Internal problem ID [1317]
Internal file name [OUTPUT/1318_Sunday_June_05_2022_02_10_02_AM_83461413/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: 28.
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]]
\[ \boxed {x^{2} \left (x +8\right ) y^{\prime \prime }+x \left (3 x +2\right ) y^{\prime }+\left (x +1\right ) y=0} \] 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. \[ \left (x^{3}+8 x^{2}\right ) y^{\prime \prime }+\left (3 x^{2}+2 x \right ) y^{\prime }+\left (x +1\right ) y = 0 \] 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 {3 x +2}{x \left (x +8\right )}\\ q(x) &= \frac {x +1}{x^{2} \left (x +8\right )}\\ \end {align*}
| \(p(x)=\frac {3 x +2}{x \left (x +8\right )}\) | |
| singularity | type |
| \(x = -8\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
| \(q(x)=\frac {x +1}{x^{2} \left (x +8\right )}\) | |
| singularity | type |
| \(x = -8\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([-8, 0, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} \left (x +8\right ) y^{\prime \prime }+\left (3 x^{2}+2 x \right ) y^{\prime }+\left (x +1\right ) y = 0 \] 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} x^{2} \left (x +8\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 (3 x^{2}+2 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (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 }}x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}8 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 }}2 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r} \\ \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 }}x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}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 =1}{\sum }}a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}8 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 }}2 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 8 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+2 x^{n +r} a_{n} \left (n +r \right )+a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 8 x^{r} a_{0} r \left (-1+r \right )+2 x^{r} a_{0} r +a_{0} x^{r} = 0 \] Or \[ \left (8 x^{r} r \left (-1+r \right )+2 x^{r} r +x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (8 r^{2}-6 r +1\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 8 r^{2}-6 r +1 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {1}{2}}\\ r_2 &= {\frac {1}{4}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (8 r^{2}-6 r +1\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {1}{2}}, {\frac {1}{4}}\right ]\).
Since \(r_1 - r_2 = {\frac {1}{4}}\) 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}{2}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +\frac {1}{4}} \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} a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+8 a_{n} \left (n +r \right ) \left (n +r -1\right )+3 a_{n -1} \left (n +r -1\right )+2 a_{n} \left (n +r \right )+a_{n -1}+a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -1} \left (n^{2}+2 n r +r^{2}\right )}{8 n^{2}+16 n r +8 r^{2}-6 n -6 r +1}\tag {4} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{n} = -\frac {a_{n -1} \left (2 n +1\right )^{2}}{32 n^{2}+8 n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {1}{2}}\) 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 \[ a_{1}=-\frac {\left (r +1\right )^{2}}{8 r^{2}+10 r +3} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{1}=-{\frac {9}{40}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {\left (r +1\right )^{2}}{8 r^{2}+10 r +3}\) | \(-{\frac {9}{40}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {\left (r +1\right )^{2} \left (2+r \right )^{2}}{64 r^{4}+288 r^{3}+452 r^{2}+288 r +63} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{2}={\frac {5}{128}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {\left (r +1\right )^{2}}{8 r^{2}+10 r +3}\) | \(-{\frac {9}{40}}\) |
| \(a_{2}\) | \(\frac {\left (r +1\right )^{2} \left (2+r \right )^{2}}{64 r^{4}+288 r^{3}+452 r^{2}+288 r +63}\) | \(\frac {5}{128}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2}}{512 r^{6}+4992 r^{5}+19232 r^{4}+37128 r^{3}+37460 r^{2}+18486 r +3465} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{3}=-{\frac {245}{39936}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {\left (r +1\right )^{2}}{8 r^{2}+10 r +3}\) | \(-{\frac {9}{40}}\) |
| \(a_{2}\) | \(\frac {\left (r +1\right )^{2} \left (2+r \right )^{2}}{64 r^{4}+288 r^{3}+452 r^{2}+288 r +63}\) | \(\frac {5}{128}\) |
| \(a_{3}\) | \(-\frac {\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2}}{512 r^{6}+4992 r^{5}+19232 r^{4}+37128 r^{3}+37460 r^{2}+18486 r +3465}\) | \(-{\frac {245}{39936}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}{\left (512 r^{6}+4992 r^{5}+19232 r^{4}+37128 r^{3}+37460 r^{2}+18486 r +3465\right ) \left (8 r^{2}+58 r +105\right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{4}={\frac {6615}{7241728}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {\left (r +1\right )^{2}}{8 r^{2}+10 r +3}\) | \(-{\frac {9}{40}}\) |
| \(a_{2}\) | \(\frac {\left (r +1\right )^{2} \left (2+r \right )^{2}}{64 r^{4}+288 r^{3}+452 r^{2}+288 r +63}\) | \(\frac {5}{128}\) |
| \(a_{3}\) | \(-\frac {\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2}}{512 r^{6}+4992 r^{5}+19232 r^{4}+37128 r^{3}+37460 r^{2}+18486 r +3465}\) | \(-{\frac {245}{39936}}\) |
| \(a_{4}\) | \(\frac {\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}{\left (512 r^{6}+4992 r^{5}+19232 r^{4}+37128 r^{3}+37460 r^{2}+18486 r +3465\right ) \left (8 r^{2}+58 r +105\right )}\) | \(\frac {6615}{7241728}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (5+r \right )^{2}}{\left (512 r^{6}+4992 r^{5}+19232 r^{4}+37128 r^{3}+37460 r^{2}+18486 r +3465\right ) \left (8 r^{2}+58 r +105\right ) \left (8 r^{2}+74 r +171\right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{5}=-{\frac {7623}{57933824}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {\left (r +1\right )^{2}}{8 r^{2}+10 r +3}\) | \(-{\frac {9}{40}}\) |
| \(a_{2}\) | \(\frac {\left (r +1\right )^{2} \left (2+r \right )^{2}}{64 r^{4}+288 r^{3}+452 r^{2}+288 r +63}\) | \(\frac {5}{128}\) |
| \(a_{3}\) | \(-\frac {\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2}}{512 r^{6}+4992 r^{5}+19232 r^{4}+37128 r^{3}+37460 r^{2}+18486 r +3465}\) | \(-{\frac {245}{39936}}\) |
| \(a_{4}\) | \(\frac {\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}{\left (512 r^{6}+4992 r^{5}+19232 r^{4}+37128 r^{3}+37460 r^{2}+18486 r +3465\right ) \left (8 r^{2}+58 r +105\right )}\) | \(\frac {6615}{7241728}\) |
| \(a_{5}\) | \(-\frac {\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (5+r \right )^{2}}{\left (512 r^{6}+4992 r^{5}+19232 r^{4}+37128 r^{3}+37460 r^{2}+18486 r +3465\right ) \left (8 r^{2}+58 r +105\right ) \left (8 r^{2}+74 r +171\right )}\) | \(-{\frac {7623}{57933824}}\) |
Using the above table, then the 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}\dots \right ) \\ &= \sqrt {x}\, \left (1-\frac {9 x}{40}+\frac {5 x^{2}}{128}-\frac {245 x^{3}}{39936}+\frac {6615 x^{4}}{7241728}-\frac {7623 x^{5}}{57933824}+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} b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+8 b_{n} \left (n +r \right ) \left (n +r -1\right )+3 b_{n -1} \left (n +r -1\right )+2 b_{n} \left (n +r \right )+b_{n -1}+b_{n} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {b_{n -1} \left (n^{2}+2 n r +r^{2}\right )}{8 n^{2}+16 n r +8 r^{2}-6 n -6 r +1}\tag {4} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ b_{n} = -\frac {b_{n -1} \left (4 n +1\right )^{2}}{128 n^{2}-32 n}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = {\frac {1}{4}}\) 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 \[ b_{1}=-\frac {\left (r +1\right )^{2}}{8 r^{2}+10 r +3} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ b_{1}=-{\frac {25}{96}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {\left (r +1\right )^{2}}{8 r^{2}+10 r +3}\) | \(-{\frac {25}{96}}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {\left (r +1\right )^{2} \left (2+r \right )^{2}}{64 r^{4}+288 r^{3}+452 r^{2}+288 r +63} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ b_{2}={\frac {675}{14336}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {\left (r +1\right )^{2}}{8 r^{2}+10 r +3}\) | \(-{\frac {25}{96}}\) |
| \(b_{2}\) | \(\frac {\left (r +1\right )^{2} \left (2+r \right )^{2}}{64 r^{4}+288 r^{3}+452 r^{2}+288 r +63}\) | \(\frac {675}{14336}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=-\frac {\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2}}{512 r^{6}+4992 r^{5}+19232 r^{4}+37128 r^{3}+37460 r^{2}+18486 r +3465} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ b_{3}=-{\frac {38025}{5046272}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {\left (r +1\right )^{2}}{8 r^{2}+10 r +3}\) | \(-{\frac {25}{96}}\) |
| \(b_{2}\) | \(\frac {\left (r +1\right )^{2} \left (2+r \right )^{2}}{64 r^{4}+288 r^{3}+452 r^{2}+288 r +63}\) | \(\frac {675}{14336}\) |
| \(b_{3}\) | \(-\frac {\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2}}{512 r^{6}+4992 r^{5}+19232 r^{4}+37128 r^{3}+37460 r^{2}+18486 r +3465}\) | \(-{\frac {38025}{5046272}}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}{\left (512 r^{6}+4992 r^{5}+19232 r^{4}+37128 r^{3}+37460 r^{2}+18486 r +3465\right ) \left (8 r^{2}+58 r +105\right )} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ b_{4}={\frac {732615}{645922816}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {\left (r +1\right )^{2}}{8 r^{2}+10 r +3}\) | \(-{\frac {25}{96}}\) |
| \(b_{2}\) | \(\frac {\left (r +1\right )^{2} \left (2+r \right )^{2}}{64 r^{4}+288 r^{3}+452 r^{2}+288 r +63}\) | \(\frac {675}{14336}\) |
| \(b_{3}\) | \(-\frac {\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2}}{512 r^{6}+4992 r^{5}+19232 r^{4}+37128 r^{3}+37460 r^{2}+18486 r +3465}\) | \(-{\frac {38025}{5046272}}\) |
| \(b_{4}\) | \(\frac {\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}{\left (512 r^{6}+4992 r^{5}+19232 r^{4}+37128 r^{3}+37460 r^{2}+18486 r +3465\right ) \left (8 r^{2}+58 r +105\right )}\) | \(\frac {732615}{645922816}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-\frac {\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (5+r \right )^{2}}{\left (512 r^{6}+4992 r^{5}+19232 r^{4}+37128 r^{3}+37460 r^{2}+18486 r +3465\right ) \left (8 r^{2}+58 r +105\right ) \left (8 r^{2}+74 r +171\right )} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ b_{5}=-{\frac {9230949}{56103010304}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {\left (r +1\right )^{2}}{8 r^{2}+10 r +3}\) | \(-{\frac {25}{96}}\) |
| \(b_{2}\) | \(\frac {\left (r +1\right )^{2} \left (2+r \right )^{2}}{64 r^{4}+288 r^{3}+452 r^{2}+288 r +63}\) | \(\frac {675}{14336}\) |
| \(b_{3}\) | \(-\frac {\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2}}{512 r^{6}+4992 r^{5}+19232 r^{4}+37128 r^{3}+37460 r^{2}+18486 r +3465}\) | \(-{\frac {38025}{5046272}}\) |
| \(b_{4}\) | \(\frac {\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}{\left (512 r^{6}+4992 r^{5}+19232 r^{4}+37128 r^{3}+37460 r^{2}+18486 r +3465\right ) \left (8 r^{2}+58 r +105\right )}\) | \(\frac {732615}{645922816}\) |
| \(b_{5}\) | \(-\frac {\left (r +1\right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (5+r \right )^{2}}{\left (512 r^{6}+4992 r^{5}+19232 r^{4}+37128 r^{3}+37460 r^{2}+18486 r +3465\right ) \left (8 r^{2}+58 r +105\right ) \left (8 r^{2}+74 r +171\right )}\) | \(-{\frac {9230949}{56103010304}}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= \sqrt {x} \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 ) \\ &= x^{\frac {1}{4}} \left (1-\frac {25 x}{96}+\frac {675 x^{2}}{14336}-\frac {38025 x^{3}}{5046272}+\frac {732615 x^{4}}{645922816}-\frac {9230949 x^{5}}{56103010304}+O\left (x^{6}\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}{40}+\frac {5 x^{2}}{128}-\frac {245 x^{3}}{39936}+\frac {6615 x^{4}}{7241728}-\frac {7623 x^{5}}{57933824}+O\left (x^{6}\right )\right ) + c_{2} x^{\frac {1}{4}} \left (1-\frac {25 x}{96}+\frac {675 x^{2}}{14336}-\frac {38025 x^{3}}{5046272}+\frac {732615 x^{4}}{645922816}-\frac {9230949 x^{5}}{56103010304}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \sqrt {x}\, \left (1-\frac {9 x}{40}+\frac {5 x^{2}}{128}-\frac {245 x^{3}}{39936}+\frac {6615 x^{4}}{7241728}-\frac {7623 x^{5}}{57933824}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {1}{4}} \left (1-\frac {25 x}{96}+\frac {675 x^{2}}{14336}-\frac {38025 x^{3}}{5046272}+\frac {732615 x^{4}}{645922816}-\frac {9230949 x^{5}}{56103010304}+O\left (x^{6}\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}{40}+\frac {5 x^{2}}{128}-\frac {245 x^{3}}{39936}+\frac {6615 x^{4}}{7241728}-\frac {7623 x^{5}}{57933824}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {1}{4}} \left (1-\frac {25 x}{96}+\frac {675 x^{2}}{14336}-\frac {38025 x^{3}}{5046272}+\frac {732615 x^{4}}{645922816}-\frac {9230949 x^{5}}{56103010304}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \sqrt {x}\, \left (1-\frac {9 x}{40}+\frac {5 x^{2}}{128}-\frac {245 x^{3}}{39936}+\frac {6615 x^{4}}{7241728}-\frac {7623 x^{5}}{57933824}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {1}{4}} \left (1-\frac {25 x}{96}+\frac {675 x^{2}}{14336}-\frac {38025 x^{3}}{5046272}+\frac {732615 x^{4}}{645922816}-\frac {9230949 x^{5}}{56103010304}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (x +8\right ) y^{\prime \prime }+\left (3 x^{2}+2 x \right ) y^{\prime }+\left (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 (x +1\right ) y}{x^{2} \left (x +8\right )}-\frac {\left (3 x +2\right ) y^{\prime }}{x \left (x +8\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 (3 x +2\right ) y^{\prime }}{x \left (x +8\right )}+\frac {\left (x +1\right ) y}{x^{2} \left (x +8\right )}=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 {3 x +2}{x \left (x +8\right )}, P_{3}\left (x \right )=\frac {x +1}{x^{2} \left (x +8\right )}\right ] \\ {} & \circ & \left (x +8\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-8 \\ {} & {} & \left (\left (x +8\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-8}}}=\frac {11}{4} \\ {} & \circ & \left (x +8\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-8 \\ {} & {} & \left (\left (x +8\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-8}}}=0 \\ {} & \circ & x =-8\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}=-8 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x^{2} \left (x +8\right ) y^{\prime \prime }+x \left (3 x +2\right ) y^{\prime }+\left (x +1\right ) y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -8\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (u^{3}-16 u^{2}+64 u \right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (3 u^{2}-46 u +176\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (u -7\right ) y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot y \left (u \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & 16 a_{0} r \left (7+4 r \right ) u^{-1+r}+\left (16 a_{1} \left (1+r \right ) \left (11+4 r \right )-a_{0} \left (16 r^{2}+30 r +7\right )\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (16 a_{k +1} \left (k +r +1\right ) \left (4 k +11+4 r \right )-a_{k} \left (16 k^{2}+32 k r +16 r^{2}+30 k +30 r +7\right )+a_{k -1} \left (k +r \right )^{2}\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 16 r \left (7+4 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -\frac {7}{4}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & 16 a_{1} \left (1+r \right ) \left (11+4 r \right )-a_{0} \left (16 r^{2}+30 r +7\right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 64 \left (k +r +1\right ) \left (k +\frac {11}{4}+r \right ) a_{k +1}-a_{k} \left (16 k^{2}+32 k r +16 r^{2}+30 k +30 r +7\right )+a_{k -1} \left (k +r \right )^{2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 64 \left (k +2+r \right ) \left (k +\frac {15}{4}+r \right ) a_{k +2}-a_{k +1} \left (16 \left (k +1\right )^{2}+32 \left (k +1\right ) r +16 r^{2}+30 k +37+30 r \right )+a_{k} \left (k +r +1\right )^{2}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-16 k^{2} a_{k +1}+2 k r a_{k}-32 k r a_{k +1}+r^{2} a_{k}-16 r^{2} a_{k +1}+2 k a_{k}-62 k a_{k +1}+2 r a_{k}-62 r a_{k +1}+a_{k}-53 a_{k +1}}{16 \left (k +2+r \right ) \left (4 k +15+4 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-16 k^{2} a_{k +1}+2 k a_{k}-62 k a_{k +1}+a_{k}-53 a_{k +1}}{16 \left (k +2\right ) \left (4 k +15\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k}, a_{k +2}=-\frac {k^{2} a_{k}-16 k^{2} a_{k +1}+2 k a_{k}-62 k a_{k +1}+a_{k}-53 a_{k +1}}{16 \left (k +2\right ) \left (4 k +15\right )}, 176 a_{1}-7 a_{0}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +8 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +8\right )^{k}, a_{k +2}=-\frac {k^{2} a_{k}-16 k^{2} a_{k +1}+2 k a_{k}-62 k a_{k +1}+a_{k}-53 a_{k +1}}{16 \left (k +2\right ) \left (4 k +15\right )}, 176 a_{1}-7 a_{0}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {7}{4} \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-16 k^{2} a_{k +1}-\frac {3}{2} k a_{k}-6 k a_{k +1}+\frac {9}{16} a_{k}+\frac {13}{2} a_{k +1}}{16 \left (k +\frac {1}{4}\right ) \left (4 k +8\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {7}{4} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k -\frac {7}{4}}, a_{k +2}=-\frac {k^{2} a_{k}-16 k^{2} a_{k +1}-\frac {3}{2} k a_{k}-6 k a_{k +1}+\frac {9}{16} a_{k}+\frac {13}{2} a_{k +1}}{16 \left (k +\frac {1}{4}\right ) \left (4 k +8\right )}, -48 a_{1}-\frac {7 a_{0}}{2}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +8 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +8\right )^{k -\frac {7}{4}}, a_{k +2}=-\frac {k^{2} a_{k}-16 k^{2} a_{k +1}-\frac {3}{2} k a_{k}-6 k a_{k +1}+\frac {9}{16} a_{k}+\frac {13}{2} a_{k +1}}{16 \left (k +\frac {1}{4}\right ) \left (4 k +8\right )}, -48 a_{1}-\frac {7 a_{0}}{2}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +8\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (x +8\right )^{k -\frac {7}{4}}\right ), a_{k +2}=-\frac {k^{2} a_{k}-16 k^{2} a_{1+k}+2 k a_{k}-62 k a_{1+k}+a_{k}-53 a_{1+k}}{16 \left (k +2\right ) \left (4 k +15\right )}, 176 a_{1}-7 a_{0}=0, b_{k +2}=-\frac {k^{2} b_{k}-16 k^{2} b_{1+k}-\frac {3}{2} k b_{k}-6 k b_{1+k}+\frac {9}{16} b_{k}+\frac {13}{2} b_{1+k}}{16 \left (k +\frac {1}{4}\right ) \left (4 k +8\right )}, -48 b_{1}-\frac {7 b_{0}}{2}=0\right ] \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> 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`
✓ Solution by Maple
Time used: 0.032 (sec). Leaf size: 47
Order:=6; dsolve(x^2*(8+x)*diff(y(x),x$2)+x*(2+3*x)*diff(y(x),x)+(1+x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{\frac {1}{4}} \left (1-\frac {25}{96} x +\frac {675}{14336} x^{2}-\frac {38025}{5046272} x^{3}+\frac {732615}{645922816} x^{4}-\frac {9230949}{56103010304} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \sqrt {x}\, \left (1-\frac {9}{40} x +\frac {5}{128} x^{2}-\frac {245}{39936} x^{3}+\frac {6615}{7241728} x^{4}-\frac {7623}{57933824} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.005 (sec). Leaf size: 90
AsymptoticDSolveValue[x^2*(8+x)*y''[x]+x*(2+3*x)*y'[x]+(1+x)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \sqrt {x} \left (-\frac {7623 x^5}{57933824}+\frac {6615 x^4}{7241728}-\frac {245 x^3}{39936}+\frac {5 x^2}{128}-\frac {9 x}{40}+1\right )+c_2 \sqrt [4]{x} \left (-\frac {9230949 x^5}{56103010304}+\frac {732615 x^4}{645922816}-\frac {38025 x^3}{5046272}+\frac {675 x^2}{14336}-\frac {25 x}{96}+1\right ) \]