Internal problem ID [6930]
Internal file name [OUTPUT/6173_Friday_August_12_2022_11_04_18_PM_66620466/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: 14.
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]]
\[ \boxed {2 x y^{\prime \prime }+\left (1+2 x \right ) y^{\prime }-5 y=0} \] 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. \[ 2 x y^{\prime \prime }+\left (1+2 x \right ) y^{\prime }-5 y = 0 \] 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 {5}{2 x}\\ \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 \[ 2 x y^{\prime \prime }+\left (1+2 x \right ) y^{\prime }-5 y = 0 \] 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 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (1+2 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )-5 \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 -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-5 a_{n} x^{n +r}\right ) = 0 \end{equation} The next step is to make all powers of \(x\) be \(n +r -1\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r -1}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-5 a_{n} x^{n +r}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-5 a_{n -1} x^{n +r -1}\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 \(n +r -1\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-5 a_{n -1} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 2 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+\left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ 2 x^{-1+r} a_{0} r \left (-1+r \right )+r a_{0} x^{-1+r} = 0 \] Or \[ \left (2 x^{-1+r} r \left (-1+r \right )+r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ r \,x^{-1+r} \left (-1+2 r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 2 r^{2}-r = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {1}{2}}\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ r \,x^{-1+r} \left (-1+2 r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {1}{2}}, 0\right ]\).
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^{n +\frac {1}{2}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n} \end {align*}
We start by finding \(y_{1}\left (x \right )\). Eq (2B) derived above is now used to find all \(a_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} 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 )-5 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -1} \left (2 n +2 r -7\right )}{2 n^{2}+4 n r +2 r^{2}-n -r}\tag {4} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{n} = -\frac {2 \left (n -3\right ) a_{n -1}}{2 n^{2}+n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {1}{2}}\) 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 \[ a_{1}=\frac {5-2 r}{2 r^{2}+3 r +1} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{1}={\frac {4}{3}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {5-2 r}{2 r^{2}+3 r +1}\) | \(\frac {4}{3}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {4 r^{2}-16 r +15}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{2}={\frac {4}{15}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {5-2 r}{2 r^{2}+3 r +1}\) | \(\frac {4}{3}\) |
| \(a_{2}\) | \(\frac {4 r^{2}-16 r +15}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6}\) | \(\frac {4}{15}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-8 r^{3}+36 r^{2}-46 r +15}{8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{3}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {5-2 r}{2 r^{2}+3 r +1}\) | \(\frac {4}{3}\) |
| \(a_{2}\) | \(\frac {4 r^{2}-16 r +15}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6}\) | \(\frac {4}{15}\) |
| \(a_{3}\) | \(\frac {-8 r^{3}+36 r^{2}-46 r +15}{8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90}\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {8 r^{3}-36 r^{2}+46 r -15}{8 r^{7}+140 r^{6}+1022 r^{5}+4025 r^{4}+9212 r^{3}+12215 r^{2}+8658 r +2520} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{4}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {5-2 r}{2 r^{2}+3 r +1}\) | \(\frac {4}{3}\) |
| \(a_{2}\) | \(\frac {4 r^{2}-16 r +15}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6}\) | \(\frac {4}{15}\) |
| \(a_{3}\) | \(\frac {-8 r^{3}+36 r^{2}-46 r +15}{8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90}\) | \(0\) |
| \(a_{4}\) | \(\frac {8 r^{3}-36 r^{2}+46 r -15}{8 r^{7}+140 r^{6}+1022 r^{5}+4025 r^{4}+9212 r^{3}+12215 r^{2}+8658 r +2520}\) | \(0\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-8 r^{3}+36 r^{2}-46 r +15}{8 r^{8}+204 r^{7}+2226 r^{6}+13545 r^{5}+50127 r^{4}+115101 r^{3}+159319 r^{2}+120630 r +37800} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{5}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {5-2 r}{2 r^{2}+3 r +1}\) | \(\frac {4}{3}\) |
| \(a_{2}\) | \(\frac {4 r^{2}-16 r +15}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6}\) | \(\frac {4}{15}\) |
| \(a_{3}\) | \(\frac {-8 r^{3}+36 r^{2}-46 r +15}{8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90}\) | \(0\) |
| \(a_{4}\) | \(\frac {8 r^{3}-36 r^{2}+46 r -15}{8 r^{7}+140 r^{6}+1022 r^{5}+4025 r^{4}+9212 r^{3}+12215 r^{2}+8658 r +2520}\) | \(0\) |
| \(a_{5}\) | \(\frac {-8 r^{3}+36 r^{2}-46 r +15}{8 r^{8}+204 r^{7}+2226 r^{6}+13545 r^{5}+50127 r^{4}+115101 r^{3}+159319 r^{2}+120630 r +37800}\) | \(0\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {8 r^{3}-36 r^{2}+46 r -15}{8 r^{9}+276 r^{8}+4146 r^{7}+35511 r^{6}+190575 r^{5}+662109 r^{4}+1481899 r^{3}+2046384 r^{2}+1566612 r +498960} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{6}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {5-2 r}{2 r^{2}+3 r +1}\) | \(\frac {4}{3}\) |
| \(a_{2}\) | \(\frac {4 r^{2}-16 r +15}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6}\) | \(\frac {4}{15}\) |
| \(a_{3}\) | \(\frac {-8 r^{3}+36 r^{2}-46 r +15}{8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90}\) | \(0\) |
| \(a_{4}\) | \(\frac {8 r^{3}-36 r^{2}+46 r -15}{8 r^{7}+140 r^{6}+1022 r^{5}+4025 r^{4}+9212 r^{3}+12215 r^{2}+8658 r +2520}\) | \(0\) |
| \(a_{5}\) | \(\frac {-8 r^{3}+36 r^{2}-46 r +15}{8 r^{8}+204 r^{7}+2226 r^{6}+13545 r^{5}+50127 r^{4}+115101 r^{3}+159319 r^{2}+120630 r +37800}\) | \(0\) |
| \(a_{6}\) | \(\frac {8 r^{3}-36 r^{2}+46 r -15}{8 r^{9}+276 r^{8}+4146 r^{7}+35511 r^{6}+190575 r^{5}+662109 r^{4}+1481899 r^{3}+2046384 r^{2}+1566612 r +498960}\) | \(0\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=\frac {-8 r^{3}+36 r^{2}-46 r +15}{8 r^{10}+356 r^{9}+6990 r^{8}+79575 r^{7}+580104 r^{6}+2820258 r^{5}+9220630 r^{4}+19905775 r^{3}+26948988 r^{2}+20437236 r +6486480} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{7}=0 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {5-2 r}{2 r^{2}+3 r +1}\) | \(\frac {4}{3}\) |
| \(a_{2}\) | \(\frac {4 r^{2}-16 r +15}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6}\) | \(\frac {4}{15}\) |
| \(a_{3}\) | \(\frac {-8 r^{3}+36 r^{2}-46 r +15}{8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90}\) | \(0\) |
| \(a_{4}\) | \(\frac {8 r^{3}-36 r^{2}+46 r -15}{8 r^{7}+140 r^{6}+1022 r^{5}+4025 r^{4}+9212 r^{3}+12215 r^{2}+8658 r +2520}\) | \(0\) |
| \(a_{5}\) | \(\frac {-8 r^{3}+36 r^{2}-46 r +15}{8 r^{8}+204 r^{7}+2226 r^{6}+13545 r^{5}+50127 r^{4}+115101 r^{3}+159319 r^{2}+120630 r +37800}\) | \(0\) |
| \(a_{6}\) | \(\frac {8 r^{3}-36 r^{2}+46 r -15}{8 r^{9}+276 r^{8}+4146 r^{7}+35511 r^{6}+190575 r^{5}+662109 r^{4}+1481899 r^{3}+2046384 r^{2}+1566612 r +498960}\) | \(0\) |
| \(a_{7}\) | \(\frac {-8 r^{3}+36 r^{2}-46 r +15}{8 r^{10}+356 r^{9}+6990 r^{8}+79575 r^{7}+580104 r^{6}+2820258 r^{5}+9220630 r^{4}+19905775 r^{3}+26948988 r^{2}+20437236 r +6486480}\) | \(0\) |
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}+a_{7} x^{7}+a_{8} x^{8}\dots \right ) \\ &= \sqrt {x}\, \left (1+\frac {4 x}{3}+\frac {4 x^{2}}{15}+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 )+\left (n +r \right ) b_{n}-5 b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {b_{n -1} \left (2 n +2 r -7\right )}{2 n^{2}+4 n r +2 r^{2}-n -r}\tag {4} \] Which for the root \(r = 0\) becomes \[ b_{n} = \frac {b_{n -1} \left (7-2 n \right )}{2 n^{2}-n}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = 0\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ b_{1}=\frac {5-2 r}{2 r^{2}+3 r +1} \] Which for the root \(r = 0\) becomes \[ b_{1}=5 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {5-2 r}{2 r^{2}+3 r +1}\) | \(5\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {4 r^{2}-16 r +15}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6} \] Which for the root \(r = 0\) becomes \[ b_{2}={\frac {5}{2}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {5-2 r}{2 r^{2}+3 r +1}\) | \(5\) |
| \(b_{2}\) | \(\frac {4 r^{2}-16 r +15}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6}\) | \(\frac {5}{2}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {-8 r^{3}+36 r^{2}-46 r +15}{8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90} \] Which for the root \(r = 0\) becomes \[ b_{3}={\frac {1}{6}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {5-2 r}{2 r^{2}+3 r +1}\) | \(5\) |
| \(b_{2}\) | \(\frac {4 r^{2}-16 r +15}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6}\) | \(\frac {5}{2}\) |
| \(b_{3}\) | \(\frac {-8 r^{3}+36 r^{2}-46 r +15}{8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90}\) | \(\frac {1}{6}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {8 r^{3}-36 r^{2}+46 r -15}{8 r^{7}+140 r^{6}+1022 r^{5}+4025 r^{4}+9212 r^{3}+12215 r^{2}+8658 r +2520} \] Which for the root \(r = 0\) becomes \[ b_{4}=-{\frac {1}{168}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {5-2 r}{2 r^{2}+3 r +1}\) | \(5\) |
| \(b_{2}\) | \(\frac {4 r^{2}-16 r +15}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6}\) | \(\frac {5}{2}\) |
| \(b_{3}\) | \(\frac {-8 r^{3}+36 r^{2}-46 r +15}{8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90}\) | \(\frac {1}{6}\) |
| \(b_{4}\) | \(\frac {8 r^{3}-36 r^{2}+46 r -15}{8 r^{7}+140 r^{6}+1022 r^{5}+4025 r^{4}+9212 r^{3}+12215 r^{2}+8658 r +2520}\) | \(-{\frac {1}{168}}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {-8 r^{3}+36 r^{2}-46 r +15}{8 r^{8}+204 r^{7}+2226 r^{6}+13545 r^{5}+50127 r^{4}+115101 r^{3}+159319 r^{2}+120630 r +37800} \] Which for the root \(r = 0\) becomes \[ b_{5}={\frac {1}{2520}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {5-2 r}{2 r^{2}+3 r +1}\) | \(5\) |
| \(b_{2}\) | \(\frac {4 r^{2}-16 r +15}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6}\) | \(\frac {5}{2}\) |
| \(b_{3}\) | \(\frac {-8 r^{3}+36 r^{2}-46 r +15}{8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90}\) | \(\frac {1}{6}\) |
| \(b_{4}\) | \(\frac {8 r^{3}-36 r^{2}+46 r -15}{8 r^{7}+140 r^{6}+1022 r^{5}+4025 r^{4}+9212 r^{3}+12215 r^{2}+8658 r +2520}\) | \(-{\frac {1}{168}}\) |
| \(b_{5}\) | \(\frac {-8 r^{3}+36 r^{2}-46 r +15}{8 r^{8}+204 r^{7}+2226 r^{6}+13545 r^{5}+50127 r^{4}+115101 r^{3}+159319 r^{2}+120630 r +37800}\) | \(\frac {1}{2520}\) |
For \(n = 6\), using the above recursive equation gives \[ b_{6}=\frac {8 r^{3}-36 r^{2}+46 r -15}{8 r^{9}+276 r^{8}+4146 r^{7}+35511 r^{6}+190575 r^{5}+662109 r^{4}+1481899 r^{3}+2046384 r^{2}+1566612 r +498960} \] Which for the root \(r = 0\) becomes \[ b_{6}=-{\frac {1}{33264}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {5-2 r}{2 r^{2}+3 r +1}\) | \(5\) |
| \(b_{2}\) | \(\frac {4 r^{2}-16 r +15}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6}\) | \(\frac {5}{2}\) |
| \(b_{3}\) | \(\frac {-8 r^{3}+36 r^{2}-46 r +15}{8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90}\) | \(\frac {1}{6}\) |
| \(b_{4}\) | \(\frac {8 r^{3}-36 r^{2}+46 r -15}{8 r^{7}+140 r^{6}+1022 r^{5}+4025 r^{4}+9212 r^{3}+12215 r^{2}+8658 r +2520}\) | \(-{\frac {1}{168}}\) |
| \(b_{5}\) | \(\frac {-8 r^{3}+36 r^{2}-46 r +15}{8 r^{8}+204 r^{7}+2226 r^{6}+13545 r^{5}+50127 r^{4}+115101 r^{3}+159319 r^{2}+120630 r +37800}\) | \(\frac {1}{2520}\) |
| \(b_{6}\) | \(\frac {8 r^{3}-36 r^{2}+46 r -15}{8 r^{9}+276 r^{8}+4146 r^{7}+35511 r^{6}+190575 r^{5}+662109 r^{4}+1481899 r^{3}+2046384 r^{2}+1566612 r +498960}\) | \(-{\frac {1}{33264}}\) |
For \(n = 7\), using the above recursive equation gives \[ b_{7}=\frac {-8 r^{3}+36 r^{2}-46 r +15}{8 r^{10}+356 r^{9}+6990 r^{8}+79575 r^{7}+580104 r^{6}+2820258 r^{5}+9220630 r^{4}+19905775 r^{3}+26948988 r^{2}+20437236 r +6486480} \] Which for the root \(r = 0\) becomes \[ b_{7}={\frac {1}{432432}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {5-2 r}{2 r^{2}+3 r +1}\) | \(5\) |
| \(b_{2}\) | \(\frac {4 r^{2}-16 r +15}{4 r^{4}+20 r^{3}+35 r^{2}+25 r +6}\) | \(\frac {5}{2}\) |
| \(b_{3}\) | \(\frac {-8 r^{3}+36 r^{2}-46 r +15}{8 r^{6}+84 r^{5}+350 r^{4}+735 r^{3}+812 r^{2}+441 r +90}\) | \(\frac {1}{6}\) |
| \(b_{4}\) | \(\frac {8 r^{3}-36 r^{2}+46 r -15}{8 r^{7}+140 r^{6}+1022 r^{5}+4025 r^{4}+9212 r^{3}+12215 r^{2}+8658 r +2520}\) | \(-{\frac {1}{168}}\) |
| \(b_{5}\) | \(\frac {-8 r^{3}+36 r^{2}-46 r +15}{8 r^{8}+204 r^{7}+2226 r^{6}+13545 r^{5}+50127 r^{4}+115101 r^{3}+159319 r^{2}+120630 r +37800}\) | \(\frac {1}{2520}\) |
| \(b_{6}\) | \(\frac {8 r^{3}-36 r^{2}+46 r -15}{8 r^{9}+276 r^{8}+4146 r^{7}+35511 r^{6}+190575 r^{5}+662109 r^{4}+1481899 r^{3}+2046384 r^{2}+1566612 r +498960}\) | \(-{\frac {1}{33264}}\) |
| \(b_{7}\) | \(\frac {-8 r^{3}+36 r^{2}-46 r +15}{8 r^{10}+356 r^{9}+6990 r^{8}+79575 r^{7}+580104 r^{6}+2820258 r^{5}+9220630 r^{4}+19905775 r^{3}+26948988 r^{2}+20437236 r +6486480}\) | \(\frac {1}{432432}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= b_{0}+b_{1} x +b_{2} x^{2}+b_{3} x^{3}+b_{4} x^{4}+b_{5} x^{5}+b_{6} x^{6}+b_{7} x^{7}+b_{8} x^{8}\dots \\ &= 1+5 x +\frac {5 x^{2}}{2}+\frac {x^{3}}{6}-\frac {x^{4}}{168}+\frac {x^{5}}{2520}-\frac {x^{6}}{33264}+\frac {x^{7}}{432432}+O\left (x^{8}\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 {4 x}{3}+\frac {4 x^{2}}{15}+O\left (x^{8}\right )\right ) + c_{2} \left (1+5 x +\frac {5 x^{2}}{2}+\frac {x^{3}}{6}-\frac {x^{4}}{168}+\frac {x^{5}}{2520}-\frac {x^{6}}{33264}+\frac {x^{7}}{432432}+O\left (x^{8}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \sqrt {x}\, \left (1+\frac {4 x}{3}+\frac {4 x^{2}}{15}+O\left (x^{8}\right )\right )+c_{2} \left (1+5 x +\frac {5 x^{2}}{2}+\frac {x^{3}}{6}-\frac {x^{4}}{168}+\frac {x^{5}}{2520}-\frac {x^{6}}{33264}+\frac {x^{7}}{432432}+O\left (x^{8}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \sqrt {x}\, \left (1+\frac {4 x}{3}+\frac {4 x^{2}}{15}+O\left (x^{8}\right )\right )+c_{2} \left (1+5 x +\frac {5 x^{2}}{2}+\frac {x^{3}}{6}-\frac {x^{4}}{168}+\frac {x^{5}}{2520}-\frac {x^{6}}{33264}+\frac {x^{7}}{432432}+O\left (x^{8}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \sqrt {x}\, \left (1+\frac {4 x}{3}+\frac {4 x^{2}}{15}+O\left (x^{8}\right )\right )+c_{2} \left (1+5 x +\frac {5 x^{2}}{2}+\frac {x^{3}}{6}-\frac {x^{4}}{168}+\frac {x^{5}}{2520}-\frac {x^{6}}{33264}+\frac {x^{7}}{432432}+O\left (x^{8}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 x \left (\frac {d}{d x}y^{\prime }\right )+\left (1+2 x \right ) y^{\prime }-5 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 {\left (1+2 x \right ) y^{\prime }}{2 x}+\frac {5 y}{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 {\left (1+2 x \right ) y^{\prime }}{2 x}-\frac {5 y}{2 x}=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 {5}{2 x}\right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=\frac {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}}}=0 \\ {} & \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 \left (\frac {d}{d x}y^{\prime }\right )+\left (1+2 x \right ) y^{\prime }-5 y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x \cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \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 -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & x \cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (-1+2 r \right ) x^{-1+r}+\left (\moverset {\infty }{\munderset {k =0}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (2 k +1+2 r \right )+a_{k} \left (2 k +2 r -5\right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (-1+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {1}{2}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 2 \left (k +r +\frac {1}{2}\right ) \left (k +1+r \right ) a_{k +1}+2 a_{k} \left (k +r -\frac {5}{2}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (2 k +2 r -5\right )}{\left (2 k +1+2 r \right ) \left (k +1+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (2 k -5\right )}{\left (2 k +1\right ) \left (k +1\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +1}=-\frac {a_{k} \left (2 k -5\right )}{\left (2 k +1\right ) \left (k +1\right )}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2}\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =2 \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (2 k -4\right )}{\left (2 k +2\right ) \left (k +\frac {3}{2}\right )} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =0 \\ {} & {} & a_{1}=\frac {4 a_{0}}{3} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =1 \\ {} & {} & a_{2}=\frac {a_{1}}{5} \\ \bullet & {} & \textrm {Express in terms of}\hspace {3pt} a_{0} \\ {} & {} & a_{2}=\frac {4 a_{0}}{15} \\ \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+\frac {4}{3} x +\frac {4}{15} x^{2}\right ) \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}\right )+b_{0}\cdot \left (1+\frac {4}{3} x +\frac {4}{15} x^{2}\right ), a_{k +1}=-\frac {a_{k} \left (2 k -5\right )}{\left (2 k +1\right ) \left (k +1\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 Solution using Kummer functions still has integrals. Trying a hypergeometric solution. -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 1F1 ODE -> 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: 42
Order:=8; dsolve(2*x*diff(y(x),x$2)+(1+2*x)*diff(y(x),x)-5*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} \sqrt {x}\, \left (1+\frac {4}{3} x +\frac {4}{15} x^{2}+\operatorname {O}\left (x^{8}\right )\right )+c_{2} \left (1+5 x +\frac {5}{2} x^{2}+\frac {1}{6} x^{3}-\frac {1}{168} x^{4}+\frac {1}{2520} x^{5}-\frac {1}{33264} x^{6}+\frac {1}{432432} x^{7}+\operatorname {O}\left (x^{8}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.003 (sec). Leaf size: 76
AsymptoticDSolveValue[2*x*y''[x]+(1+2*x)*y'[x]-5*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_1 \sqrt {x} \left (\frac {4 x^2}{15}+\frac {4 x}{3}+1\right )+c_2 \left (\frac {x^7}{432432}-\frac {x^6}{33264}+\frac {x^5}{2520}-\frac {x^4}{168}+\frac {x^3}{6}+\frac {5 x^2}{2}+5 x+1\right ) \]