Internal problem ID [14886]
Internal file name [OUTPUT/14566_Monday_April_08_2024_06_27_25_AM_48665625/index.tex]
Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton.
Fourth edition 2014. ElScAe. 2014
Section: Chapter 4. Higher Order Equations. Chapter 4 review exercises, page 219
Problem number: 66.
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
[[_Emden, _Fowler]]
\[ \boxed {3 x y^{\prime \prime }+11 y^{\prime }-y=0} \] 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. \[ 3 x y^{\prime \prime }+11 y^{\prime }-y = 0 \] 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 {11}{3 x}\\ q(x) &= -\frac {1}{3 x}\\ \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 \[ 3 x y^{\prime \prime }+11 y^{\prime }-y = 0 \] 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} 3 x \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+11 \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 }}3 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}11 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} x^{n +r}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r -1\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r -1}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} x^{n +r}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} x^{n +r -1}\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 \(n +r -1\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}11 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 3 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+11 \left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ 3 x^{-1+r} a_{0} r \left (-1+r \right )+11 r a_{0} x^{-1+r} = 0 \] Or \[ \left (3 x^{-1+r} r \left (-1+r \right )+11 r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ r \,x^{-1+r} \left (8+3 r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 3 r^{2}+8 r = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 0\\ r_2 &= -{\frac {8}{3}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ r \,x^{-1+r} \left (8+3 r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [0, -{\frac {8}{3}}\right ]\).
Since \(r_1 - r_2 = {\frac {8}{3}}\) 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}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -\frac {8}{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} 3 a_{n} \left (n +r \right ) \left (n +r -1\right )+11 a_{n} \left (n +r \right )-a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {a_{n -1}}{3 n^{2}+6 n r +3 r^{2}+8 n +8 r}\tag {4} \] Which for the root \(r = 0\) becomes \[ a_{n} = \frac {a_{n -1}}{n \left (3 n +8\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 0\) 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 \[ a_{1}=\frac {1}{3 r^{2}+14 r +11} \] Which for the root \(r = 0\) becomes \[ a_{1}={\frac {1}{11}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{3 r^{2}+14 r +11}\) | \(\frac {1}{11}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {1}{9 r^{4}+102 r^{3}+397 r^{2}+612 r +308} \] Which for the root \(r = 0\) becomes \[ a_{2}={\frac {1}{308}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{3 r^{2}+14 r +11}\) | \(\frac {1}{11}\) |
| \(a_{2}\) | \(\frac {1}{9 r^{4}+102 r^{3}+397 r^{2}+612 r +308}\) | \(\frac {1}{308}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {1}{27 r^{6}+540 r^{5}+4302 r^{4}+17360 r^{3}+37083 r^{2}+39220 r +15708} \] Which for the root \(r = 0\) becomes \[ a_{3}={\frac {1}{15708}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{3 r^{2}+14 r +11}\) | \(\frac {1}{11}\) |
| \(a_{2}\) | \(\frac {1}{9 r^{4}+102 r^{3}+397 r^{2}+612 r +308}\) | \(\frac {1}{308}\) |
| \(a_{3}\) | \(\frac {1}{27 r^{6}+540 r^{5}+4302 r^{4}+17360 r^{3}+37083 r^{2}+39220 r +15708}\) | \(\frac {1}{15708}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {1}{81 r^{8}+2484 r^{7}+32346 r^{6}+232944 r^{5}+1010929 r^{4}+2693116 r^{3}+4268804 r^{2}+3640256 r +1256640} \] Which for the root \(r = 0\) becomes \[ a_{4}={\frac {1}{1256640}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{3 r^{2}+14 r +11}\) | \(\frac {1}{11}\) |
| \(a_{2}\) | \(\frac {1}{9 r^{4}+102 r^{3}+397 r^{2}+612 r +308}\) | \(\frac {1}{308}\) |
| \(a_{3}\) | \(\frac {1}{27 r^{6}+540 r^{5}+4302 r^{4}+17360 r^{3}+37083 r^{2}+39220 r +15708}\) | \(\frac {1}{15708}\) |
| \(a_{4}\) | \(\frac {1}{81 r^{8}+2484 r^{7}+32346 r^{6}+232944 r^{5}+1010929 r^{4}+2693116 r^{3}+4268804 r^{2}+3640256 r +1256640}\) | \(\frac {1}{1256640}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {1}{\left (81 r^{8}+2484 r^{7}+32346 r^{6}+232944 r^{5}+1010929 r^{4}+2693116 r^{3}+4268804 r^{2}+3640256 r +1256640\right ) \left (3 r^{2}+38 r +115\right )} \] Which for the root \(r = 0\) becomes \[ a_{5}={\frac {1}{144513600}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{3 r^{2}+14 r +11}\) | \(\frac {1}{11}\) |
| \(a_{2}\) | \(\frac {1}{9 r^{4}+102 r^{3}+397 r^{2}+612 r +308}\) | \(\frac {1}{308}\) |
| \(a_{3}\) | \(\frac {1}{27 r^{6}+540 r^{5}+4302 r^{4}+17360 r^{3}+37083 r^{2}+39220 r +15708}\) | \(\frac {1}{15708}\) |
| \(a_{4}\) | \(\frac {1}{81 r^{8}+2484 r^{7}+32346 r^{6}+232944 r^{5}+1010929 r^{4}+2693116 r^{3}+4268804 r^{2}+3640256 r +1256640}\) | \(\frac {1}{1256640}\) |
| \(a_{5}\) | \(\frac {1}{\left (81 r^{8}+2484 r^{7}+32346 r^{6}+232944 r^{5}+1010929 r^{4}+2693116 r^{3}+4268804 r^{2}+3640256 r +1256640\right ) \left (3 r^{2}+38 r +115\right )}\) | \(\frac {1}{144513600}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= 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 \\ &= 1+\frac {x}{11}+\frac {x^{2}}{308}+\frac {x^{3}}{15708}+\frac {x^{4}}{1256640}+\frac {x^{5}}{144513600}+O\left (x^{6}\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} 3 b_{n} \left (n +r \right ) \left (n +r -1\right )+11 \left (n +r \right ) b_{n}-b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = \frac {b_{n -1}}{3 n^{2}+6 n r +3 r^{2}+8 n +8 r}\tag {4} \] Which for the root \(r = -{\frac {8}{3}}\) becomes \[ b_{n} = \frac {b_{n -1}}{n \left (3 n -8\right )}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -{\frac {8}{3}}\) 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 \[ b_{1}=\frac {1}{3 r^{2}+14 r +11} \] Which for the root \(r = -{\frac {8}{3}}\) becomes \[ b_{1}=-{\frac {1}{5}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{3 r^{2}+14 r +11}\) | \(-{\frac {1}{5}}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {1}{9 r^{4}+102 r^{3}+397 r^{2}+612 r +308} \] Which for the root \(r = -{\frac {8}{3}}\) becomes \[ b_{2}={\frac {1}{20}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{3 r^{2}+14 r +11}\) | \(-{\frac {1}{5}}\) |
| \(b_{2}\) | \(\frac {1}{9 r^{4}+102 r^{3}+397 r^{2}+612 r +308}\) | \(\frac {1}{20}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {1}{27 r^{6}+540 r^{5}+4302 r^{4}+17360 r^{3}+37083 r^{2}+39220 r +15708} \] Which for the root \(r = -{\frac {8}{3}}\) becomes \[ b_{3}={\frac {1}{60}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{3 r^{2}+14 r +11}\) | \(-{\frac {1}{5}}\) |
| \(b_{2}\) | \(\frac {1}{9 r^{4}+102 r^{3}+397 r^{2}+612 r +308}\) | \(\frac {1}{20}\) |
| \(b_{3}\) | \(\frac {1}{27 r^{6}+540 r^{5}+4302 r^{4}+17360 r^{3}+37083 r^{2}+39220 r +15708}\) | \(\frac {1}{60}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {1}{81 r^{8}+2484 r^{7}+32346 r^{6}+232944 r^{5}+1010929 r^{4}+2693116 r^{3}+4268804 r^{2}+3640256 r +1256640} \] Which for the root \(r = -{\frac {8}{3}}\) becomes \[ b_{4}={\frac {1}{960}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{3 r^{2}+14 r +11}\) | \(-{\frac {1}{5}}\) |
| \(b_{2}\) | \(\frac {1}{9 r^{4}+102 r^{3}+397 r^{2}+612 r +308}\) | \(\frac {1}{20}\) |
| \(b_{3}\) | \(\frac {1}{27 r^{6}+540 r^{5}+4302 r^{4}+17360 r^{3}+37083 r^{2}+39220 r +15708}\) | \(\frac {1}{60}\) |
| \(b_{4}\) | \(\frac {1}{81 r^{8}+2484 r^{7}+32346 r^{6}+232944 r^{5}+1010929 r^{4}+2693116 r^{3}+4268804 r^{2}+3640256 r +1256640}\) | \(\frac {1}{960}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {1}{\left (81 r^{8}+2484 r^{7}+32346 r^{6}+232944 r^{5}+1010929 r^{4}+2693116 r^{3}+4268804 r^{2}+3640256 r +1256640\right ) \left (3 r^{2}+38 r +115\right )} \] Which for the root \(r = -{\frac {8}{3}}\) becomes \[ b_{5}={\frac {1}{33600}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{3 r^{2}+14 r +11}\) | \(-{\frac {1}{5}}\) |
| \(b_{2}\) | \(\frac {1}{9 r^{4}+102 r^{3}+397 r^{2}+612 r +308}\) | \(\frac {1}{20}\) |
| \(b_{3}\) | \(\frac {1}{27 r^{6}+540 r^{5}+4302 r^{4}+17360 r^{3}+37083 r^{2}+39220 r +15708}\) | \(\frac {1}{60}\) |
| \(b_{4}\) | \(\frac {1}{81 r^{8}+2484 r^{7}+32346 r^{6}+232944 r^{5}+1010929 r^{4}+2693116 r^{3}+4268804 r^{2}+3640256 r +1256640}\) | \(\frac {1}{960}\) |
| \(b_{5}\) | \(\frac {1}{\left (81 r^{8}+2484 r^{7}+32346 r^{6}+232944 r^{5}+1010929 r^{4}+2693116 r^{3}+4268804 r^{2}+3640256 r +1256640\right ) \left (3 r^{2}+38 r +115\right )}\) | \(\frac {1}{33600}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= 1 \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 {x}{5}+\frac {x^{2}}{20}+\frac {x^{3}}{60}+\frac {x^{4}}{960}+\frac {x^{5}}{33600}+O\left (x^{6}\right )}{x^{\frac {8}{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} \left (1+\frac {x}{11}+\frac {x^{2}}{308}+\frac {x^{3}}{15708}+\frac {x^{4}}{1256640}+\frac {x^{5}}{144513600}+O\left (x^{6}\right )\right ) + \frac {c_{2} \left (1-\frac {x}{5}+\frac {x^{2}}{20}+\frac {x^{3}}{60}+\frac {x^{4}}{960}+\frac {x^{5}}{33600}+O\left (x^{6}\right )\right )}{x^{\frac {8}{3}}} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \left (1+\frac {x}{11}+\frac {x^{2}}{308}+\frac {x^{3}}{15708}+\frac {x^{4}}{1256640}+\frac {x^{5}}{144513600}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-\frac {x}{5}+\frac {x^{2}}{20}+\frac {x^{3}}{60}+\frac {x^{4}}{960}+\frac {x^{5}}{33600}+O\left (x^{6}\right )\right )}{x^{\frac {8}{3}}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (1+\frac {x}{11}+\frac {x^{2}}{308}+\frac {x^{3}}{15708}+\frac {x^{4}}{1256640}+\frac {x^{5}}{144513600}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-\frac {x}{5}+\frac {x^{2}}{20}+\frac {x^{3}}{60}+\frac {x^{4}}{960}+\frac {x^{5}}{33600}+O\left (x^{6}\right )\right )}{x^{\frac {8}{3}}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (1+\frac {x}{11}+\frac {x^{2}}{308}+\frac {x^{3}}{15708}+\frac {x^{4}}{1256640}+\frac {x^{5}}{144513600}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-\frac {x}{5}+\frac {x^{2}}{20}+\frac {x^{3}}{60}+\frac {x^{4}}{960}+\frac {x^{5}}{33600}+O\left (x^{6}\right )\right )}{x^{\frac {8}{3}}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 3 x \left (\frac {d}{d x}y^{\prime }\right )+11 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}{3 x}-\frac {11 y^{\prime }}{3 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 {11 y^{\prime }}{3 x}-\frac {y}{3 x}=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 {11}{3 x}, P_{3}\left (x \right )=-\frac {1}{3 x}\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}}}=0 \\ {} & \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} \\ {} & {} & 3 x \left (\frac {d}{d x}y^{\prime }\right )+11 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} y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & 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 \\ {} & {} & 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 \cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \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 -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & x \cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (8+3 r \right ) x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (3 k +11+3 r \right )-a_{k}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (8+3 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -\frac {8}{3}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 3 \left (k +1+r \right ) \left (k +\frac {11}{3}+r \right ) a_{k +1}-a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k}}{\left (k +1+r \right ) \left (3 k +11+3 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {a_{k}}{\left (k +1\right ) \left (3 k +11\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +1}=\frac {a_{k}}{\left (k +1\right ) \left (3 k +11\right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {8}{3} \\ {} & {} & a_{k +1}=\frac {a_{k}}{\left (k -\frac {5}{3}\right ) \left (3 k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {8}{3} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {8}{3}}, a_{k +1}=\frac {a_{k}}{\left (k -\frac {5}{3}\right ) \left (3 k +3\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k -\frac {8}{3}}\right ), a_{k +1}=\frac {a_{k}}{\left (k +1\right ) \left (3 k +11\right )}, b_{k +1}=\frac {b_{k}}{\left (k -\frac {5}{3}\right ) \left (3 k +3\right )}\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 <- Bessel successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.062 (sec). Leaf size: 44
Order:=6; dsolve(3*x*diff(y(x),x$2)+11*diff(y(x),x)-y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{1} \left (1-\frac {1}{5} x +\frac {1}{20} x^{2}+\frac {1}{60} x^{3}+\frac {1}{960} x^{4}+\frac {1}{33600} x^{5}+\operatorname {O}\left (x^{6}\right )\right )}{x^{\frac {8}{3}}}+c_{2} \left (1+\frac {1}{11} x +\frac {1}{308} x^{2}+\frac {1}{15708} x^{3}+\frac {1}{1256640} x^{4}+\frac {1}{144513600} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.002 (sec). Leaf size: 85
AsymptoticDSolveValue[3*x*y''[x]+11*y'[x]-y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (\frac {x^5}{144513600}+\frac {x^4}{1256640}+\frac {x^3}{15708}+\frac {x^2}{308}+\frac {x}{11}+1\right )+\frac {c_2 \left (\frac {x^5}{33600}+\frac {x^4}{960}+\frac {x^3}{60}+\frac {x^2}{20}-\frac {x}{5}+1\right )}{x^{8/3}} \]