Internal problem ID [6962]
Internal file name [OUTPUT/6205_Friday_August_12_2022_11_05_19_PM_40367264/index.tex]
Book: Elementary differential equations. Rainville, Bedient, Bedient. Prentice Hall. NJ. 8th
edition. 1997.
Section: CHAPTER 18. Power series solutions. 18.6. Indicial Equation with Equal Roots.
Exercises page 373
Problem number: 15.
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]]
\[ \boxed {4 x^{2} y^{\prime \prime }+8 x \left (x +1\right ) y^{\prime }+y=0} \] 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. \[ 4 x^{2} y^{\prime \prime }+\left (8 x^{2}+8 x \right ) y^{\prime }+y = 0 \] 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}\\ q(x) &= \frac {1}{4 x^{2}}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ 4 x^{2} y^{\prime \prime }+\left (8 x^{2}+8 x \right ) y^{\prime }+y = 0 \] 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 (8 x^{2}+8 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -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 }}8 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 }}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^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}8 a_{n -1} \left (n +r -1\right ) 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 =1}{\sum }}8 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 =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 \[ \left (2 r +1\right )^{2} x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ \left (2 r +1\right )^{2} = 0 \] 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 \[ \left (2 r +1\right )^{2} x^{r} = 0 \] 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 \(a_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} 4 a_{n} \left (n +r \right ) \left (n +r -1\right )+8 a_{n -1} \left (n +r -1\right )+8 a_{n} \left (n +r \right )+a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {8 a_{n -1} \left (n +r -1\right )}{4 n^{2}+8 n r +4 r^{2}+4 n +4 r +1}\tag {4} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ a_{n} = \frac {a_{n -1} \left (3-2 n \right )}{n^{2}}\tag {5} \] 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.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ a_{1}=-\frac {8 r}{\left (3+2 r \right )^{2}} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ a_{1}=1 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {8 r}{\left (3+2 r \right )^{2}}\) | \(1\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {64 r \left (1+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2}} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ a_{2}=-{\frac {1}{4}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {8 r}{\left (3+2 r \right )^{2}}\) | \(1\) |
| \(a_{2}\) | \(\frac {64 r \left (1+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2}}\) | \(-{\frac {1}{4}}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {512 r \left (1+r \right ) \left (2+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2}} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ a_{3}={\frac {1}{12}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {8 r}{\left (3+2 r \right )^{2}}\) | \(1\) |
| \(a_{2}\) | \(\frac {64 r \left (1+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2}}\) | \(-{\frac {1}{4}}\) |
| \(a_{3}\) | \(-\frac {512 r \left (1+r \right ) \left (2+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2}}\) | \(\frac {1}{12}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {4096 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (9+2 r \right )^{2}} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ a_{4}=-{\frac {5}{192}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {8 r}{\left (3+2 r \right )^{2}}\) | \(1\) |
| \(a_{2}\) | \(\frac {64 r \left (1+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2}}\) | \(-{\frac {1}{4}}\) |
| \(a_{3}\) | \(-\frac {512 r \left (1+r \right ) \left (2+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2}}\) | \(\frac {1}{12}\) |
| \(a_{4}\) | \(\frac {4096 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (9+2 r \right )^{2}}\) | \(-{\frac {5}{192}}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {32768 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (9+2 r \right )^{2} \left (11+2 r \right )^{2}} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ a_{5}={\frac {7}{960}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {8 r}{\left (3+2 r \right )^{2}}\) | \(1\) |
| \(a_{2}\) | \(\frac {64 r \left (1+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2}}\) | \(-{\frac {1}{4}}\) |
| \(a_{3}\) | \(-\frac {512 r \left (1+r \right ) \left (2+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2}}\) | \(\frac {1}{12}\) |
| \(a_{4}\) | \(\frac {4096 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (9+2 r \right )^{2}}\) | \(-{\frac {5}{192}}\) |
| \(a_{5}\) | \(-\frac {32768 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (9+2 r \right )^{2} \left (11+2 r \right )^{2}}\) | \(\frac {7}{960}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {262144 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (9+2 r \right )^{2} \left (11+2 r \right )^{2} \left (13+2 r \right )^{2}} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ a_{6}=-{\frac {7}{3840}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {8 r}{\left (3+2 r \right )^{2}}\) | \(1\) |
| \(a_{2}\) | \(\frac {64 r \left (1+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2}}\) | \(-{\frac {1}{4}}\) |
| \(a_{3}\) | \(-\frac {512 r \left (1+r \right ) \left (2+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2}}\) | \(\frac {1}{12}\) |
| \(a_{4}\) | \(\frac {4096 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (9+2 r \right )^{2}}\) | \(-{\frac {5}{192}}\) |
| \(a_{5}\) | \(-\frac {32768 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (9+2 r \right )^{2} \left (11+2 r \right )^{2}}\) | \(\frac {7}{960}\) |
| \(a_{6}\) | \(\frac {262144 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (9+2 r \right )^{2} \left (11+2 r \right )^{2} \left (13+2 r \right )^{2}}\) | \(-{\frac {7}{3840}}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=-\frac {2097152 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right ) \left (6+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (9+2 r \right )^{2} \left (11+2 r \right )^{2} \left (13+2 r \right )^{2} \left (15+2 r \right )^{2}} \] Which for the root \(r = -{\frac {1}{2}}\) becomes \[ a_{7}={\frac {11}{26880}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {8 r}{\left (3+2 r \right )^{2}}\) | \(1\) |
| \(a_{2}\) | \(\frac {64 r \left (1+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2}}\) | \(-{\frac {1}{4}}\) |
| \(a_{3}\) | \(-\frac {512 r \left (1+r \right ) \left (2+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2}}\) | \(\frac {1}{12}\) |
| \(a_{4}\) | \(\frac {4096 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (9+2 r \right )^{2}}\) | \(-{\frac {5}{192}}\) |
| \(a_{5}\) | \(-\frac {32768 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (9+2 r \right )^{2} \left (11+2 r \right )^{2}}\) | \(\frac {7}{960}\) |
| \(a_{6}\) | \(\frac {262144 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (9+2 r \right )^{2} \left (11+2 r \right )^{2} \left (13+2 r \right )^{2}}\) | \(-{\frac {7}{3840}}\) |
| \(a_{7}\) | \(-\frac {2097152 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right ) \left (6+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (9+2 r \right )^{2} \left (11+2 r \right )^{2} \left (13+2 r \right )^{2} \left (15+2 r \right )^{2}}\) | \(\frac {11}{26880}\) |
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 {x +1-\frac {x^{2}}{4}+\frac {x^{3}}{12}-\frac {5 x^{4}}{192}+\frac {7 x^{5}}{960}-\frac {7 x^{6}}{3840}+\frac {11 x^{7}}{26880}+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 |
| \(b_{1}\) | \(-\frac {8 r}{\left (3+2 r \right )^{2}}\) | \(1\) | \(\frac {-24+16 r}{\left (3+2 r \right )^{3}}\) | \(-4\) |
| \(b_{2}\) | \(\frac {64 r \left (1+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2}}\) | \(-{\frac {1}{4}}\) | \(\frac {-512 r^{3}-768 r^{2}+896 r +960}{\left (3+2 r \right )^{3} \left (5+2 r \right )^{3}}\) | \(\frac {3}{4}\) |
| \(b_{3}\) | \(-\frac {512 r \left (1+r \right ) \left (2+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2}}\) | \(\frac {1}{12}\) | \(\frac {12288 r^{5}+79872 r^{4}+152576 r^{3}+23040 r^{2}-177152 r -107520}{\left (3+2 r \right )^{3} \left (5+2 r \right )^{3} \left (7+2 r \right )^{3}}\) | \(-{\frac {1}{4}}\) |
| \(b_{4}\) | \(\frac {4096 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (9+2 r \right )^{2}}\) | \(-{\frac {5}{192}}\) | \(-\frac {8192 \left (32 r^{7}+432 r^{6}+2256 r^{5}+5544 r^{4}+5590 r^{3}-1089 r^{2}-5931 r -2835\right )}{\left (3+2 r \right )^{3} \left (5+2 r \right )^{3} \left (7+2 r \right )^{3} \left (9+2 r \right )^{3}}\) | \(\frac {31}{384}\) |
| \(b_{5}\) | \(-\frac {32768 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (9+2 r \right )^{2} \left (11+2 r \right )^{2}}\) | \(\frac {7}{960}\) | \(\frac {5242880 r^{9}+117964800 r^{8}+1114112000 r^{7}+5700321280 r^{6}+16873488384 r^{5}+27864104960 r^{4}+19672924160 r^{3}-7359528960 r^{2}-19703660544 r -8174960640}{\left (3+2 r \right )^{3} \left (5+2 r \right )^{3} \left (7+2 r \right )^{3} \left (9+2 r \right )^{3} \left (11+2 r \right )^{3}}\) | \(-{\frac {3}{128}}\) |
| \(b_{6}\) | \(\frac {262144 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (9+2 r \right )^{2} \left (11+2 r \right )^{2} \left (13+2 r \right )^{2}}\) | \(-{\frac {7}{3840}}\) | \(-\frac {262144 \left (384 r^{11}+12864 r^{10}+188320 r^{9}+1578960 r^{8}+8334888 r^{7}+28576476 r^{6}+62810790 r^{5}+81809655 r^{4}+45531428 r^{3}-21723165 r^{2}-43076700 r -16216200\right )}{\left (3+2 r \right )^{3} \left (5+2 r \right )^{3} \left (7+2 r \right )^{3} \left (9+2 r \right )^{3} \left (11+2 r \right )^{3} \left (13+2 r \right )^{3}}\) | \(\frac {419}{69120}\) |
| \(b_{7}\) | \(-\frac {2097152 r \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right ) \left (6+r \right )}{\left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (9+2 r \right )^{2} \left (11+2 r \right )^{2} \left (13+2 r \right )^{2} \left (15+2 r \right )^{2}}\) | \(\frac {11}{26880}\) | \(\frac {1879048192 r^{13}+87375740928 r^{12}+1819388411904 r^{11}+22375940751360 r^{10}+180411527331840 r^{9}+999725218136064 r^{8}+3874472704606208 r^{7}+10442496100270080 r^{6}+18911804604284928 r^{5}+20926047304286208 r^{4}+9814395718729728 r^{3}-5255708949872640 r^{2}-8742595932979200 r -3060705263616000}{\left (3+2 r \right )^{3} \left (5+2 r \right )^{3} \left (7+2 r \right )^{3} \left (9+2 r \right )^{3} \left (11+2 r \right )^{3} \left (13+2 r \right )^{3} \left (15+2 r \right )^{3}}\) | \(-{\frac {97}{69120}}\) |
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 (x +1-\frac {x^{2}}{4}+\frac {x^{3}}{12}-\frac {5 x^{4}}{192}+\frac {7 x^{5}}{960}-\frac {7 x^{6}}{3840}+\frac {11 x^{7}}{26880}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{\sqrt {x}}+\frac {-4 x +\frac {3 x^{2}}{4}-\frac {x^{3}}{4}+\frac {31 x^{4}}{384}-\frac {3 x^{5}}{128}+\frac {419 x^{6}}{69120}-\frac {97 x^{7}}{69120}+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 (x +1-\frac {x^{2}}{4}+\frac {x^{3}}{12}-\frac {5 x^{4}}{192}+\frac {7 x^{5}}{960}-\frac {7 x^{6}}{3840}+\frac {11 x^{7}}{26880}+O\left (x^{8}\right )\right )}{\sqrt {x}} + c_{2} \left (\frac {\left (x +1-\frac {x^{2}}{4}+\frac {x^{3}}{12}-\frac {5 x^{4}}{192}+\frac {7 x^{5}}{960}-\frac {7 x^{6}}{3840}+\frac {11 x^{7}}{26880}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{\sqrt {x}}+\frac {-4 x +\frac {3 x^{2}}{4}-\frac {x^{3}}{4}+\frac {31 x^{4}}{384}-\frac {3 x^{5}}{128}+\frac {419 x^{6}}{69120}-\frac {97 x^{7}}{69120}+O\left (x^{8}\right )}{\sqrt {x}}\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= \frac {c_{1} \left (x +1-\frac {x^{2}}{4}+\frac {x^{3}}{12}-\frac {5 x^{4}}{192}+\frac {7 x^{5}}{960}-\frac {7 x^{6}}{3840}+\frac {11 x^{7}}{26880}+O\left (x^{8}\right )\right )}{\sqrt {x}}+c_{2} \left (\frac {\left (x +1-\frac {x^{2}}{4}+\frac {x^{3}}{12}-\frac {5 x^{4}}{192}+\frac {7 x^{5}}{960}-\frac {7 x^{6}}{3840}+\frac {11 x^{7}}{26880}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{\sqrt {x}}+\frac {-4 x +\frac {3 x^{2}}{4}-\frac {x^{3}}{4}+\frac {31 x^{4}}{384}-\frac {3 x^{5}}{128}+\frac {419 x^{6}}{69120}-\frac {97 x^{7}}{69120}+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 (x +1-\frac {x^{2}}{4}+\frac {x^{3}}{12}-\frac {5 x^{4}}{192}+\frac {7 x^{5}}{960}-\frac {7 x^{6}}{3840}+\frac {11 x^{7}}{26880}+O\left (x^{8}\right )\right )}{\sqrt {x}}+c_{2} \left (\frac {\left (x +1-\frac {x^{2}}{4}+\frac {x^{3}}{12}-\frac {5 x^{4}}{192}+\frac {7 x^{5}}{960}-\frac {7 x^{6}}{3840}+\frac {11 x^{7}}{26880}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{\sqrt {x}}+\frac {-4 x +\frac {3 x^{2}}{4}-\frac {x^{3}}{4}+\frac {31 x^{4}}{384}-\frac {3 x^{5}}{128}+\frac {419 x^{6}}{69120}-\frac {97 x^{7}}{69120}+O\left (x^{8}\right )}{\sqrt {x}}\right ) \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1} \left (x +1-\frac {x^{2}}{4}+\frac {x^{3}}{12}-\frac {5 x^{4}}{192}+\frac {7 x^{5}}{960}-\frac {7 x^{6}}{3840}+\frac {11 x^{7}}{26880}+O\left (x^{8}\right )\right )}{\sqrt {x}}+c_{2} \left (\frac {\left (x +1-\frac {x^{2}}{4}+\frac {x^{3}}{12}-\frac {5 x^{4}}{192}+\frac {7 x^{5}}{960}-\frac {7 x^{6}}{3840}+\frac {11 x^{7}}{26880}+O\left (x^{8}\right )\right ) \ln \left (x \right )}{\sqrt {x}}+\frac {-4 x +\frac {3 x^{2}}{4}-\frac {x^{3}}{4}+\frac {31 x^{4}}{384}-\frac {3 x^{5}}{128}+\frac {419 x^{6}}{69120}-\frac {97 x^{7}}{69120}+O\left (x^{8}\right )}{\sqrt {x}}\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 x^{2} \left (\frac {d}{d x}y^{\prime }\right )+\left (8 x^{2}+8 x \right ) y^{\prime }+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 {y}{4 x^{2}}-\frac {2 \left (x +1\right ) y^{\prime }}{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 {2 \left (x +1\right ) y^{\prime }}{x}+\frac {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 {2 \left (x +1\right )}{x}, P_{3}\left (x \right )=\frac {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} \left (\frac {d}{d x}y^{\prime }\right )+8 x \left (x +1\right ) y^{\prime }+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^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..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^{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+2 r \right )^{2} x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (2 k +2 r +1\right )^{2}+8 a_{k -1} \left (k +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+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 in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (2 k +2 r +1\right )^{2}+8 a_{k -1} \left (k +r -1\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +1} \left (2 k +3+2 r \right )^{2}+8 a_{k} \left (k +r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {8 a_{k} \left (k +r \right )}{\left (2 k +3+2 r \right )^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {1}{2} \\ {} & {} & a_{k +1}=-\frac {8 a_{k} \left (k -\frac {1}{2}\right )}{\left (2 k +2\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 +1}=-\frac {8 a_{k} \left (k -\frac {1}{2}\right )}{\left (2 k +2\right )^{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 <- Kummer successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 81
Order:=8; dsolve(4*x^2*diff(y(x),x$2)+8*x*(x+1)*diff(y(x),x)+y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {\left (c_{2} \ln \left (x \right )+c_{1} \right ) \left (1+x -\frac {1}{4} x^{2}+\frac {1}{12} x^{3}-\frac {5}{192} x^{4}+\frac {7}{960} x^{5}-\frac {7}{3840} x^{6}+\frac {11}{26880} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+\left (\left (-4\right ) x +\frac {3}{4} x^{2}-\frac {1}{4} x^{3}+\frac {31}{384} x^{4}-\frac {3}{128} x^{5}+\frac {419}{69120} x^{6}-\frac {97}{69120} x^{7}+\operatorname {O}\left (x^{8}\right )\right ) c_{2}}{\sqrt {x}} \]
✓ Solution by Mathematica
Time used: 0.006 (sec). Leaf size: 166
AsymptoticDSolveValue[4*x^2*y''[x]+8*x*(x+1)*y'[x]+y[x]==0,y[x],{x,0,7}]
\[ y(x)\to \frac {c_1 \left (\frac {11 x^7}{26880}-\frac {7 x^6}{3840}+\frac {7 x^5}{960}-\frac {5 x^4}{192}+\frac {x^3}{12}-\frac {x^2}{4}+x+1\right )}{\sqrt {x}}+c_2 \left (\frac {-\frac {97 x^7}{69120}+\frac {419 x^6}{69120}-\frac {3 x^5}{128}+\frac {31 x^4}{384}-\frac {x^3}{4}+\frac {3 x^2}{4}-4 x}{\sqrt {x}}+\frac {\left (\frac {11 x^7}{26880}-\frac {7 x^6}{3840}+\frac {7 x^5}{960}-\frac {5 x^4}{192}+\frac {x^3}{12}-\frac {x^2}{4}+x+1\right ) \log (x)}{\sqrt {x}}\right ) \]