Internal problem ID [1318]
Internal file name [OUTPUT/1319_Sunday_June_05_2022_02_10_05_AM_13927065/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: 29.
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]]
\[ \boxed {x^{2} \left (4 x +3\right ) y^{\prime \prime }+x \left (11+4 x \right ) y^{\prime }-\left (4 x +3\right ) y=0} \] 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. \[ \left (4 x^{3}+3 x^{2}\right ) y^{\prime \prime }+\left (4 x^{2}+11 x \right ) y^{\prime }+\left (-4 x -3\right ) y = 0 \] 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 {11+4 x}{x \left (4 x +3\right )}\\ q(x) &= -\frac {1}{x^{2}}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \(\left [0, -{\frac {3}{4}}, \infty \right ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} \left (4 x +3\right ) y^{\prime \prime }+\left (4 x^{2}+11 x \right ) y^{\prime }+\left (-4 x -3\right ) y = 0 \] 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} x^{2} \left (4 x +3\right ) \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^{2}+11 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (-4 x -3\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^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}11 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 x^{1+n +r} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 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^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}4 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 x^{1+n +r} a_{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 a_{n -1} x^{n +r}\right ) \\ \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 =1}{\sum }}4 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}11 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-4 a_{n -1} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 3 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+11 x^{n +r} a_{n} \left (n +r \right )-3 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 3 x^{r} a_{0} r \left (-1+r \right )+11 x^{r} a_{0} r -3 a_{0} x^{r} = 0 \] Or \[ \left (3 x^{r} r \left (-1+r \right )+11 x^{r} r -3 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (3 r^{2}+8 r -3\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 3 r^{2}+8 r -3 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {1}{3}}\\ r_2 &= -3 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (3 r^{2}+8 r -3\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {1}{3}}, -3\right ]\).
Since \(r_1 - r_2 = {\frac {10}{3}}\) 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}{3}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -3} \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 \(a_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} 4 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+3 a_{n} \left (n +r \right ) \left (n +r -1\right )+4 a_{n -1} \left (n +r -1\right )+11 a_{n} \left (n +r \right )-4 a_{n -1}-3 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {4 a_{n -1} \left (n^{2}+2 n r +r^{2}-2 n -2 r \right )}{3 n^{2}+6 n r +3 r^{2}+8 n +8 r -3}\tag {4} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ a_{n} = -\frac {4 a_{n -1} \left (9 n^{2}-12 n -5\right )}{27 n^{2}+90 n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {1}{3}}\) 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 \[ a_{1}=\frac {-4 r^{2}+4}{3 r^{2}+14 r +8} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ a_{1}={\frac {32}{117}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-4 r^{2}+4}{3 r^{2}+14 r +8}\) | \(\frac {32}{117}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {16 \left (r^{2}-1\right ) r \left (r +2\right )}{\left (3 r^{2}+14 r +8\right ) \left (3 r^{2}+20 r +25\right )} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ a_{2}=-{\frac {28}{1053}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-4 r^{2}+4}{3 r^{2}+14 r +8}\) | \(\frac {32}{117}\) |
| \(a_{2}\) | \(\frac {16 \left (r^{2}-1\right ) r \left (r +2\right )}{\left (3 r^{2}+14 r +8\right ) \left (3 r^{2}+20 r +25\right )}\) | \(-{\frac {28}{1053}}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {64 \left (r^{2}-1\right ) r \left (r +2\right ) \left (r^{2}+4 r +3\right )}{\left (3 r^{2}+14 r +8\right ) \left (3 r^{2}+20 r +25\right ) \left (3 r^{2}+26 r +48\right )} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ a_{3}={\frac {4480}{540189}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-4 r^{2}+4}{3 r^{2}+14 r +8}\) | \(\frac {32}{117}\) |
| \(a_{2}\) | \(\frac {16 \left (r^{2}-1\right ) r \left (r +2\right )}{\left (3 r^{2}+14 r +8\right ) \left (3 r^{2}+20 r +25\right )}\) | \(-{\frac {28}{1053}}\) |
| \(a_{3}\) | \(-\frac {64 \left (r^{2}-1\right ) r \left (r +2\right ) \left (r^{2}+4 r +3\right )}{\left (3 r^{2}+14 r +8\right ) \left (3 r^{2}+20 r +25\right ) \left (3 r^{2}+26 r +48\right )}\) | \(\frac {4480}{540189}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {256 \left (r +2\right )^{2} \left (-1+r \right ) \left (r +1\right )^{2} r \left (r +3\right )}{\left (3 r^{2}+32 r +77\right ) \left (3 r^{2}+26 r +48\right ) \left (3 r^{2}+20 r +25\right ) \left (3 r +2\right )} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ a_{4}=-{\frac {15680}{4113747}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-4 r^{2}+4}{3 r^{2}+14 r +8}\) | \(\frac {32}{117}\) |
| \(a_{2}\) | \(\frac {16 \left (r^{2}-1\right ) r \left (r +2\right )}{\left (3 r^{2}+14 r +8\right ) \left (3 r^{2}+20 r +25\right )}\) | \(-{\frac {28}{1053}}\) |
| \(a_{3}\) | \(-\frac {64 \left (r^{2}-1\right ) r \left (r +2\right ) \left (r^{2}+4 r +3\right )}{\left (3 r^{2}+14 r +8\right ) \left (3 r^{2}+20 r +25\right ) \left (3 r^{2}+26 r +48\right )}\) | \(\frac {4480}{540189}\) |
| \(a_{4}\) | \(\frac {256 \left (r +2\right )^{2} \left (-1+r \right ) \left (r +1\right )^{2} r \left (r +3\right )}{\left (3 r^{2}+32 r +77\right ) \left (3 r^{2}+26 r +48\right ) \left (3 r^{2}+20 r +25\right ) \left (3 r +2\right )}\) | \(-{\frac {15680}{4113747}}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {1024 \left (r +3\right )^{2} \left (r +2\right )^{2} \left (-1+r \right ) \left (r +1\right )^{2} r}{\left (3 r^{2}+38 r +112\right ) \left (3 r +2\right ) \left (3 r +5\right ) \left (3 r^{2}+26 r +48\right ) \left (3 r^{2}+32 r +77\right )} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ a_{5}={\frac {401408}{185118615}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-4 r^{2}+4}{3 r^{2}+14 r +8}\) | \(\frac {32}{117}\) |
| \(a_{2}\) | \(\frac {16 \left (r^{2}-1\right ) r \left (r +2\right )}{\left (3 r^{2}+14 r +8\right ) \left (3 r^{2}+20 r +25\right )}\) | \(-{\frac {28}{1053}}\) |
| \(a_{3}\) | \(-\frac {64 \left (r^{2}-1\right ) r \left (r +2\right ) \left (r^{2}+4 r +3\right )}{\left (3 r^{2}+14 r +8\right ) \left (3 r^{2}+20 r +25\right ) \left (3 r^{2}+26 r +48\right )}\) | \(\frac {4480}{540189}\) |
| \(a_{4}\) | \(\frac {256 \left (r +2\right )^{2} \left (-1+r \right ) \left (r +1\right )^{2} r \left (r +3\right )}{\left (3 r^{2}+32 r +77\right ) \left (3 r^{2}+26 r +48\right ) \left (3 r^{2}+20 r +25\right ) \left (3 r +2\right )}\) | \(-{\frac {15680}{4113747}}\) |
| \(a_{5}\) | \(-\frac {1024 \left (r +3\right )^{2} \left (r +2\right )^{2} \left (-1+r \right ) \left (r +1\right )^{2} r}{\left (3 r^{2}+38 r +112\right ) \left (3 r +2\right ) \left (3 r +5\right ) \left (3 r^{2}+26 r +48\right ) \left (3 r^{2}+32 r +77\right )}\) | \(\frac {401408}{185118615}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{\frac {1}{3}} \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}{3}} \left (1+\frac {32 x}{117}-\frac {28 x^{2}}{1053}+\frac {4480 x^{3}}{540189}-\frac {15680 x^{4}}{4113747}+\frac {401408 x^{5}}{185118615}+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 \(b_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} 4 b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+3 b_{n} \left (n +r \right ) \left (n +r -1\right )+4 b_{n -1} \left (n +r -1\right )+11 b_{n} \left (n +r \right )-4 b_{n -1}-3 b_{n} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {4 b_{n -1} \left (n^{2}+2 n r +r^{2}-2 n -2 r \right )}{3 n^{2}+6 n r +3 r^{2}+8 n +8 r -3}\tag {4} \] Which for the root \(r = -3\) becomes \[ b_{n} = -\frac {4 b_{n -1} \left (n^{2}-8 n +15\right )}{n \left (3 n -10\right )}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -3\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ b_{1}=\frac {-4 r^{2}+4}{3 r^{2}+14 r +8} \] Which for the root \(r = -3\) becomes \[ b_{1}={\frac {32}{7}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-4 r^{2}+4}{3 r^{2}+14 r +8}\) | \(\frac {32}{7}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {16 \left (r^{2}-1\right ) r \left (r +2\right )}{\left (3 r^{2}+14 r +8\right ) \left (3 r^{2}+20 r +25\right )} \] Which for the root \(r = -3\) becomes \[ b_{2}={\frac {48}{7}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-4 r^{2}+4}{3 r^{2}+14 r +8}\) | \(\frac {32}{7}\) |
| \(b_{2}\) | \(\frac {16 \left (r^{2}-1\right ) r \left (r +2\right )}{\left (3 r^{2}+14 r +8\right ) \left (3 r^{2}+20 r +25\right )}\) | \(\frac {48}{7}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=-\frac {64 \left (r^{2}-1\right ) r \left (r +2\right ) \left (r^{2}+4 r +3\right )}{\left (3 r^{2}+14 r +8\right ) \left (3 r^{2}+20 r +25\right ) \left (3 r^{2}+26 r +48\right )} \] Which for the root \(r = -3\) becomes \[ b_{3}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-4 r^{2}+4}{3 r^{2}+14 r +8}\) | \(\frac {32}{7}\) |
| \(b_{2}\) | \(\frac {16 \left (r^{2}-1\right ) r \left (r +2\right )}{\left (3 r^{2}+14 r +8\right ) \left (3 r^{2}+20 r +25\right )}\) | \(\frac {48}{7}\) |
| \(b_{3}\) | \(-\frac {64 \left (r^{2}-1\right ) r \left (r +2\right ) \left (r^{2}+4 r +3\right )}{\left (3 r^{2}+14 r +8\right ) \left (3 r^{2}+20 r +25\right ) \left (3 r^{2}+26 r +48\right )}\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {256 \left (r +2\right )^{2} \left (-1+r \right ) \left (r +1\right )^{2} r \left (r +3\right )}{\left (3 r^{2}+32 r +77\right ) \left (3 r^{2}+26 r +48\right ) \left (3 r^{2}+20 r +25\right ) \left (3 r +2\right )} \] Which for the root \(r = -3\) becomes \[ b_{4}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-4 r^{2}+4}{3 r^{2}+14 r +8}\) | \(\frac {32}{7}\) |
| \(b_{2}\) | \(\frac {16 \left (r^{2}-1\right ) r \left (r +2\right )}{\left (3 r^{2}+14 r +8\right ) \left (3 r^{2}+20 r +25\right )}\) | \(\frac {48}{7}\) |
| \(b_{3}\) | \(-\frac {64 \left (r^{2}-1\right ) r \left (r +2\right ) \left (r^{2}+4 r +3\right )}{\left (3 r^{2}+14 r +8\right ) \left (3 r^{2}+20 r +25\right ) \left (3 r^{2}+26 r +48\right )}\) | \(0\) |
| \(b_{4}\) | \(\frac {256 \left (r +2\right )^{2} \left (-1+r \right ) \left (r +1\right )^{2} r \left (r +3\right )}{\left (3 r^{2}+32 r +77\right ) \left (3 r^{2}+26 r +48\right ) \left (3 r^{2}+20 r +25\right ) \left (3 r +2\right )}\) | \(0\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-\frac {1024 \left (r +3\right )^{2} \left (r +2\right )^{2} \left (-1+r \right ) \left (r +1\right )^{2} r}{\left (3 r^{2}+38 r +112\right ) \left (3 r +2\right ) \left (3 r +5\right ) \left (3 r^{2}+26 r +48\right ) \left (3 r^{2}+32 r +77\right )} \] Which for the root \(r = -3\) becomes \[ b_{5}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-4 r^{2}+4}{3 r^{2}+14 r +8}\) | \(\frac {32}{7}\) |
| \(b_{2}\) | \(\frac {16 \left (r^{2}-1\right ) r \left (r +2\right )}{\left (3 r^{2}+14 r +8\right ) \left (3 r^{2}+20 r +25\right )}\) | \(\frac {48}{7}\) |
| \(b_{3}\) | \(-\frac {64 \left (r^{2}-1\right ) r \left (r +2\right ) \left (r^{2}+4 r +3\right )}{\left (3 r^{2}+14 r +8\right ) \left (3 r^{2}+20 r +25\right ) \left (3 r^{2}+26 r +48\right )}\) | \(0\) |
| \(b_{4}\) | \(\frac {256 \left (r +2\right )^{2} \left (-1+r \right ) \left (r +1\right )^{2} r \left (r +3\right )}{\left (3 r^{2}+32 r +77\right ) \left (3 r^{2}+26 r +48\right ) \left (3 r^{2}+20 r +25\right ) \left (3 r +2\right )}\) | \(0\) |
| \(b_{5}\) | \(-\frac {1024 \left (r +3\right )^{2} \left (r +2\right )^{2} \left (-1+r \right ) \left (r +1\right )^{2} r}{\left (3 r^{2}+38 r +112\right ) \left (3 r +2\right ) \left (3 r +5\right ) \left (3 r^{2}+26 r +48\right ) \left (3 r^{2}+32 r +77\right )}\) | \(0\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x^{\frac {1}{3}} \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+\frac {32 x}{7}+\frac {48 x^{2}}{7}+O\left (x^{6}\right )}{x^{3}} \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}{3}} \left (1+\frac {32 x}{117}-\frac {28 x^{2}}{1053}+\frac {4480 x^{3}}{540189}-\frac {15680 x^{4}}{4113747}+\frac {401408 x^{5}}{185118615}+O\left (x^{6}\right )\right ) + \frac {c_{2} \left (1+\frac {32 x}{7}+\frac {48 x^{2}}{7}+O\left (x^{6}\right )\right )}{x^{3}} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {1}{3}} \left (1+\frac {32 x}{117}-\frac {28 x^{2}}{1053}+\frac {4480 x^{3}}{540189}-\frac {15680 x^{4}}{4113747}+\frac {401408 x^{5}}{185118615}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+\frac {32 x}{7}+\frac {48 x^{2}}{7}+O\left (x^{6}\right )\right )}{x^{3}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {1}{3}} \left (1+\frac {32 x}{117}-\frac {28 x^{2}}{1053}+\frac {4480 x^{3}}{540189}-\frac {15680 x^{4}}{4113747}+\frac {401408 x^{5}}{185118615}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+\frac {32 x}{7}+\frac {48 x^{2}}{7}+O\left (x^{6}\right )\right )}{x^{3}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {1}{3}} \left (1+\frac {32 x}{117}-\frac {28 x^{2}}{1053}+\frac {4480 x^{3}}{540189}-\frac {15680 x^{4}}{4113747}+\frac {401408 x^{5}}{185118615}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+\frac {32 x}{7}+\frac {48 x^{2}}{7}+O\left (x^{6}\right )\right )}{x^{3}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (4 x +3\right ) y^{\prime \prime }+\left (4 x^{2}+11 x \right ) y^{\prime }+\left (-4 x -3\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 {y}{x^{2}}-\frac {\left (11+4 x \right ) y^{\prime }}{x \left (4 x +3\right )} \\ \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 (11+4 x \right ) y^{\prime }}{x \left (4 x +3\right )}-\frac {y}{x^{2}}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {11+4 x}{x \left (4 x +3\right )}, P_{3}\left (x \right )=-\frac {1}{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 {11}{3} \\ {} & \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}}}=-1 \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x^{2} \left (4 x +3\right ) y^{\prime \prime }+x \left (11+4 x \right ) y^{\prime }+\left (-4 x -3\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^{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^{m}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..3 \\ {} & {} & x^{m}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & x^{m}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (3+r \right ) \left (-1+3 r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (k +r +3\right ) \left (3 k +3 r -1\right )+4 a_{k -1} \left (k +r \right ) \left (k -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 (3+r \right ) \left (-1+3 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-3, \frac {1}{3}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 3 \left (k +r +3\right ) \left (k +r -\frac {1}{3}\right ) a_{k}+4 a_{k -1} \left (k +r \right ) \left (k -2+r \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 3 \left (k +4+r \right ) \left (k +\frac {2}{3}+r \right ) a_{k +1}+4 a_{k} \left (k +r +1\right ) \left (k +r -1\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {4 a_{k} \left (k +r +1\right ) \left (k +r -1\right )}{\left (k +4+r \right ) \left (3 k +2+3 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-3\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =2 \\ {} & {} & a_{k +1}=-\frac {4 a_{k} \left (k -2\right ) \left (k -4\right )}{\left (k +1\right ) \left (3 k -7\right )} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =0 \\ {} & {} & a_{1}=\frac {32 a_{0}}{7} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =1 \\ {} & {} & a_{2}=\frac {3 a_{1}}{2} \\ \bullet & {} & \textrm {Express in terms of}\hspace {3pt} a_{0} \\ {} & {} & a_{2}=\frac {48 a_{0}}{7} \\ \bullet & {} & \textrm {Terminating series solution of the ODE for}\hspace {3pt} r =-3\hspace {3pt}\textrm {. Use reduction of order to find the second linearly independent solution}\hspace {3pt} \\ {} & {} & y=a_{0}\cdot \left (1+\frac {32}{7} x +\frac {48}{7} x^{2}\right ) \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{3} \\ {} & {} & a_{k +1}=-\frac {4 a_{k} \left (k +\frac {4}{3}\right ) \left (k -\frac {2}{3}\right )}{\left (k +\frac {13}{3}\right ) \left (3 k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{3} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{3}}, a_{k +1}=-\frac {4 a_{k} \left (k +\frac {4}{3}\right ) \left (k -\frac {2}{3}\right )}{\left (k +\frac {13}{3}\right ) \left (3 k +3\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=a_{0}\cdot \left (1+\frac {32}{7} x +\frac {48}{7} x^{2}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {1}{3}}\right ), b_{1+k}=-\frac {4 b_{k} \left (k +\frac {4}{3}\right ) \left (k -\frac {2}{3}\right )}{\left (k +\frac {13}{3}\right ) \left (3 k +3\right )}\right ] \end {array} \]
Maple trace Kovacic algorithm successful
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm A Liouvillian solution exists Reducible group (found an exponential solution) Group is reducible, not completely reducible Solution has integrals. Trying a special function solution free of integrals... -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 2F1 ODE <- hypergeometric successful <- special function solution successful -> Trying to convert hypergeometric functions to elementary form... <- elementary form for at least one hypergeometric solution is achieved - returning with no uncomputed integrals <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 41
Order:=6; dsolve(x^2*(3+4*x)*diff(y(x),x$2)+x*(11+4*x)*diff(y(x),x)-(3+4*x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{1} \left (1+\frac {32}{7} x +\frac {48}{7} x^{2}+\operatorname {O}\left (x^{6}\right )\right )}{x^{3}}+c_{2} x^{\frac {1}{3}} \left (1+\frac {32}{117} x -\frac {28}{1053} x^{2}+\frac {4480}{540189} x^{3}-\frac {15680}{4113747} x^{4}+\frac {401408}{185118615} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.007 (sec). Leaf size: 67
AsymptoticDSolveValue[x^2*(3+4*x)*y''[x]+x*(11+4*x)*y'[x]-(3+4*x)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to \frac {c_2 \left (\frac {48 x^2}{7}+\frac {32 x}{7}+1\right )}{x^3}+c_1 \sqrt [3]{x} \left (\frac {401408 x^5}{185118615}-\frac {15680 x^4}{4113747}+\frac {4480 x^3}{540189}-\frac {28 x^2}{1053}+\frac {32 x}{117}+1\right ) \]