Internal problem ID [6931]
Internal file name [OUTPUT/6174_Friday_August_12_2022_11_04_20_PM_14734510/index.tex]
Book: Elementary differential equations. Rainville, Bedient, Bedient. Prentice Hall. NJ. 8th
edition. 1997.
Section: CHAPTER 18. Power series solutions. 18.4 Indicial Equation with Difference of Roots
Nonintegral. Exercises page 365
Problem number: 15.
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]]
\[ \boxed {2 x^{2} y^{\prime \prime }-3 x \left (1-x \right ) y^{\prime }+2 y=0} \] 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. \[ 2 x^{2} y^{\prime \prime }+\left (3 x^{2}-3 x \right ) y^{\prime }+2 y = 0 \] 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 {\frac {3 x}{2}-\frac {3}{2}}{x}\\ q(x) &= \frac {1}{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 \[ 2 x^{2} y^{\prime \prime }+\left (3 x^{2}-3 x \right ) y^{\prime }+2 y = 0 \] 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 (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (3 x^{2}-3 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+2 \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 )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{1+n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 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 }}3 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}3 a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \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 )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}3 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-3 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 2 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-3 x^{n +r} a_{n} \left (n +r \right )+2 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 2 x^{r} a_{0} r \left (-1+r \right )-3 x^{r} a_{0} r +2 a_{0} x^{r} = 0 \] Or \[ \left (2 x^{r} r \left (-1+r \right )-3 x^{r} r +2 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (2 r^{2}-5 r +2\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 2 r^{2}-5 r +2 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 2\\ r_2 &= {\frac {1}{2}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (2 r^{2}-5 r +2\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [2, {\frac {1}{2}}\right ]\).
Since \(r_1 - r_2 = {\frac {3}{2}}\) 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^{n +2}\\ 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 )+3 a_{n -1} \left (n +r -1\right )-3 a_{n} \left (n +r \right )+2 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {3 a_{n -1} \left (n +r -1\right )}{2 n^{2}+4 n r +2 r^{2}-5 n -5 r +2}\tag {4} \] Which for the root \(r = 2\) becomes \[ a_{n} = -\frac {3 a_{n -1} \left (1+n \right )}{n \left (2 n +3\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 2\) 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 \[ a_{1}=-\frac {3 r}{2 r^{2}-r -1} \] Which for the root \(r = 2\) becomes \[ a_{1}=-{\frac {6}{5}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3 r}{2 r^{2}-r -1}\) | \(-{\frac {6}{5}}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {9+9 r}{4 r^{3}+4 r^{2}-5 r -3} \] Which for the root \(r = 2\) becomes \[ a_{2}={\frac {27}{35}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3 r}{2 r^{2}-r -1}\) | \(-{\frac {6}{5}}\) |
| \(a_{2}\) | \(\frac {9+9 r}{4 r^{3}+4 r^{2}-5 r -3}\) | \(\frac {27}{35}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-54-27 r}{8 r^{4}+28 r^{3}+10 r^{2}-31 r -15} \] Which for the root \(r = 2\) becomes \[ a_{3}=-{\frac {12}{35}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3 r}{2 r^{2}-r -1}\) | \(-{\frac {6}{5}}\) |
| \(a_{2}\) | \(\frac {9+9 r}{4 r^{3}+4 r^{2}-5 r -3}\) | \(\frac {27}{35}\) |
| \(a_{3}\) | \(\frac {-54-27 r}{8 r^{4}+28 r^{3}+10 r^{2}-31 r -15}\) | \(-{\frac {12}{35}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {243+81 r}{16 r^{5}+112 r^{4}+216 r^{3}+8 r^{2}-247 r -105} \] Which for the root \(r = 2\) becomes \[ a_{4}={\frac {9}{77}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3 r}{2 r^{2}-r -1}\) | \(-{\frac {6}{5}}\) |
| \(a_{2}\) | \(\frac {9+9 r}{4 r^{3}+4 r^{2}-5 r -3}\) | \(\frac {27}{35}\) |
| \(a_{3}\) | \(\frac {-54-27 r}{8 r^{4}+28 r^{3}+10 r^{2}-31 r -15}\) | \(-{\frac {12}{35}}\) |
| \(a_{4}\) | \(\frac {243+81 r}{16 r^{5}+112 r^{4}+216 r^{3}+8 r^{2}-247 r -105}\) | \(\frac {9}{77}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-972-243 r}{32 r^{6}+368 r^{5}+1440 r^{4}+1960 r^{3}-422 r^{2}-2433 r -945} \] Which for the root \(r = 2\) becomes \[ a_{5}=-{\frac {162}{5005}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3 r}{2 r^{2}-r -1}\) | \(-{\frac {6}{5}}\) |
| \(a_{2}\) | \(\frac {9+9 r}{4 r^{3}+4 r^{2}-5 r -3}\) | \(\frac {27}{35}\) |
| \(a_{3}\) | \(\frac {-54-27 r}{8 r^{4}+28 r^{3}+10 r^{2}-31 r -15}\) | \(-{\frac {12}{35}}\) |
| \(a_{4}\) | \(\frac {243+81 r}{16 r^{5}+112 r^{4}+216 r^{3}+8 r^{2}-247 r -105}\) | \(\frac {9}{77}\) |
| \(a_{5}\) | \(\frac {-972-243 r}{32 r^{6}+368 r^{5}+1440 r^{4}+1960 r^{3}-422 r^{2}-2433 r -945}\) | \(-{\frac {162}{5005}}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {3645+729 r}{64 r^{7}+1088 r^{6}+6928 r^{5}+19760 r^{4}+20716 r^{3}-9508 r^{2}-28653 r -10395} \] Which for the root \(r = 2\) becomes \[ a_{6}={\frac {27}{3575}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3 r}{2 r^{2}-r -1}\) | \(-{\frac {6}{5}}\) |
| \(a_{2}\) | \(\frac {9+9 r}{4 r^{3}+4 r^{2}-5 r -3}\) | \(\frac {27}{35}\) |
| \(a_{3}\) | \(\frac {-54-27 r}{8 r^{4}+28 r^{3}+10 r^{2}-31 r -15}\) | \(-{\frac {12}{35}}\) |
| \(a_{4}\) | \(\frac {243+81 r}{16 r^{5}+112 r^{4}+216 r^{3}+8 r^{2}-247 r -105}\) | \(\frac {9}{77}\) |
| \(a_{5}\) | \(\frac {-972-243 r}{32 r^{6}+368 r^{5}+1440 r^{4}+1960 r^{3}-422 r^{2}-2433 r -945}\) | \(-{\frac {162}{5005}}\) |
| \(a_{6}\) | \(\frac {3645+729 r}{64 r^{7}+1088 r^{6}+6928 r^{5}+19760 r^{4}+20716 r^{3}-9508 r^{2}-28653 r -10395}\) | \(\frac {27}{3575}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=\frac {-13122-2187 r}{128 r^{8}+3008 r^{7}+28000 r^{6}+129584 r^{5}+298312 r^{4}+250292 r^{3}-180910 r^{2}-393279 r -135135} \] Which for the root \(r = 2\) becomes \[ a_{7}=-{\frac {648}{425425}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {3 r}{2 r^{2}-r -1}\) | \(-{\frac {6}{5}}\) |
| \(a_{2}\) | \(\frac {9+9 r}{4 r^{3}+4 r^{2}-5 r -3}\) | \(\frac {27}{35}\) |
| \(a_{3}\) | \(\frac {-54-27 r}{8 r^{4}+28 r^{3}+10 r^{2}-31 r -15}\) | \(-{\frac {12}{35}}\) |
| \(a_{4}\) | \(\frac {243+81 r}{16 r^{5}+112 r^{4}+216 r^{3}+8 r^{2}-247 r -105}\) | \(\frac {9}{77}\) |
| \(a_{5}\) | \(\frac {-972-243 r}{32 r^{6}+368 r^{5}+1440 r^{4}+1960 r^{3}-422 r^{2}-2433 r -945}\) | \(-{\frac {162}{5005}}\) |
| \(a_{6}\) | \(\frac {3645+729 r}{64 r^{7}+1088 r^{6}+6928 r^{5}+19760 r^{4}+20716 r^{3}-9508 r^{2}-28653 r -10395}\) | \(\frac {27}{3575}\) |
| \(a_{7}\) | \(\frac {-13122-2187 r}{128 r^{8}+3008 r^{7}+28000 r^{6}+129584 r^{5}+298312 r^{4}+250292 r^{3}-180910 r^{2}-393279 r -135135}\) | \(-{\frac {648}{425425}}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{2} \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^{2} \left (1-\frac {6 x}{5}+\frac {27 x^{2}}{35}-\frac {12 x^{3}}{35}+\frac {9 x^{4}}{77}-\frac {162 x^{5}}{5005}+\frac {27 x^{6}}{3575}-\frac {648 x^{7}}{425425}+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 )+3 b_{n -1} \left (n +r -1\right )-3 b_{n} \left (n +r \right )+2 b_{n} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {3 b_{n -1} \left (n +r -1\right )}{2 n^{2}+4 n r +2 r^{2}-5 n -5 r +2}\tag {4} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{n} = \frac {-6 n b_{n -1}+3 b_{n -1}}{4 n^{2}-6 n}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = {\frac {1}{2}}\) 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 \[ b_{1}=-\frac {3 r}{2 r^{2}-r -1} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{1}={\frac {3}{2}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {3 r}{2 r^{2}-r -1}\) | \(\frac {3}{2}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {9+9 r}{4 r^{3}+4 r^{2}-5 r -3} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{2}=-{\frac {27}{8}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {3 r}{2 r^{2}-r -1}\) | \(\frac {3}{2}\) |
| \(b_{2}\) | \(\frac {9+9 r}{4 r^{3}+4 r^{2}-5 r -3}\) | \(-{\frac {27}{8}}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {-54-27 r}{8 r^{4}+28 r^{3}+10 r^{2}-31 r -15} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{3}={\frac {45}{16}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {3 r}{2 r^{2}-r -1}\) | \(\frac {3}{2}\) |
| \(b_{2}\) | \(\frac {9+9 r}{4 r^{3}+4 r^{2}-5 r -3}\) | \(-{\frac {27}{8}}\) |
| \(b_{3}\) | \(\frac {-54-27 r}{8 r^{4}+28 r^{3}+10 r^{2}-31 r -15}\) | \(\frac {45}{16}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {243+81 r}{16 r^{5}+112 r^{4}+216 r^{3}+8 r^{2}-247 r -105} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{4}=-{\frac {189}{128}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {3 r}{2 r^{2}-r -1}\) | \(\frac {3}{2}\) |
| \(b_{2}\) | \(\frac {9+9 r}{4 r^{3}+4 r^{2}-5 r -3}\) | \(-{\frac {27}{8}}\) |
| \(b_{3}\) | \(\frac {-54-27 r}{8 r^{4}+28 r^{3}+10 r^{2}-31 r -15}\) | \(\frac {45}{16}\) |
| \(b_{4}\) | \(\frac {243+81 r}{16 r^{5}+112 r^{4}+216 r^{3}+8 r^{2}-247 r -105}\) | \(-{\frac {189}{128}}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {-972-243 r}{32 r^{6}+368 r^{5}+1440 r^{4}+1960 r^{3}-422 r^{2}-2433 r -945} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{5}={\frac {729}{1280}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {3 r}{2 r^{2}-r -1}\) | \(\frac {3}{2}\) |
| \(b_{2}\) | \(\frac {9+9 r}{4 r^{3}+4 r^{2}-5 r -3}\) | \(-{\frac {27}{8}}\) |
| \(b_{3}\) | \(\frac {-54-27 r}{8 r^{4}+28 r^{3}+10 r^{2}-31 r -15}\) | \(\frac {45}{16}\) |
| \(b_{4}\) | \(\frac {243+81 r}{16 r^{5}+112 r^{4}+216 r^{3}+8 r^{2}-247 r -105}\) | \(-{\frac {189}{128}}\) |
| \(b_{5}\) | \(\frac {-972-243 r}{32 r^{6}+368 r^{5}+1440 r^{4}+1960 r^{3}-422 r^{2}-2433 r -945}\) | \(\frac {729}{1280}\) |
For \(n = 6\), using the above recursive equation gives \[ b_{6}=\frac {3645+729 r}{64 r^{7}+1088 r^{6}+6928 r^{5}+19760 r^{4}+20716 r^{3}-9508 r^{2}-28653 r -10395} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{6}=-{\frac {891}{5120}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {3 r}{2 r^{2}-r -1}\) | \(\frac {3}{2}\) |
| \(b_{2}\) | \(\frac {9+9 r}{4 r^{3}+4 r^{2}-5 r -3}\) | \(-{\frac {27}{8}}\) |
| \(b_{3}\) | \(\frac {-54-27 r}{8 r^{4}+28 r^{3}+10 r^{2}-31 r -15}\) | \(\frac {45}{16}\) |
| \(b_{4}\) | \(\frac {243+81 r}{16 r^{5}+112 r^{4}+216 r^{3}+8 r^{2}-247 r -105}\) | \(-{\frac {189}{128}}\) |
| \(b_{5}\) | \(\frac {-972-243 r}{32 r^{6}+368 r^{5}+1440 r^{4}+1960 r^{3}-422 r^{2}-2433 r -945}\) | \(\frac {729}{1280}\) |
| \(b_{6}\) | \(\frac {3645+729 r}{64 r^{7}+1088 r^{6}+6928 r^{5}+19760 r^{4}+20716 r^{3}-9508 r^{2}-28653 r -10395}\) | \(-{\frac {891}{5120}}\) |
For \(n = 7\), using the above recursive equation gives \[ b_{7}=\frac {-13122-2187 r}{128 r^{8}+3008 r^{7}+28000 r^{6}+129584 r^{5}+298312 r^{4}+250292 r^{3}-180910 r^{2}-393279 r -135135} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{7}={\frac {3159}{71680}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {3 r}{2 r^{2}-r -1}\) | \(\frac {3}{2}\) |
| \(b_{2}\) | \(\frac {9+9 r}{4 r^{3}+4 r^{2}-5 r -3}\) | \(-{\frac {27}{8}}\) |
| \(b_{3}\) | \(\frac {-54-27 r}{8 r^{4}+28 r^{3}+10 r^{2}-31 r -15}\) | \(\frac {45}{16}\) |
| \(b_{4}\) | \(\frac {243+81 r}{16 r^{5}+112 r^{4}+216 r^{3}+8 r^{2}-247 r -105}\) | \(-{\frac {189}{128}}\) |
| \(b_{5}\) | \(\frac {-972-243 r}{32 r^{6}+368 r^{5}+1440 r^{4}+1960 r^{3}-422 r^{2}-2433 r -945}\) | \(\frac {729}{1280}\) |
| \(b_{6}\) | \(\frac {3645+729 r}{64 r^{7}+1088 r^{6}+6928 r^{5}+19760 r^{4}+20716 r^{3}-9508 r^{2}-28653 r -10395}\) | \(-{\frac {891}{5120}}\) |
| \(b_{7}\) | \(\frac {-13122-2187 r}{128 r^{8}+3008 r^{7}+28000 r^{6}+129584 r^{5}+298312 r^{4}+250292 r^{3}-180910 r^{2}-393279 r -135135}\) | \(\frac {3159}{71680}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x^{2} \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+\frac {3 x}{2}-\frac {27 x^{2}}{8}+\frac {45 x^{3}}{16}-\frac {189 x^{4}}{128}+\frac {729 x^{5}}{1280}-\frac {891 x^{6}}{5120}+\frac {3159 x^{7}}{71680}+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^{2} \left (1-\frac {6 x}{5}+\frac {27 x^{2}}{35}-\frac {12 x^{3}}{35}+\frac {9 x^{4}}{77}-\frac {162 x^{5}}{5005}+\frac {27 x^{6}}{3575}-\frac {648 x^{7}}{425425}+O\left (x^{8}\right )\right ) + c_{2} \sqrt {x}\, \left (1+\frac {3 x}{2}-\frac {27 x^{2}}{8}+\frac {45 x^{3}}{16}-\frac {189 x^{4}}{128}+\frac {729 x^{5}}{1280}-\frac {891 x^{6}}{5120}+\frac {3159 x^{7}}{71680}+O\left (x^{8}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{2} \left (1-\frac {6 x}{5}+\frac {27 x^{2}}{35}-\frac {12 x^{3}}{35}+\frac {9 x^{4}}{77}-\frac {162 x^{5}}{5005}+\frac {27 x^{6}}{3575}-\frac {648 x^{7}}{425425}+O\left (x^{8}\right )\right )+c_{2} \sqrt {x}\, \left (1+\frac {3 x}{2}-\frac {27 x^{2}}{8}+\frac {45 x^{3}}{16}-\frac {189 x^{4}}{128}+\frac {729 x^{5}}{1280}-\frac {891 x^{6}}{5120}+\frac {3159 x^{7}}{71680}+O\left (x^{8}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{2} \left (1-\frac {6 x}{5}+\frac {27 x^{2}}{35}-\frac {12 x^{3}}{35}+\frac {9 x^{4}}{77}-\frac {162 x^{5}}{5005}+\frac {27 x^{6}}{3575}-\frac {648 x^{7}}{425425}+O\left (x^{8}\right )\right )+c_{2} \sqrt {x}\, \left (1+\frac {3 x}{2}-\frac {27 x^{2}}{8}+\frac {45 x^{3}}{16}-\frac {189 x^{4}}{128}+\frac {729 x^{5}}{1280}-\frac {891 x^{6}}{5120}+\frac {3159 x^{7}}{71680}+O\left (x^{8}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{2} \left (1-\frac {6 x}{5}+\frac {27 x^{2}}{35}-\frac {12 x^{3}}{35}+\frac {9 x^{4}}{77}-\frac {162 x^{5}}{5005}+\frac {27 x^{6}}{3575}-\frac {648 x^{7}}{425425}+O\left (x^{8}\right )\right )+c_{2} \sqrt {x}\, \left (1+\frac {3 x}{2}-\frac {27 x^{2}}{8}+\frac {45 x^{3}}{16}-\frac {189 x^{4}}{128}+\frac {729 x^{5}}{1280}-\frac {891 x^{6}}{5120}+\frac {3159 x^{7}}{71680}+O\left (x^{8}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 x^{2} \left (\frac {d}{d x}y^{\prime }\right )+\left (3 x^{2}-3 x \right ) y^{\prime }+2 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}{x^{2}}-\frac {3 \left (x -1\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} \\ {} & {} & \frac {d}{d x}y^{\prime }+\frac {3 \left (x -1\right ) y^{\prime }}{2 x}+\frac {y}{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 {3 \left (x -1\right )}{2 x}, P_{3}\left (x \right )=\frac {1}{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 {3}{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}}}=1 \\ {} & \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 x^{2} \left (\frac {d}{d x}y^{\prime }\right )+3 x \left (x -1\right ) y^{\prime }+2 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 =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 \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\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} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (-1+2 r \right ) \left (-2+r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (2 k +2 r -1\right ) \left (k +r -2\right )+3 a_{k -1} \left (k +r -1\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 (-2+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{2, \frac {1}{2}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 2 \left (k +r -2\right ) \left (k +r -\frac {1}{2}\right ) a_{k}+3 a_{k -1} \left (k +r -1\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 2 \left (k +r -1\right ) \left (k +\frac {1}{2}+r \right ) a_{k +1}+3 a_{k} \left (k +r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {3 a_{k} \left (k +r \right )}{\left (k +r -1\right ) \left (2 k +1+2 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +1}=-\frac {3 a_{k} \left (k +2\right )}{\left (k +1\right ) \left (2 k +5\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +2}, a_{k +1}=-\frac {3 a_{k} \left (k +2\right )}{\left (k +1\right ) \left (2 k +5\right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & a_{k +1}=-\frac {3 a_{k} \left (k +\frac {1}{2}\right )}{\left (k -\frac {1}{2}\right ) \left (2 k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}}, a_{k +1}=-\frac {3 a_{k} \left (k +\frac {1}{2}\right )}{\left (k -\frac {1}{2}\right ) \left (2 k +2\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +2}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {1}{2}}\right ), a_{k +1}=-\frac {3 a_{k} \left (k +2\right )}{\left (k +1\right ) \left (2 k +5\right )}, b_{k +1}=-\frac {3 b_{k} \left (k +\frac {1}{2}\right )}{\left (k -\frac {1}{2}\right ) \left (2 k +2\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: 0.015 (sec). Leaf size: 55
Order:=8; dsolve(2*x^2*diff(y(x),x$2)-3*x*(1-x)*diff(y(x),x)+2*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} \sqrt {x}\, \left (1+\frac {3}{2} x -\frac {27}{8} x^{2}+\frac {45}{16} x^{3}-\frac {189}{128} x^{4}+\frac {729}{1280} x^{5}-\frac {891}{5120} x^{6}+\frac {3159}{71680} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+c_{2} x^{2} \left (1-\frac {6}{5} x +\frac {27}{35} x^{2}-\frac {12}{35} x^{3}+\frac {9}{77} x^{4}-\frac {162}{5005} x^{5}+\frac {27}{3575} x^{6}-\frac {648}{425425} x^{7}+\operatorname {O}\left (x^{8}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.007 (sec). Leaf size: 116
AsymptoticDSolveValue[2*x^2*y''[x]-3*x*(1-x)*y'[x]+2*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_1 \left (-\frac {648 x^7}{425425}+\frac {27 x^6}{3575}-\frac {162 x^5}{5005}+\frac {9 x^4}{77}-\frac {12 x^3}{35}+\frac {27 x^2}{35}-\frac {6 x}{5}+1\right ) x^2+c_2 \left (\frac {3159 x^7}{71680}-\frac {891 x^6}{5120}+\frac {729 x^5}{1280}-\frac {189 x^4}{128}+\frac {45 x^3}{16}-\frac {27 x^2}{8}+\frac {3 x}{2}+1\right ) \sqrt {x} \]