Internal problem ID [1356]
Internal file name [OUTPUT/1357_Sunday_June_05_2022_02_12_43_AM_60657887/index.tex]
Book: Elementary differential equations with boundary value problems. William F. Trench.
Brooks/Cole 2001
Section: Chapter 7 Series Solutions of Linear Second Equations. 7.6 THE METHOD OF
FROBENIUS II. Exercises 7.6. Page 374
Problem number: 4.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Repeated root"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {4 x^{2} \left (x^{2}+x +1\right ) y^{\prime \prime }+12 x^{2} \left (x +1\right ) y^{\prime }+\left (3 x^{2}+3 x +1\right ) y=0} \] 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. \[ \left (4 x^{4}+4 x^{3}+4 x^{2}\right ) y^{\prime \prime }+\left (12 x^{3}+12 x^{2}\right ) y^{\prime }+\left (3 x^{2}+3 x +1\right ) y = 0 \] 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 {3 x +3}{x^{2}+x +1}\\ q(x) &= \frac {3 x^{2}+3 x +1}{4 x^{2} \left (x^{2}+x +1\right )}\\ \end {align*}
| \(p(x)=\frac {3 x +3}{x^{2}+x +1}\) | |
| singularity | type |
| \(x = -\frac {1}{2}-\frac {i \sqrt {3}}{2}\) | \(\text {``regular''}\) |
| \(x = -\frac {1}{2}+\frac {i \sqrt {3}}{2}\) | \(\text {``regular''}\) |
| \(q(x)=\frac {3 x^{2}+3 x +1}{4 x^{2} \left (x^{2}+x +1\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -\frac {1}{2}-\frac {i \sqrt {3}}{2}\) | \(\text {``regular''}\) |
| \(x = -\frac {1}{2}+\frac {i \sqrt {3}}{2}\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \(\left [-\frac {1}{2}-\frac {i \sqrt {3}}{2}, -\frac {1}{2}+\frac {i \sqrt {3}}{2}, 0, \infty \right ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ 4 x^{2} \left (x^{2}+x +1\right ) y^{\prime \prime }+\left (12 x^{3}+12 x^{2}\right ) y^{\prime }+\left (3 x^{2}+3 x +1\right ) y = 0 \] 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} 4 x^{2} \left (x^{2}+x +1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right )+\left (12 x^{3}+12 x^{2}\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (3 x^{2}+3 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 }}4 x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}12 x^{n +r +2} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}12 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r +2} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{1+n +r} a_{n}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}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 }}4 x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}4 a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}4 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}12 x^{n +r +2} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}12 a_{n -2} \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}12 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}12 a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}3 x^{n +r +2} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}3 a_{n -2} x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}3 x^{1+n +r} a_{n} &= \moverset {\infty }{\munderset {n =1}{\sum }}3 a_{n -1} x^{n +r} \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =2}{\sum }}4 a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}12 a_{n -2} \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}12 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}3 a_{n -2} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}3 a_{n -1} x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 4 x^{r} a_{0} r \left (-1+r \right )+a_{0} x^{r} = 0 \] Or \[ \left (4 x^{r} r \left (-1+r \right )+x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ x^{r} \left (2 r -1\right )^{2} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ \left (2 r -1\right )^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {1}{2}}\\ r_2 &= {\frac {1}{2}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ x^{r} \left (2 r -1\right )^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {1}{2}}, {\frac {1}{2}}\right ]\).
Since the root of the indicial equation is repeated, then we can construct two linearly independent solutions. The first solution has the form \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\tag {1A} \end {align*}
Now the second solution \(y_{2}\) is found using \begin {align*} y_{2}\left (x \right ) &= y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +r}\right )\tag {1B} \end {align*}
Then the general solution will be \[ y = c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \] In Eq (1B) the sum starts from 1 and not zero. In Eq (1A), \(a_{0}\) is never zero, and is arbitrary and is typically taken as \(a_{0} = 1\), and \(\{c_{1}, c_{2}\}\) are two arbitray constants of integration which can be found from initial conditions. Using the value of the indicial root found earlier, \(r = {\frac {1}{2}}\), Eqs (1A,1B) become \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 ) &= y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +\frac {1}{2}}\right ) \end {align*}
We start by finding the first solution \(y_{1}\left (x \right )\). 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 \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = \frac {-3-2 r}{2 r +1} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} 4 a_{n -2} \left (n +r -2\right ) \left (n -3+r \right )+4 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+4 a_{n} \left (n +r \right ) \left (n +r -1\right )+12 a_{n -2} \left (n +r -2\right )+12 a_{n -1} \left (n +r -1\right )+3 a_{n -2}+3 a_{n -1}+a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {2 n a_{n -2}+2 n a_{n -1}+2 r a_{n -2}+2 r a_{n -1}-3 a_{n -2}+a_{n -1}}{-1+2 n +2 r}\tag {4} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{n} = \frac {\left (-a_{n -2}-a_{n -1}\right ) n +a_{n -2}-a_{n -1}}{n}\tag {5} \] 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.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-2 r}{2 r +1}\) | \(-2\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {12 r +14}{4 r^{2}+8 r +3} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{2}={\frac {5}{2}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-2 r}{2 r +1}\) | \(-2\) |
| \(a_{2}\) | \(\frac {12 r +14}{4 r^{2}+8 r +3}\) | \(\frac {5}{2}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {8 r^{3}+12 r^{2}-58 r -71}{8 r^{3}+36 r^{2}+46 r +15} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{3}=-2 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-2 r}{2 r +1}\) | \(-2\) |
| \(a_{2}\) | \(\frac {12 r +14}{4 r^{2}+8 r +3}\) | \(\frac {5}{2}\) |
| \(a_{3}\) | \(\frac {8 r^{3}+12 r^{2}-58 r -71}{8 r^{3}+36 r^{2}+46 r +15}\) | \(-2\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {-16 r^{4}-144 r^{3}-288 r^{2}+84 r +289}{16 r^{4}+128 r^{3}+344 r^{2}+352 r +105} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{4}={\frac {5}{8}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-2 r}{2 r +1}\) | \(-2\) |
| \(a_{2}\) | \(\frac {12 r +14}{4 r^{2}+8 r +3}\) | \(\frac {5}{2}\) |
| \(a_{3}\) | \(\frac {8 r^{3}+12 r^{2}-58 r -71}{8 r^{3}+36 r^{2}+46 r +15}\) | \(-2\) |
| \(a_{4}\) | \(\frac {-16 r^{4}-144 r^{3}-288 r^{2}+84 r +289}{16 r^{4}+128 r^{3}+344 r^{2}+352 r +105}\) | \(\frac {5}{8}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {192 r^{4}+1664 r^{3}+4320 r^{2}+3328 r +300}{32 r^{5}+400 r^{4}+1840 r^{3}+3800 r^{2}+3378 r +945} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{5}={\frac {17}{20}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-2 r}{2 r +1}\) | \(-2\) |
| \(a_{2}\) | \(\frac {12 r +14}{4 r^{2}+8 r +3}\) | \(\frac {5}{2}\) |
| \(a_{3}\) | \(\frac {8 r^{3}+12 r^{2}-58 r -71}{8 r^{3}+36 r^{2}+46 r +15}\) | \(-2\) |
| \(a_{4}\) | \(\frac {-16 r^{4}-144 r^{3}-288 r^{2}+84 r +289}{16 r^{4}+128 r^{3}+344 r^{2}+352 r +105}\) | \(\frac {5}{8}\) |
| \(a_{5}\) | \(\frac {192 r^{4}+1664 r^{3}+4320 r^{2}+3328 r +300}{32 r^{5}+400 r^{4}+1840 r^{3}+3800 r^{2}+3378 r +945}\) | \(\frac {17}{20}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {64 r^{6}+768 r^{5}+1808 r^{4}-8576 r^{3}-43668 r^{2}-61072 r -27309}{64 r^{6}+1152 r^{5}+8080 r^{4}+27840 r^{3}+48556 r^{2}+39048 r +10395} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{6}=-{\frac {121}{80}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-2 r}{2 r +1}\) | \(-2\) |
| \(a_{2}\) | \(\frac {12 r +14}{4 r^{2}+8 r +3}\) | \(\frac {5}{2}\) |
| \(a_{3}\) | \(\frac {8 r^{3}+12 r^{2}-58 r -71}{8 r^{3}+36 r^{2}+46 r +15}\) | \(-2\) |
| \(a_{4}\) | \(\frac {-16 r^{4}-144 r^{3}-288 r^{2}+84 r +289}{16 r^{4}+128 r^{3}+344 r^{2}+352 r +105}\) | \(\frac {5}{8}\) |
| \(a_{5}\) | \(\frac {192 r^{4}+1664 r^{3}+4320 r^{2}+3328 r +300}{32 r^{5}+400 r^{4}+1840 r^{3}+3800 r^{2}+3378 r +945}\) | \(\frac {17}{20}\) |
| \(a_{6}\) | \(\frac {64 r^{6}+768 r^{5}+1808 r^{4}-8576 r^{3}-43668 r^{2}-61072 r -27309}{64 r^{6}+1152 r^{5}+8080 r^{4}+27840 r^{3}+48556 r^{2}+39048 r +10395}\) | \(-{\frac {121}{80}}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=\frac {-128 r^{7}-3264 r^{6}-30240 r^{5}-123696 r^{4}-188760 r^{3}+106812 r^{2}+554810 r +373335}{128 r^{7}+3136 r^{6}+31136 r^{5}+160720 r^{4}+459032 r^{3}+709324 r^{2}+528414 r +135135} \] Which for the root \(r = {\frac {1}{2}}\) becomes \[ a_{7}=1 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-3-2 r}{2 r +1}\) | \(-2\) |
| \(a_{2}\) | \(\frac {12 r +14}{4 r^{2}+8 r +3}\) | \(\frac {5}{2}\) |
| \(a_{3}\) | \(\frac {8 r^{3}+12 r^{2}-58 r -71}{8 r^{3}+36 r^{2}+46 r +15}\) | \(-2\) |
| \(a_{4}\) | \(\frac {-16 r^{4}-144 r^{3}-288 r^{2}+84 r +289}{16 r^{4}+128 r^{3}+344 r^{2}+352 r +105}\) | \(\frac {5}{8}\) |
| \(a_{5}\) | \(\frac {192 r^{4}+1664 r^{3}+4320 r^{2}+3328 r +300}{32 r^{5}+400 r^{4}+1840 r^{3}+3800 r^{2}+3378 r +945}\) | \(\frac {17}{20}\) |
| \(a_{6}\) | \(\frac {64 r^{6}+768 r^{5}+1808 r^{4}-8576 r^{3}-43668 r^{2}-61072 r -27309}{64 r^{6}+1152 r^{5}+8080 r^{4}+27840 r^{3}+48556 r^{2}+39048 r +10395}\) | \(-{\frac {121}{80}}\) |
| \(a_{7}\) | \(\frac {-128 r^{7}-3264 r^{6}-30240 r^{5}-123696 r^{4}-188760 r^{3}+106812 r^{2}+554810 r +373335}{128 r^{7}+3136 r^{6}+31136 r^{5}+160720 r^{4}+459032 r^{3}+709324 r^{2}+528414 r +135135}\) | \(1\) |
Using the above table, then the first 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 (1-2 x +\frac {5 x^{2}}{2}-2 x^{3}+\frac {5 x^{4}}{8}+\frac {17 x^{5}}{20}-\frac {121 x^{6}}{80}+x^{7}+O\left (x^{8}\right )\right ) \\ \end{align*} Now the second solution is found. The second solution is given by \[ y_{2}\left (x \right ) = y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +r}\right ) \] Where \(b_{n}\) is found using \[ b_{n} = \frac {d}{d r}a_{n ,r} \] And the above is then evaluated at \(r = {\frac {1}{2}}\). The above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table
| \(n\) | \(b_{n ,r}\) | \(a_{n}\) | \(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) | \(b_{n}\left (r =\frac {1}{2}\right )\) |
| \(b_{0}\) | \(1\) | \(1\) | N/A since \(b_{n}\) starts from 1 | N/A |
| \(b_{1}\) | \(\frac {-3-2 r}{2 r +1}\) | \(-2\) | \(\frac {4}{\left (2 r +1\right )^{2}}\) | \(1\) |
| \(b_{2}\) | \(\frac {12 r +14}{4 r^{2}+8 r +3}\) | \(\frac {5}{2}\) | \(\frac {-48 r^{2}-112 r -76}{\left (4 r^{2}+8 r +3\right )^{2}}\) | \(-{\frac {9}{4}}\) |
| \(b_{3}\) | \(\frac {8 r^{3}+12 r^{2}-58 r -71}{8 r^{3}+36 r^{2}+46 r +15}\) | \(-2\) | \(\frac {192 r^{4}+1664 r^{3}+4704 r^{2}+5472 r +2396}{\left (8 r^{3}+36 r^{2}+46 r +15\right )^{2}}\) | \(\frac {17}{6}\) |
| \(b_{4}\) | \(\frac {-16 r^{4}-144 r^{3}-288 r^{2}+84 r +289}{16 r^{4}+128 r^{3}+344 r^{2}+352 r +105}\) | \(\frac {5}{8}\) | \(\frac {256 r^{6}-1792 r^{5}-33600 r^{4}-148096 r^{3}-286608 r^{2}-259312 r -92908}{\left (16 r^{4}+128 r^{3}+344 r^{2}+352 r +105\right )^{2}}\) | \(-{\frac {205}{96}}\) |
| \(b_{5}\) | \(\frac {192 r^{4}+1664 r^{3}+4320 r^{2}+3328 r +300}{32 r^{5}+400 r^{4}+1840 r^{3}+3800 r^{2}+3378 r +945}\) | \(\frac {17}{20}\) | \(-\frac {8 \left (768 r^{8}+13312 r^{7}+90880 r^{6}+302848 r^{5}+465184 r^{4}+94912 r^{3}-626000 r^{2}-735600 r -266445\right )}{\left (32 r^{5}+400 r^{4}+1840 r^{3}+3800 r^{2}+3378 r +945\right )^{2}}\) | \(\frac {481}{1200}\) |
| \(b_{6}\) | \(\frac {64 r^{6}+768 r^{5}+1808 r^{4}-8576 r^{3}-43668 r^{2}-61072 r -27309}{64 r^{6}+1152 r^{5}+8080 r^{4}+27840 r^{3}+48556 r^{2}+39048 r +10395}\) | \(-{\frac {121}{80}}\) | \(\frac {24576 r^{10}+802816 r^{9}+11114496 r^{8}+86130688 r^{7}+414456832 r^{6}+1297106944 r^{5}+2688699136 r^{4}+3688541184 r^{3}+3273669088 r^{2}+1744173888 r +431518392}{\left (64 r^{6}+1152 r^{5}+8080 r^{4}+27840 r^{3}+48556 r^{2}+39048 r +10395\right )^{2}}\) | \(\frac {2109}{1600}\) |
| \(b_{7}\) | \(\frac {-128 r^{7}-3264 r^{6}-30240 r^{5}-123696 r^{4}-188760 r^{3}+106812 r^{2}+554810 r +373335}{128 r^{7}+3136 r^{6}+31136 r^{5}+160720 r^{4}+459032 r^{3}+709324 r^{2}+528414 r +135135}\) | \(1\) | \(\frac {16384 r^{12}-229376 r^{11}-21012480 r^{10}-411738112 r^{9}-4250088448 r^{8}-27440492544 r^{7}-118548610048 r^{6}-352501101568 r^{5}-725070613056 r^{4}-1015710181760 r^{3}-927741475232 r^{2}-500762871840 r -122301191340}{\left (128 r^{7}+3136 r^{6}+31136 r^{5}+160720 r^{4}+459032 r^{3}+709324 r^{2}+528414 r +135135\right )^{2}}\) | \(-{\frac {1063}{560}}\) |
The above table gives all values of \(b_{n}\) needed. Hence the second solution is \begin{align*} y_{2}\left (x \right )&=y_{1}\left (x \right ) \ln \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 \\ &= \sqrt {x}\, \left (1-2 x +\frac {5 x^{2}}{2}-2 x^{3}+\frac {5 x^{4}}{8}+\frac {17 x^{5}}{20}-\frac {121 x^{6}}{80}+x^{7}+O\left (x^{8}\right )\right ) \ln \left (x \right )+\sqrt {x}\, \left (x -\frac {9 x^{2}}{4}+\frac {17 x^{3}}{6}-\frac {205 x^{4}}{96}+\frac {481 x^{5}}{1200}+\frac {2109 x^{6}}{1600}-\frac {1063 x^{7}}{560}+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} \sqrt {x}\, \left (1-2 x +\frac {5 x^{2}}{2}-2 x^{3}+\frac {5 x^{4}}{8}+\frac {17 x^{5}}{20}-\frac {121 x^{6}}{80}+x^{7}+O\left (x^{8}\right )\right ) + c_{2} \left (\sqrt {x}\, \left (1-2 x +\frac {5 x^{2}}{2}-2 x^{3}+\frac {5 x^{4}}{8}+\frac {17 x^{5}}{20}-\frac {121 x^{6}}{80}+x^{7}+O\left (x^{8}\right )\right ) \ln \left (x \right )+\sqrt {x}\, \left (x -\frac {9 x^{2}}{4}+\frac {17 x^{3}}{6}-\frac {205 x^{4}}{96}+\frac {481 x^{5}}{1200}+\frac {2109 x^{6}}{1600}-\frac {1063 x^{7}}{560}+O\left (x^{8}\right )\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \sqrt {x}\, \left (1-2 x +\frac {5 x^{2}}{2}-2 x^{3}+\frac {5 x^{4}}{8}+\frac {17 x^{5}}{20}-\frac {121 x^{6}}{80}+x^{7}+O\left (x^{8}\right )\right )+c_{2} \left (\sqrt {x}\, \left (1-2 x +\frac {5 x^{2}}{2}-2 x^{3}+\frac {5 x^{4}}{8}+\frac {17 x^{5}}{20}-\frac {121 x^{6}}{80}+x^{7}+O\left (x^{8}\right )\right ) \ln \left (x \right )+\sqrt {x}\, \left (x -\frac {9 x^{2}}{4}+\frac {17 x^{3}}{6}-\frac {205 x^{4}}{96}+\frac {481 x^{5}}{1200}+\frac {2109 x^{6}}{1600}-\frac {1063 x^{7}}{560}+O\left (x^{8}\right )\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \sqrt {x}\, \left (1-2 x +\frac {5 x^{2}}{2}-2 x^{3}+\frac {5 x^{4}}{8}+\frac {17 x^{5}}{20}-\frac {121 x^{6}}{80}+x^{7}+O\left (x^{8}\right )\right )+c_{2} \left (\sqrt {x}\, \left (1-2 x +\frac {5 x^{2}}{2}-2 x^{3}+\frac {5 x^{4}}{8}+\frac {17 x^{5}}{20}-\frac {121 x^{6}}{80}+x^{7}+O\left (x^{8}\right )\right ) \ln \left (x \right )+\sqrt {x}\, \left (x -\frac {9 x^{2}}{4}+\frac {17 x^{3}}{6}-\frac {205 x^{4}}{96}+\frac {481 x^{5}}{1200}+\frac {2109 x^{6}}{1600}-\frac {1063 x^{7}}{560}+O\left (x^{8}\right )\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \sqrt {x}\, \left (1-2 x +\frac {5 x^{2}}{2}-2 x^{3}+\frac {5 x^{4}}{8}+\frac {17 x^{5}}{20}-\frac {121 x^{6}}{80}+x^{7}+O\left (x^{8}\right )\right )+c_{2} \left (\sqrt {x}\, \left (1-2 x +\frac {5 x^{2}}{2}-2 x^{3}+\frac {5 x^{4}}{8}+\frac {17 x^{5}}{20}-\frac {121 x^{6}}{80}+x^{7}+O\left (x^{8}\right )\right ) \ln \left (x \right )+\sqrt {x}\, \left (x -\frac {9 x^{2}}{4}+\frac {17 x^{3}}{6}-\frac {205 x^{4}}{96}+\frac {481 x^{5}}{1200}+\frac {2109 x^{6}}{1600}-\frac {1063 x^{7}}{560}+O\left (x^{8}\right )\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 x^{2} \left (x^{2}+x +1\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (12 x^{3}+12 x^{2}\right ) y^{\prime }+\left (3 x^{2}+3 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 {\left (3 x^{2}+3 x +1\right ) y}{4 x^{2} \left (x^{2}+x +1\right )}-\frac {3 \left (x +1\right ) y^{\prime }}{x^{2}+x +1} \\ \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 }}{x^{2}+x +1}+\frac {\left (3 x^{2}+3 x +1\right ) y}{4 x^{2} \left (x^{2}+x +1\right )}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{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 )}{x^{2}+x +1}, P_{3}\left (x \right )=\frac {3 x^{2}+3 x +1}{4 x^{2} \left (x^{2}+x +1\right )}\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}}}=0 \\ {} & \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}{4} \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 4 x^{2} \left (x^{2}+x +1\right ) \left (\frac {d}{d x}y^{\prime }\right )+12 x^{2} \left (x +1\right ) y^{\prime }+\left (3 x^{2}+3 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..2 \\ {} & {} & 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 =2..3 \\ {} & {} & 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^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..4 \\ {} & {} & x^{m}\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 -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (-1+2 r \right )^{2} x^{r}+\left (a_{1} \left (1+2 r \right )^{2}+a_{0} \left (3+2 r \right ) \left (1+2 r \right )\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (2 k +2 r -1\right )^{2}+a_{k -1} \left (2 k +2 r +1\right ) \left (2 k +2 r -1\right )+a_{k -2} \left (2 k +2 r -1\right ) \left (2 k -3+2 r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (-1+2 r \right )^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =\frac {1}{2} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+2 r \right )^{2}+a_{0} \left (3+2 r \right ) \left (1+2 r \right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=-\frac {\left (3+2 r \right ) a_{0}}{1+2 r} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 4 \left (k +r -\frac {1}{2}\right ) \left (\left (a_{k}+a_{k -2}+a_{k -1}\right ) k +\left (a_{k}+a_{k -2}+a_{k -1}\right ) r -\frac {a_{k}}{2}-\frac {3 a_{k -2}}{2}+\frac {a_{k -1}}{2}\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & 4 \left (k +\frac {3}{2}+r \right ) \left (\left (a_{k +2}+a_{k}+a_{k +1}\right ) \left (k +2\right )+\left (a_{k +2}+a_{k}+a_{k +1}\right ) r -\frac {a_{k +2}}{2}-\frac {3 a_{k}}{2}+\frac {a_{k +1}}{2}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {2 k a_{k}+2 k a_{k +1}+2 r a_{k}+2 r a_{k +1}+a_{k}+5 a_{k +1}}{2 k +2 r +3} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {1}{2} \\ {} & {} & a_{k +2}=-\frac {2 k a_{k}+2 k a_{k +1}+2 a_{k}+6 a_{k +1}}{2 k +4} \\ \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 +2}=-\frac {2 k a_{k}+2 k a_{k +1}+2 a_{k}+6 a_{k +1}}{2 k +4}, a_{1}=-2 a_{0}\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 -> Whittaker -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 2F1 ODE <- hypergeometric 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: 0.031 (sec). Leaf size: 81
Order:=8; dsolve(4*x^2*(1+x+x^2)*diff(y(x),x$2)+12*x^2*(1+x)*diff(y(x),x)+(1+3*x+3*x^2)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \sqrt {x}\, \left (\left (c_{2} \ln \left (x \right )+c_{1} \right ) \left (1-2 x +\frac {5}{2} x^{2}-2 x^{3}+\frac {5}{8} x^{4}+\frac {17}{20} x^{5}-\frac {121}{80} x^{6}+x^{7}+\operatorname {O}\left (x^{8}\right )\right )+\left (x -\frac {9}{4} x^{2}+\frac {17}{6} x^{3}-\frac {205}{96} x^{4}+\frac {481}{1200} x^{5}+\frac {2109}{1600} x^{6}-\frac {1063}{560} x^{7}+\operatorname {O}\left (x^{8}\right )\right ) c_{2} \right ) \]
✓ Solution by Mathematica
Time used: 0.011 (sec). Leaf size: 156
AsymptoticDSolveValue[4*x^2*(1+x+x^2)*y''[x]+12*x^2*(1+x)*y'[x]+(1+3*x+3*x^2)*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_1 \sqrt {x} \left (x^7-\frac {121 x^6}{80}+\frac {17 x^5}{20}+\frac {5 x^4}{8}-2 x^3+\frac {5 x^2}{2}-2 x+1\right )+c_2 \left (\sqrt {x} \left (-\frac {1063 x^7}{560}+\frac {2109 x^6}{1600}+\frac {481 x^5}{1200}-\frac {205 x^4}{96}+\frac {17 x^3}{6}-\frac {9 x^2}{4}+x\right )+\sqrt {x} \left (x^7-\frac {121 x^6}{80}+\frac {17 x^5}{20}+\frac {5 x^4}{8}-2 x^3+\frac {5 x^2}{2}-2 x+1\right ) \log (x)\right ) \]