Internal problem ID [7020]
Internal file name [OUTPUT/6263_Thursday_August_18_2022_07_12_13_AM_96052526/index.tex]
Book: Elementary differential equations. Rainville, Bedient, Bedient. Prentice Hall. NJ. 8th
edition. 1997.
Section: CHAPTER 18. Power series solutions. Miscellaneous Exercises. page 394
Problem number: 19.
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]]
\[ \boxed {4 x^{2} y^{\prime \prime }+3 y^{\prime } x^{2}+\left (1+3 x \right ) y=0} \] 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. \[ 4 x^{2} y^{\prime \prime }+3 y^{\prime } x^{2}+\left (1+3 x \right ) y = 0 \] 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 {3}{4}}\\ q(x) &= \frac {1+3 x}{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 \[ 4 x^{2} y^{\prime \prime }+3 y^{\prime } x^{2}+\left (1+3 x \right ) y = 0 \] 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 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right ) x^{2}+3 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right ) x^{2}+\left (1+3 x \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 }}3 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{1+n +r} a_{n}\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 }}3 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}3 a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}3 x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}3 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 =1}{\sum }}3 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}3 a_{n -1} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 4 x^{r} a_{0} r \left (-1+r \right )+a_{0} x^{r} = 0 \] Or \[ \left (4 x^{r} r \left (-1+r \right )+x^{r}\right ) a_{0} = 0 \] 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 \(a_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} 4 a_{n} \left (n +r \right ) \left (n +r -1\right )+3 a_{n -1} \left (n +r -1\right )+a_{n}+3 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {3 a_{n -1} \left (n +r \right )}{4 n^{2}+8 n r +4 r^{2}-4 n -4 r +1}\tag {4} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{n} = -\frac {3 a_{n -1} \left (2 n +1\right )}{8 n^{2}}\tag {5} \] 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 \[ a_{1}=\frac {-3-3 r}{\left (2 r +1\right )^{2}} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{1}=-{\frac {9}{8}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-3 r}{\left (2 r +1\right )^{2}}\) | \(-{\frac {9}{8}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {9 \left (1+r \right ) \left (2+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2}} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{2}={\frac {135}{256}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-3 r}{\left (2 r +1\right )^{2}}\) | \(-{\frac {9}{8}}\) |
| \(a_{2}\) | \(\frac {9 \left (1+r \right ) \left (2+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2}}\) | \(\frac {135}{256}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {27 \left (3+r \right ) \left (1+r \right ) \left (2+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2}} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{3}=-{\frac {315}{2048}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-3 r}{\left (2 r +1\right )^{2}}\) | \(-{\frac {9}{8}}\) |
| \(a_{2}\) | \(\frac {9 \left (1+r \right ) \left (2+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2}}\) | \(\frac {135}{256}\) |
| \(a_{3}\) | \(-\frac {27 \left (3+r \right ) \left (1+r \right ) \left (2+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2}}\) | \(-{\frac {315}{2048}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {81 \left (3+r \right ) \left (1+r \right ) \left (2+r \right ) \left (4+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2}} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{4}={\frac {8505}{262144}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-3 r}{\left (2 r +1\right )^{2}}\) | \(-{\frac {9}{8}}\) |
| \(a_{2}\) | \(\frac {9 \left (1+r \right ) \left (2+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2}}\) | \(\frac {135}{256}\) |
| \(a_{3}\) | \(-\frac {27 \left (3+r \right ) \left (1+r \right ) \left (2+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2}}\) | \(-{\frac {315}{2048}}\) |
| \(a_{4}\) | \(\frac {81 \left (3+r \right ) \left (1+r \right ) \left (2+r \right ) \left (4+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2}}\) | \(\frac {8505}{262144}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {243 \left (3+r \right ) \left (1+r \right ) \left (2+r \right ) \left (4+r \right ) \left (5+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (2 r +9\right )^{2}} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{5}=-{\frac {56133}{10485760}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-3 r}{\left (2 r +1\right )^{2}}\) | \(-{\frac {9}{8}}\) |
| \(a_{2}\) | \(\frac {9 \left (1+r \right ) \left (2+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2}}\) | \(\frac {135}{256}\) |
| \(a_{3}\) | \(-\frac {27 \left (3+r \right ) \left (1+r \right ) \left (2+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2}}\) | \(-{\frac {315}{2048}}\) |
| \(a_{4}\) | \(\frac {81 \left (3+r \right ) \left (1+r \right ) \left (2+r \right ) \left (4+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2}}\) | \(\frac {8505}{262144}\) |
| \(a_{5}\) | \(-\frac {243 \left (3+r \right ) \left (1+r \right ) \left (2+r \right ) \left (4+r \right ) \left (5+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (2 r +9\right )^{2}}\) | \(-{\frac {56133}{10485760}}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {729 \left (3+r \right ) \left (1+r \right ) \left (2+r \right ) \left (4+r \right ) \left (5+r \right ) \left (6+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (2 r +9\right )^{2} \left (2 r +11\right )^{2}} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{6}={\frac {243243}{335544320}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-3 r}{\left (2 r +1\right )^{2}}\) | \(-{\frac {9}{8}}\) |
| \(a_{2}\) | \(\frac {9 \left (1+r \right ) \left (2+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2}}\) | \(\frac {135}{256}\) |
| \(a_{3}\) | \(-\frac {27 \left (3+r \right ) \left (1+r \right ) \left (2+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2}}\) | \(-{\frac {315}{2048}}\) |
| \(a_{4}\) | \(\frac {81 \left (3+r \right ) \left (1+r \right ) \left (2+r \right ) \left (4+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2}}\) | \(\frac {8505}{262144}\) |
| \(a_{5}\) | \(-\frac {243 \left (3+r \right ) \left (1+r \right ) \left (2+r \right ) \left (4+r \right ) \left (5+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (2 r +9\right )^{2}}\) | \(-{\frac {56133}{10485760}}\) |
| \(a_{6}\) | \(\frac {729 \left (3+r \right ) \left (1+r \right ) \left (2+r \right ) \left (4+r \right ) \left (5+r \right ) \left (6+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (2 r +9\right )^{2} \left (2 r +11\right )^{2}}\) | \(\frac {243243}{335544320}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=-\frac {2187 \left (3+r \right ) \left (1+r \right ) \left (2+r \right ) \left (4+r \right ) \left (5+r \right ) \left (6+r \right ) \left (7+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (2 r +9\right )^{2} \left (2 r +11\right )^{2} \left (2 r +13\right )^{2}} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{7}=-{\frac {312741}{3758096384}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-3 r}{\left (2 r +1\right )^{2}}\) | \(-{\frac {9}{8}}\) |
| \(a_{2}\) | \(\frac {9 \left (1+r \right ) \left (2+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2}}\) | \(\frac {135}{256}\) |
| \(a_{3}\) | \(-\frac {27 \left (3+r \right ) \left (1+r \right ) \left (2+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2}}\) | \(-{\frac {315}{2048}}\) |
| \(a_{4}\) | \(\frac {81 \left (3+r \right ) \left (1+r \right ) \left (2+r \right ) \left (4+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2}}\) | \(\frac {8505}{262144}\) |
| \(a_{5}\) | \(-\frac {243 \left (3+r \right ) \left (1+r \right ) \left (2+r \right ) \left (4+r \right ) \left (5+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (2 r +9\right )^{2}}\) | \(-{\frac {56133}{10485760}}\) |
| \(a_{6}\) | \(\frac {729 \left (3+r \right ) \left (1+r \right ) \left (2+r \right ) \left (4+r \right ) \left (5+r \right ) \left (6+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (2 r +9\right )^{2} \left (2 r +11\right )^{2}}\) | \(\frac {243243}{335544320}\) |
| \(a_{7}\) | \(-\frac {2187 \left (3+r \right ) \left (1+r \right ) \left (2+r \right ) \left (4+r \right ) \left (5+r \right ) \left (6+r \right ) \left (7+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (2 r +9\right )^{2} \left (2 r +11\right )^{2} \left (2 r +13\right )^{2}}\) | \(-{\frac {312741}{3758096384}}\) |
Using the above table, then the first solution \(y_{1}\left (x \right )\) is \begin{align*} y_{1}\left (x \right )&= \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 ) \\ &= \sqrt {x}\, \left (1-\frac {9 x}{8}+\frac {135 x^{2}}{256}-\frac {315 x^{3}}{2048}+\frac {8505 x^{4}}{262144}-\frac {56133 x^{5}}{10485760}+\frac {243243 x^{6}}{335544320}-\frac {312741 x^{7}}{3758096384}+O\left (x^{8}\right )\right ) \\ \end{align*} Now the second solution is found. The second solution is given by \[ y_{2}\left (x \right ) = y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +r}\right ) \] Where \(b_{n}\) is found using \[ b_{n} = \frac {d}{d r}a_{n ,r} \] And the above is then evaluated at \(r = {\frac {1}{2}}\). 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 |
| \(b_{1}\) | \(\frac {-3-3 r}{\left (2 r +1\right )^{2}}\) | \(-{\frac {9}{8}}\) | \(\frac {6 r +9}{\left (2 r +1\right )^{3}}\) | \(\frac {3}{2}\) |
| \(b_{2}\) | \(\frac {9 \left (1+r \right ) \left (2+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2}}\) | \(\frac {135}{256}\) | \(\frac {-72 r^{3}-324 r^{2}-450 r -207}{\left (3+2 r \right )^{3} \left (2 r +1\right )^{3}}\) | \(-{\frac {261}{256}}\) |
| \(b_{3}\) | \(-\frac {27 \left (3+r \right ) \left (1+r \right ) \left (2+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2}}\) | \(-{\frac {315}{2048}}\) | \(\frac {648 r^{5}+6156 r^{4}+22302 r^{3}+38637 r^{2}+32130 r +10449}{\left (5+2 r \right )^{3} \left (3+2 r \right )^{3} \left (2 r +1\right )^{3}}\) | \(\frac {729}{2048}\) |
| \(b_{4}\) | \(\frac {81 \left (3+r \right ) \left (1+r \right ) \left (2+r \right ) \left (4+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2}}\) | \(\frac {8505}{262144}\) | \(-\frac {162 \left (32 r^{7}+528 r^{6}+3600 r^{5}+13128 r^{4}+27606 r^{3}+33441 r^{2}+21637 r +5823\right )}{\left (7+2 r \right )^{3} \left (5+2 r \right )^{3} \left (3+2 r \right )^{3} \left (2 r +1\right )^{3}}\) | \(-{\frac {44091}{524288}}\) |
| \(b_{5}\) | \(-\frac {243 \left (3+r \right ) \left (1+r \right ) \left (2+r \right ) \left (4+r \right ) \left (5+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (2 r +9\right )^{2}}\) | \(-{\frac {56133}{10485760}}\) | \(\frac {38880 r^{9}+991440 r^{8}+10905840 r^{7}+67797000 r^{6}+261948654 r^{5}+650884005 r^{4}+1037845710 r^{3}+1022404275 r^{2}+564810246 r +134084970}{\left (2 r +9\right )^{3} \left (7+2 r \right )^{3} \left (5+2 r \right )^{3} \left (3+2 r \right )^{3} \left (2 r +1\right )^{3}}\) | \(\frac {63099}{4194304}\) |
| \(b_{6}\) | \(\frac {729 \left (3+r \right ) \left (1+r \right ) \left (2+r \right ) \left (4+r \right ) \left (5+r \right ) \left (6+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (2 r +9\right )^{2} \left (2 r +11\right )^{2}}\) | \(\frac {243243}{335544320}\) | \(-\frac {729 \left (384 r^{11}+14016 r^{10}+226720 r^{9}+2142000 r^{8}+13109928 r^{7}+54470772 r^{6}+156438438 r^{5}+309845265 r^{4}+413822708 r^{3}+354306177 r^{2}+174958992 r +37892340\right )}{\left (2 r +1\right )^{3} \left (3+2 r \right )^{3} \left (5+2 r \right )^{3} \left (7+2 r \right )^{3} \left (2 r +9\right )^{3} \left (2 r +11\right )^{3}}\) | \(-{\frac {1454463}{671088640}}\) |
| \(b_{7}\) | \(-\frac {2187 \left (3+r \right ) \left (1+r \right ) \left (2+r \right ) \left (4+r \right ) \left (5+r \right ) \left (6+r \right ) \left (7+r \right )}{\left (2 r +1\right )^{2} \left (3+2 r \right )^{2} \left (5+2 r \right )^{2} \left (7+2 r \right )^{2} \left (2 r +9\right )^{2} \left (2 r +11\right )^{2} \left (2 r +13\right )^{2}}\) | \(-{\frac {312741}{3758096384}}\) | \(\frac {1959552 r^{13}+96997824 r^{12}+2167754400 r^{11}+28923722352 r^{10}+256695800760 r^{9}+1597616438292 r^{8}+7161048611550 r^{7}+23359986091701 r^{6}+55338743503764 r^{5}+93813057611154 r^{4}+110370834847350 r^{3}+85173325896897 r^{2}+38613891469824 r +7786745571780}{\left (2 r +1\right )^{3} \left (3+2 r \right )^{3} \left (5+2 r \right )^{3} \left (7+2 r \right )^{3} \left (2 r +9\right )^{3} \left (2 r +11\right )^{3} \left (2 r +13\right )^{3}}\) | \(\frac {1403811}{5368709120}\) |
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 \\ &= \sqrt {x}\, \left (1-\frac {9 x}{8}+\frac {135 x^{2}}{256}-\frac {315 x^{3}}{2048}+\frac {8505 x^{4}}{262144}-\frac {56133 x^{5}}{10485760}+\frac {243243 x^{6}}{335544320}-\frac {312741 x^{7}}{3758096384}+O\left (x^{8}\right )\right ) \ln \left (x \right )+\sqrt {x}\, \left (\frac {3 x}{2}-\frac {261 x^{2}}{256}+\frac {729 x^{3}}{2048}-\frac {44091 x^{4}}{524288}+\frac {63099 x^{5}}{4194304}-\frac {1454463 x^{6}}{671088640}+\frac {1403811 x^{7}}{5368709120}+O\left (x^{8}\right )\right ) \\ \end{align*} Therefore the homogeneous solution is \begin{align*} y_h(x) &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= c_{1} \sqrt {x}\, \left (1-\frac {9 x}{8}+\frac {135 x^{2}}{256}-\frac {315 x^{3}}{2048}+\frac {8505 x^{4}}{262144}-\frac {56133 x^{5}}{10485760}+\frac {243243 x^{6}}{335544320}-\frac {312741 x^{7}}{3758096384}+O\left (x^{8}\right )\right ) + c_{2} \left (\sqrt {x}\, \left (1-\frac {9 x}{8}+\frac {135 x^{2}}{256}-\frac {315 x^{3}}{2048}+\frac {8505 x^{4}}{262144}-\frac {56133 x^{5}}{10485760}+\frac {243243 x^{6}}{335544320}-\frac {312741 x^{7}}{3758096384}+O\left (x^{8}\right )\right ) \ln \left (x \right )+\sqrt {x}\, \left (\frac {3 x}{2}-\frac {261 x^{2}}{256}+\frac {729 x^{3}}{2048}-\frac {44091 x^{4}}{524288}+\frac {63099 x^{5}}{4194304}-\frac {1454463 x^{6}}{671088640}+\frac {1403811 x^{7}}{5368709120}+O\left (x^{8}\right )\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \sqrt {x}\, \left (1-\frac {9 x}{8}+\frac {135 x^{2}}{256}-\frac {315 x^{3}}{2048}+\frac {8505 x^{4}}{262144}-\frac {56133 x^{5}}{10485760}+\frac {243243 x^{6}}{335544320}-\frac {312741 x^{7}}{3758096384}+O\left (x^{8}\right )\right )+c_{2} \left (\sqrt {x}\, \left (1-\frac {9 x}{8}+\frac {135 x^{2}}{256}-\frac {315 x^{3}}{2048}+\frac {8505 x^{4}}{262144}-\frac {56133 x^{5}}{10485760}+\frac {243243 x^{6}}{335544320}-\frac {312741 x^{7}}{3758096384}+O\left (x^{8}\right )\right ) \ln \left (x \right )+\sqrt {x}\, \left (\frac {3 x}{2}-\frac {261 x^{2}}{256}+\frac {729 x^{3}}{2048}-\frac {44091 x^{4}}{524288}+\frac {63099 x^{5}}{4194304}-\frac {1454463 x^{6}}{671088640}+\frac {1403811 x^{7}}{5368709120}+O\left (x^{8}\right )\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \sqrt {x}\, \left (1-\frac {9 x}{8}+\frac {135 x^{2}}{256}-\frac {315 x^{3}}{2048}+\frac {8505 x^{4}}{262144}-\frac {56133 x^{5}}{10485760}+\frac {243243 x^{6}}{335544320}-\frac {312741 x^{7}}{3758096384}+O\left (x^{8}\right )\right )+c_{2} \left (\sqrt {x}\, \left (1-\frac {9 x}{8}+\frac {135 x^{2}}{256}-\frac {315 x^{3}}{2048}+\frac {8505 x^{4}}{262144}-\frac {56133 x^{5}}{10485760}+\frac {243243 x^{6}}{335544320}-\frac {312741 x^{7}}{3758096384}+O\left (x^{8}\right )\right ) \ln \left (x \right )+\sqrt {x}\, \left (\frac {3 x}{2}-\frac {261 x^{2}}{256}+\frac {729 x^{3}}{2048}-\frac {44091 x^{4}}{524288}+\frac {63099 x^{5}}{4194304}-\frac {1454463 x^{6}}{671088640}+\frac {1403811 x^{7}}{5368709120}+O\left (x^{8}\right )\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \sqrt {x}\, \left (1-\frac {9 x}{8}+\frac {135 x^{2}}{256}-\frac {315 x^{3}}{2048}+\frac {8505 x^{4}}{262144}-\frac {56133 x^{5}}{10485760}+\frac {243243 x^{6}}{335544320}-\frac {312741 x^{7}}{3758096384}+O\left (x^{8}\right )\right )+c_{2} \left (\sqrt {x}\, \left (1-\frac {9 x}{8}+\frac {135 x^{2}}{256}-\frac {315 x^{3}}{2048}+\frac {8505 x^{4}}{262144}-\frac {56133 x^{5}}{10485760}+\frac {243243 x^{6}}{335544320}-\frac {312741 x^{7}}{3758096384}+O\left (x^{8}\right )\right ) \ln \left (x \right )+\sqrt {x}\, \left (\frac {3 x}{2}-\frac {261 x^{2}}{256}+\frac {729 x^{3}}{2048}-\frac {44091 x^{4}}{524288}+\frac {63099 x^{5}}{4194304}-\frac {1454463 x^{6}}{671088640}+\frac {1403811 x^{7}}{5368709120}+O\left (x^{8}\right )\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 y^{\prime \prime } x^{2}+3 y^{\prime } x^{2}+\left (1+3 x \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 {3 y^{\prime }}{4}-\frac {\left (1+3 x \right ) y}{4 x^{2}} \\ \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 {3 y^{\prime }}{4}+\frac {\left (1+3 x \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 {3}{4}, P_{3}\left (x \right )=\frac {1+3 x}{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}}}=0 \\ {} & \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 y^{\prime \prime } x^{2}+3 y^{\prime } x^{2}+\left (1+3 x \right ) y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{2}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r +1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -1 \\ {} & {} & x^{2}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =1}{\sum }}a_{k -1} \left (k -1+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 -1+r \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}+3 a_{k -1} \left (k +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 in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (2 k +2 r -1\right )^{2}+3 a_{k -1} \left (k +r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +1} \left (2 k +1+2 r \right )^{2}+3 a_{k} \left (k +r +1\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {3 a_{k} \left (k +r +1\right )}{\left (2 k +1+2 r \right )^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & a_{k +1}=-\frac {3 a_{k} \left (k +\frac {3}{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 {3 a_{k} \left (k +\frac {3}{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)+3*x^2*diff(y(x),x)+(1+3*x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \sqrt {x}\, \left (\left (c_{2} \ln \left (x \right )+c_{1} \right ) \left (1-\frac {9}{8} x +\frac {135}{256} x^{2}-\frac {315}{2048} x^{3}+\frac {8505}{262144} x^{4}-\frac {56133}{10485760} x^{5}+\frac {243243}{335544320} x^{6}-\frac {312741}{3758096384} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+\left (\frac {3}{2} x -\frac {261}{256} x^{2}+\frac {729}{2048} x^{3}-\frac {44091}{524288} x^{4}+\frac {63099}{4194304} x^{5}-\frac {1454463}{671088640} x^{6}+\frac {1403811}{5368709120} x^{7}+\operatorname {O}\left (x^{8}\right )\right ) c_{2} \right ) \]
✓ Solution by Mathematica
Time used: 0.006 (sec). Leaf size: 176
AsymptoticDSolveValue[4*x^2*y''[x]+3*x^2*y'[x]+(1+3*x)*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_1 \sqrt {x} \left (-\frac {312741 x^7}{3758096384}+\frac {243243 x^6}{335544320}-\frac {56133 x^5}{10485760}+\frac {8505 x^4}{262144}-\frac {315 x^3}{2048}+\frac {135 x^2}{256}-\frac {9 x}{8}+1\right )+c_2 \left (\sqrt {x} \left (\frac {1403811 x^7}{5368709120}-\frac {1454463 x^6}{671088640}+\frac {63099 x^5}{4194304}-\frac {44091 x^4}{524288}+\frac {729 x^3}{2048}-\frac {261 x^2}{256}+\frac {3 x}{2}\right )+\sqrt {x} \left (-\frac {312741 x^7}{3758096384}+\frac {243243 x^6}{335544320}-\frac {56133 x^5}{10485760}+\frac {8505 x^4}{262144}-\frac {315 x^3}{2048}+\frac {135 x^2}{256}-\frac {9 x}{8}+1\right ) \log (x)\right ) \]