Internal problem ID [11901]
Internal file name [OUTPUT/11911_Saturday_April_13_2024_01_14_37_AM_31651573/index.tex]
Book: Differential Equations by Shepley L. Ross. Third edition. John Willey. New Delhi.
2004.
Section: Chapter 6, Series solutions of linear differential equations. Section 6.2 (Frobenius).
Exercises page 251
Problem number: 1.
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]]
\[ \boxed {\left (x^{2}-3 x \right ) y^{\prime \prime }+\left (x +2\right ) y^{\prime }+y=0} \] With the expansion point for the power series method at \(x = 0\).
The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ \left (x^{2}-3 x \right ) y^{\prime \prime }+\left (x +2\right ) y^{\prime }+y = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} y^{\prime \prime }+p(x) y^{\prime } + q(x) y &=0 \end {align*}
Where \begin {align*} p(x) &= \frac {x +2}{x \left (x -3\right )}\\ q(x) &= \frac {1}{\left (x -3\right ) x}\\ \end {align*}
| \(p(x)=\frac {x +2}{x \left (x -3\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = 3\) | \(\text {``regular''}\) |
| \(q(x)=\frac {1}{\left (x -3\right ) x}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = 3\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, 3, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ y^{\prime \prime } x \left (x -3\right )+\left (x +2\right ) y^{\prime }+y = 0 \] Let the solution be represented as Frobenius power series of the form \[ y = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} \] Then \begin{align*} y^{\prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1} \\ y^{\prime \prime } &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2} \\ \end{align*} Substituting the above back into the ode gives \begin{equation} \tag{1} \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right ) x \left (x -3\right )+\left (x +2\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 \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} 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 }}x^{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 -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r -1} \\ \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 =1}{\sum }}a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ -3 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+2 \left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ -3 x^{-1+r} a_{0} r \left (-1+r \right )+2 r a_{0} x^{-1+r} = 0 \] Or \[ \left (-3 x^{-1+r} r \left (-1+r \right )+2 r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (-3 r^{2}+5 r \right ) x^{-1+r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ -3 r^{2}+5 r = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {5}{3}}\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (-3 r^{2}+5 r \right ) x^{-1+r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {5}{3}}, 0\right ]\).
Since \(r_1 - r_2 = {\frac {5}{3}}\) 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 {5}{3}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n} \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 )-3 a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n -1} \left (n +r -1\right )+2 a_{n} \left (n +r \right )+a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {a_{n -1} \left (n^{2}+2 n r +r^{2}-2 n -2 r +2\right )}{3 n^{2}+6 n r +3 r^{2}-5 n -5 r}\tag {4} \] Which for the root \(r = {\frac {5}{3}}\) becomes \[ a_{n} = \frac {a_{n -1} \left (9 n^{2}+12 n +13\right )}{27 n^{2}+45 n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {5}{3}}\) 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 \[ a_{1}=\frac {r^{2}+1}{3 r^{2}+r -2} \] Which for the root \(r = {\frac {5}{3}}\) becomes \[ a_{1}={\frac {17}{36}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r^{2}+1}{3 r^{2}+r -2}\) | \(\frac {17}{36}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right )}{9 r^{4}+24 r^{3}+7 r^{2}-12 r -4} \] Which for the root \(r = {\frac {5}{3}}\) becomes \[ a_{2}={\frac {1241}{7128}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r^{2}+1}{3 r^{2}+r -2}\) | \(\frac {17}{36}\) |
| \(a_{2}\) | \(\frac {\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right )}{9 r^{4}+24 r^{3}+7 r^{2}-12 r -4}\) | \(\frac {1241}{7128}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right )}{27 r^{6}+189 r^{5}+441 r^{4}+343 r^{3}-84 r^{2}-196 r -48} \] Which for the root \(r = {\frac {5}{3}}\) becomes \[ a_{3}={\frac {80665}{1347192}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r^{2}+1}{3 r^{2}+r -2}\) | \(\frac {17}{36}\) |
| \(a_{2}\) | \(\frac {\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right )}{9 r^{4}+24 r^{3}+7 r^{2}-12 r -4}\) | \(\frac {1241}{7128}\) |
| \(a_{3}\) | \(\frac {\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right )}{27 r^{6}+189 r^{5}+441 r^{4}+343 r^{3}-84 r^{2}-196 r -48}\) | \(\frac {80665}{1347192}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right )}{81 r^{8}+1080 r^{7}+5670 r^{6}+14700 r^{5}+18613 r^{4}+7420 r^{3}-6220 r^{2}-6400 r -1344} \] Which for the root \(r = {\frac {5}{3}}\) becomes \[ a_{4}={\frac {972725}{48498912}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r^{2}+1}{3 r^{2}+r -2}\) | \(\frac {17}{36}\) |
| \(a_{2}\) | \(\frac {\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right )}{9 r^{4}+24 r^{3}+7 r^{2}-12 r -4}\) | \(\frac {1241}{7128}\) |
| \(a_{3}\) | \(\frac {\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right )}{27 r^{6}+189 r^{5}+441 r^{4}+343 r^{3}-84 r^{2}-196 r -48}\) | \(\frac {80665}{1347192}\) |
| \(a_{4}\) | \(\frac {\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right )}{81 r^{8}+1080 r^{7}+5670 r^{6}+14700 r^{5}+18613 r^{4}+7420 r^{3}-6220 r^{2}-6400 r -1344}\) | \(\frac {972725}{48498912}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right )}{\left (81 r^{8}+1080 r^{7}+5670 r^{6}+14700 r^{5}+18613 r^{4}+7420 r^{3}-6220 r^{2}-6400 r -1344\right ) \left (3 r^{2}+25 r +50\right )} \] Which for the root \(r = {\frac {5}{3}}\) becomes \[ a_{5}={\frac {5797441}{872980416}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r^{2}+1}{3 r^{2}+r -2}\) | \(\frac {17}{36}\) |
| \(a_{2}\) | \(\frac {\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right )}{9 r^{4}+24 r^{3}+7 r^{2}-12 r -4}\) | \(\frac {1241}{7128}\) |
| \(a_{3}\) | \(\frac {\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right )}{27 r^{6}+189 r^{5}+441 r^{4}+343 r^{3}-84 r^{2}-196 r -48}\) | \(\frac {80665}{1347192}\) |
| \(a_{4}\) | \(\frac {\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right )}{81 r^{8}+1080 r^{7}+5670 r^{6}+14700 r^{5}+18613 r^{4}+7420 r^{3}-6220 r^{2}-6400 r -1344}\) | \(\frac {972725}{48498912}\) |
| \(a_{5}\) | \(\frac {\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right )}{\left (81 r^{8}+1080 r^{7}+5670 r^{6}+14700 r^{5}+18613 r^{4}+7420 r^{3}-6220 r^{2}-6400 r -1344\right ) \left (3 r^{2}+25 r +50\right )}\) | \(\frac {5797441}{872980416}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{\frac {5}{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 {5}{3}} \left (1+\frac {17 x}{36}+\frac {1241 x^{2}}{7128}+\frac {80665 x^{3}}{1347192}+\frac {972725 x^{4}}{48498912}+\frac {5797441 x^{5}}{872980416}+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 )-3 b_{n} \left (n +r \right ) \left (n +r -1\right )+b_{n -1} \left (n +r -1\right )+2 \left (n +r \right ) b_{n}+b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = \frac {b_{n -1} \left (n^{2}+2 n r +r^{2}-2 n -2 r +2\right )}{3 n^{2}+6 n r +3 r^{2}-5 n -5 r}\tag {4} \] Which for the root \(r = 0\) becomes \[ b_{n} = \frac {b_{n -1} \left (n^{2}-2 n +2\right )}{n \left (3 n -5\right )}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = 0\) 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 \[ b_{1}=\frac {r^{2}+1}{3 r^{2}+r -2} \] Which for the root \(r = 0\) becomes \[ b_{1}=-{\frac {1}{2}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {r^{2}+1}{3 r^{2}+r -2}\) | \(-{\frac {1}{2}}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right )}{9 r^{4}+24 r^{3}+7 r^{2}-12 r -4} \] Which for the root \(r = 0\) becomes \[ b_{2}=-{\frac {1}{2}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {r^{2}+1}{3 r^{2}+r -2}\) | \(-{\frac {1}{2}}\) |
| \(b_{2}\) | \(\frac {\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right )}{9 r^{4}+24 r^{3}+7 r^{2}-12 r -4}\) | \(-{\frac {1}{2}}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right )}{27 r^{6}+189 r^{5}+441 r^{4}+343 r^{3}-84 r^{2}-196 r -48} \] Which for the root \(r = 0\) becomes \[ b_{3}=-{\frac {5}{24}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {r^{2}+1}{3 r^{2}+r -2}\) | \(-{\frac {1}{2}}\) |
| \(b_{2}\) | \(\frac {\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right )}{9 r^{4}+24 r^{3}+7 r^{2}-12 r -4}\) | \(-{\frac {1}{2}}\) |
| \(b_{3}\) | \(\frac {\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right )}{27 r^{6}+189 r^{5}+441 r^{4}+343 r^{3}-84 r^{2}-196 r -48}\) | \(-{\frac {5}{24}}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right )}{81 r^{8}+1080 r^{7}+5670 r^{6}+14700 r^{5}+18613 r^{4}+7420 r^{3}-6220 r^{2}-6400 r -1344} \] Which for the root \(r = 0\) becomes \[ b_{4}=-{\frac {25}{336}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {r^{2}+1}{3 r^{2}+r -2}\) | \(-{\frac {1}{2}}\) |
| \(b_{2}\) | \(\frac {\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right )}{9 r^{4}+24 r^{3}+7 r^{2}-12 r -4}\) | \(-{\frac {1}{2}}\) |
| \(b_{3}\) | \(\frac {\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right )}{27 r^{6}+189 r^{5}+441 r^{4}+343 r^{3}-84 r^{2}-196 r -48}\) | \(-{\frac {5}{24}}\) |
| \(b_{4}\) | \(\frac {\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right )}{81 r^{8}+1080 r^{7}+5670 r^{6}+14700 r^{5}+18613 r^{4}+7420 r^{3}-6220 r^{2}-6400 r -1344}\) | \(-{\frac {25}{336}}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right )}{\left (81 r^{8}+1080 r^{7}+5670 r^{6}+14700 r^{5}+18613 r^{4}+7420 r^{3}-6220 r^{2}-6400 r -1344\right ) \left (3 r^{2}+25 r +50\right )} \] Which for the root \(r = 0\) becomes \[ b_{5}=-{\frac {17}{672}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {r^{2}+1}{3 r^{2}+r -2}\) | \(-{\frac {1}{2}}\) |
| \(b_{2}\) | \(\frac {\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right )}{9 r^{4}+24 r^{3}+7 r^{2}-12 r -4}\) | \(-{\frac {1}{2}}\) |
| \(b_{3}\) | \(\frac {\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right )}{27 r^{6}+189 r^{5}+441 r^{4}+343 r^{3}-84 r^{2}-196 r -48}\) | \(-{\frac {5}{24}}\) |
| \(b_{4}\) | \(\frac {\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right )}{81 r^{8}+1080 r^{7}+5670 r^{6}+14700 r^{5}+18613 r^{4}+7420 r^{3}-6220 r^{2}-6400 r -1344}\) | \(-{\frac {25}{336}}\) |
| \(b_{5}\) | \(\frac {\left (r^{2}+1\right ) \left (r^{2}+2 r +2\right ) \left (r^{2}+4 r +5\right ) \left (r^{2}+6 r +10\right ) \left (r^{2}+8 r +17\right )}{\left (81 r^{8}+1080 r^{7}+5670 r^{6}+14700 r^{5}+18613 r^{4}+7420 r^{3}-6220 r^{2}-6400 r -1344\right ) \left (3 r^{2}+25 r +50\right )}\) | \(-{\frac {17}{672}}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\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}\dots \\ &= 1-\frac {x}{2}-\frac {x^{2}}{2}-\frac {5 x^{3}}{24}-\frac {25 x^{4}}{336}-\frac {17 x^{5}}{672}+O\left (x^{6}\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} x^{\frac {5}{3}} \left (1+\frac {17 x}{36}+\frac {1241 x^{2}}{7128}+\frac {80665 x^{3}}{1347192}+\frac {972725 x^{4}}{48498912}+\frac {5797441 x^{5}}{872980416}+O\left (x^{6}\right )\right ) + c_{2} \left (1-\frac {x}{2}-\frac {x^{2}}{2}-\frac {5 x^{3}}{24}-\frac {25 x^{4}}{336}-\frac {17 x^{5}}{672}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {5}{3}} \left (1+\frac {17 x}{36}+\frac {1241 x^{2}}{7128}+\frac {80665 x^{3}}{1347192}+\frac {972725 x^{4}}{48498912}+\frac {5797441 x^{5}}{872980416}+O\left (x^{6}\right )\right )+c_{2} \left (1-\frac {x}{2}-\frac {x^{2}}{2}-\frac {5 x^{3}}{24}-\frac {25 x^{4}}{336}-\frac {17 x^{5}}{672}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {5}{3}} \left (1+\frac {17 x}{36}+\frac {1241 x^{2}}{7128}+\frac {80665 x^{3}}{1347192}+\frac {972725 x^{4}}{48498912}+\frac {5797441 x^{5}}{872980416}+O\left (x^{6}\right )\right )+c_{2} \left (1-\frac {x}{2}-\frac {x^{2}}{2}-\frac {5 x^{3}}{24}-\frac {25 x^{4}}{336}-\frac {17 x^{5}}{672}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {5}{3}} \left (1+\frac {17 x}{36}+\frac {1241 x^{2}}{7128}+\frac {80665 x^{3}}{1347192}+\frac {972725 x^{4}}{48498912}+\frac {5797441 x^{5}}{872980416}+O\left (x^{6}\right )\right )+c_{2} \left (1-\frac {x}{2}-\frac {x^{2}}{2}-\frac {5 x^{3}}{24}-\frac {25 x^{4}}{336}-\frac {17 x^{5}}{672}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime } x \left (x -3\right )+\left (x +2\right ) y^{\prime }+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 \left (x -3\right )}-\frac {\left (x +2\right ) y^{\prime }}{x \left (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 (x +2\right ) y^{\prime }}{x \left (x -3\right )}+\frac {y}{x \left (x -3\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 {x +2}{x \left (x -3\right )}, P_{3}\left (x \right )=\frac {1}{\left (x -3\right ) 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 {2}{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}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & y^{\prime \prime } x \left (x -3\right )+\left (x +2\right ) y^{\prime }+y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & 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 =1..2 \\ {} & {} & 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} r \left (-5+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 -2+3 r \right )+a_{k} \left (k^{2}+2 k r +r^{2}+1\right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -r \left (-5+3 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {5}{3}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -3 \left (k -\frac {2}{3}+r \right ) \left (k +1+r \right ) a_{k +1}+a_{k} \left (k^{2}+2 k r +r^{2}+1\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k^{2}+2 k r +r^{2}+1\right )}{\left (3 k -2+3 r \right ) \left (k +1+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k^{2}+1\right )}{\left (3 k -2\right ) \left (k +1\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^{2}+1\right )}{\left (3 k -2\right ) \left (k +1\right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {5}{3} \\ {} & {} & a_{k +1}=\frac {a_{k} \left (k^{2}+\frac {10}{3} k +\frac {34}{9}\right )}{\left (3 k +3\right ) \left (k +\frac {8}{3}\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {5}{3} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {5}{3}}, a_{k +1}=\frac {a_{k} \left (k^{2}+\frac {10}{3} k +\frac {34}{9}\right )}{\left (3 k +3\right ) \left (k +\frac {8}{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 {5}{3}}\right ), a_{k +1}=\frac {a_{k} \left (k^{2}+1\right )}{\left (3 k -2\right ) \left (k +1\right )}, b_{k +1}=\frac {b_{k} \left (k^{2}+\frac {10}{3} k +\frac {34}{9}\right )}{\left (3 k +3\right ) \left (k +\frac {8}{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 -> elliptic -> Legendre -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach <- heuristic approach successful <- hypergeometric successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.032 (sec). Leaf size: 44
Order:=6; dsolve((x^2-3*x)*diff(y(x),x$2)+(x+2)*diff(y(x),x)+y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{\frac {5}{3}} \left (1+\frac {17}{36} x +\frac {1241}{7128} x^{2}+\frac {80665}{1347192} x^{3}+\frac {972725}{48498912} x^{4}+\frac {5797441}{872980416} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (1-\frac {1}{2} x -\frac {1}{2} x^{2}-\frac {5}{24} x^{3}-\frac {25}{336} x^{4}-\frac {17}{672} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.011 (sec). Leaf size: 85
AsymptoticDSolveValue[(x^2-3*x)*y''[x]+(x+2)*y'[x]+y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_2 \left (-\frac {17 x^5}{672}-\frac {25 x^4}{336}-\frac {5 x^3}{24}-\frac {x^2}{2}-\frac {x}{2}+1\right )+c_1 \left (\frac {5797441 x^5}{872980416}+\frac {972725 x^4}{48498912}+\frac {80665 x^3}{1347192}+\frac {1241 x^2}{7128}+\frac {17 x}{36}+1\right ) x^{5/3} \]