Internal problem ID [7019]
Internal file name [OUTPUT/6262_Thursday_August_18_2022_07_12_11_AM_55985488/index.tex]
Book: Elementary differential equations. Rainville, Bedient, Bedient. Prentice Hall. NJ. 8th
edition. 1997.
Section: CHAPTER 18. Power series solutions. Miscellaneous Exercises. page 394
Problem number: 17.
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]]
\[ \boxed {2 x^{2} y^{\prime \prime }-x \left (1+2 x \right ) y^{\prime }+\left (1+3 x \right ) y=0} \] 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. \[ 2 x^{2} y^{\prime \prime }+\left (-2 x^{2}-x \right ) y^{\prime }+\left (1+3 x \right ) y = 0 \] 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+2 x}{2 x}\\ q(x) &= \frac {1+3 x}{2 x^{2}}\\ \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 \[ 2 x^{2} y^{\prime \prime }+\left (-2 x^{2}-x \right ) y^{\prime }+\left (1+3 x \right ) y = 0 \] 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 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right ) x^{2}+\left (-2 x^{2}-x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (1+3 x \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^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{1+n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{1+n +r} a_{n}\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 }}\left (-2 x^{1+n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} \left (n +r -1\right ) x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}3 x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}3 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 =0}{\sum }}2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}3 a_{n -1} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-x^{n +r} a_{n} \left (n +r \right )+a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 2 x^{r} a_{0} r \left (-1+r \right )-x^{r} a_{0} r +a_{0} x^{r} = 0 \] Or \[ \left (2 x^{r} r \left (-1+r \right )-x^{r} r +x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (2 r^{2}-3 r +1\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 2 r^{2}-3 r +1 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 1\\ r_2 &= {\frac {1}{2}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (2 r^{2}-3 r +1\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [1, {\frac {1}{2}}\right ]\).
Since \(r_1 - r_2 = {\frac {1}{2}}\) 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^{1+n}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +\frac {1}{2}} \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} \left (n +r \right ) \left (n +r -1\right )-2 a_{n -1} \left (n +r -1\right )-a_{n} \left (n +r \right )+a_{n}+3 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {a_{n -1} \left (2 n +2 r -5\right )}{2 n^{2}+4 n r +2 r^{2}-3 n -3 r +1}\tag {4} \] Which for the root \(r = 1\) becomes \[ a_{n} = \frac {a_{n -1} \left (2 n -3\right )}{n \left (2 n +1\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 1\) 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 \[ a_{1}=\frac {-3+2 r}{r \left (2 r +1\right )} \] Which for the root \(r = 1\) becomes \[ a_{1}=-{\frac {1}{3}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3+2 r}{r \left (2 r +1\right )}\) | \(-{\frac {1}{3}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {4 r^{2}-8 r +3}{4 r^{4}+12 r^{3}+11 r^{2}+3 r} \] Which for the root \(r = 1\) becomes \[ a_{2}=-{\frac {1}{30}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3+2 r}{r \left (2 r +1\right )}\) | \(-{\frac {1}{3}}\) |
| \(a_{2}\) | \(\frac {4 r^{2}-8 r +3}{4 r^{4}+12 r^{3}+11 r^{2}+3 r}\) | \(-{\frac {1}{30}}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {4 r^{2}-8 r +3}{4 r^{5}+28 r^{4}+71 r^{3}+77 r^{2}+30 r} \] Which for the root \(r = 1\) becomes \[ a_{3}=-{\frac {1}{210}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3+2 r}{r \left (2 r +1\right )}\) | \(-{\frac {1}{3}}\) |
| \(a_{2}\) | \(\frac {4 r^{2}-8 r +3}{4 r^{4}+12 r^{3}+11 r^{2}+3 r}\) | \(-{\frac {1}{30}}\) |
| \(a_{3}\) | \(\frac {4 r^{2}-8 r +3}{4 r^{5}+28 r^{4}+71 r^{3}+77 r^{2}+30 r}\) | \(-{\frac {1}{210}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {4 r^{2}-8 r +3}{\left (2 r^{2}+13 r +21\right ) \left (2 r^{3}+11 r^{2}+19 r +10\right ) r} \] Which for the root \(r = 1\) becomes \[ a_{4}=-{\frac {1}{1512}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3+2 r}{r \left (2 r +1\right )}\) | \(-{\frac {1}{3}}\) |
| \(a_{2}\) | \(\frac {4 r^{2}-8 r +3}{4 r^{4}+12 r^{3}+11 r^{2}+3 r}\) | \(-{\frac {1}{30}}\) |
| \(a_{3}\) | \(\frac {4 r^{2}-8 r +3}{4 r^{5}+28 r^{4}+71 r^{3}+77 r^{2}+30 r}\) | \(-{\frac {1}{210}}\) |
| \(a_{4}\) | \(\frac {4 r^{2}-8 r +3}{\left (2 r^{2}+13 r +21\right ) \left (2 r^{3}+11 r^{2}+19 r +10\right ) r}\) | \(-{\frac {1}{1512}}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {4 r^{2}-8 r +3}{\left (2 r^{2}+17 r +36\right ) r \left (r^{2}+3 r +2\right ) \left (2 r^{2}+13 r +21\right )} \] Which for the root \(r = 1\) becomes \[ a_{5}=-{\frac {1}{11880}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3+2 r}{r \left (2 r +1\right )}\) | \(-{\frac {1}{3}}\) |
| \(a_{2}\) | \(\frac {4 r^{2}-8 r +3}{4 r^{4}+12 r^{3}+11 r^{2}+3 r}\) | \(-{\frac {1}{30}}\) |
| \(a_{3}\) | \(\frac {4 r^{2}-8 r +3}{4 r^{5}+28 r^{4}+71 r^{3}+77 r^{2}+30 r}\) | \(-{\frac {1}{210}}\) |
| \(a_{4}\) | \(\frac {4 r^{2}-8 r +3}{\left (2 r^{2}+13 r +21\right ) \left (2 r^{3}+11 r^{2}+19 r +10\right ) r}\) | \(-{\frac {1}{1512}}\) |
| \(a_{5}\) | \(\frac {4 r^{2}-8 r +3}{\left (2 r^{2}+17 r +36\right ) r \left (r^{2}+3 r +2\right ) \left (2 r^{2}+13 r +21\right )}\) | \(-{\frac {1}{11880}}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {4 r^{2}-8 r +3}{\left (2 r^{2}+21 r +55\right ) \left (r +3\right ) \left (r^{2}+3 r +2\right ) r \left (2 r^{2}+17 r +36\right )} \] Which for the root \(r = 1\) becomes \[ a_{6}=-{\frac {1}{102960}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3+2 r}{r \left (2 r +1\right )}\) | \(-{\frac {1}{3}}\) |
| \(a_{2}\) | \(\frac {4 r^{2}-8 r +3}{4 r^{4}+12 r^{3}+11 r^{2}+3 r}\) | \(-{\frac {1}{30}}\) |
| \(a_{3}\) | \(\frac {4 r^{2}-8 r +3}{4 r^{5}+28 r^{4}+71 r^{3}+77 r^{2}+30 r}\) | \(-{\frac {1}{210}}\) |
| \(a_{4}\) | \(\frac {4 r^{2}-8 r +3}{\left (2 r^{2}+13 r +21\right ) \left (2 r^{3}+11 r^{2}+19 r +10\right ) r}\) | \(-{\frac {1}{1512}}\) |
| \(a_{5}\) | \(\frac {4 r^{2}-8 r +3}{\left (2 r^{2}+17 r +36\right ) r \left (r^{2}+3 r +2\right ) \left (2 r^{2}+13 r +21\right )}\) | \(-{\frac {1}{11880}}\) |
| \(a_{6}\) | \(\frac {4 r^{2}-8 r +3}{\left (2 r^{2}+21 r +55\right ) \left (r +3\right ) \left (r^{2}+3 r +2\right ) r \left (2 r^{2}+17 r +36\right )}\) | \(-{\frac {1}{102960}}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=\frac {4 r^{2}-8 r +3}{\left (r +4\right ) r \left (r^{2}+3 r +2\right ) \left (r +3\right ) \left (2 r^{2}+21 r +55\right ) \left (2 r^{2}+25 r +78\right )} \] Which for the root \(r = 1\) becomes \[ a_{7}=-{\frac {1}{982800}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3+2 r}{r \left (2 r +1\right )}\) | \(-{\frac {1}{3}}\) |
| \(a_{2}\) | \(\frac {4 r^{2}-8 r +3}{4 r^{4}+12 r^{3}+11 r^{2}+3 r}\) | \(-{\frac {1}{30}}\) |
| \(a_{3}\) | \(\frac {4 r^{2}-8 r +3}{4 r^{5}+28 r^{4}+71 r^{3}+77 r^{2}+30 r}\) | \(-{\frac {1}{210}}\) |
| \(a_{4}\) | \(\frac {4 r^{2}-8 r +3}{\left (2 r^{2}+13 r +21\right ) \left (2 r^{3}+11 r^{2}+19 r +10\right ) r}\) | \(-{\frac {1}{1512}}\) |
| \(a_{5}\) | \(\frac {4 r^{2}-8 r +3}{\left (2 r^{2}+17 r +36\right ) r \left (r^{2}+3 r +2\right ) \left (2 r^{2}+13 r +21\right )}\) | \(-{\frac {1}{11880}}\) |
| \(a_{6}\) | \(\frac {4 r^{2}-8 r +3}{\left (2 r^{2}+21 r +55\right ) \left (r +3\right ) \left (r^{2}+3 r +2\right ) r \left (2 r^{2}+17 r +36\right )}\) | \(-{\frac {1}{102960}}\) |
| \(a_{7}\) | \(\frac {4 r^{2}-8 r +3}{\left (r +4\right ) r \left (r^{2}+3 r +2\right ) \left (r +3\right ) \left (2 r^{2}+21 r +55\right ) \left (2 r^{2}+25 r +78\right )}\) | \(-{\frac {1}{982800}}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= 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}+a_{7} x^{7}+a_{8} x^{8}\dots \right ) \\ &= x \left (1-\frac {x}{3}-\frac {x^{2}}{30}-\frac {x^{3}}{210}-\frac {x^{4}}{1512}-\frac {x^{5}}{11880}-\frac {x^{6}}{102960}-\frac {x^{7}}{982800}+O\left (x^{8}\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} \left (n +r \right ) \left (n +r -1\right )-2 b_{n -1} \left (n +r -1\right )-b_{n} \left (n +r \right )+b_{n}+3 b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = \frac {b_{n -1} \left (2 n +2 r -5\right )}{2 n^{2}+4 n r +2 r^{2}-3 n -3 r +1}\tag {4} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{n} = \frac {2 b_{n -1} \left (n -2\right )}{n \left (2 n -1\right )}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = {\frac {1}{2}}\) 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 \[ b_{1}=\frac {-3+2 r}{r \left (2 r +1\right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{1}=-2 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-3+2 r}{r \left (2 r +1\right )}\) | \(-2\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {4 r^{2}-8 r +3}{4 r^{4}+12 r^{3}+11 r^{2}+3 r} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{2}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-3+2 r}{r \left (2 r +1\right )}\) | \(-2\) |
| \(b_{2}\) | \(\frac {4 r^{2}-8 r +3}{4 r^{4}+12 r^{3}+11 r^{2}+3 r}\) | \(0\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {4 r^{2}-8 r +3}{4 r^{5}+28 r^{4}+71 r^{3}+77 r^{2}+30 r} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{3}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-3+2 r}{r \left (2 r +1\right )}\) | \(-2\) |
| \(b_{2}\) | \(\frac {4 r^{2}-8 r +3}{4 r^{4}+12 r^{3}+11 r^{2}+3 r}\) | \(0\) |
| \(b_{3}\) | \(\frac {4 r^{2}-8 r +3}{4 r^{5}+28 r^{4}+71 r^{3}+77 r^{2}+30 r}\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {4 r^{2}-8 r +3}{\left (2 r^{2}+13 r +21\right ) \left (2 r^{3}+11 r^{2}+19 r +10\right ) r} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{4}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-3+2 r}{r \left (2 r +1\right )}\) | \(-2\) |
| \(b_{2}\) | \(\frac {4 r^{2}-8 r +3}{4 r^{4}+12 r^{3}+11 r^{2}+3 r}\) | \(0\) |
| \(b_{3}\) | \(\frac {4 r^{2}-8 r +3}{4 r^{5}+28 r^{4}+71 r^{3}+77 r^{2}+30 r}\) | \(0\) |
| \(b_{4}\) | \(\frac {4 r^{2}-8 r +3}{\left (2 r^{2}+13 r +21\right ) \left (2 r^{3}+11 r^{2}+19 r +10\right ) r}\) | \(0\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {4 r^{2}-8 r +3}{\left (2 r^{2}+17 r +36\right ) r \left (r^{2}+3 r +2\right ) \left (2 r^{2}+13 r +21\right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{5}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-3+2 r}{r \left (2 r +1\right )}\) | \(-2\) |
| \(b_{2}\) | \(\frac {4 r^{2}-8 r +3}{4 r^{4}+12 r^{3}+11 r^{2}+3 r}\) | \(0\) |
| \(b_{3}\) | \(\frac {4 r^{2}-8 r +3}{4 r^{5}+28 r^{4}+71 r^{3}+77 r^{2}+30 r}\) | \(0\) |
| \(b_{4}\) | \(\frac {4 r^{2}-8 r +3}{\left (2 r^{2}+13 r +21\right ) \left (2 r^{3}+11 r^{2}+19 r +10\right ) r}\) | \(0\) |
| \(b_{5}\) | \(\frac {4 r^{2}-8 r +3}{\left (2 r^{2}+17 r +36\right ) r \left (r^{2}+3 r +2\right ) \left (2 r^{2}+13 r +21\right )}\) | \(0\) |
For \(n = 6\), using the above recursive equation gives \[ b_{6}=\frac {4 r^{2}-8 r +3}{\left (2 r^{2}+21 r +55\right ) \left (r +3\right ) \left (r^{2}+3 r +2\right ) r \left (2 r^{2}+17 r +36\right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{6}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-3+2 r}{r \left (2 r +1\right )}\) | \(-2\) |
| \(b_{2}\) | \(\frac {4 r^{2}-8 r +3}{4 r^{4}+12 r^{3}+11 r^{2}+3 r}\) | \(0\) |
| \(b_{3}\) | \(\frac {4 r^{2}-8 r +3}{4 r^{5}+28 r^{4}+71 r^{3}+77 r^{2}+30 r}\) | \(0\) |
| \(b_{4}\) | \(\frac {4 r^{2}-8 r +3}{\left (2 r^{2}+13 r +21\right ) \left (2 r^{3}+11 r^{2}+19 r +10\right ) r}\) | \(0\) |
| \(b_{5}\) | \(\frac {4 r^{2}-8 r +3}{\left (2 r^{2}+17 r +36\right ) r \left (r^{2}+3 r +2\right ) \left (2 r^{2}+13 r +21\right )}\) | \(0\) |
| \(b_{6}\) | \(\frac {4 r^{2}-8 r +3}{\left (2 r^{2}+21 r +55\right ) \left (r +3\right ) \left (r^{2}+3 r +2\right ) r \left (2 r^{2}+17 r +36\right )}\) | \(0\) |
For \(n = 7\), using the above recursive equation gives \[ b_{7}=\frac {4 r^{2}-8 r +3}{\left (r +4\right ) r \left (r^{2}+3 r +2\right ) \left (r +3\right ) \left (2 r^{2}+21 r +55\right ) \left (2 r^{2}+25 r +78\right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{7}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {-3+2 r}{r \left (2 r +1\right )}\) | \(-2\) |
| \(b_{2}\) | \(\frac {4 r^{2}-8 r +3}{4 r^{4}+12 r^{3}+11 r^{2}+3 r}\) | \(0\) |
| \(b_{3}\) | \(\frac {4 r^{2}-8 r +3}{4 r^{5}+28 r^{4}+71 r^{3}+77 r^{2}+30 r}\) | \(0\) |
| \(b_{4}\) | \(\frac {4 r^{2}-8 r +3}{\left (2 r^{2}+13 r +21\right ) \left (2 r^{3}+11 r^{2}+19 r +10\right ) r}\) | \(0\) |
| \(b_{5}\) | \(\frac {4 r^{2}-8 r +3}{\left (2 r^{2}+17 r +36\right ) r \left (r^{2}+3 r +2\right ) \left (2 r^{2}+13 r +21\right )}\) | \(0\) |
| \(b_{6}\) | \(\frac {4 r^{2}-8 r +3}{\left (2 r^{2}+21 r +55\right ) \left (r +3\right ) \left (r^{2}+3 r +2\right ) r \left (2 r^{2}+17 r +36\right )}\) | \(0\) |
| \(b_{7}\) | \(\frac {4 r^{2}-8 r +3}{\left (r +4\right ) r \left (r^{2}+3 r +2\right ) \left (r +3\right ) \left (2 r^{2}+21 r +55\right ) \left (2 r^{2}+25 r +78\right )}\) | \(0\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= 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}+b_{7} x^{7}+b_{8} x^{8}\dots \right ) \\ &= \sqrt {x}\, \left (1-2 x +O\left (x^{8}\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} x \left (1-\frac {x}{3}-\frac {x^{2}}{30}-\frac {x^{3}}{210}-\frac {x^{4}}{1512}-\frac {x^{5}}{11880}-\frac {x^{6}}{102960}-\frac {x^{7}}{982800}+O\left (x^{8}\right )\right ) + c_{2} \sqrt {x}\, \left (1-2 x +O\left (x^{8}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x \left (1-\frac {x}{3}-\frac {x^{2}}{30}-\frac {x^{3}}{210}-\frac {x^{4}}{1512}-\frac {x^{5}}{11880}-\frac {x^{6}}{102960}-\frac {x^{7}}{982800}+O\left (x^{8}\right )\right )+c_{2} \sqrt {x}\, \left (1-2 x +O\left (x^{8}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x \left (1-\frac {x}{3}-\frac {x^{2}}{30}-\frac {x^{3}}{210}-\frac {x^{4}}{1512}-\frac {x^{5}}{11880}-\frac {x^{6}}{102960}-\frac {x^{7}}{982800}+O\left (x^{8}\right )\right )+c_{2} \sqrt {x}\, \left (1-2 x +O\left (x^{8}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x \left (1-\frac {x}{3}-\frac {x^{2}}{30}-\frac {x^{3}}{210}-\frac {x^{4}}{1512}-\frac {x^{5}}{11880}-\frac {x^{6}}{102960}-\frac {x^{7}}{982800}+O\left (x^{8}\right )\right )+c_{2} \sqrt {x}\, \left (1-2 x +O\left (x^{8}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 y^{\prime \prime } x^{2}+\left (-2 x^{2}-x \right ) y^{\prime }+\left (1+3 x \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 (1+3 x \right ) y}{2 x^{2}}+\frac {\left (1+2 x \right ) y^{\prime }}{2 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} \\ {} & {} & y^{\prime \prime }-\frac {\left (1+2 x \right ) y^{\prime }}{2 x}+\frac {\left (1+3 x \right ) y}{2 x^{2}}=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 {1+2 x}{2 x}, P_{3}\left (x \right )=\frac {1+3 x}{2 x^{2}}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=-\frac {1}{2} \\ {} & \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}}}=\frac {1}{2} \\ {} & \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} \\ {} & {} & 2 y^{\prime \prime } x^{2}-x \left (1+2 x \right ) y^{\prime }+\left (1+3 x \right ) y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..2 \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{2}\cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (-1+2 r \right ) \left (-1+r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (2 k +2 r -1\right ) \left (k +r -1\right )-a_{k -1} \left (2 k -5+2 r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (-1+2 r \right ) \left (-1+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{1, \frac {1}{2}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 2 \left (k +r -1\right ) \left (k +r -\frac {1}{2}\right ) a_{k}-2 \left (k -\frac {5}{2}+r \right ) a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 2 \left (k +r \right ) \left (k +\frac {1}{2}+r \right ) a_{k +1}-2 \left (k -\frac {3}{2}+r \right ) a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {\left (2 k +2 r -3\right ) a_{k}}{\left (k +r \right ) \left (2 k +1+2 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +1}=\frac {\left (2 k -1\right ) a_{k}}{\left (k +1\right ) \left (2 k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1}, a_{k +1}=\frac {\left (2 k -1\right ) a_{k}}{\left (k +1\right ) \left (2 k +3\right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2}\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =1 \\ {} & {} & a_{k +1}=\frac {\left (2 k -2\right ) a_{k}}{\left (k +\frac {1}{2}\right ) \left (2 k +2\right )} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =0 \\ {} & {} & a_{1}=-2 a_{0} \\ \bullet & {} & \textrm {Terminating series solution of the ODE for}\hspace {3pt} r =\frac {1}{2}\hspace {3pt}\textrm {. Use reduction of order to find the second linearly independent solution}\hspace {3pt} \\ {} & {} & y=a_{0}\cdot \left (1-2 x \right ) \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1}\right )+b_{0}\cdot \left (1-2 x \right ), a_{k +1}=\frac {\left (2 k -1\right ) a_{k}}{\left (k +1\right ) \left (2 k +3\right )}\right ] \end {array} \]
Maple trace Kovacic algorithm successful
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm A Liouvillian solution exists Reducible group (found an exponential solution) Group is reducible, not completely reducible Solution has integrals. Trying a special function solution free of integrals... -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre <- Kummer successful <- special function solution successful -> Trying to convert hypergeometric functions to elementary form... <- elementary form is not straightforward to achieve - returning special function solution free of uncomputed integrals <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 41
Order:=8; dsolve(2*x^2*diff(y(x),x$2)-x*(1+2*x)*diff(y(x),x)+(1+3*x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} \sqrt {x}\, \left (1-2 x +\operatorname {O}\left (x^{8}\right )\right )+c_{2} x \left (1-\frac {1}{3} x -\frac {1}{30} x^{2}-\frac {1}{210} x^{3}-\frac {1}{1512} x^{4}-\frac {1}{11880} x^{5}-\frac {1}{102960} x^{6}-\frac {1}{982800} x^{7}+\operatorname {O}\left (x^{8}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 70
AsymptoticDSolveValue[2*x^2*y''[x]-x*(1+2*x)*y'[x]+(1+3*x)*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_1 x \left (-\frac {x^7}{982800}-\frac {x^6}{102960}-\frac {x^5}{11880}-\frac {x^4}{1512}-\frac {x^3}{210}-\frac {x^2}{30}-\frac {x}{3}+1\right )+c_2 (1-2 x) \sqrt {x} \]