Internal problem ID [1361]
Internal file name [OUTPUT/1362_Sunday_June_05_2022_02_13_10_AM_25547940/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: 9.
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]]
\[ \boxed {4 x^{2} y^{\prime \prime }+2 x \left (x^{2}+x +4\right ) y^{\prime }+\left (3 x^{2}+5 x +1\right ) y=0} \] 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. \[ 4 x^{2} y^{\prime \prime }+\left (2 x^{3}+2 x^{2}+8 x \right ) y^{\prime }+\left (3 x^{2}+5 x +1\right ) y = 0 \] 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 {x^{2}+x +4}{2 x}\\ q(x) &= \frac {3 x^{2}+5 x +1}{4 x^{2}}\\ \end {align*}
| \(p(x)=\frac {x^{2}+x +4}{2 x}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = \infty \) | \(\text {``regular''}\) |
| \(x = -\infty \) | \(\text {``regular''}\) |
| \(q(x)=\frac {3 x^{2}+5 x +1}{4 x^{2}}\) | |
| 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 \[ 4 x^{2} y^{\prime \prime }+\left (2 x^{3}+2 x^{2}+8 x \right ) y^{\prime }+\left (3 x^{2}+5 x +1\right ) y = 0 \] 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} 4 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 (2 x^{3}+2 x^{2}+8 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (3 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 }}4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r +2} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}8 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 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 }}2 x^{n +r +2} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}2 a_{n -2} \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r +2} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}3 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 }}4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}2 a_{n -2} \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}8 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}3 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 \[ 4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+8 x^{n +r} a_{n} \left (n +r \right )+a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 4 x^{r} a_{0} r \left (-1+r \right )+8 x^{r} a_{0} r +a_{0} x^{r} = 0 \] Or \[ \left (4 x^{r} r \left (-1+r \right )+8 x^{r} r +x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ x^{r} \left (1+2 r \right )^{2} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ \left (1+2 r \right )^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= -{\frac {1}{2}}\\ r_2 &= -{\frac {1}{2}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ x^{r} \left (1+2 r \right )^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [-{\frac {1}{2}}, -{\frac {1}{2}}\right ]\).
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, \(r = -{\frac {1}{2}}\), Eqs (1A,1B) become \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n -\frac {1}{2}}\\ 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}{2}}\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 \[ a_{1} = \frac {-2 r -5}{\left (2 r +3\right )^{2}} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} 4 a_{n} \left (n +r \right ) \left (n +r -1\right )+2 a_{n -2} \left (n +r -2\right )+2 a_{n -1} \left (n +r -1\right )+8 a_{n} \left (n +r \right )+3 a_{n -2}+5 a_{n -1}+a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {2 n a_{n -2}+2 n a_{n -1}+2 r a_{n -2}+2 r a_{n -1}-a_{n -2}+3 a_{n -1}}{4 n^{2}+8 n r +4 r^{2}+4 n +4 r +1}\tag {4} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ a_{n} = \frac {\left (-a_{n -2}-a_{n -1}\right ) n +a_{n -2}-a_{n -1}}{2 n^{2}}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = -{\frac {1}{2}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2 r -5}{\left (2 r +3\right )^{2}}\) | \(-1\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {-4 r^{3}-16 r^{2}-15 r +4}{8 \left (r +\frac {3}{2}\right )^{2} \left (r +\frac {5}{2}\right )^{2}} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ a_{2}={\frac {1}{4}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2 r -5}{\left (2 r +3\right )^{2}}\) | \(-1\) |
| \(a_{2}\) | \(\frac {-4 r^{3}-16 r^{2}-15 r +4}{8 \left (r +\frac {3}{2}\right )^{2} \left (r +\frac {5}{2}\right )^{2}}\) | \(\frac {1}{4}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {32 r^{4}+296 r^{3}+948 r^{2}+1254 r +553}{64 \left (r +\frac {7}{2}\right )^{2} \left (r +\frac {3}{2}\right )^{2} \left (r +\frac {5}{2}\right )^{2}} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ a_{3}={\frac {1}{18}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2 r -5}{\left (2 r +3\right )^{2}}\) | \(-1\) |
| \(a_{2}\) | \(\frac {-4 r^{3}-16 r^{2}-15 r +4}{8 \left (r +\frac {3}{2}\right )^{2} \left (r +\frac {5}{2}\right )^{2}}\) | \(\frac {1}{4}\) |
| \(a_{3}\) | \(\frac {32 r^{4}+296 r^{3}+948 r^{2}+1254 r +553}{64 \left (r +\frac {7}{2}\right )^{2} \left (r +\frac {3}{2}\right )^{2} \left (r +\frac {5}{2}\right )^{2}}\) | \(\frac {1}{18}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {64 r^{6}+864 r^{5}+4336 r^{4}+9456 r^{3}+6188 r^{2}-6962 r -8827}{\left (2 r +7\right )^{2} \left (2 r +3\right )^{2} \left (2 r +5\right )^{2} \left (4 r^{2}+36 r +81\right )} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ a_{4}=-{\frac {37}{1152}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2 r -5}{\left (2 r +3\right )^{2}}\) | \(-1\) |
| \(a_{2}\) | \(\frac {-4 r^{3}-16 r^{2}-15 r +4}{8 \left (r +\frac {3}{2}\right )^{2} \left (r +\frac {5}{2}\right )^{2}}\) | \(\frac {1}{4}\) |
| \(a_{3}\) | \(\frac {32 r^{4}+296 r^{3}+948 r^{2}+1254 r +553}{64 \left (r +\frac {7}{2}\right )^{2} \left (r +\frac {3}{2}\right )^{2} \left (r +\frac {5}{2}\right )^{2}}\) | \(\frac {1}{18}\) |
| \(a_{4}\) | \(\frac {64 r^{6}+864 r^{5}+4336 r^{4}+9456 r^{3}+6188 r^{2}-6962 r -8827}{\left (2 r +7\right )^{2} \left (2 r +3\right )^{2} \left (2 r +5\right )^{2} \left (4 r^{2}+36 r +81\right )}\) | \(-{\frac {37}{1152}}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-192 r^{7}-4192 r^{6}-37504 r^{5}-177440 r^{4}-475836 r^{3}-713390 r^{2}-537382 r -144193}{512 \left (r +\frac {7}{2}\right )^{2} \left (r +\frac {3}{2}\right )^{2} \left (r +\frac {5}{2}\right )^{2} \left (r +\frac {9}{2}\right )^{2} \left (r +\frac {11}{2}\right )^{2}} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ a_{5}=-{\frac {17}{28800}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2 r -5}{\left (2 r +3\right )^{2}}\) | \(-1\) |
| \(a_{2}\) | \(\frac {-4 r^{3}-16 r^{2}-15 r +4}{8 \left (r +\frac {3}{2}\right )^{2} \left (r +\frac {5}{2}\right )^{2}}\) | \(\frac {1}{4}\) |
| \(a_{3}\) | \(\frac {32 r^{4}+296 r^{3}+948 r^{2}+1254 r +553}{64 \left (r +\frac {7}{2}\right )^{2} \left (r +\frac {3}{2}\right )^{2} \left (r +\frac {5}{2}\right )^{2}}\) | \(\frac {1}{18}\) |
| \(a_{4}\) | \(\frac {64 r^{6}+864 r^{5}+4336 r^{4}+9456 r^{3}+6188 r^{2}-6962 r -8827}{\left (2 r +7\right )^{2} \left (2 r +3\right )^{2} \left (2 r +5\right )^{2} \left (4 r^{2}+36 r +81\right )}\) | \(-{\frac {37}{1152}}\) |
| \(a_{5}\) | \(\frac {-192 r^{7}-4192 r^{6}-37504 r^{5}-177440 r^{4}-475836 r^{3}-713390 r^{2}-537382 r -144193}{512 \left (r +\frac {7}{2}\right )^{2} \left (r +\frac {3}{2}\right )^{2} \left (r +\frac {5}{2}\right )^{2} \left (r +\frac {9}{2}\right )^{2} \left (r +\frac {11}{2}\right )^{2}}\) | \(-{\frac {17}{28800}}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {-512 r^{9}-14592 r^{8}-172672 r^{7}-1084672 r^{6}-3760736 r^{5}-6170848 r^{4}+1039816 r^{3}+21534576 r^{2}+32373056 r +16074527}{\left (4 r^{2}+52 r +169\right ) \left (2 r +7\right )^{2} \left (2 r +3\right )^{2} \left (2 r +5\right )^{2} \left (2 r +9\right )^{2} \left (2 r +11\right )^{2}} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ a_{6}={\frac {593}{259200}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2 r -5}{\left (2 r +3\right )^{2}}\) | \(-1\) |
| \(a_{2}\) | \(\frac {-4 r^{3}-16 r^{2}-15 r +4}{8 \left (r +\frac {3}{2}\right )^{2} \left (r +\frac {5}{2}\right )^{2}}\) | \(\frac {1}{4}\) |
| \(a_{3}\) | \(\frac {32 r^{4}+296 r^{3}+948 r^{2}+1254 r +553}{64 \left (r +\frac {7}{2}\right )^{2} \left (r +\frac {3}{2}\right )^{2} \left (r +\frac {5}{2}\right )^{2}}\) | \(\frac {1}{18}\) |
| \(a_{4}\) | \(\frac {64 r^{6}+864 r^{5}+4336 r^{4}+9456 r^{3}+6188 r^{2}-6962 r -8827}{\left (2 r +7\right )^{2} \left (2 r +3\right )^{2} \left (2 r +5\right )^{2} \left (4 r^{2}+36 r +81\right )}\) | \(-{\frac {37}{1152}}\) |
| \(a_{5}\) | \(\frac {-192 r^{7}-4192 r^{6}-37504 r^{5}-177440 r^{4}-475836 r^{3}-713390 r^{2}-537382 r -144193}{512 \left (r +\frac {7}{2}\right )^{2} \left (r +\frac {3}{2}\right )^{2} \left (r +\frac {5}{2}\right )^{2} \left (r +\frac {9}{2}\right )^{2} \left (r +\frac {11}{2}\right )^{2}}\) | \(-{\frac {17}{28800}}\) |
| \(a_{6}\) | \(\frac {-512 r^{9}-14592 r^{8}-172672 r^{7}-1084672 r^{6}-3760736 r^{5}-6170848 r^{4}+1039816 r^{3}+21534576 r^{2}+32373056 r +16074527}{\left (4 r^{2}+52 r +169\right ) \left (2 r +7\right )^{2} \left (2 r +3\right )^{2} \left (2 r +5\right )^{2} \left (2 r +9\right )^{2} \left (2 r +11\right )^{2}}\) | \(\frac {593}{259200}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=\frac {4096 r^{10}+164864 r^{9}+2890752 r^{8}+28990080 r^{7}+183413312 r^{6}+760790176 r^{5}+2078667344 r^{4}+3646802552 r^{3}+3838600668 r^{2}+2071188906 r +360317083}{16384 \left (r +\frac {7}{2}\right )^{2} \left (r +\frac {13}{2}\right )^{2} \left (r +\frac {3}{2}\right )^{2} \left (r +\frac {5}{2}\right )^{2} \left (r +\frac {9}{2}\right )^{2} \left (r +\frac {15}{2}\right )^{2} \left (r +\frac {11}{2}\right )^{2}} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ a_{7}=-{\frac {1913}{12700800}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2 r -5}{\left (2 r +3\right )^{2}}\) | \(-1\) |
| \(a_{2}\) | \(\frac {-4 r^{3}-16 r^{2}-15 r +4}{8 \left (r +\frac {3}{2}\right )^{2} \left (r +\frac {5}{2}\right )^{2}}\) | \(\frac {1}{4}\) |
| \(a_{3}\) | \(\frac {32 r^{4}+296 r^{3}+948 r^{2}+1254 r +553}{64 \left (r +\frac {7}{2}\right )^{2} \left (r +\frac {3}{2}\right )^{2} \left (r +\frac {5}{2}\right )^{2}}\) | \(\frac {1}{18}\) |
| \(a_{4}\) | \(\frac {64 r^{6}+864 r^{5}+4336 r^{4}+9456 r^{3}+6188 r^{2}-6962 r -8827}{\left (2 r +7\right )^{2} \left (2 r +3\right )^{2} \left (2 r +5\right )^{2} \left (4 r^{2}+36 r +81\right )}\) | \(-{\frac {37}{1152}}\) |
| \(a_{5}\) | \(\frac {-192 r^{7}-4192 r^{6}-37504 r^{5}-177440 r^{4}-475836 r^{3}-713390 r^{2}-537382 r -144193}{512 \left (r +\frac {7}{2}\right )^{2} \left (r +\frac {3}{2}\right )^{2} \left (r +\frac {5}{2}\right )^{2} \left (r +\frac {9}{2}\right )^{2} \left (r +\frac {11}{2}\right )^{2}}\) | \(-{\frac {17}{28800}}\) |
| \(a_{6}\) | \(\frac {-512 r^{9}-14592 r^{8}-172672 r^{7}-1084672 r^{6}-3760736 r^{5}-6170848 r^{4}+1039816 r^{3}+21534576 r^{2}+32373056 r +16074527}{\left (4 r^{2}+52 r +169\right ) \left (2 r +7\right )^{2} \left (2 r +3\right )^{2} \left (2 r +5\right )^{2} \left (2 r +9\right )^{2} \left (2 r +11\right )^{2}}\) | \(\frac {593}{259200}\) |
| \(a_{7}\) | \(\frac {4096 r^{10}+164864 r^{9}+2890752 r^{8}+28990080 r^{7}+183413312 r^{6}+760790176 r^{5}+2078667344 r^{4}+3646802552 r^{3}+3838600668 r^{2}+2071188906 r +360317083}{16384 \left (r +\frac {7}{2}\right )^{2} \left (r +\frac {13}{2}\right )^{2} \left (r +\frac {3}{2}\right )^{2} \left (r +\frac {5}{2}\right )^{2} \left (r +\frac {9}{2}\right )^{2} \left (r +\frac {15}{2}\right )^{2} \left (r +\frac {11}{2}\right )^{2}}\) | \(-{\frac {1913}{12700800}}\) |
Using the above table, then the first solution \(y_{1}\left (x \right )\) is \begin{align*} y_{1}\left (x \right )&= \frac {1}{\sqrt {x}} \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-x +\frac {x^{2}}{4}+\frac {x^{3}}{18}-\frac {37 x^{4}}{1152}-\frac {17 x^{5}}{28800}+\frac {593 x^{6}}{259200}-\frac {1913 x^{7}}{12700800}+O\left (x^{8}\right )}{\sqrt {x}} \\ \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 \(r = -{\frac {1}{2}}\). The above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table
| \(n\) | \(b_{n ,r}\) | \(a_{n}\) | \(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) | \(b_{n}\left (r =-\frac {1}{2}\right )\) |
| \(b_{0}\) | \(1\) | \(1\) | N/A since \(b_{n}\) starts from 1 | N/A |
| \(b_{1}\) | \(\frac {-2 r -5}{\left (2 r +3\right )^{2}}\) | \(-1\) | \(\frac {4 r +14}{\left (2 r +3\right )^{3}}\) | \(\frac {3}{2}\) |
| \(b_{2}\) | \(\frac {-4 r^{3}-16 r^{2}-15 r +4}{8 \left (r +\frac {3}{2}\right )^{2} \left (r +\frac {5}{2}\right )^{2}}\) | \(\frac {1}{4}\) | \(\frac {32 r^{4}+128 r^{3}-608 r -706}{\left (2 r +3\right )^{3} \left (2 r +5\right )^{3}}\) | \(-{\frac {13}{16}}\) |
| \(b_{3}\) | \(\frac {32 r^{4}+296 r^{3}+948 r^{2}+1254 r +553}{64 \left (r +\frac {7}{2}\right )^{2} \left (r +\frac {3}{2}\right )^{2} \left (r +\frac {5}{2}\right )^{2}}\) | \(\frac {1}{18}\) | \(\frac {-512 r^{6}-7104 r^{5}-39008 r^{4}-108448 r^{3}-159024 r^{2}-111708 r -25382}{\left (2 r +7\right )^{3} \left (2 r +3\right )^{3} \left (2 r +5\right )^{3}}\) | \(\frac {1}{54}\) |
| \(b_{4}\) | \(\frac {64 r^{6}+864 r^{5}+4336 r^{4}+9456 r^{3}+6188 r^{2}-6962 r -8827}{\left (2 r +7\right )^{2} \left (2 r +3\right )^{2} \left (2 r +5\right )^{2} \left (4 r^{2}+36 r +81\right )}\) | \(-{\frac {37}{1152}}\) | \(\frac {-2048 r^{9}-41472 r^{8}-337920 r^{7}-1328640 r^{6}-1820928 r^{5}+5221952 r^{4}+28076160 r^{3}+54186528 r^{2}+51148568 r +19690062}{\left (2 r +7\right )^{3} \left (2 r +3\right )^{3} \left (2 r +5\right )^{3} \left (2 r +9\right )^{3}}\) | \(\frac {1103}{13824}\) |
| \(b_{5}\) | \(\frac {-192 r^{7}-4192 r^{6}-37504 r^{5}-177440 r^{4}-475836 r^{3}-713390 r^{2}-537382 r -144193}{512 \left (r +\frac {7}{2}\right )^{2} \left (r +\frac {3}{2}\right )^{2} \left (r +\frac {5}{2}\right )^{2} \left (r +\frac {9}{2}\right )^{2} \left (r +\frac {11}{2}\right )^{2}}\) | \(-{\frac {17}{28800}}\) | \(\frac {36864 r^{11}+1288192 r^{10}+19947520 r^{9}+180280320 r^{8}+1053193728 r^{7}+4155296512 r^{6}+11206974720 r^{5}+20364921600 r^{4}+23722857104 r^{3}+15648572520 r^{2}+3848954672 r -641468604}{\left (2 r +7\right )^{3} \left (2 r +3\right )^{3} \left (2 r +5\right )^{3} \left (2 r +9\right )^{3} \left (2 r +11\right )^{3}}\) | \(-{\frac {19507}{1728000}}\) |
| \(b_{6}\) | \(\frac {-512 r^{9}-14592 r^{8}-172672 r^{7}-1084672 r^{6}-3760736 r^{5}-6170848 r^{4}+1039816 r^{3}+21534576 r^{2}+32373056 r +16074527}{\left (4 r^{2}+52 r +169\right ) \left (2 r +7\right )^{2} \left (2 r +3\right )^{2} \left (2 r +5\right )^{2} \left (2 r +9\right )^{2} \left (2 r +11\right )^{2}}\) | \(\frac {593}{259200}\) | \(\frac {12288 r^{14}+565248 r^{13}+11563008 r^{12}+137515008 r^{11}+1035142912 r^{10}+4919719424 r^{9}+12503602176 r^{8}-5334067712 r^{7}-184150708720 r^{6}-798954105504 r^{5}-1976636998320 r^{4}-3136747445936 r^{3}-3170341756011 r^{2}-1868576236054 r -490544059152}{32768 \left (r +\frac {7}{2}\right )^{3} \left (r +\frac {13}{2}\right )^{3} \left (r +\frac {3}{2}\right )^{3} \left (r +\frac {5}{2}\right )^{3} \left (r +\frac {9}{2}\right )^{3} \left (r +\frac {11}{2}\right )^{3}}\) | \(-{\frac {98531}{20736000}}\) |
| \(b_{7}\) | \(\frac {4096 r^{10}+164864 r^{9}+2890752 r^{8}+28990080 r^{7}+183413312 r^{6}+760790176 r^{5}+2078667344 r^{4}+3646802552 r^{3}+3838600668 r^{2}+2071188906 r +360317083}{16384 \left (r +\frac {7}{2}\right )^{2} \left (r +\frac {13}{2}\right )^{2} \left (r +\frac {3}{2}\right )^{2} \left (r +\frac {5}{2}\right )^{2} \left (r +\frac {9}{2}\right )^{2} \left (r +\frac {15}{2}\right )^{2} \left (r +\frac {11}{2}\right )^{2}}\) | \(-{\frac {1913}{12700800}}\) | \(-\frac {2 \left (1048576 r^{16}+69271552 r^{15}+2107146240 r^{14}+39130800128 r^{13}+495840399360 r^{12}+4537799933952 r^{11}+30955620671488 r^{10}+160076715744768 r^{9}+631451871167232 r^{8}+1894290553466112 r^{7}+4263178189826688 r^{6}+6992087314461792 r^{5}+7882874247211120 r^{4}+5312426146035744 r^{3}+1111888597228080 r^{2}-1032382163796462 r -606587551960635\right )}{\left (2 r +7\right )^{3} \left (2 r +13\right )^{3} \left (2 r +3\right )^{3} \left (2 r +5\right )^{3} \left (2 r +9\right )^{3} \left (2 r +15\right )^{3} \left (2 r +11\right )^{3}}\) | \(\frac {982189}{889056000}\) |
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-x +\frac {x^{2}}{4}+\frac {x^{3}}{18}-\frac {37 x^{4}}{1152}-\frac {17 x^{5}}{28800}+\frac {593 x^{6}}{259200}-\frac {1913 x^{7}}{12700800}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{\sqrt {x}}+\frac {\frac {3 x}{2}-\frac {13 x^{2}}{16}+\frac {x^{3}}{54}+\frac {1103 x^{4}}{13824}-\frac {19507 x^{5}}{1728000}-\frac {98531 x^{6}}{20736000}+\frac {982189 x^{7}}{889056000}+O\left (x^{8}\right )}{\sqrt {x}} \\ \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-x +\frac {x^{2}}{4}+\frac {x^{3}}{18}-\frac {37 x^{4}}{1152}-\frac {17 x^{5}}{28800}+\frac {593 x^{6}}{259200}-\frac {1913 x^{7}}{12700800}+O\left (x^{8}\right )\right )}{\sqrt {x}} + c_{2} \left (\frac {\left (1-x +\frac {x^{2}}{4}+\frac {x^{3}}{18}-\frac {37 x^{4}}{1152}-\frac {17 x^{5}}{28800}+\frac {593 x^{6}}{259200}-\frac {1913 x^{7}}{12700800}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{\sqrt {x}}+\frac {\frac {3 x}{2}-\frac {13 x^{2}}{16}+\frac {x^{3}}{54}+\frac {1103 x^{4}}{13824}-\frac {19507 x^{5}}{1728000}-\frac {98531 x^{6}}{20736000}+\frac {982189 x^{7}}{889056000}+O\left (x^{8}\right )}{\sqrt {x}}\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= \frac {c_{1} \left (1-x +\frac {x^{2}}{4}+\frac {x^{3}}{18}-\frac {37 x^{4}}{1152}-\frac {17 x^{5}}{28800}+\frac {593 x^{6}}{259200}-\frac {1913 x^{7}}{12700800}+O\left (x^{8}\right )\right )}{\sqrt {x}}+c_{2} \left (\frac {\left (1-x +\frac {x^{2}}{4}+\frac {x^{3}}{18}-\frac {37 x^{4}}{1152}-\frac {17 x^{5}}{28800}+\frac {593 x^{6}}{259200}-\frac {1913 x^{7}}{12700800}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{\sqrt {x}}+\frac {\frac {3 x}{2}-\frac {13 x^{2}}{16}+\frac {x^{3}}{54}+\frac {1103 x^{4}}{13824}-\frac {19507 x^{5}}{1728000}-\frac {98531 x^{6}}{20736000}+\frac {982189 x^{7}}{889056000}+O\left (x^{8}\right )}{\sqrt {x}}\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1} \left (1-x +\frac {x^{2}}{4}+\frac {x^{3}}{18}-\frac {37 x^{4}}{1152}-\frac {17 x^{5}}{28800}+\frac {593 x^{6}}{259200}-\frac {1913 x^{7}}{12700800}+O\left (x^{8}\right )\right )}{\sqrt {x}}+c_{2} \left (\frac {\left (1-x +\frac {x^{2}}{4}+\frac {x^{3}}{18}-\frac {37 x^{4}}{1152}-\frac {17 x^{5}}{28800}+\frac {593 x^{6}}{259200}-\frac {1913 x^{7}}{12700800}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{\sqrt {x}}+\frac {\frac {3 x}{2}-\frac {13 x^{2}}{16}+\frac {x^{3}}{54}+\frac {1103 x^{4}}{13824}-\frac {19507 x^{5}}{1728000}-\frac {98531 x^{6}}{20736000}+\frac {982189 x^{7}}{889056000}+O\left (x^{8}\right )}{\sqrt {x}}\right ) \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1} \left (1-x +\frac {x^{2}}{4}+\frac {x^{3}}{18}-\frac {37 x^{4}}{1152}-\frac {17 x^{5}}{28800}+\frac {593 x^{6}}{259200}-\frac {1913 x^{7}}{12700800}+O\left (x^{8}\right )\right )}{\sqrt {x}}+c_{2} \left (\frac {\left (1-x +\frac {x^{2}}{4}+\frac {x^{3}}{18}-\frac {37 x^{4}}{1152}-\frac {17 x^{5}}{28800}+\frac {593 x^{6}}{259200}-\frac {1913 x^{7}}{12700800}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{\sqrt {x}}+\frac {\frac {3 x}{2}-\frac {13 x^{2}}{16}+\frac {x^{3}}{54}+\frac {1103 x^{4}}{13824}-\frac {19507 x^{5}}{1728000}-\frac {98531 x^{6}}{20736000}+\frac {982189 x^{7}}{889056000}+O\left (x^{8}\right )}{\sqrt {x}}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 x^{2} y^{\prime \prime }+\left (2 x^{3}+2 x^{2}+8 x \right ) y^{\prime }+\left (3 x^{2}+5 x +1\right ) 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 {\left (3 x^{2}+5 x +1\right ) y}{4 x^{2}}-\frac {\left (x^{2}+x +4\right ) y^{\prime }}{2 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 {\left (x^{2}+x +4\right ) y^{\prime }}{2 x}+\frac {\left (3 x^{2}+5 x +1\right ) y}{4 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 {x^{2}+x +4}{2 x}, P_{3}\left (x \right )=\frac {3 x^{2}+5 x +1}{4 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}}}=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}{4} \\ {} & \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} \\ {} & {} & 4 x^{2} y^{\prime \prime }+2 x \left (x^{2}+x +4\right ) y^{\prime }+\left (3 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 y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot y^{\prime \prime }=\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+2 r \right )^{2} x^{r}+\left (a_{1} \left (3+2 r \right )^{2}+a_{0} \left (5+2 r \right )\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (2 k +2 r +1\right )^{2}+a_{k -1} \left (2 k +2 r +3\right )+a_{k -2} \left (2 k -1+2 r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (1+2 r \right )^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =-\frac {1}{2} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (3+2 r \right )^{2}+a_{0} \left (5+2 r \right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=-\frac {a_{0} \left (5+2 r \right )}{4 r^{2}+12 r +9} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (2 k +2 r +1\right )^{2}+a_{k -1} \left (2 k +2 r +3\right )+a_{k -2} \left (2 k -1+2 r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +2} \left (2 k +2 r +5\right )^{2}+a_{k +1} \left (2 k +7+2 r \right )+a_{k} \left (2 k +2 r +3\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {2 k a_{k}+2 k a_{k +1}+2 r a_{k}+2 r a_{k +1}+3 a_{k}+7 a_{k +1}}{\left (2 k +2 r +5\right )^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {1}{2} \\ {} & {} & a_{k +2}=-\frac {2 k a_{k}+2 k a_{k +1}+2 a_{k}+6 a_{k +1}}{\left (2 k +4\right )^{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {1}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {1}{2}}, a_{k +2}=-\frac {2 k a_{k}+2 k a_{k +1}+2 a_{k}+6 a_{k +1}}{\left (2 k +4\right )^{2}}, a_{1}=-a_{0}\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)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm <- No Liouvillian solutions exists -> 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 trying a solution in terms of MeijerG functions -> 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 `
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 81
Order:=8; dsolve(4*x^2*diff(y(x),x$2)+2*x*(4+x+x^2)*diff(y(x),x)+(1+5*x+3*x^2)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {\left (c_{2} \ln \left (x \right )+c_{1} \right ) \left (1-x +\frac {1}{4} x^{2}+\frac {1}{18} x^{3}-\frac {37}{1152} x^{4}-\frac {17}{28800} x^{5}+\frac {593}{259200} x^{6}-\frac {1913}{12700800} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+\left (\frac {3}{2} x -\frac {13}{16} x^{2}+\frac {1}{54} x^{3}+\frac {1103}{13824} x^{4}-\frac {19507}{1728000} x^{5}-\frac {98531}{20736000} x^{6}+\frac {982189}{889056000} x^{7}+\operatorname {O}\left (x^{8}\right )\right ) c_{2}}{\sqrt {x}} \]
✓ Solution by Mathematica
Time used: 0.01 (sec). Leaf size: 172
AsymptoticDSolveValue[4*x^2*y''[x]+2*x*(4+x+x^2)*y'[x]+(1+5*x+3*x^2)*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to \frac {c_1 \left (-\frac {1913 x^7}{12700800}+\frac {593 x^6}{259200}-\frac {17 x^5}{28800}-\frac {37 x^4}{1152}+\frac {x^3}{18}+\frac {x^2}{4}-x+1\right )}{\sqrt {x}}+c_2 \left (\frac {\frac {982189 x^7}{889056000}-\frac {98531 x^6}{20736000}-\frac {19507 x^5}{1728000}+\frac {1103 x^4}{13824}+\frac {x^3}{54}-\frac {13 x^2}{16}+\frac {3 x}{2}}{\sqrt {x}}+\frac {\left (-\frac {1913 x^7}{12700800}+\frac {593 x^6}{259200}-\frac {17 x^5}{28800}-\frac {37 x^4}{1152}+\frac {x^3}{18}+\frac {x^2}{4}-x+1\right ) \log (x)}{\sqrt {x}}\right ) \]