}+56 r +45}\) \(-{\frac {7}{32}}\)
\(a_{3}\) \(\frac {16 r +27}{64 r^{3}+432 r^{2}+908 r +585}\) \(\frac {23}{384}\)
\(a_{4}\) \(\frac {64 r^{2}+264 r +207}{256 r^{4}+2816 r^{3}+10976 r^{2}+17776 r +9945}\) \(\frac {145}{6144}\)

For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-192 r^{2}-1024 r -1125}{1024 r^{5}+16640 r^{4}+103040 r^{3}+301600 r^{2}+413076 r +208845} \] Which for the root \(r = -{\frac {1}{4}}\) becomes \[ a_{5}=-{\frac {881}{122880}} \] And the table now becomes

\(n\) \(a_{n ,r}\) \(a_{n}\)
\(a_{0}\) \(1\) \(1\)
\(a_{1}\) \(-\frac {1}{4 r +5}\) \(-{\frac {1}{4}}\)
\(a_{2}\) \(\frac {-9-8 r}{16 r^{2}+56 r +45}\) \(-{\frac {7}{32}}\)
\(a_{3}\) \(\frac {16 r +27}{64 r^{3}+432 r^{2}+908 r +585}\) \(\frac {23}{384}\)
\(a_{4}\) \(\frac {64 r^{2}+264 r +207}{256 r^{4}+2816 r^{3}+10976 r^{2}+17776 r +9945}\) \(\frac {145}{6144}\)
\(a_{5}\) \(\frac {-192 r^{2}-1024 r -1125}{1024 r^{5}+16640 r^{4}+103040 r^{3}+301600 r^{2}+413076 r +208845}\) \(-{\frac {881}{122880}}\)

For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {-512 r^{3}-4608 r^{2}-11720 r -7569}{4096 r^{6}+92160 r^{5}+828160 r^{4}+3782400 r^{3}+9192304 r^{2}+11162280 r +5221125} \] Which for the root \(r = -{\frac {1}{4}}\) becomes \[ a_{6}=-{\frac {4919}{2949120}} \] And the table now becomes

\(n\) \(a_{n ,r}\) \(a_{n}\)
\(a_{0}\) \(1\) \(1\)
\(a_{1}\) \(-\frac {1}{4 r +5}\) \(-{\frac {1}{4}}\)
\(a_{2}\) \(\frac {-9-8 r}{16 r^{2}+56 r +45}\) \(-{\frac {7}{32}}\)
\(a_{3}\) \(\frac {16 r +27}{64 r^{3}+432 r^{2}+908 r +585}\) \(\frac {23}{384}\)
\(a_{4}\) \(\frac {64 r^{2}+264 r +207}{256 r^{4}+2816 r^{3}+10976 r^{2}+17776 r +9945}\) \(\frac {145}{6144}\)
\(a_{5}\) \(\frac {-192 r^{2}-1024 r -1125}{1024 r^{5}+16640 r^{4}+103040 r^{3}+301600 r^{2}+413076 r +208845}\) \(-{\frac {881}{122880}}\)
\(a_{6}\) \(\frac {-512 r^{3}-4608 r^{2}-11720 r -7569}{4096 r^{6}+92160 r^{5}+828160 r^{4}+3782400 r^{3}+9192304 r^{2}+11162280 r +5221125}\) \(-{\frac {4919}{2949120}}\)

For \(n = 7\), using the above recursive equation gives \[ a_{7}=\frac {2048 r^{3}+22400 r^{2}+71920 r +63819}{16384 r^{7}+487424 r^{6}+5985280 r^{5}+39146240 r^{4}+146458816 r^{3}+311225936 r^{2}+344590620 r +151412625} \] Which for the root \(r = -{\frac {1}{4}}\) becomes \[ a_{7}={\frac {47207}{82575360}} \] And the table now becomes

\(n\) \(a_{n ,r}\) \(a_{n}\)
\(a_{0}\) \(1\) \(1\)
\(a_{1}\) \(-\frac {1}{4 r +5}\) \(-{\frac {1}{4}}\)
\(a_{2}\) \(\frac {-9-8 r}{16 r^{2}+56 r +45}\) \(-{\frac {7}{32}}\)
\(a_{3}\) \(\frac {16 r +27}{64 r^{3}+432 r^{2}+908 r +585}\) \(\frac {23}{384}\)
\(a_{4}\) \(\frac {64 r^{2}+264 r +207}{256 r^{4}+2816 r^{3}+10976 r^{2}+17776 r +9945}\) \(\frac {145}{6144}\)
\(a_{5}\) \(\frac {-192 r^{2}-1024 r -1125}{1024 r^{5}+16640 r^{4}+103040 r^{3}+301600 r^{2}+413076 r +208845}\) \(-{\frac {881}{122880}}\)
\(a_{6}\) \(\frac {-512 r^{3}-4608 r^{2}-11720 r -7569}{4096 r^{6}+92160 r^{5}+828160 r^{4}+3782400 r^{3}+9192304 r^{2}+11162280 r +5221125}\) \(-{\frac {4919}{2949120}}\)
\(a_{7}\) \(\frac {2048 r^{3}+22400 r^{2}+71920 r +63819}{16384 r^{7}+487424 r^{6}+5985280 r^{5}+39146240 r^{4}+146458816 r^{3}+311225936 r^{2}+344590620 r +151412625}\) \(\frac {47207}{82575360}\)

Using the above table, then the first solution \(y_{1}\left (x \right )\) is \begin{align*} y_{1}\left (x \right )&= \frac {1}{x^{\frac {1}{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}+a_{7} x^{7}+a_{8} x^{8}\dots \right ) \\ &= \frac {1-\frac {x}{4}-\frac {7 x^{2}}{32}+\frac {23 x^{3}}{384}+\frac {145 x^{4}}{6144}-\frac {881 x^{5}}{122880}-\frac {4919 x^{6}}{2949120}+\frac {47207 x^{7}}{82575360}+O\left (x^{8}\right )}{x^{\frac {1}{4}}} \\ \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 \(r = -{\frac {1}{4}}\). The above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table

\(n\) \(b_{n ,r}\) \(a_{n}\) \(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) \(b_{n}\left (r =-\frac {1}{4}\right )\)
\(b_{0}\) \(1\) \(1\) N/A since \(b_{n}\) starts from 1 N/A
\(b_{1}\) \(-\frac {1}{4 r +5}\) \(-{\frac {1}{4}}\) \(\frac {4}{\left (4 r +5\right )^{2}}\) \(\frac {1}{4}\)
\(b_{2}\) \(\frac {-9-8 r}{16 r^{2}+56 r +45}\) \(-{\frac {7}{32}}\) \(\frac {128 r^{2}+288 r +144}{\left (16 r^{2}+56 r +45\right )^{2}}\) \(\frac {5}{64}\)
\(b_{3}\) \(\frac {16 r +27}{64 r^{3}+432 r^{2}+908 r +585}\) \(\frac {23}{384}\) \(\frac {-2048 r^{3}-12096 r^{2}-23328 r -15156}{\left (64 r^{3}+432 r^{2}+908 r +585\right )^{2}}\) \(-{\frac {157}{2304}}\)
\(b_{4}\) \(\frac {64 r^{2}+264 r +207}{256 r^{4}+2816 r^{3}+10976 r^{2}+17776 r +9945}\) \(\frac {145}{6144}\) \(-\frac {8 \left (4096 r^{5}+47872 r^{4}+212352 r^{3}+438592 r^{2}+408888 r +131769\right )}{\left (256 r^{4}+2816 r^{3}+10976 r^{2}+17776 r +9945\right )^{2}}\) \(-{\frac {841}{73728}}\)
\(b_{5}\) \(\frac {-192 r^{2}-1024 r -1125}{1024 r^{5}+16640 r^{4}+103040 r^{3}+301600 r^{2}+413076 r +208845}\) \(-{\frac {881}{122880}}\) \(\frac {589824 r^{6}+10584064 r^{5}+76661760 r^{4}+285905920 r^{3}+577287808 r^{2}+598403520 r +250853220}{\left (1024 r^{5}+16640 r^{4}+103040 r^{3}+301600 r^{2}+413076 r +208845\right )^{2}}\) \(\frac {65017}{7372800}\)
\(b_{6}\) \(\frac {-512 r^{3}-4608 r^{2}-11720 r -7569}{4096 r^{6}+92160 r^{5}+828160 r^{4}+3782400 r^{3}+9192304 r^{2}+11162280 r +5221125}\) \(-{\frac {4919}{2949120}}\) \(\frac {6291456 r^{8}+169869312 r^{7}+1938063360 r^{6}+12138799104 r^{5}+45328740352 r^{4}+102302653440 r^{3}+134165325440 r^{2}+91035209952 r +23295712320}{\left (4096 r^{6}+92160 r^{5}+828160 r^{4}+3782400 r^{3}+9192304 r^{2}+11162280 r +5221125\right )^{2}}\) \(\frac {50791}{58982400}\)
\(b_{7}\) \(\frac {2048 r^{3}+22400 r^{2}+71920 r +63819}{16384 r^{7}+487424 r^{6}+5985280 r^{5}+39146240 r^{4}+146458816 r^{3}+311225936 r^{2}+344590620 r +151412625}\) \(\frac {47207}{82575360}\) \(-\frac {4 \left (33554432 r^{9}+1207435264 r^{8}+18814730240 r^{7}+166244814848 r^{6}+915559593984 r^{5}+3249838106368 r^{4}+7412072119040 r^{3}+10443706449008 r^{2}+8235242604792 r +2775458196945\right )}{\left (16384 r^{7}+487424 r^{6}+5985280 r^{5}+39146240 r^{4}+146458816 r^{3}+311225936 r^{2}+344590620 r +151412625\right )^{2}}\) \(-{\frac {953509}{1284505600}}\)

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 \\ &= \frac {\left (1-\frac {x}{4}-\frac {7 x^{2}}{32}+\frac {23 x^{3}}{384}+\frac {145 x^{4}}{6144}-\frac {881 x^{5}}{122880}-\frac {4919 x^{6}}{2949120}+\frac {47207 x^{7}}{82575360}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{x^{\frac {1}{4}}}+\frac {\frac {x}{4}+\frac {5 x^{2}}{64}-\frac {157 x^{3}}{2304}-\frac {841 x^{4}}{73728}+\frac {65017 x^{5}}{7372800}+\frac {50791 x^{6}}{58982400}-\frac {953509 x^{7}}{1284505600}+O\left (x^{8}\right )}{x^{\frac {1}{4}}} \\ \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 ) \\ &= \frac {c_{1} \left (1-\frac {x}{4}-\frac {7 x^{2}}{32}+\frac {23 x^{3}}{384}+\frac {145 x^{4}}{6144}-\frac {881 x^{5}}{122880}-\frac {4919 x^{6}}{2949120}+\frac {47207 x^{7}}{82575360}+O\left (x^{8}\right )\right )}{x^{\frac {1}{4}}} + c_{2} \left (\frac {\left (1-\frac {x}{4}-\frac {7 x^{2}}{32}+\frac {23 x^{3}}{384}+\frac {145 x^{4}}{6144}-\frac {881 x^{5}}{122880}-\frac {4919 x^{6}}{2949120}+\frac {47207 x^{7}}{82575360}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{x^{\frac {1}{4}}}+\frac {\frac {x}{4}+\frac {5 x^{2}}{64}-\frac {157 x^{3}}{2304}-\frac {841 x^{4}}{73728}+\frac {65017 x^{5}}{7372800}+\frac {50791 x^{6}}{58982400}-\frac {953509 x^{7}}{1284505600}+O\left (x^{8}\right )}{x^{\frac {1}{4}}}\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= \frac {c_{1} \left (1-\frac {x}{4}-\frac {7 x^{2}}{32}+\frac {23 x^{3}}{384}+\frac {145 x^{4}}{6144}-\frac {881 x^{5}}{122880}-\frac {4919 x^{6}}{2949120}+\frac {47207 x^{7}}{82575360}+O\left (x^{8}\right )\right )}{x^{\frac {1}{4}}}+c_{2} \left (\frac {\left (1-\frac {x}{4}-\frac {7 x^{2}}{32}+\frac {23 x^{3}}{384}+\frac {145 x^{4}}{6144}-\frac {881 x^{5}}{122880}-\frac {4919 x^{6}}{2949120}+\frac {47207 x^{7}}{82575360}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{x^{\frac {1}{4}}}+\frac {\frac {x}{4}+\frac {5 x^{2}}{64}-\frac {157 x^{3}}{2304}-\frac {841 x^{4}}{73728}+\frac {65017 x^{5}}{7372800}+\frac {50791 x^{6}}{58982400}-\frac {953509 x^{7}}{1284505600}+O\left (x^{8}\right )}{x^{\frac {1}{4}}}\right ) \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1} \left (1-\frac {x}{4}-\frac {7 x^{2}}{32}+\frac {23 x^{3}}{384}+\frac {145 x^{4}}{6144}-\frac {881 x^{5}}{122880}-\frac {4919 x^{6}}{2949120}+\frac {47207 x^{7}}{82575360}+O\left (x^{8}\right )\right )}{x^{\frac {1}{4}}}+c_{2} \left (\frac {\left (1-\frac {x}{4}-\frac {7 x^{2}}{32}+\frac {23 x^{3}}{384}+\frac {145 x^{4}}{6144}-\frac {881 x^{5}}{122880}-\frac {4919 x^{6}}{2949120}+\frac {47207 x^{7}}{82575360}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{x^{\frac {1}{4}}}+\frac {\frac {x}{4}+\frac {5 x^{2}}{64}-\frac {157 x^{3}}{2304}-\frac {841 x^{4}}{73728}+\frac {65017 x^{5}}{7372800}+\frac {50791 x^{6}}{58982400}-\frac {953509 x^{7}}{1284505600}+O\left (x^{8}\right )}{x^{\frac {1}{4}}}\right ) \\ \end{align*}

Verification of solutions

\[ y = \frac {c_{1} \left (1-\frac {x}{4}-\frac {7 x^{2}}{32}+\frac {23 x^{3}}{384}+\frac {145 x^{4}}{6144}-\frac {881 x^{5}}{122880}-\frac {4919 x^{6}}{2949120}+\frac {47207 x^{7}}{82575360}+O\left (x^{8}\right )\right )}{x^{\frac {1}{4}}}+c_{2} \left (\frac {\left (1-\frac {x}{4}-\frac {7 x^{2}}{32}+\frac {23 x^{3}}{384}+\frac {145 x^{4}}{6144}-\frac {881 x^{5}}{122880}-\frac {4919 x^{6}}{2949120}+\frac {47207 x^{7}}{82575360}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{x^{\frac {1}{4}}}+\frac {\frac {x}{4}+\frac {5 x^{2}}{64}-\frac {157 x^{3}}{2304}-\frac {841 x^{4}}{73728}+\frac {65017 x^{5}}{7372800}+\frac {50791 x^{6}}{58982400}-\frac {953509 x^{7}}{1284505600}+O\left (x^{8}\right )}{x^{\frac {1}{4}}}\right ) \] Verified OK.

15.14.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 16 x^{2} \left (\frac {d}{d x}y^{\prime }\right )+\left (8 x^{3}+4 x^{2}+24 x \right ) y^{\prime }+\left (18 x^{2}+5 x +1\right ) 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 {\left (18 x^{2}+5 x +1\right ) y}{16 x^{2}}-\frac {\left (2 x^{2}+x +6\right ) y^{\prime }}{4 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} \\ {} & {} & \frac {d}{d x}y^{\prime }+\frac {\left (2 x^{2}+x +6\right ) y^{\prime }}{4 x}+\frac {\left (18 x^{2}+5 x +1\right ) y}{16 x^{2}}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=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^{2}+x +6}{4 x}, P_{3}\left (x \right )=\frac {18 x^{2}+5 x +1}{16 x^{2}}\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}}}=\frac {3}{2} \\ {} & \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}}}=\frac {1}{16} \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 16 x^{2} \left (\frac {d}{d x}y^{\prime }\right )+4 x \left (2 x^{2}+x +6\right ) y^{\prime }+\left (18 x^{2}+5 x +1\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..2 \\ {} & {} & 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 =1..3 \\ {} & {} & 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^{2}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (1+4 r \right )^{2} x^{r}+\left (a_{1} \left (5+4 r \right )^{2}+a_{0} \left (5+4 r \right )\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (4 k +4 r +1\right )^{2}+a_{k -1} \left (4 k +4 r +1\right )+2 a_{k -2} \left (4 k +4 r +1\right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (1+4 r \right )^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =-\frac {1}{4} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (5+4 r \right )^{2}+a_{0} \left (5+4 r \right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=-\frac {a_{0}}{5+4 r} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (4 k +4 r +1\right )^{2}+\left (4 k +4 r +1\right ) \left (2 a_{k -2}+a_{k -1}\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +2} \left (4 k +4 r +9\right )^{2}+\left (4 k +4 r +9\right ) \left (2 a_{k}+a_{k +1}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {2 a_{k}+a_{k +1}}{4 k +4 r +9} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {1}{4} \\ {} & {} & a_{k +2}=-\frac {2 a_{k}+a_{k +1}}{4 k +8} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {1}{4} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {1}{4}}, a_{k +2}=-\frac {2 a_{k}+a_{k +1}}{4 k +8}, a_{1}=-\frac {a_{0}}{4}\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 
   Solution has integrals. Trying a special function solution free of integrals... 
   -> Trying a solution in terms of special functions: 
      -> Bessel 
      -> elliptic 
      -> Legendre 
      -> Whittaker 
         -> hyper3: Equivalence to 1F1 under a power @ Moebius 
      -> hypergeometric 
         -> heuristic approach 
         -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius 
      -> Mathieu 
         -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius 
      -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius 
      <- Heun successful: received ODE is equivalent to the  HeunB  ODE, case  c = 0 
   Special function solution also has integrals. Returning default Liouvillian solution. 
<- Kovacics algorithm successful`
 

Solution by Maple

Time used: 0.046 (sec). Leaf size: 81

Order:=8; 
dsolve(16*x^2*diff(y(x),x$2)+4*x*(6+x+2*x^2)*diff(y(x),x)+(1+5*x+18*x^2)*y(x)=0,y(x),type='series',x=0);
 

\[ y \left (x \right ) = \frac {\left (c_{2} \ln \left (x \right )+c_{1} \right ) \left (1-\frac {1}{4} x -\frac {7}{32} x^{2}+\frac {23}{384} x^{3}+\frac {145}{6144} x^{4}-\frac {881}{122880} x^{5}-\frac {4919}{2949120} x^{6}+\frac {47207}{82575360} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+\left (\frac {1}{4} x +\frac {5}{64} x^{2}-\frac {157}{2304} x^{3}-\frac {841}{73728} x^{4}+\frac {65017}{7372800} x^{5}+\frac {50791}{58982400} x^{6}-\frac {953509}{1284505600} x^{7}+\operatorname {O}\left (x^{8}\right )\right ) c_{2}}{x^{\frac {1}{4}}} \]

Solution by Mathematica

Time used: 0.011 (sec). Leaf size: 176

AsymptoticDSolveValue[16*x^2*y''[x]+4*x*(6+x+2*x^2)*y'[x]+(1+5*x+18*x^2)*y[x]==0,y[x],{x,0,7}]
 

\[ y(x)\to \frac {c_1 \left (\frac {47207 x^7}{82575360}-\frac {4919 x^6}{2949120}-\frac {881 x^5}{122880}+\frac {145 x^4}{6144}+\frac {23 x^3}{384}-\frac {7 x^2}{32}-\frac {x}{4}+1\right )}{\sqrt [4]{x}}+c_2 \left (\frac {-\frac {953509 x^7}{1284505600}+\frac {50791 x^6}{58982400}+\frac {65017 x^5}{7372800}-\frac {841 x^4}{73728}-\frac {157 x^3}{2304}+\frac {5 x^2}{64}+\frac {x}{4}}{\sqrt [4]{x}}+\frac {\left (\frac {47207 x^7}{82575360}-\frac {4919 x^6}{2949120}-\frac {881 x^5}{122880}+\frac {145 x^4}{6144}+\frac {23 x^3}{384}-\frac {7 x^2}{32}-\frac {x}{4}+1\right ) \log (x)}{\sqrt [4]{x}}\right ) \]