Internal problem ID [6928]
Internal file name [OUTPUT/6171_Friday_August_12_2022_11_04_13_PM_5854370/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: 12.
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]]
\[ \boxed {3 x^{2} y^{\prime \prime }+y^{\prime } x -\left (1+x \right ) y=0} \] 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. \[ 3 x^{2} y^{\prime \prime }+y^{\prime } x +\left (-x -1\right ) y = 0 \] 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}{3 x}\\ q(x) &= -\frac {1+x}{3 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 \[ 3 x^{2} y^{\prime \prime }+y^{\prime } x +\left (-x -1\right ) y = 0 \] 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} 3 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 (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right ) x +\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 }}3 x^{n +r} 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 )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{1+n +r} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} x^{n +r}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r\) 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 (-x^{1+n +r} a_{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} x^{n +r}\right ) \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r} 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 )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 3 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+x^{n +r} a_{n} \left (n +r \right )-a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 3 x^{r} a_{0} r \left (-1+r \right )+x^{r} a_{0} r -a_{0} x^{r} = 0 \] Or \[ \left (3 x^{r} r \left (-1+r \right )+x^{r} r -x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (3 r^{2}-2 r -1\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 3 r^{2}-2 r -1 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 1\\ r_2 &= -{\frac {1}{3}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (3 r^{2}-2 r -1\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [1, -{\frac {1}{3}}\right ]\).
Since \(r_1 - r_2 = {\frac {4}{3}}\) 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^{1+n}\\ 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} 3 a_{n} \left (n +r \right ) \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 \[ a_{n} = \frac {a_{n -1}}{3 n^{2}+6 n r +3 r^{2}-2 n -2 r -1}\tag {4} \] Which for the root \(r = 1\) becomes \[ a_{n} = \frac {a_{n -1}}{n \left (3 n +4\right )}\tag {5} \] 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 \[ a_{1}=\frac {1}{r \left (3 r +4\right )} \] Which for the root \(r = 1\) becomes \[ a_{1}={\frac {1}{7}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{r \left (3 r +4\right )}\) | \(\frac {1}{7}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {1}{9 r^{4}+42 r^{3}+61 r^{2}+28 r} \] Which for the root \(r = 1\) becomes \[ a_{2}={\frac {1}{140}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{r \left (3 r +4\right )}\) | \(\frac {1}{7}\) |
| \(a_{2}\) | \(\frac {1}{9 r^{4}+42 r^{3}+61 r^{2}+28 r}\) | \(\frac {1}{140}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {1}{27 r^{6}+270 r^{5}+1035 r^{4}+1900 r^{3}+1668 r^{2}+560 r} \] Which for the root \(r = 1\) becomes \[ a_{3}={\frac {1}{5460}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{r \left (3 r +4\right )}\) | \(\frac {1}{7}\) |
| \(a_{2}\) | \(\frac {1}{9 r^{4}+42 r^{3}+61 r^{2}+28 r}\) | \(\frac {1}{140}\) |
| \(a_{3}\) | \(\frac {1}{27 r^{6}+270 r^{5}+1035 r^{4}+1900 r^{3}+1668 r^{2}+560 r}\) | \(\frac {1}{5460}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {1}{81 r^{8}+1404 r^{7}+10098 r^{6}+39000 r^{5}+87169 r^{4}+112476 r^{3}+77372 r^{2}+21840 r} \] Which for the root \(r = 1\) becomes \[ a_{4}={\frac {1}{349440}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{r \left (3 r +4\right )}\) | \(\frac {1}{7}\) |
| \(a_{2}\) | \(\frac {1}{9 r^{4}+42 r^{3}+61 r^{2}+28 r}\) | \(\frac {1}{140}\) |
| \(a_{3}\) | \(\frac {1}{27 r^{6}+270 r^{5}+1035 r^{4}+1900 r^{3}+1668 r^{2}+560 r}\) | \(\frac {1}{5460}\) |
| \(a_{4}\) | \(\frac {1}{81 r^{8}+1404 r^{7}+10098 r^{6}+39000 r^{5}+87169 r^{4}+112476 r^{3}+77372 r^{2}+21840 r}\) | \(\frac {1}{349440}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {1}{243 r^{10}+6480 r^{9}+74790 r^{8}+489600 r^{7}+1999779 r^{6}+5274160 r^{5}+8960260 r^{4}+9430400 r^{3}+5563328 r^{2}+1397760 r} \] Which for the root \(r = 1\) becomes \[ a_{5}={\frac {1}{33196800}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{r \left (3 r +4\right )}\) | \(\frac {1}{7}\) |
| \(a_{2}\) | \(\frac {1}{9 r^{4}+42 r^{3}+61 r^{2}+28 r}\) | \(\frac {1}{140}\) |
| \(a_{3}\) | \(\frac {1}{27 r^{6}+270 r^{5}+1035 r^{4}+1900 r^{3}+1668 r^{2}+560 r}\) | \(\frac {1}{5460}\) |
| \(a_{4}\) | \(\frac {1}{81 r^{8}+1404 r^{7}+10098 r^{6}+39000 r^{5}+87169 r^{4}+112476 r^{3}+77372 r^{2}+21840 r}\) | \(\frac {1}{349440}\) |
| \(a_{5}\) | \(\frac {1}{243 r^{10}+6480 r^{9}+74790 r^{8}+489600 r^{7}+1999779 r^{6}+5274160 r^{5}+8960260 r^{4}+9430400 r^{3}+5563328 r^{2}+1397760 r}\) | \(\frac {1}{33196800}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {1}{r \left (243 r^{9}+6480 r^{8}+74790 r^{7}+489600 r^{6}+1999779 r^{5}+5274160 r^{4}+8960260 r^{3}+9430400 r^{2}+5563328 r +1397760\right ) \left (3 r^{2}+34 r +95\right )} \] Which for the root \(r = 1\) becomes \[ a_{6}={\frac {1}{4381977600}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{r \left (3 r +4\right )}\) | \(\frac {1}{7}\) |
| \(a_{2}\) | \(\frac {1}{9 r^{4}+42 r^{3}+61 r^{2}+28 r}\) | \(\frac {1}{140}\) |
| \(a_{3}\) | \(\frac {1}{27 r^{6}+270 r^{5}+1035 r^{4}+1900 r^{3}+1668 r^{2}+560 r}\) | \(\frac {1}{5460}\) |
| \(a_{4}\) | \(\frac {1}{81 r^{8}+1404 r^{7}+10098 r^{6}+39000 r^{5}+87169 r^{4}+112476 r^{3}+77372 r^{2}+21840 r}\) | \(\frac {1}{349440}\) |
| \(a_{5}\) | \(\frac {1}{243 r^{10}+6480 r^{9}+74790 r^{8}+489600 r^{7}+1999779 r^{6}+5274160 r^{5}+8960260 r^{4}+9430400 r^{3}+5563328 r^{2}+1397760 r}\) | \(\frac {1}{33196800}\) |
| \(a_{6}\) | \(\frac {1}{r \left (243 r^{9}+6480 r^{8}+74790 r^{7}+489600 r^{6}+1999779 r^{5}+5274160 r^{4}+8960260 r^{3}+9430400 r^{2}+5563328 r +1397760\right ) \left (3 r^{2}+34 r +95\right )}\) | \(\frac {1}{4381977600}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=\frac {1}{r \left (243 r^{9}+6480 r^{8}+74790 r^{7}+489600 r^{6}+1999779 r^{5}+5274160 r^{4}+8960260 r^{3}+9430400 r^{2}+5563328 r +1397760\right ) \left (3 r^{2}+34 r +95\right ) \left (3 r^{2}+40 r +132\right )} \] Which for the root \(r = 1\) becomes \[ a_{7}={\frac {1}{766846080000}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{r \left (3 r +4\right )}\) | \(\frac {1}{7}\) |
| \(a_{2}\) | \(\frac {1}{9 r^{4}+42 r^{3}+61 r^{2}+28 r}\) | \(\frac {1}{140}\) |
| \(a_{3}\) | \(\frac {1}{27 r^{6}+270 r^{5}+1035 r^{4}+1900 r^{3}+1668 r^{2}+560 r}\) | \(\frac {1}{5460}\) |
| \(a_{4}\) | \(\frac {1}{81 r^{8}+1404 r^{7}+10098 r^{6}+39000 r^{5}+87169 r^{4}+112476 r^{3}+77372 r^{2}+21840 r}\) | \(\frac {1}{349440}\) |
| \(a_{5}\) | \(\frac {1}{243 r^{10}+6480 r^{9}+74790 r^{8}+489600 r^{7}+1999779 r^{6}+5274160 r^{5}+8960260 r^{4}+9430400 r^{3}+5563328 r^{2}+1397760 r}\) | \(\frac {1}{33196800}\) |
| \(a_{6}\) | \(\frac {1}{r \left (243 r^{9}+6480 r^{8}+74790 r^{7}+489600 r^{6}+1999779 r^{5}+5274160 r^{4}+8960260 r^{3}+9430400 r^{2}+5563328 r +1397760\right ) \left (3 r^{2}+34 r +95\right )}\) | \(\frac {1}{4381977600}\) |
| \(a_{7}\) | \(\frac {1}{r \left (243 r^{9}+6480 r^{8}+74790 r^{7}+489600 r^{6}+1999779 r^{5}+5274160 r^{4}+8960260 r^{3}+9430400 r^{2}+5563328 r +1397760\right ) \left (3 r^{2}+34 r +95\right ) \left (3 r^{2}+40 r +132\right )}\) | \(\frac {1}{766846080000}\) |
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}{7}+\frac {x^{2}}{140}+\frac {x^{3}}{5460}+\frac {x^{4}}{349440}+\frac {x^{5}}{33196800}+\frac {x^{6}}{4381977600}+\frac {x^{7}}{766846080000}+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} 3 b_{n} \left (n +r \right ) \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 \[ b_{n} = \frac {b_{n -1}}{3 n^{2}+6 n r +3 r^{2}-2 n -2 r -1}\tag {4} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ b_{n} = \frac {b_{n -1}}{n \left (3 n -4\right )}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -{\frac {1}{3}}\) 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 \[ b_{1}=\frac {1}{r \left (3 r +4\right )} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ b_{1}=-1 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{r \left (3 r +4\right )}\) | \(-1\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {1}{9 r^{4}+42 r^{3}+61 r^{2}+28 r} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ b_{2}=-{\frac {1}{4}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{r \left (3 r +4\right )}\) | \(-1\) |
| \(b_{2}\) | \(\frac {1}{9 r^{4}+42 r^{3}+61 r^{2}+28 r}\) | \(-{\frac {1}{4}}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {1}{27 r^{6}+270 r^{5}+1035 r^{4}+1900 r^{3}+1668 r^{2}+560 r} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ b_{3}=-{\frac {1}{60}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{r \left (3 r +4\right )}\) | \(-1\) |
| \(b_{2}\) | \(\frac {1}{9 r^{4}+42 r^{3}+61 r^{2}+28 r}\) | \(-{\frac {1}{4}}\) |
| \(b_{3}\) | \(\frac {1}{27 r^{6}+270 r^{5}+1035 r^{4}+1900 r^{3}+1668 r^{2}+560 r}\) | \(-{\frac {1}{60}}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {1}{81 r^{8}+1404 r^{7}+10098 r^{6}+39000 r^{5}+87169 r^{4}+112476 r^{3}+77372 r^{2}+21840 r} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ b_{4}=-{\frac {1}{1920}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{r \left (3 r +4\right )}\) | \(-1\) |
| \(b_{2}\) | \(\frac {1}{9 r^{4}+42 r^{3}+61 r^{2}+28 r}\) | \(-{\frac {1}{4}}\) |
| \(b_{3}\) | \(\frac {1}{27 r^{6}+270 r^{5}+1035 r^{4}+1900 r^{3}+1668 r^{2}+560 r}\) | \(-{\frac {1}{60}}\) |
| \(b_{4}\) | \(\frac {1}{81 r^{8}+1404 r^{7}+10098 r^{6}+39000 r^{5}+87169 r^{4}+112476 r^{3}+77372 r^{2}+21840 r}\) | \(-{\frac {1}{1920}}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {1}{243 r^{10}+6480 r^{9}+74790 r^{8}+489600 r^{7}+1999779 r^{6}+5274160 r^{5}+8960260 r^{4}+9430400 r^{3}+5563328 r^{2}+1397760 r} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ b_{5}=-{\frac {1}{105600}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{r \left (3 r +4\right )}\) | \(-1\) |
| \(b_{2}\) | \(\frac {1}{9 r^{4}+42 r^{3}+61 r^{2}+28 r}\) | \(-{\frac {1}{4}}\) |
| \(b_{3}\) | \(\frac {1}{27 r^{6}+270 r^{5}+1035 r^{4}+1900 r^{3}+1668 r^{2}+560 r}\) | \(-{\frac {1}{60}}\) |
| \(b_{4}\) | \(\frac {1}{81 r^{8}+1404 r^{7}+10098 r^{6}+39000 r^{5}+87169 r^{4}+112476 r^{3}+77372 r^{2}+21840 r}\) | \(-{\frac {1}{1920}}\) |
| \(b_{5}\) | \(\frac {1}{243 r^{10}+6480 r^{9}+74790 r^{8}+489600 r^{7}+1999779 r^{6}+5274160 r^{5}+8960260 r^{4}+9430400 r^{3}+5563328 r^{2}+1397760 r}\) | \(-{\frac {1}{105600}}\) |
For \(n = 6\), using the above recursive equation gives \[ b_{6}=\frac {1}{r \left (243 r^{9}+6480 r^{8}+74790 r^{7}+489600 r^{6}+1999779 r^{5}+5274160 r^{4}+8960260 r^{3}+9430400 r^{2}+5563328 r +1397760\right ) \left (3 r^{2}+34 r +95\right )} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ b_{6}=-{\frac {1}{8870400}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{r \left (3 r +4\right )}\) | \(-1\) |
| \(b_{2}\) | \(\frac {1}{9 r^{4}+42 r^{3}+61 r^{2}+28 r}\) | \(-{\frac {1}{4}}\) |
| \(b_{3}\) | \(\frac {1}{27 r^{6}+270 r^{5}+1035 r^{4}+1900 r^{3}+1668 r^{2}+560 r}\) | \(-{\frac {1}{60}}\) |
| \(b_{4}\) | \(\frac {1}{81 r^{8}+1404 r^{7}+10098 r^{6}+39000 r^{5}+87169 r^{4}+112476 r^{3}+77372 r^{2}+21840 r}\) | \(-{\frac {1}{1920}}\) |
| \(b_{5}\) | \(\frac {1}{243 r^{10}+6480 r^{9}+74790 r^{8}+489600 r^{7}+1999779 r^{6}+5274160 r^{5}+8960260 r^{4}+9430400 r^{3}+5563328 r^{2}+1397760 r}\) | \(-{\frac {1}{105600}}\) |
| \(b_{6}\) | \(\frac {1}{r \left (243 r^{9}+6480 r^{8}+74790 r^{7}+489600 r^{6}+1999779 r^{5}+5274160 r^{4}+8960260 r^{3}+9430400 r^{2}+5563328 r +1397760\right ) \left (3 r^{2}+34 r +95\right )}\) | \(-{\frac {1}{8870400}}\) |
For \(n = 7\), using the above recursive equation gives \[ b_{7}=\frac {1}{r \left (243 r^{9}+6480 r^{8}+74790 r^{7}+489600 r^{6}+1999779 r^{5}+5274160 r^{4}+8960260 r^{3}+9430400 r^{2}+5563328 r +1397760\right ) \left (3 r^{2}+34 r +95\right ) \left (3 r^{2}+40 r +132\right )} \] Which for the root \(r = -{\frac {1}{3}}\) becomes \[ b_{7}=-{\frac {1}{1055577600}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{r \left (3 r +4\right )}\) | \(-1\) |
| \(b_{2}\) | \(\frac {1}{9 r^{4}+42 r^{3}+61 r^{2}+28 r}\) | \(-{\frac {1}{4}}\) |
| \(b_{3}\) | \(\frac {1}{27 r^{6}+270 r^{5}+1035 r^{4}+1900 r^{3}+1668 r^{2}+560 r}\) | \(-{\frac {1}{60}}\) |
| \(b_{4}\) | \(\frac {1}{81 r^{8}+1404 r^{7}+10098 r^{6}+39000 r^{5}+87169 r^{4}+112476 r^{3}+77372 r^{2}+21840 r}\) | \(-{\frac {1}{1920}}\) |
| \(b_{5}\) | \(\frac {1}{243 r^{10}+6480 r^{9}+74790 r^{8}+489600 r^{7}+1999779 r^{6}+5274160 r^{5}+8960260 r^{4}+9430400 r^{3}+5563328 r^{2}+1397760 r}\) | \(-{\frac {1}{105600}}\) |
| \(b_{6}\) | \(\frac {1}{r \left (243 r^{9}+6480 r^{8}+74790 r^{7}+489600 r^{6}+1999779 r^{5}+5274160 r^{4}+8960260 r^{3}+9430400 r^{2}+5563328 r +1397760\right ) \left (3 r^{2}+34 r +95\right )}\) | \(-{\frac {1}{8870400}}\) |
| \(b_{7}\) | \(\frac {1}{r \left (243 r^{9}+6480 r^{8}+74790 r^{7}+489600 r^{6}+1999779 r^{5}+5274160 r^{4}+8960260 r^{3}+9430400 r^{2}+5563328 r +1397760\right ) \left (3 r^{2}+34 r +95\right ) \left (3 r^{2}+40 r +132\right )}\) | \(-{\frac {1}{1055577600}}\) |
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 ) \\ &= \frac {1-x -\frac {x^{2}}{4}-\frac {x^{3}}{60}-\frac {x^{4}}{1920}-\frac {x^{5}}{105600}-\frac {x^{6}}{8870400}-\frac {x^{7}}{1055577600}+O\left (x^{8}\right )}{x^{\frac {1}{3}}} \end {align*}
Therefore the homogeneous solution is \begin{align*} y_h(x) &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= c_{1} x \left (1+\frac {x}{7}+\frac {x^{2}}{140}+\frac {x^{3}}{5460}+\frac {x^{4}}{349440}+\frac {x^{5}}{33196800}+\frac {x^{6}}{4381977600}+\frac {x^{7}}{766846080000}+O\left (x^{8}\right )\right ) + \frac {c_{2} \left (1-x -\frac {x^{2}}{4}-\frac {x^{3}}{60}-\frac {x^{4}}{1920}-\frac {x^{5}}{105600}-\frac {x^{6}}{8870400}-\frac {x^{7}}{1055577600}+O\left (x^{8}\right )\right )}{x^{\frac {1}{3}}} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x \left (1+\frac {x}{7}+\frac {x^{2}}{140}+\frac {x^{3}}{5460}+\frac {x^{4}}{349440}+\frac {x^{5}}{33196800}+\frac {x^{6}}{4381977600}+\frac {x^{7}}{766846080000}+O\left (x^{8}\right )\right )+\frac {c_{2} \left (1-x -\frac {x^{2}}{4}-\frac {x^{3}}{60}-\frac {x^{4}}{1920}-\frac {x^{5}}{105600}-\frac {x^{6}}{8870400}-\frac {x^{7}}{1055577600}+O\left (x^{8}\right )\right )}{x^{\frac {1}{3}}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x \left (1+\frac {x}{7}+\frac {x^{2}}{140}+\frac {x^{3}}{5460}+\frac {x^{4}}{349440}+\frac {x^{5}}{33196800}+\frac {x^{6}}{4381977600}+\frac {x^{7}}{766846080000}+O\left (x^{8}\right )\right )+\frac {c_{2} \left (1-x -\frac {x^{2}}{4}-\frac {x^{3}}{60}-\frac {x^{4}}{1920}-\frac {x^{5}}{105600}-\frac {x^{6}}{8870400}-\frac {x^{7}}{1055577600}+O\left (x^{8}\right )\right )}{x^{\frac {1}{3}}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} x \left (1+\frac {x}{7}+\frac {x^{2}}{140}+\frac {x^{3}}{5460}+\frac {x^{4}}{349440}+\frac {x^{5}}{33196800}+\frac {x^{6}}{4381977600}+\frac {x^{7}}{766846080000}+O\left (x^{8}\right )\right )+\frac {c_{2} \left (1-x -\frac {x^{2}}{4}-\frac {x^{3}}{60}-\frac {x^{4}}{1920}-\frac {x^{5}}{105600}-\frac {x^{6}}{8870400}-\frac {x^{7}}{1055577600}+O\left (x^{8}\right )\right )}{x^{\frac {1}{3}}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 3 x^{2} \left (\frac {d}{d x}y^{\prime }\right )+y^{\prime } x +\left (-x -1\right ) 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^{\prime }}{3 x}+\frac {\left (1+x \right ) y}{3 x^{2}} \\ \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 {y^{\prime }}{3 x}-\frac {\left (1+x \right ) y}{3 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}{3 x}, P_{3}\left (x \right )=-\frac {1+x}{3 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}{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}}}=-\frac {1}{3} \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 3 x^{2} \left (\frac {d}{d x}y^{\prime }\right )+y^{\prime } x +\left (-x -1\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 \cdot y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +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+3 r \right ) \left (-1+r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (3 k +3 r +1\right ) \left (k +r -1\right )-a_{k -1}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (1+3 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}{3}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 3 \left (k +r -1\right ) \left (k +r +\frac {1}{3}\right ) a_{k}-a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 3 \left (k +r \right ) \left (k +\frac {4}{3}+r \right ) a_{k +1}-a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k}}{\left (k +r \right ) \left (3 k +4+3 r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +1}=\frac {a_{k}}{\left (k +1\right ) \left (3 k +7\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 {a_{k}}{\left (k +1\right ) \left (3 k +7\right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {1}{3} \\ {} & {} & a_{k +1}=\frac {a_{k}}{\left (k -\frac {1}{3}\right ) \left (3 k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {1}{3} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {1}{3}}, a_{k +1}=\frac {a_{k}}{\left (k -\frac {1}{3}\right ) \left (3 k +3\right )}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k -\frac {1}{3}}\right ), a_{k +1}=\frac {a_{k}}{\left (k +1\right ) \left (3 k +7\right )}, b_{k +1}=\frac {b_{k}}{\left (k -\frac {1}{3}\right ) \left (3 k +3\right )}\right ] \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel <- Bessel successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 53
Order:=8; dsolve(3*x^2*diff(y(x),x$2)+x*diff(y(x),x)-(1+x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{1} \left (1-x -\frac {1}{4} x^{2}-\frac {1}{60} x^{3}-\frac {1}{1920} x^{4}-\frac {1}{105600} x^{5}-\frac {1}{8870400} x^{6}-\frac {1}{1055577600} x^{7}+\operatorname {O}\left (x^{8}\right )\right )}{x^{\frac {1}{3}}}+c_{2} x \left (1+\frac {1}{7} x +\frac {1}{140} x^{2}+\frac {1}{5460} x^{3}+\frac {1}{349440} x^{4}+\frac {1}{33196800} x^{5}+\frac {1}{4381977600} x^{6}+\frac {1}{766846080000} x^{7}+\operatorname {O}\left (x^{8}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.003 (sec). Leaf size: 112
AsymptoticDSolveValue[3*x^2*y''[x]+x*y'[x]-(1+x)*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_1 x \left (\frac {x^7}{766846080000}+\frac {x^6}{4381977600}+\frac {x^5}{33196800}+\frac {x^4}{349440}+\frac {x^3}{5460}+\frac {x^2}{140}+\frac {x}{7}+1\right )+\frac {c_2 \left (-\frac {x^7}{1055577600}-\frac {x^6}{8870400}-\frac {x^5}{105600}-\frac {x^4}{1920}-\frac {x^3}{60}-\frac {x^2}{4}-x+1\right )}{\sqrt [3]{x}} \]