Internal problem ID [1298]
Internal file name [OUTPUT/1299_Sunday_June_05_2022_02_08_58_AM_10712497/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. 7.5 THE METHOD OF
FROBENIUS I. Exercises 7.5. Page 358
Problem number: 4.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference not integer"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {4 x^{2} y^{\prime \prime }+x \left (4 x^{2}+2 x +7\right ) y^{\prime }-\left (-7 x^{2}-4 x +1\right ) y=0} \] 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. \[ 4 x^{2} y^{\prime \prime }+\left (4 x^{3}+2 x^{2}+7 x \right ) y^{\prime }+\left (7 x^{2}+4 x -1\right ) y = 0 \] 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 {4 x^{2}+2 x +7}{4 x}\\ q(x) &= \frac {7 x^{2}+4 x -1}{4 x^{2}}\\ \end {align*}
| \(p(x)=\frac {4 x^{2}+2 x +7}{4 x}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = \infty \) | \(\text {``regular''}\) |
| \(x = -\infty \) | \(\text {``regular''}\) |
| \(q(x)=\frac {7 x^{2}+4 x -1}{4 x^{2}}\) | |
| 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 \[ 4 x^{2} y^{\prime \prime }+\left (4 x^{3}+2 x^{2}+7 x \right ) y^{\prime }+\left (7 x^{2}+4 x -1\right ) y = 0 \] 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 (4 x^{3}+2 x^{2}+7 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (7 x^{2}+4 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 }}4 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 }}7 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}7 x^{n +r +2} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{1+n +r} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-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 }}4 x^{n +r +2} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}4 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 }}7 x^{n +r +2} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}7 a_{n -2} x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}4 x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}4 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 }}4 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 }}7 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}7 a_{n -2} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+7 x^{n +r} a_{n} \left (n +r \right )-a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 4 x^{r} a_{0} r \left (-1+r \right )+7 x^{r} a_{0} r -a_{0} x^{r} = 0 \] Or \[ \left (4 x^{r} r \left (-1+r \right )+7 x^{r} r -x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (4 r^{2}+3 r -1\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 4 r^{2}+3 r -1 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {1}{4}}\\ r_2 &= -1 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (4 r^{2}+3 r -1\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {1}{4}}, -1\right ]\).
Since \(r_1 - r_2 = {\frac {5}{4}}\) is not 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 ) &= 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 ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +\frac {1}{4}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -1} \end {align*}
We start by finding \(y_{1}\left (x \right )\). 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 \[ a_{1} = -\frac {2}{4 r +3} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} 4 a_{n} \left (n +r \right ) \left (n +r -1\right )+4 a_{n -2} \left (n +r -2\right )+2 a_{n -1} \left (n +r -1\right )+7 a_{n} \left (n +r \right )+7 a_{n -2}+4 a_{n -1}-a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {4 n a_{n -2}+2 n a_{n -1}+4 r a_{n -2}+2 r a_{n -1}-a_{n -2}+2 a_{n -1}}{4 n^{2}+8 n r +4 r^{2}+3 n +3 r -1}\tag {4} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ a_{n} = \frac {\left (-8 a_{n -2}-4 a_{n -1}\right ) n -5 a_{n -1}}{8 n^{2}+10 n}\tag {5} \] 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.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2}{4 r +3}\) | \(-{\frac {1}{2}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {-16 r^{2}-36 r -9}{16 r^{3}+88 r^{2}+141 r +63} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ a_{2}=-{\frac {19}{104}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2}{4 r +3}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {-16 r^{2}-36 r -9}{16 r^{3}+88 r^{2}+141 r +63}\) | \(-{\frac {19}{104}}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {64 r^{3}+440 r^{2}+892 r +534}{64 r^{5}+784 r^{4}+3644 r^{3}+7931 r^{2}+7905 r +2772} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ a_{3}={\frac {1571}{10608}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2}{4 r +3}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {-16 r^{2}-36 r -9}{16 r^{3}+88 r^{2}+141 r +63}\) | \(-{\frac {19}{104}}\) |
| \(a_{3}\) | \(\frac {64 r^{3}+440 r^{2}+892 r +534}{64 r^{5}+784 r^{4}+3644 r^{3}+7931 r^{2}+7905 r +2772}\) | \(\frac {1571}{10608}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {256 r^{5}+3136 r^{4}+13968 r^{3}+26804 r^{2}+19001 r +600}{256 r^{7}+5376 r^{6}+46816 r^{5}+218064 r^{4}+582505 r^{3}+882588 r^{2}+689895 r +207900} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ a_{4}={\frac {3225}{198016}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2}{4 r +3}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {-16 r^{2}-36 r -9}{16 r^{3}+88 r^{2}+141 r +63}\) | \(-{\frac {19}{104}}\) |
| \(a_{3}\) | \(\frac {64 r^{3}+440 r^{2}+892 r +534}{64 r^{5}+784 r^{4}+3644 r^{3}+7931 r^{2}+7905 r +2772}\) | \(\frac {1571}{10608}\) |
| \(a_{4}\) | \(\frac {256 r^{5}+3136 r^{4}+13968 r^{3}+26804 r^{2}+19001 r +600}{256 r^{7}+5376 r^{6}+46816 r^{5}+218064 r^{4}+582505 r^{3}+882588 r^{2}+689895 r +207900}\) | \(\frac {3225}{198016}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-1536 r^{6}-30208 r^{5}-236640 r^{4}-938240 r^{3}-1962774 r^{2}-2015622 r -768150}{\left (4 r^{2}+43 r +114\right ) \left (256 r^{7}+5376 r^{6}+46816 r^{5}+218064 r^{4}+582505 r^{3}+882588 r^{2}+689895 r +207900\right )} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ a_{5}=-{\frac {752183}{29702400}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {2}{4 r +3}\) | \(-{\frac {1}{2}}\) |
| \(a_{2}\) | \(\frac {-16 r^{2}-36 r -9}{16 r^{3}+88 r^{2}+141 r +63}\) | \(-{\frac {19}{104}}\) |
| \(a_{3}\) | \(\frac {64 r^{3}+440 r^{2}+892 r +534}{64 r^{5}+784 r^{4}+3644 r^{3}+7931 r^{2}+7905 r +2772}\) | \(\frac {1571}{10608}\) |
| \(a_{4}\) | \(\frac {256 r^{5}+3136 r^{4}+13968 r^{3}+26804 r^{2}+19001 r +600}{256 r^{7}+5376 r^{6}+46816 r^{5}+218064 r^{4}+582505 r^{3}+882588 r^{2}+689895 r +207900}\) | \(\frac {3225}{198016}\) |
| \(a_{5}\) | \(\frac {-1536 r^{6}-30208 r^{5}-236640 r^{4}-938240 r^{3}-1962774 r^{2}-2015622 r -768150}{\left (4 r^{2}+43 r +114\right ) \left (256 r^{7}+5376 r^{6}+46816 r^{5}+218064 r^{4}+582505 r^{3}+882588 r^{2}+689895 r +207900\right )}\) | \(-{\frac {752183}{29702400}}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= 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}\dots \right ) \\ &= x^{\frac {1}{4}} \left (1-\frac {x}{2}-\frac {19 x^{2}}{104}+\frac {1571 x^{3}}{10608}+\frac {3225 x^{4}}{198016}-\frac {752183 x^{5}}{29702400}+O\left (x^{6}\right )\right ) \end {align*}
Now the second solution \(y_{2}\left (x \right )\) is found. Eq (2B) derived above is now 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 \(b_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ b_{1} = -\frac {2}{4 r +3} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} 4 b_{n} \left (n +r \right ) \left (n +r -1\right )+4 b_{n -2} \left (n +r -2\right )+2 b_{n -1} \left (n +r -1\right )+7 b_{n} \left (n +r \right )+7 b_{n -2}+4 b_{n -1}-b_{n} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {4 n b_{n -2}+2 n b_{n -1}+4 r b_{n -2}+2 r b_{n -1}-b_{n -2}+2 b_{n -1}}{4 n^{2}+8 n r +4 r^{2}+3 n +3 r -1}\tag {4} \] Which for the root \(r = -1\) becomes \[ b_{n} = \frac {\left (-4 b_{n -2}-2 b_{n -1}\right ) n +5 b_{n -2}}{4 n^{2}-5 n}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -1\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {2}{4 r +3}\) | \(2\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {-16 r^{2}-36 r -9}{16 r^{3}+88 r^{2}+141 r +63} \] Which for the root \(r = -1\) becomes \[ b_{2}=-{\frac {11}{6}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {2}{4 r +3}\) | \(2\) |
| \(b_{2}\) | \(\frac {-16 r^{2}-36 r -9}{16 r^{3}+88 r^{2}+141 r +63}\) | \(-{\frac {11}{6}}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {64 r^{3}+440 r^{2}+892 r +534}{64 r^{5}+784 r^{4}+3644 r^{3}+7931 r^{2}+7905 r +2772} \] Which for the root \(r = -1\) becomes \[ b_{3}=-{\frac {1}{7}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {2}{4 r +3}\) | \(2\) |
| \(b_{2}\) | \(\frac {-16 r^{2}-36 r -9}{16 r^{3}+88 r^{2}+141 r +63}\) | \(-{\frac {11}{6}}\) |
| \(b_{3}\) | \(\frac {64 r^{3}+440 r^{2}+892 r +534}{64 r^{5}+784 r^{4}+3644 r^{3}+7931 r^{2}+7905 r +2772}\) | \(-{\frac {1}{7}}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {256 r^{5}+3136 r^{4}+13968 r^{3}+26804 r^{2}+19001 r +600}{256 r^{7}+5376 r^{6}+46816 r^{5}+218064 r^{4}+582505 r^{3}+882588 r^{2}+689895 r +207900} \] Which for the root \(r = -1\) becomes \[ b_{4}={\frac {895}{1848}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {2}{4 r +3}\) | \(2\) |
| \(b_{2}\) | \(\frac {-16 r^{2}-36 r -9}{16 r^{3}+88 r^{2}+141 r +63}\) | \(-{\frac {11}{6}}\) |
| \(b_{3}\) | \(\frac {64 r^{3}+440 r^{2}+892 r +534}{64 r^{5}+784 r^{4}+3644 r^{3}+7931 r^{2}+7905 r +2772}\) | \(-{\frac {1}{7}}\) |
| \(b_{4}\) | \(\frac {256 r^{5}+3136 r^{4}+13968 r^{3}+26804 r^{2}+19001 r +600}{256 r^{7}+5376 r^{6}+46816 r^{5}+218064 r^{4}+582505 r^{3}+882588 r^{2}+689895 r +207900}\) | \(\frac {895}{1848}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {-1536 r^{6}-30208 r^{5}-236640 r^{4}-938240 r^{3}-1962774 r^{2}-2015622 r -768150}{\left (4 r^{2}+43 r +114\right ) \left (256 r^{7}+5376 r^{6}+46816 r^{5}+218064 r^{4}+582505 r^{3}+882588 r^{2}+689895 r +207900\right )} \] Which for the root \(r = -1\) becomes \[ b_{5}=-{\frac {499}{13860}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {2}{4 r +3}\) | \(2\) |
| \(b_{2}\) | \(\frac {-16 r^{2}-36 r -9}{16 r^{3}+88 r^{2}+141 r +63}\) | \(-{\frac {11}{6}}\) |
| \(b_{3}\) | \(\frac {64 r^{3}+440 r^{2}+892 r +534}{64 r^{5}+784 r^{4}+3644 r^{3}+7931 r^{2}+7905 r +2772}\) | \(-{\frac {1}{7}}\) |
| \(b_{4}\) | \(\frac {256 r^{5}+3136 r^{4}+13968 r^{3}+26804 r^{2}+19001 r +600}{256 r^{7}+5376 r^{6}+46816 r^{5}+218064 r^{4}+582505 r^{3}+882588 r^{2}+689895 r +207900}\) | \(\frac {895}{1848}\) |
| \(b_{5}\) | \(\frac {-1536 r^{6}-30208 r^{5}-236640 r^{4}-938240 r^{3}-1962774 r^{2}-2015622 r -768150}{\left (4 r^{2}+43 r +114\right ) \left (256 r^{7}+5376 r^{6}+46816 r^{5}+218064 r^{4}+582505 r^{3}+882588 r^{2}+689895 r +207900\right )}\) | \(-{\frac {499}{13860}}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x^{\frac {1}{4}} \left (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 \right ) \\ &= \frac {1+2 x -\frac {11 x^{2}}{6}-\frac {x^{3}}{7}+\frac {895 x^{4}}{1848}-\frac {499 x^{5}}{13860}+O\left (x^{6}\right )}{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 ) \\ &= c_{1} x^{\frac {1}{4}} \left (1-\frac {x}{2}-\frac {19 x^{2}}{104}+\frac {1571 x^{3}}{10608}+\frac {3225 x^{4}}{198016}-\frac {752183 x^{5}}{29702400}+O\left (x^{6}\right )\right ) + \frac {c_{2} \left (1+2 x -\frac {11 x^{2}}{6}-\frac {x^{3}}{7}+\frac {895 x^{4}}{1848}-\frac {499 x^{5}}{13860}+O\left (x^{6}\right )\right )}{x} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {1}{4}} \left (1-\frac {x}{2}-\frac {19 x^{2}}{104}+\frac {1571 x^{3}}{10608}+\frac {3225 x^{4}}{198016}-\frac {752183 x^{5}}{29702400}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+2 x -\frac {11 x^{2}}{6}-\frac {x^{3}}{7}+\frac {895 x^{4}}{1848}-\frac {499 x^{5}}{13860}+O\left (x^{6}\right )\right )}{x} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {1}{4}} \left (1-\frac {x}{2}-\frac {19 x^{2}}{104}+\frac {1571 x^{3}}{10608}+\frac {3225 x^{4}}{198016}-\frac {752183 x^{5}}{29702400}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+2 x -\frac {11 x^{2}}{6}-\frac {x^{3}}{7}+\frac {895 x^{4}}{1848}-\frac {499 x^{5}}{13860}+O\left (x^{6}\right )\right )}{x} \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {1}{4}} \left (1-\frac {x}{2}-\frac {19 x^{2}}{104}+\frac {1571 x^{3}}{10608}+\frac {3225 x^{4}}{198016}-\frac {752183 x^{5}}{29702400}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+2 x -\frac {11 x^{2}}{6}-\frac {x^{3}}{7}+\frac {895 x^{4}}{1848}-\frac {499 x^{5}}{13860}+O\left (x^{6}\right )\right )}{x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 x^{2} y^{\prime \prime }+\left (4 x^{3}+2 x^{2}+7 x \right ) y^{\prime }+\left (7 x^{2}+4 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 (7 x^{2}+4 x -1\right ) y}{4 x^{2}}-\frac {\left (4 x^{2}+2 x +7\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} \\ {} & {} & y^{\prime \prime }+\frac {\left (4 x^{2}+2 x +7\right ) y^{\prime }}{4 x}+\frac {\left (7 x^{2}+4 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 {4 x^{2}+2 x +7}{4 x}, P_{3}\left (x \right )=\frac {7 x^{2}+4 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}}}=\frac {7}{4} \\ {} & \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 }+x \left (4 x^{2}+2 x +7\right ) y^{\prime }+\left (7 x^{2}+4 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+r \right ) \left (-1+4 r \right ) x^{r}+\left (a_{1} \left (2+r \right ) \left (3+4 r \right )+2 a_{0} \left (2+r \right )\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (k +r +1\right ) \left (4 k +4 r -1\right )+2 a_{k -1} \left (k +r +1\right )+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+r \right ) \left (-1+4 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-1, \frac {1}{4}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (2+r \right ) \left (3+4 r \right )+2 a_{0} \left (2+r \right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=-\frac {2 a_{0}}{3+4 r} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 4 \left (k +r +1\right ) \left (k +r -\frac {1}{4}\right ) a_{k}+\left (4 a_{k -2}+2 a_{k -1}\right ) k +\left (4 a_{k -2}+2 a_{k -1}\right ) r -a_{k -2}+2 a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & 4 \left (k +3+r \right ) \left (k +\frac {7}{4}+r \right ) a_{k +2}+\left (4 a_{k}+2 a_{k +1}\right ) \left (k +2\right )+\left (4 a_{k}+2 a_{k +1}\right ) r -a_{k}+2 a_{k +1}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {4 k a_{k}+2 k a_{k +1}+4 r a_{k}+2 r a_{k +1}+7 a_{k}+6 a_{k +1}}{\left (k +3+r \right ) \left (4 k +4 r +7\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-1 \\ {} & {} & a_{k +2}=-\frac {4 k a_{k}+2 k a_{k +1}+3 a_{k}+4 a_{k +1}}{\left (k +2\right ) \left (4 k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -1}, a_{k +2}=-\frac {4 k a_{k}+2 k a_{k +1}+3 a_{k}+4 a_{k +1}}{\left (k +2\right ) \left (4 k +3\right )}, a_{1}=2 a_{0}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{4} \\ {} & {} & a_{k +2}=-\frac {4 k a_{k}+2 k a_{k +1}+8 a_{k}+\frac {13}{2} a_{k +1}}{\left (k +\frac {13}{4}\right ) \left (4 k +8\right )} \\ \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 {4 k a_{k}+2 k a_{k +1}+8 a_{k}+\frac {13}{2} a_{k +1}}{\left (k +\frac {13}{4}\right ) \left (4 k +8\right )}, a_{1}=-\frac {a_{0}}{2}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {1}{4}}\right ), a_{k +2}=-\frac {4 k a_{k}+2 k a_{1+k}+3 a_{k}+4 a_{1+k}}{\left (k +2\right ) \left (4 k +3\right )}, a_{1}=2 a_{0}, b_{k +2}=-\frac {4 k b_{k}+2 k b_{1+k}+8 b_{k}+\frac {13}{2} b_{1+k}}{\left (k +\frac {13}{4}\right ) \left (4 k +8\right )}, b_{1}=-\frac {b_{0}}{2}\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: 47
Order:=6; dsolve(4*x^2*diff(y(x),x$2)+x*(7+2*x+4*x^2)*diff(y(x),x)-(1-4*x-7*x^2)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{2} x^{\frac {5}{4}} \left (1-\frac {1}{2} x -\frac {19}{104} x^{2}+\frac {1571}{10608} x^{3}+\frac {3225}{198016} x^{4}-\frac {752183}{29702400} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{1} \left (1+2 x -\frac {11}{6} x^{2}-\frac {1}{7} x^{3}+\frac {895}{1848} x^{4}-\frac {499}{13860} x^{5}+\operatorname {O}\left (x^{6}\right )\right )}{x} \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 86
AsymptoticDSolveValue[4*x^2*y''[x]+x*(7+2*x+4*x^2)*y'[x]-(1-4*x-7*x^2)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \sqrt [4]{x} \left (-\frac {752183 x^5}{29702400}+\frac {3225 x^4}{198016}+\frac {1571 x^3}{10608}-\frac {19 x^2}{104}-\frac {x}{2}+1\right )+\frac {c_2 \left (-\frac {499 x^5}{13860}+\frac {895 x^4}{1848}-\frac {x^3}{7}-\frac {11 x^2}{6}+2 x+1\right )}{x} \]