Internal problem ID [1312]
Internal file name [OUTPUT/1313_Sunday_June_05_2022_02_09_46_AM_32578457/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: 21.
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]]
\[ \boxed {2 x^{2} \left (x +3\right ) y^{\prime \prime }+x \left (1+5 x \right ) y^{\prime }+\left (x +1\right ) y=0} \] 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. \[ \left (2 x^{3}+6 x^{2}\right ) y^{\prime \prime }+\left (5 x^{2}+x \right ) y^{\prime }+\left (x +1\right ) y = 0 \] 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 {1+5 x}{2 x \left (x +3\right )}\\ q(x) &= \frac {x +1}{2 x^{2} \left (x +3\right )}\\ \end {align*}
| \(p(x)=\frac {1+5 x}{2 x \left (x +3\right )}\) | |
| singularity | type |
| \(x = -3\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
| \(q(x)=\frac {x +1}{2 x^{2} \left (x +3\right )}\) | |
| singularity | type |
| \(x = -3\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([-3, 0, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ 2 x^{2} \left (x +3\right ) y^{\prime \prime }+\left (5 x^{2}+x \right ) y^{\prime }+\left (x +1\right ) y = 0 \] 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} 2 x^{2} \left (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 (5 x^{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 }}2 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}6 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}5 x^{1+n +r} a_{n} \left (n +r \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 }}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 }}2 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}5 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}5 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 }}2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}6 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}5 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}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 \[ 6 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+x^{n +r} a_{n} \left (n +r \right )+a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 6 x^{r} a_{0} r \left (-1+r \right )+x^{r} a_{0} r +a_{0} x^{r} = 0 \] Or \[ \left (6 x^{r} r \left (-1+r \right )+x^{r} r +x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (6 r^{2}-5 r +1\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 6 r^{2}-5 r +1 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {1}{2}}\\ r_2 &= {\frac {1}{3}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (6 r^{2}-5 r +1\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {1}{2}}, {\frac {1}{3}}\right ]\).
Since \(r_1 - r_2 = {\frac {1}{6}}\) 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}{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} 2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+6 a_{n} \left (n +r \right ) \left (n +r -1\right )+5 a_{n -1} \left (n +r -1\right )+a_{n} \left (n +r \right )+a_{n -1}+a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {\left (n +r \right ) a_{n -1}}{3 n +3 r -1}\tag {4} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{n} = \frac {a_{n -1} \left (-1-2 n \right )}{6 n +1}\tag {5} \] 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 \[ a_{1}=\frac {-1-r}{2+3 r} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{1}=-{\frac {3}{7}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-1-r}{2+3 r}\) | \(-{\frac {3}{7}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {r^{2}+3 r +2}{9 r^{2}+21 r +10} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{2}={\frac {15}{91}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-1-r}{2+3 r}\) | \(-{\frac {3}{7}}\) |
| \(a_{2}\) | \(\frac {r^{2}+3 r +2}{9 r^{2}+21 r +10}\) | \(\frac {15}{91}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-r^{3}-6 r^{2}-11 r -6}{27 r^{3}+135 r^{2}+198 r +80} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{3}=-{\frac {15}{247}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-1-r}{2+3 r}\) | \(-{\frac {3}{7}}\) |
| \(a_{2}\) | \(\frac {r^{2}+3 r +2}{9 r^{2}+21 r +10}\) | \(\frac {15}{91}\) |
| \(a_{3}\) | \(\frac {-r^{3}-6 r^{2}-11 r -6}{27 r^{3}+135 r^{2}+198 r +80}\) | \(-{\frac {15}{247}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {r^{4}+10 r^{3}+35 r^{2}+50 r +24}{81 r^{4}+702 r^{3}+2079 r^{2}+2418 r +880} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{4}={\frac {27}{1235}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-1-r}{2+3 r}\) | \(-{\frac {3}{7}}\) |
| \(a_{2}\) | \(\frac {r^{2}+3 r +2}{9 r^{2}+21 r +10}\) | \(\frac {15}{91}\) |
| \(a_{3}\) | \(\frac {-r^{3}-6 r^{2}-11 r -6}{27 r^{3}+135 r^{2}+198 r +80}\) | \(-{\frac {15}{247}}\) |
| \(a_{4}\) | \(\frac {r^{4}+10 r^{3}+35 r^{2}+50 r +24}{81 r^{4}+702 r^{3}+2079 r^{2}+2418 r +880}\) | \(\frac {27}{1235}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-r^{5}-15 r^{4}-85 r^{3}-225 r^{2}-274 r -120}{243 r^{5}+3240 r^{4}+16065 r^{3}+36360 r^{2}+36492 r +12320} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{5}=-{\frac {297}{38285}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-1-r}{2+3 r}\) | \(-{\frac {3}{7}}\) |
| \(a_{2}\) | \(\frac {r^{2}+3 r +2}{9 r^{2}+21 r +10}\) | \(\frac {15}{91}\) |
| \(a_{3}\) | \(\frac {-r^{3}-6 r^{2}-11 r -6}{27 r^{3}+135 r^{2}+198 r +80}\) | \(-{\frac {15}{247}}\) |
| \(a_{4}\) | \(\frac {r^{4}+10 r^{3}+35 r^{2}+50 r +24}{81 r^{4}+702 r^{3}+2079 r^{2}+2418 r +880}\) | \(\frac {27}{1235}\) |
| \(a_{5}\) | \(\frac {-r^{5}-15 r^{4}-85 r^{3}-225 r^{2}-274 r -120}{243 r^{5}+3240 r^{4}+16065 r^{3}+36360 r^{2}+36492 r +12320}\) | \(-{\frac {297}{38285}}\) |
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 {3 x}{7}+\frac {15 x^{2}}{91}-\frac {15 x^{3}}{247}+\frac {27 x^{4}}{1235}-\frac {297 x^{5}}{38285}+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} 2 b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+6 b_{n} \left (n +r \right ) \left (n +r -1\right )+5 b_{n -1} \left (n +r -1\right )+b_{n} \left (n +r \right )+b_{n -1}+b_{n} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {\left (n +r \right ) b_{n -1}}{3 n +3 r -1}\tag {4} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ b_{n} = -\frac {\left (3 n +1\right ) b_{n -1}}{9 n}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = {\frac {1}{3}}\) 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 \[ b_{1}=\frac {-1-r}{2+3 r} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ b_{1}=-{\frac {4}{9}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-1-r}{2+3 r}\) | \(-{\frac {4}{9}}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {r^{2}+3 r +2}{9 r^{2}+21 r +10} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ b_{2}={\frac {14}{81}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-1-r}{2+3 r}\) | \(-{\frac {4}{9}}\) |
| \(b_{2}\) | \(\frac {r^{2}+3 r +2}{9 r^{2}+21 r +10}\) | \(\frac {14}{81}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {-r^{3}-6 r^{2}-11 r -6}{27 r^{3}+135 r^{2}+198 r +80} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ b_{3}=-{\frac {140}{2187}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-1-r}{2+3 r}\) | \(-{\frac {4}{9}}\) |
| \(b_{2}\) | \(\frac {r^{2}+3 r +2}{9 r^{2}+21 r +10}\) | \(\frac {14}{81}\) |
| \(b_{3}\) | \(\frac {-r^{3}-6 r^{2}-11 r -6}{27 r^{3}+135 r^{2}+198 r +80}\) | \(-{\frac {140}{2187}}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {r^{4}+10 r^{3}+35 r^{2}+50 r +24}{81 r^{4}+702 r^{3}+2079 r^{2}+2418 r +880} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ b_{4}={\frac {455}{19683}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-1-r}{2+3 r}\) | \(-{\frac {4}{9}}\) |
| \(b_{2}\) | \(\frac {r^{2}+3 r +2}{9 r^{2}+21 r +10}\) | \(\frac {14}{81}\) |
| \(b_{3}\) | \(\frac {-r^{3}-6 r^{2}-11 r -6}{27 r^{3}+135 r^{2}+198 r +80}\) | \(-{\frac {140}{2187}}\) |
| \(b_{4}\) | \(\frac {r^{4}+10 r^{3}+35 r^{2}+50 r +24}{81 r^{4}+702 r^{3}+2079 r^{2}+2418 r +880}\) | \(\frac {455}{19683}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {-r^{5}-15 r^{4}-85 r^{3}-225 r^{2}-274 r -120}{243 r^{5}+3240 r^{4}+16065 r^{3}+36360 r^{2}+36492 r +12320} \] Which for the root \(r = {\frac {1}{3}}\) becomes \[ b_{5}=-{\frac {1456}{177147}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-1-r}{2+3 r}\) | \(-{\frac {4}{9}}\) |
| \(b_{2}\) | \(\frac {r^{2}+3 r +2}{9 r^{2}+21 r +10}\) | \(\frac {14}{81}\) |
| \(b_{3}\) | \(\frac {-r^{3}-6 r^{2}-11 r -6}{27 r^{3}+135 r^{2}+198 r +80}\) | \(-{\frac {140}{2187}}\) |
| \(b_{4}\) | \(\frac {r^{4}+10 r^{3}+35 r^{2}+50 r +24}{81 r^{4}+702 r^{3}+2079 r^{2}+2418 r +880}\) | \(\frac {455}{19683}\) |
| \(b_{5}\) | \(\frac {-r^{5}-15 r^{4}-85 r^{3}-225 r^{2}-274 r -120}{243 r^{5}+3240 r^{4}+16065 r^{3}+36360 r^{2}+36492 r +12320}\) | \(-{\frac {1456}{177147}}\) |
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}{3}} \left (1-\frac {4 x}{9}+\frac {14 x^{2}}{81}-\frac {140 x^{3}}{2187}+\frac {455 x^{4}}{19683}-\frac {1456 x^{5}}{177147}+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 {3 x}{7}+\frac {15 x^{2}}{91}-\frac {15 x^{3}}{247}+\frac {27 x^{4}}{1235}-\frac {297 x^{5}}{38285}+O\left (x^{6}\right )\right ) + c_{2} x^{\frac {1}{3}} \left (1-\frac {4 x}{9}+\frac {14 x^{2}}{81}-\frac {140 x^{3}}{2187}+\frac {455 x^{4}}{19683}-\frac {1456 x^{5}}{177147}+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 {3 x}{7}+\frac {15 x^{2}}{91}-\frac {15 x^{3}}{247}+\frac {27 x^{4}}{1235}-\frac {297 x^{5}}{38285}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {1}{3}} \left (1-\frac {4 x}{9}+\frac {14 x^{2}}{81}-\frac {140 x^{3}}{2187}+\frac {455 x^{4}}{19683}-\frac {1456 x^{5}}{177147}+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 {3 x}{7}+\frac {15 x^{2}}{91}-\frac {15 x^{3}}{247}+\frac {27 x^{4}}{1235}-\frac {297 x^{5}}{38285}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {1}{3}} \left (1-\frac {4 x}{9}+\frac {14 x^{2}}{81}-\frac {140 x^{3}}{2187}+\frac {455 x^{4}}{19683}-\frac {1456 x^{5}}{177147}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \sqrt {x}\, \left (1-\frac {3 x}{7}+\frac {15 x^{2}}{91}-\frac {15 x^{3}}{247}+\frac {27 x^{4}}{1235}-\frac {297 x^{5}}{38285}+O\left (x^{6}\right )\right )+c_{2} x^{\frac {1}{3}} \left (1-\frac {4 x}{9}+\frac {14 x^{2}}{81}-\frac {140 x^{3}}{2187}+\frac {455 x^{4}}{19683}-\frac {1456 x^{5}}{177147}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 x^{2} \left (x +3\right ) y^{\prime \prime }+\left (5 x^{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}{2 x^{2} \left (x +3\right )}-\frac {\left (1+5 x \right ) y^{\prime }}{2 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 (1+5 x \right ) y^{\prime }}{2 x \left (x +3\right )}+\frac {\left (x +1\right ) y}{2 x^{2} \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 {1+5 x}{2 x \left (x +3\right )}, P_{3}\left (x \right )=\frac {x +1}{2 x^{2} \left (x +3\right )}\right ] \\ {} & \circ & \left (x +3\right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-3 \\ {} & {} & \left (\left (x +3\right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-3}}}=\frac {7}{3} \\ {} & \circ & \left (x +3\right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-3 \\ {} & {} & \left (\left (x +3\right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-3}}}=0 \\ {} & \circ & x =-3\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}=-3 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 2 x^{2} \left (x +3\right ) y^{\prime \prime }+x \left (1+5 x \right ) y^{\prime }+\left (x +1\right ) y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -3\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (2 u^{3}-12 u^{2}+18 u \right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (5 u^{2}-29 u +42\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (u -2\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} \\ {} & {} & 6 a_{0} r \left (4+3 r \right ) u^{-1+r}+\left (6 a_{1} \left (1+r \right ) \left (7+3 r \right )-a_{0} \left (12 r^{2}+17 r +2\right )\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (6 a_{k +1} \left (k +r +1\right ) \left (3 k +7+3 r \right )-a_{k} \left (12 k^{2}+24 k r +12 r^{2}+17 k +17 r +2\right )+a_{k -1} \left (k +r \right ) \left (2 k -1+2 r \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 6 r \left (4+3 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -\frac {4}{3}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & 6 a_{1} \left (1+r \right ) \left (7+3 r \right )-a_{0} \left (12 r^{2}+17 r +2\right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 2 \left (-6 a_{k}+a_{k -1}+9 a_{k +1}\right ) k^{2}+\left (4 \left (-6 a_{k}+a_{k -1}+9 a_{k +1}\right ) r -17 a_{k}-a_{k -1}+60 a_{k +1}\right ) k +2 \left (-6 a_{k}+a_{k -1}+9 a_{k +1}\right ) r^{2}+\left (-17 a_{k}-a_{k -1}+60 a_{k +1}\right ) r -2 a_{k}+42 a_{k +1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 2 \left (-6 a_{k +1}+a_{k}+9 a_{k +2}\right ) \left (k +1\right )^{2}+\left (4 \left (-6 a_{k +1}+a_{k}+9 a_{k +2}\right ) r -17 a_{k +1}-a_{k}+60 a_{k +2}\right ) \left (k +1\right )+2 \left (-6 a_{k +1}+a_{k}+9 a_{k +2}\right ) r^{2}+\left (-17 a_{k +1}-a_{k}+60 a_{k +2}\right ) r -2 a_{k +1}+42 a_{k +2}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {2 k^{2} a_{k}-12 k^{2} a_{k +1}+4 k r a_{k}-24 k r a_{k +1}+2 r^{2} a_{k}-12 r^{2} a_{k +1}+3 k a_{k}-41 k a_{k +1}+3 r a_{k}-41 r a_{k +1}+a_{k}-31 a_{k +1}}{6 \left (3 k^{2}+6 k r +3 r^{2}+16 k +16 r +20\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {2 k^{2} a_{k}-12 k^{2} a_{k +1}+3 k a_{k}-41 k a_{k +1}+a_{k}-31 a_{k +1}}{6 \left (3 k^{2}+16 k +20\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 {2 k^{2} a_{k}-12 k^{2} a_{k +1}+3 k a_{k}-41 k a_{k +1}+a_{k}-31 a_{k +1}}{6 \left (3 k^{2}+16 k +20\right )}, 42 a_{1}-2 a_{0}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +3 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +3\right )^{k}, a_{k +2}=-\frac {2 k^{2} a_{k}-12 k^{2} a_{k +1}+3 k a_{k}-41 k a_{k +1}+a_{k}-31 a_{k +1}}{6 \left (3 k^{2}+16 k +20\right )}, 42 a_{1}-2 a_{0}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {4}{3} \\ {} & {} & a_{k +2}=-\frac {2 k^{2} a_{k}-12 k^{2} a_{k +1}-\frac {7}{3} k a_{k}-9 k a_{k +1}+\frac {5}{9} a_{k}+\frac {7}{3} a_{k +1}}{6 \left (3 k^{2}+8 k +4\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {4}{3} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k -\frac {4}{3}}, a_{k +2}=-\frac {2 k^{2} a_{k}-12 k^{2} a_{k +1}-\frac {7}{3} k a_{k}-9 k a_{k +1}+\frac {5}{9} a_{k}+\frac {7}{3} a_{k +1}}{6 \left (3 k^{2}+8 k +4\right )}, -6 a_{1}-\frac {2 a_{0}}{3}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =x +3 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +3\right )^{k -\frac {4}{3}}, a_{k +2}=-\frac {2 k^{2} a_{k}-12 k^{2} a_{k +1}-\frac {7}{3} k a_{k}-9 k a_{k +1}+\frac {5}{9} a_{k}+\frac {7}{3} a_{k +1}}{6 \left (3 k^{2}+8 k +4\right )}, -6 a_{1}-\frac {2 a_{0}}{3}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (x +3\right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (x +3\right )^{k -\frac {4}{3}}\right ), a_{k +2}=-\frac {2 k^{2} a_{k}-12 k^{2} a_{1+k}+3 k a_{k}-41 k a_{1+k}+a_{k}-31 a_{1+k}}{6 \left (3 k^{2}+16 k +20\right )}, 42 a_{1}-2 a_{0}=0, b_{k +2}=-\frac {2 k^{2} b_{k}-12 k^{2} b_{1+k}-\frac {7}{3} k b_{k}-9 k b_{1+k}+\frac {5}{9} b_{k}+\frac {7}{3} b_{1+k}}{6 \left (3 k^{2}+8 k +4\right )}, -6 b_{1}-\frac {2 b_{0}}{3}=0\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 <- heuristic approach successful <- 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: 0.016 (sec). Leaf size: 47
Order:=6; dsolve(2*x^2*(3+x)*diff(y(x),x$2)+x*(1+5*x)*diff(y(x),x)+(1+x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{\frac {1}{3}} \left (1-\frac {4}{9} x +\frac {14}{81} x^{2}-\frac {140}{2187} x^{3}+\frac {455}{19683} x^{4}-\frac {1456}{177147} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \sqrt {x}\, \left (1-\frac {3}{7} x +\frac {15}{91} x^{2}-\frac {15}{247} x^{3}+\frac {27}{1235} x^{4}-\frac {297}{38285} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.006 (sec). Leaf size: 90
AsymptoticDSolveValue[2*x^2*(3+x)*y''[x]+x*(1+5*x)*y'[x]+(1+x)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \sqrt {x} \left (-\frac {297 x^5}{38285}+\frac {27 x^4}{1235}-\frac {15 x^3}{247}+\frac {15 x^2}{91}-\frac {3 x}{7}+1\right )+c_2 \sqrt [3]{x} \left (-\frac {1456 x^5}{177147}+\frac {455 x^4}{19683}-\frac {140 x^3}{2187}+\frac {14 x^2}{81}-\frac {4 x}{9}+1\right ) \]