Internal problem ID [7013]
Internal file name [OUTPUT/6256_Thursday_August_18_2022_07_11_58_AM_54582995/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: 11.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference is integer"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {4 x^{2} y^{\prime \prime }-2 x \left (x +2\right ) y^{\prime }+\left (x +3\right ) y=0} \] 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. \[ 4 x^{2} y^{\prime \prime }+\left (-2 x^{2}-4 x \right ) y^{\prime }+\left (x +3\right ) y = 0 \] 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 {x +2}{2 x}\\ q(x) &= \frac {x +3}{4 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 \[ 4 x^{2} y^{\prime \prime }+\left (-2 x^{2}-4 x \right ) y^{\prime }+\left (x +3\right ) y = 0 \] 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} 4 \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}-4 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (x +3\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 }}4 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 (-4 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 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 (-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 }}x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r} \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}4 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 (-4 x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-4 x^{n +r} a_{n} \left (n +r \right )+3 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 4 x^{r} a_{0} r \left (-1+r \right )-4 x^{r} a_{0} r +3 a_{0} x^{r} = 0 \] Or \[ \left (4 x^{r} r \left (-1+r \right )-4 x^{r} r +3 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (4 r^{2}-8 r +3\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 4 r^{2}-8 r +3 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {3}{2}}\\ r_2 &= {\frac {1}{2}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (4 r^{2}-8 r +3\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {3}{2}}, {\frac {1}{2}}\right ]\).
Since \(r_1 - r_2 = 1\) is 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 ) &= C y_{1}\left (x \right ) \ln \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 ) &= x^{\frac {3}{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\sqrt {x}\, \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 {3}{2}}\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +\frac {1}{2}}\right ) \end {align*}
Where \(C\) above can be zero. We start by finding \(y_{1}\). 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} 4 a_{n} \left (n +r \right ) \left (n +r -1\right )-2 a_{n -1} \left (n +r -1\right )-4 a_{n} \left (n +r \right )+a_{n -1}+3 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {a_{n -1}}{2 n +2 r -1}\tag {4} \] Which for the root \(r = {\frac {3}{2}}\) becomes \[ a_{n} = \frac {a_{n -1}}{2 n +2}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {3}{2}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ a_{1}=\frac {1}{1+2 r} \] Which for the root \(r = {\frac {3}{2}}\) becomes \[ a_{1}={\frac {1}{4}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{1+2 r}\) | \(\frac {1}{4}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {1}{4 r^{2}+8 r +3} \] Which for the root \(r = {\frac {3}{2}}\) becomes \[ a_{2}={\frac {1}{24}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{1+2 r}\) | \(\frac {1}{4}\) |
| \(a_{2}\) | \(\frac {1}{4 r^{2}+8 r +3}\) | \(\frac {1}{24}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {1}{8 r^{3}+36 r^{2}+46 r +15} \] Which for the root \(r = {\frac {3}{2}}\) becomes \[ a_{3}={\frac {1}{192}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{1+2 r}\) | \(\frac {1}{4}\) |
| \(a_{2}\) | \(\frac {1}{4 r^{2}+8 r +3}\) | \(\frac {1}{24}\) |
| \(a_{3}\) | \(\frac {1}{8 r^{3}+36 r^{2}+46 r +15}\) | \(\frac {1}{192}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {1}{16 r^{4}+128 r^{3}+344 r^{2}+352 r +105} \] Which for the root \(r = {\frac {3}{2}}\) becomes \[ a_{4}={\frac {1}{1920}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{1+2 r}\) | \(\frac {1}{4}\) |
| \(a_{2}\) | \(\frac {1}{4 r^{2}+8 r +3}\) | \(\frac {1}{24}\) |
| \(a_{3}\) | \(\frac {1}{8 r^{3}+36 r^{2}+46 r +15}\) | \(\frac {1}{192}\) |
| \(a_{4}\) | \(\frac {1}{16 r^{4}+128 r^{3}+344 r^{2}+352 r +105}\) | \(\frac {1}{1920}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {1}{32 r^{5}+400 r^{4}+1840 r^{3}+3800 r^{2}+3378 r +945} \] Which for the root \(r = {\frac {3}{2}}\) becomes \[ a_{5}={\frac {1}{23040}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{1+2 r}\) | \(\frac {1}{4}\) |
| \(a_{2}\) | \(\frac {1}{4 r^{2}+8 r +3}\) | \(\frac {1}{24}\) |
| \(a_{3}\) | \(\frac {1}{8 r^{3}+36 r^{2}+46 r +15}\) | \(\frac {1}{192}\) |
| \(a_{4}\) | \(\frac {1}{16 r^{4}+128 r^{3}+344 r^{2}+352 r +105}\) | \(\frac {1}{1920}\) |
| \(a_{5}\) | \(\frac {1}{32 r^{5}+400 r^{4}+1840 r^{3}+3800 r^{2}+3378 r +945}\) | \(\frac {1}{23040}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {1}{64 r^{6}+1152 r^{5}+8080 r^{4}+27840 r^{3}+48556 r^{2}+39048 r +10395} \] Which for the root \(r = {\frac {3}{2}}\) becomes \[ a_{6}={\frac {1}{322560}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{1+2 r}\) | \(\frac {1}{4}\) |
| \(a_{2}\) | \(\frac {1}{4 r^{2}+8 r +3}\) | \(\frac {1}{24}\) |
| \(a_{3}\) | \(\frac {1}{8 r^{3}+36 r^{2}+46 r +15}\) | \(\frac {1}{192}\) |
| \(a_{4}\) | \(\frac {1}{16 r^{4}+128 r^{3}+344 r^{2}+352 r +105}\) | \(\frac {1}{1920}\) |
| \(a_{5}\) | \(\frac {1}{32 r^{5}+400 r^{4}+1840 r^{3}+3800 r^{2}+3378 r +945}\) | \(\frac {1}{23040}\) |
| \(a_{6}\) | \(\frac {1}{64 r^{6}+1152 r^{5}+8080 r^{4}+27840 r^{3}+48556 r^{2}+39048 r +10395}\) | \(\frac {1}{322560}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=\frac {1}{128 r^{7}+3136 r^{6}+31136 r^{5}+160720 r^{4}+459032 r^{3}+709324 r^{2}+528414 r +135135} \] Which for the root \(r = {\frac {3}{2}}\) becomes \[ a_{7}={\frac {1}{5160960}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{1+2 r}\) | \(\frac {1}{4}\) |
| \(a_{2}\) | \(\frac {1}{4 r^{2}+8 r +3}\) | \(\frac {1}{24}\) |
| \(a_{3}\) | \(\frac {1}{8 r^{3}+36 r^{2}+46 r +15}\) | \(\frac {1}{192}\) |
| \(a_{4}\) | \(\frac {1}{16 r^{4}+128 r^{3}+344 r^{2}+352 r +105}\) | \(\frac {1}{1920}\) |
| \(a_{5}\) | \(\frac {1}{32 r^{5}+400 r^{4}+1840 r^{3}+3800 r^{2}+3378 r +945}\) | \(\frac {1}{23040}\) |
| \(a_{6}\) | \(\frac {1}{64 r^{6}+1152 r^{5}+8080 r^{4}+27840 r^{3}+48556 r^{2}+39048 r +10395}\) | \(\frac {1}{322560}\) |
| \(a_{7}\) | \(\frac {1}{128 r^{7}+3136 r^{6}+31136 r^{5}+160720 r^{4}+459032 r^{3}+709324 r^{2}+528414 r +135135}\) | \(\frac {1}{5160960}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{\frac {3}{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^{\frac {3}{2}} \left (1+\frac {x}{4}+\frac {x^{2}}{24}+\frac {x^{3}}{192}+\frac {x^{4}}{1920}+\frac {x^{5}}{23040}+\frac {x^{6}}{322560}+\frac {x^{7}}{5160960}+O\left (x^{8}\right )\right ) \end {align*}
Now the second solution \(y_{2}\left (x \right )\) is found. Let \[ r_{1}-r_{2} = N \] Where \(N\) is positive integer which is the difference between the two roots. \(r_{1}\) is taken as the larger root. Hence for this problem we have \(N=1\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{1}\left (r \right )\). If this limit exists, then \(C = 0\), else we need to keep the log term and \(C \neq 0\). The above table shows that \begin {align*} a_N &= a_{1} \\ &= \frac {1}{1+2 r} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {1}{1+2 r}&= \lim _{r\rightarrow {\frac {1}{2}}}\frac {1}{1+2 r}\\ &= {\frac {1}{2}} \end {align*}
The limit is \(\frac {1}{2}\). Since the limit exists then the log term is not needed and we can set \(C = 0\). Therefore the second solution has the form \begin {align*} y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r}\\ &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +\frac {1}{2}} \end {align*}
Eq (3) derived above is 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{4} 4 b_{n} \left (n +r \right ) \left (n +r -1\right )-2 b_{n -1} \left (n +r -1\right )-4 b_{n} \left (n +r \right )+b_{n -1}+3 b_{n} = 0 \end{equation} Which for for the root \(r = {\frac {1}{2}}\) becomes \begin{equation} \tag{4A} 4 b_{n} \left (n +\frac {1}{2}\right ) \left (n -\frac {1}{2}\right )-2 b_{n -1} \left (n -\frac {1}{2}\right )-4 b_{n} \left (n +\frac {1}{2}\right )+b_{n -1}+3 b_{n} = 0 \end{equation} Solving for \(b_{n}\) from the recursive equation (4) gives \[ b_{n} = \frac {b_{n -1}}{2 n +2 r -1}\tag {5} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{n} = \frac {b_{n -1}}{2 n}\tag {6} \] 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 \[ b_{1}=\frac {1}{1+2 r} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{1}={\frac {1}{2}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{1+2 r}\) | \(\frac {1}{2}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {1}{\left (1+2 r \right ) \left (3+2 r \right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{2}={\frac {1}{8}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{1+2 r}\) | \(\frac {1}{2}\) |
| \(b_{2}\) | \(\frac {1}{4 r^{2}+8 r +3}\) | \(\frac {1}{8}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {1}{\left (1+2 r \right ) \left (5+2 r \right ) \left (3+2 r \right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{3}={\frac {1}{48}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{1+2 r}\) | \(\frac {1}{2}\) |
| \(b_{2}\) | \(\frac {1}{4 r^{2}+8 r +3}\) | \(\frac {1}{8}\) |
| \(b_{3}\) | \(\frac {1}{8 r^{3}+36 r^{2}+46 r +15}\) | \(\frac {1}{48}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {1}{\left (1+2 r \right ) \left (3+2 r \right ) \left (7+2 r \right ) \left (5+2 r \right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{4}={\frac {1}{384}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{1+2 r}\) | \(\frac {1}{2}\) |
| \(b_{2}\) | \(\frac {1}{4 r^{2}+8 r +3}\) | \(\frac {1}{8}\) |
| \(b_{3}\) | \(\frac {1}{8 r^{3}+36 r^{2}+46 r +15}\) | \(\frac {1}{48}\) |
| \(b_{4}\) | \(\frac {1}{16 r^{4}+128 r^{3}+344 r^{2}+352 r +105}\) | \(\frac {1}{384}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {1}{\left (1+2 r \right ) \left (5+2 r \right ) \left (3+2 r \right ) \left (9+2 r \right ) \left (7+2 r \right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{5}={\frac {1}{3840}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{1+2 r}\) | \(\frac {1}{2}\) |
| \(b_{2}\) | \(\frac {1}{4 r^{2}+8 r +3}\) | \(\frac {1}{8}\) |
| \(b_{3}\) | \(\frac {1}{8 r^{3}+36 r^{2}+46 r +15}\) | \(\frac {1}{48}\) |
| \(b_{4}\) | \(\frac {1}{16 r^{4}+128 r^{3}+344 r^{2}+352 r +105}\) | \(\frac {1}{384}\) |
| \(b_{5}\) | \(\frac {1}{32 r^{5}+400 r^{4}+1840 r^{3}+3800 r^{2}+3378 r +945}\) | \(\frac {1}{3840}\) |
For \(n = 6\), using the above recursive equation gives \[ b_{6}=\frac {1}{\left (1+2 r \right ) \left (3+2 r \right ) \left (7+2 r \right ) \left (5+2 r \right ) \left (9+2 r \right ) \left (11+2 r \right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{6}={\frac {1}{46080}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{1+2 r}\) | \(\frac {1}{2}\) |
| \(b_{2}\) | \(\frac {1}{4 r^{2}+8 r +3}\) | \(\frac {1}{8}\) |
| \(b_{3}\) | \(\frac {1}{8 r^{3}+36 r^{2}+46 r +15}\) | \(\frac {1}{48}\) |
| \(b_{4}\) | \(\frac {1}{16 r^{4}+128 r^{3}+344 r^{2}+352 r +105}\) | \(\frac {1}{384}\) |
| \(b_{5}\) | \(\frac {1}{32 r^{5}+400 r^{4}+1840 r^{3}+3800 r^{2}+3378 r +945}\) | \(\frac {1}{3840}\) |
| \(b_{6}\) | \(\frac {1}{\left (1+2 r \right ) \left (3+2 r \right ) \left (7+2 r \right ) \left (5+2 r \right ) \left (9+2 r \right ) \left (11+2 r \right )}\) | \(\frac {1}{46080}\) |
For \(n = 7\), using the above recursive equation gives \[ b_{7}=\frac {1}{\left (1+2 r \right ) \left (5+2 r \right ) \left (3+2 r \right ) \left (9+2 r \right ) \left (7+2 r \right ) \left (11+2 r \right ) \left (13+2 r \right )} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ b_{7}={\frac {1}{645120}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{1+2 r}\) | \(\frac {1}{2}\) |
| \(b_{2}\) | \(\frac {1}{4 r^{2}+8 r +3}\) | \(\frac {1}{8}\) |
| \(b_{3}\) | \(\frac {1}{8 r^{3}+36 r^{2}+46 r +15}\) | \(\frac {1}{48}\) |
| \(b_{4}\) | \(\frac {1}{16 r^{4}+128 r^{3}+344 r^{2}+352 r +105}\) | \(\frac {1}{384}\) |
| \(b_{5}\) | \(\frac {1}{32 r^{5}+400 r^{4}+1840 r^{3}+3800 r^{2}+3378 r +945}\) | \(\frac {1}{3840}\) |
| \(b_{6}\) | \(\frac {1}{\left (1+2 r \right ) \left (3+2 r \right ) \left (7+2 r \right ) \left (5+2 r \right ) \left (9+2 r \right ) \left (11+2 r \right )}\) | \(\frac {1}{46080}\) |
| \(b_{7}\) | \(\frac {1}{\left (1+2 r \right ) \left (5+2 r \right ) \left (3+2 r \right ) \left (9+2 r \right ) \left (7+2 r \right ) \left (11+2 r \right ) \left (13+2 r \right )}\) | \(\frac {1}{645120}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x^{\frac {3}{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 {x}{2}+\frac {x^{2}}{8}+\frac {x^{3}}{48}+\frac {x^{4}}{384}+\frac {x^{5}}{3840}+\frac {x^{6}}{46080}+\frac {x^{7}}{645120}+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^{\frac {3}{2}} \left (1+\frac {x}{4}+\frac {x^{2}}{24}+\frac {x^{3}}{192}+\frac {x^{4}}{1920}+\frac {x^{5}}{23040}+\frac {x^{6}}{322560}+\frac {x^{7}}{5160960}+O\left (x^{8}\right )\right ) + c_{2} \sqrt {x}\, \left (1+\frac {x}{2}+\frac {x^{2}}{8}+\frac {x^{3}}{48}+\frac {x^{4}}{384}+\frac {x^{5}}{3840}+\frac {x^{6}}{46080}+\frac {x^{7}}{645120}+O\left (x^{8}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {3}{2}} \left (1+\frac {x}{4}+\frac {x^{2}}{24}+\frac {x^{3}}{192}+\frac {x^{4}}{1920}+\frac {x^{5}}{23040}+\frac {x^{6}}{322560}+\frac {x^{7}}{5160960}+O\left (x^{8}\right )\right )+c_{2} \sqrt {x}\, \left (1+\frac {x}{2}+\frac {x^{2}}{8}+\frac {x^{3}}{48}+\frac {x^{4}}{384}+\frac {x^{5}}{3840}+\frac {x^{6}}{46080}+\frac {x^{7}}{645120}+O\left (x^{8}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {3}{2}} \left (1+\frac {x}{4}+\frac {x^{2}}{24}+\frac {x^{3}}{192}+\frac {x^{4}}{1920}+\frac {x^{5}}{23040}+\frac {x^{6}}{322560}+\frac {x^{7}}{5160960}+O\left (x^{8}\right )\right )+c_{2} \sqrt {x}\, \left (1+\frac {x}{2}+\frac {x^{2}}{8}+\frac {x^{3}}{48}+\frac {x^{4}}{384}+\frac {x^{5}}{3840}+\frac {x^{6}}{46080}+\frac {x^{7}}{645120}+O\left (x^{8}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {3}{2}} \left (1+\frac {x}{4}+\frac {x^{2}}{24}+\frac {x^{3}}{192}+\frac {x^{4}}{1920}+\frac {x^{5}}{23040}+\frac {x^{6}}{322560}+\frac {x^{7}}{5160960}+O\left (x^{8}\right )\right )+c_{2} \sqrt {x}\, \left (1+\frac {x}{2}+\frac {x^{2}}{8}+\frac {x^{3}}{48}+\frac {x^{4}}{384}+\frac {x^{5}}{3840}+\frac {x^{6}}{46080}+\frac {x^{7}}{645120}+O\left (x^{8}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 y^{\prime \prime } x^{2}+\left (-2 x^{2}-4 x \right ) y^{\prime }+\left (x +3\right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=-\frac {\left (x +3\right ) y}{4 x^{2}}+\frac {\left (x +2\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 (x +2\right ) y^{\prime }}{2 x}+\frac {\left (x +3\right ) y}{4 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 {x +2}{2 x}, P_{3}\left (x \right )=\frac {x +3}{4 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}}}=-1 \\ {} & \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 {3}{4} \\ {} & \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} \\ {} & {} & 4 y^{\prime \prime } x^{2}-2 x \left (x +2\right ) y^{\prime }+\left (x +3\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 (-3+2 r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (2 k +2 r -1\right ) \left (2 k +2 r -3\right )-a_{k -1} \left (2 k +2 r -3\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 (-3+2 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{\frac {1}{2}, \frac {3}{2}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (2 k +2 r -3\right ) \left (2 k a_{k}+2 r a_{k}-a_{k}-a_{k -1}\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (2 k +2 r -1\right ) \left (2 \left (k +1\right ) a_{k +1}+2 r a_{k +1}-a_{k +1}-a_{k}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {a_{k}}{2 k +1+2 r} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & a_{k +1}=\frac {a_{k}}{2 k +2} \\ \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 {a_{k}}{2 k +2}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {3}{2} \\ {} & {} & a_{k +1}=\frac {a_{k}}{2 k +4} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {3}{2} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {3}{2}}, a_{k +1}=\frac {a_{k}}{2 k +4}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +\frac {1}{2}}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +\frac {3}{2}}\right ), a_{k +1}=\frac {a_{k}}{2 k +2}, b_{k +1}=\frac {b_{k}}{2 k +4}\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) Reducible group (found another exponential solution) <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 55
Order:=8; dsolve(4*x^2*diff(y(x),x$2)-2*x*(2+x)*diff(y(x),x)+(3+x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \sqrt {x}\, \left (x \left (1+\frac {1}{4} x +\frac {1}{24} x^{2}+\frac {1}{192} x^{3}+\frac {1}{1920} x^{4}+\frac {1}{23040} x^{5}+\frac {1}{322560} x^{6}+\frac {1}{5160960} x^{7}+\operatorname {O}\left (x^{8}\right )\right ) c_{1} +\left (1+\frac {1}{2} x +\frac {1}{8} x^{2}+\frac {1}{48} x^{3}+\frac {1}{384} x^{4}+\frac {1}{3840} x^{5}+\frac {1}{46080} x^{6}+\frac {1}{645120} x^{7}+\operatorname {O}\left (x^{8}\right )\right ) c_{2} \right ) \]
✓ Solution by Mathematica
Time used: 0.114 (sec). Leaf size: 130
AsymptoticDSolveValue[4*x^2*y''[x]-2*x*(2+x)*y'[x]+(3+x)*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_1 \left (\frac {x^{13/2}}{46080}+\frac {x^{11/2}}{3840}+\frac {x^{9/2}}{384}+\frac {x^{7/2}}{48}+\frac {x^{5/2}}{8}+\frac {x^{3/2}}{2}+\sqrt {x}\right )+c_2 \left (\frac {x^{15/2}}{322560}+\frac {x^{13/2}}{23040}+\frac {x^{11/2}}{1920}+\frac {x^{9/2}}{192}+\frac {x^{7/2}}{24}+\frac {x^{5/2}}{4}+x^{3/2}\right ) \]