Internal problem ID [1357]
Internal file name [OUTPUT/1358_Sunday_June_05_2022_02_12_49_AM_75467743/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: 5.
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]]
\[ \boxed {x^{2} \left (x^{2}+x +1\right ) y^{\prime \prime }-x \left (-2 x^{2}-4 x +1\right ) y^{\prime }+y=0} \] 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. \[ \left (x^{4}+x^{3}+x^{2}\right ) y^{\prime \prime }+\left (2 x^{3}+4 x^{2}-x \right ) y^{\prime }+y = 0 \] 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 {2 x^{2}+4 x -1}{x \left (x^{2}+x +1\right )}\\ q(x) &= \frac {1}{x^{2} \left (x^{2}+x +1\right )}\\ \end {align*}
| \(p(x)=\frac {2 x^{2}+4 x -1}{x \left (x^{2}+x +1\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -\frac {1}{2}-\frac {i \sqrt {3}}{2}\) | \(\text {``regular''}\) |
| \(x = -\frac {1}{2}+\frac {i \sqrt {3}}{2}\) | \(\text {``regular''}\) |
| \(q(x)=\frac {1}{x^{2} \left (x^{2}+x +1\right )}\) | |
| 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 : \(\left [0, -\frac {1}{2}-\frac {i \sqrt {3}}{2}, -\frac {1}{2}+\frac {i \sqrt {3}}{2}, \infty \right ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} \left (x^{2}+x +1\right ) y^{\prime \prime }+\left (2 x^{3}+4 x^{2}-x \right ) y^{\prime }+y = 0 \] 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} 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 (2 x^{3}+4 x^{2}-x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -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 }}x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{1+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 ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r +2} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{1+n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} a_{n} \left (n +r \right )\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 }}x^{n +r +2} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r +2} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}2 a_{n -2} \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}4 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =2}{\sum }}a_{n -2} \left (n +r -2\right ) \left (n -3+r \right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}2 a_{n -2} \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}4 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r} a_{n} \left (n +r \right )\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 \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-x^{n +r} a_{n} \left (n +r \right )+a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )-x^{r} a_{0} r +a_{0} x^{r} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )-x^{r} r +x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (-1+r \right )^{2} x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ \left (-1+r \right )^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 1\\ r_2 &= 1 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (-1+r \right )^{2} x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([1, 1]\).
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, \(r = 1\), Eqs (1A,1B) become \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{1+n}\\ y_{2}\left (x \right ) &= y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{1+n}\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 \[ a_{1} = \frac {-r -3}{r} \] For \(2\le n\) the recursive equation is \begin{equation} \tag{3} a_{n -2} \left (n +r -2\right ) \left (n -3+r \right )+a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+a_{n} \left (n +r \right ) \left (n +r -1\right )+2 a_{n -2} \left (n +r -2\right )+4 a_{n -1} \left (n +r -1\right )-a_{n} \left (n +r \right )+a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {n a_{n -2}+n a_{n -1}+r a_{n -2}+r a_{n -1}-2 a_{n -2}+2 a_{n -1}}{n +r -1}\tag {4} \] Which for the root \(r = 1\) becomes \[ a_{n} = \frac {\left (-a_{n -2}-a_{n -1}\right ) n +a_{n -2}-3 a_{n -1}}{n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 1\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r -3}{r}\) | \(-4\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {7 r +12}{\left (1+r \right ) r} \] Which for the root \(r = 1\) becomes \[ a_{2}={\frac {19}{2}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r -3}{r}\) | \(-4\) |
| \(a_{2}\) | \(\frac {7 r +12}{\left (1+r \right ) r}\) | \(\frac {19}{2}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {r^{3}-2 r^{2}-40 r -57}{\left (2+r \right ) r \left (1+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{3}=-{\frac {49}{3}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r -3}{r}\) | \(-4\) |
| \(a_{2}\) | \(\frac {7 r +12}{\left (1+r \right ) r}\) | \(\frac {19}{2}\) |
| \(a_{3}\) | \(\frac {r^{3}-2 r^{2}-40 r -57}{\left (2+r \right ) r \left (1+r \right )}\) | \(-{\frac {49}{3}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {-r^{4}-11 r^{3}+12 r^{2}+221 r +294}{r \left (1+r \right ) \left (2+r \right ) \left (r +3\right )} \] Which for the root \(r = 1\) becomes \[ a_{4}={\frac {515}{24}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r -3}{r}\) | \(-4\) |
| \(a_{2}\) | \(\frac {7 r +12}{\left (1+r \right ) r}\) | \(\frac {19}{2}\) |
| \(a_{3}\) | \(\frac {r^{3}-2 r^{2}-40 r -57}{\left (2+r \right ) r \left (1+r \right )}\) | \(-{\frac {49}{3}}\) |
| \(a_{4}\) | \(\frac {-r^{4}-11 r^{3}+12 r^{2}+221 r +294}{r \left (1+r \right ) \left (2+r \right ) \left (r +3\right )}\) | \(\frac {515}{24}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {14 r^{4}+108 r^{3}+10 r^{2}-1139 r -1545}{\left (4+r \right ) r \left (1+r \right ) \left (2+r \right ) \left (r +3\right )} \] Which for the root \(r = 1\) becomes \[ a_{5}=-{\frac {319}{15}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r -3}{r}\) | \(-4\) |
| \(a_{2}\) | \(\frac {7 r +12}{\left (1+r \right ) r}\) | \(\frac {19}{2}\) |
| \(a_{3}\) | \(\frac {r^{3}-2 r^{2}-40 r -57}{\left (2+r \right ) r \left (1+r \right )}\) | \(-{\frac {49}{3}}\) |
| \(a_{4}\) | \(\frac {-r^{4}-11 r^{3}+12 r^{2}+221 r +294}{r \left (1+r \right ) \left (2+r \right ) \left (r +3\right )}\) | \(\frac {515}{24}\) |
| \(a_{5}\) | \(\frac {14 r^{4}+108 r^{3}+10 r^{2}-1139 r -1545}{\left (4+r \right ) r \left (1+r \right ) \left (2+r \right ) \left (r +3\right )}\) | \(-{\frac {319}{15}}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {r^{6}+5 r^{5}-128 r^{4}-1015 r^{3}-1195 r^{2}+4769 r +7656}{\left (5+r \right ) r \left (1+r \right ) \left (2+r \right ) \left (r +3\right ) \left (4+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{6}={\frac {10093}{720}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r -3}{r}\) | \(-4\) |
| \(a_{2}\) | \(\frac {7 r +12}{\left (1+r \right ) r}\) | \(\frac {19}{2}\) |
| \(a_{3}\) | \(\frac {r^{3}-2 r^{2}-40 r -57}{\left (2+r \right ) r \left (1+r \right )}\) | \(-{\frac {49}{3}}\) |
| \(a_{4}\) | \(\frac {-r^{4}-11 r^{3}+12 r^{2}+221 r +294}{r \left (1+r \right ) \left (2+r \right ) \left (r +3\right )}\) | \(\frac {515}{24}\) |
| \(a_{5}\) | \(\frac {14 r^{4}+108 r^{3}+10 r^{2}-1139 r -1545}{\left (4+r \right ) r \left (1+r \right ) \left (2+r \right ) \left (r +3\right )}\) | \(-{\frac {319}{15}}\) |
| \(a_{6}\) | \(\frac {r^{6}+5 r^{5}-128 r^{4}-1015 r^{3}-1195 r^{2}+4769 r +7656}{\left (5+r \right ) r \left (1+r \right ) \left (2+r \right ) \left (r +3\right ) \left (4+r \right )}\) | \(\frac {10093}{720}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=\frac {-r^{7}-28 r^{6}-165 r^{5}+727 r^{4}+8669 r^{3}+18671 r^{2}-6652 r -30279}{\left (6+r \right ) \left (5+r \right ) r \left (1+r \right ) \left (2+r \right ) \left (r +3\right ) \left (4+r \right )} \] Which for the root \(r = 1\) becomes \[ a_{7}=-{\frac {647}{360}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-r -3}{r}\) | \(-4\) |
| \(a_{2}\) | \(\frac {7 r +12}{\left (1+r \right ) r}\) | \(\frac {19}{2}\) |
| \(a_{3}\) | \(\frac {r^{3}-2 r^{2}-40 r -57}{\left (2+r \right ) r \left (1+r \right )}\) | \(-{\frac {49}{3}}\) |
| \(a_{4}\) | \(\frac {-r^{4}-11 r^{3}+12 r^{2}+221 r +294}{r \left (1+r \right ) \left (2+r \right ) \left (r +3\right )}\) | \(\frac {515}{24}\) |
| \(a_{5}\) | \(\frac {14 r^{4}+108 r^{3}+10 r^{2}-1139 r -1545}{\left (4+r \right ) r \left (1+r \right ) \left (2+r \right ) \left (r +3\right )}\) | \(-{\frac {319}{15}}\) |
| \(a_{6}\) | \(\frac {r^{6}+5 r^{5}-128 r^{4}-1015 r^{3}-1195 r^{2}+4769 r +7656}{\left (5+r \right ) r \left (1+r \right ) \left (2+r \right ) \left (r +3\right ) \left (4+r \right )}\) | \(\frac {10093}{720}\) |
| \(a_{7}\) | \(\frac {-r^{7}-28 r^{6}-165 r^{5}+727 r^{4}+8669 r^{3}+18671 r^{2}-6652 r -30279}{\left (6+r \right ) \left (5+r \right ) r \left (1+r \right ) \left (2+r \right ) \left (r +3\right ) \left (4+r \right )}\) | \(-{\frac {647}{360}}\) |
Using the above table, then the first 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 (1-4 x +\frac {19 x^{2}}{2}-\frac {49 x^{3}}{3}+\frac {515 x^{4}}{24}-\frac {319 x^{5}}{15}+\frac {10093 x^{6}}{720}-\frac {647 x^{7}}{360}+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 \(r = 1\). The above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table
| \(n\) | \(b_{n ,r}\) | \(a_{n}\) | \(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) | \(b_{n}\left (r =1\right )\) |
| \(b_{0}\) | \(1\) | \(1\) | N/A since \(b_{n}\) starts from 1 | N/A |
| \(b_{1}\) | \(\frac {-r -3}{r}\) | \(-4\) | \(\frac {3}{r^{2}}\) | \(3\) |
| \(b_{2}\) | \(\frac {7 r +12}{\left (1+r \right ) r}\) | \(\frac {19}{2}\) | \(\frac {-7 r^{2}-24 r -12}{\left (1+r \right )^{2} r^{2}}\) | \(-{\frac {43}{4}}\) |
| \(b_{3}\) | \(\frac {r^{3}-2 r^{2}-40 r -57}{\left (2+r \right ) r \left (1+r \right )}\) | \(-{\frac {49}{3}}\) | \(\frac {5 r^{4}+84 r^{3}+287 r^{2}+342 r +114}{\left (2+r \right )^{2} r^{2} \left (1+r \right )^{2}}\) | \(\frac {208}{9}\) |
| \(b_{4}\) | \(\frac {-r^{4}-11 r^{3}+12 r^{2}+221 r +294}{r \left (1+r \right ) \left (2+r \right ) \left (r +3\right )}\) | \(\frac {515}{24}\) | \(\frac {5 r^{6}-46 r^{5}-874 r^{4}-3960 r^{3}-7651 r^{2}-6468 r -1764}{r^{2} \left (1+r \right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2}}\) | \(-{\frac {10379}{288}}\) |
| \(b_{5}\) | \(\frac {14 r^{4}+108 r^{3}+10 r^{2}-1139 r -1545}{\left (4+r \right ) r \left (1+r \right ) \left (2+r \right ) \left (r +3\right )}\) | \(-{\frac {319}{15}}\) | \(\frac {-14 r^{8}-216 r^{7}-620 r^{6}+5756 r^{5}+47953 r^{4}+146714 r^{3}+219415 r^{2}+154500 r +37080}{\left (4+r \right )^{2} r^{2} \left (1+r \right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2}}\) | \(\frac {76321}{1800}\) |
| \(b_{6}\) | \(\frac {r^{6}+5 r^{5}-128 r^{4}-1015 r^{3}-1195 r^{2}+4769 r +7656}{\left (5+r \right ) r \left (1+r \right ) \left (2+r \right ) \left (r +3\right ) \left (4+r \right )}\) | \(\frac {10093}{720}\) | \(\frac {10 r^{10}+426 r^{9}+6065 r^{8}+38576 r^{7}+92115 r^{6}-196670 r^{5}-1845610 r^{4}-4992690 r^{3}-6617906 r^{2}-4195488 r -918720}{\left (5+r \right )^{2} r^{2} \left (1+r \right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (4+r \right )^{2}}\) | \(-{\frac {172499}{4800}}\) |
| \(b_{7}\) | \(\frac {-r^{7}-28 r^{6}-165 r^{5}+727 r^{4}+8669 r^{3}+18671 r^{2}-6652 r -30279}{\left (6+r \right ) \left (5+r \right ) r \left (1+r \right ) \left (2+r \right ) \left (r +3\right ) \left (4+r \right )}\) | \(-{\frac {647}{360}}\) | \(\frac {7 r^{12}-20 r^{11}-5821 r^{10}-112866 r^{9}-1033238 r^{8}-5300410 r^{7}-15056909 r^{6}-16885160 r^{5}+27702517 r^{4}+123109316 r^{3}+172696536 r^{2}+106824312 r +21800880}{\left (6+r \right )^{2} \left (5+r \right )^{2} r^{2} \left (1+r \right )^{2} \left (2+r \right )^{2} \left (r +3\right )^{2} \left (4+r \right )^{2}}\) | \(\frac {39091}{2400}\) |
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 \\ &= x \left (1-4 x +\frac {19 x^{2}}{2}-\frac {49 x^{3}}{3}+\frac {515 x^{4}}{24}-\frac {319 x^{5}}{15}+\frac {10093 x^{6}}{720}-\frac {647 x^{7}}{360}+O\left (x^{8}\right )\right ) \ln \left (x \right )+x \left (3 x -\frac {43 x^{2}}{4}+\frac {208 x^{3}}{9}-\frac {10379 x^{4}}{288}+\frac {76321 x^{5}}{1800}-\frac {172499 x^{6}}{4800}+\frac {39091 x^{7}}{2400}+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 \left (1-4 x +\frac {19 x^{2}}{2}-\frac {49 x^{3}}{3}+\frac {515 x^{4}}{24}-\frac {319 x^{5}}{15}+\frac {10093 x^{6}}{720}-\frac {647 x^{7}}{360}+O\left (x^{8}\right )\right ) + c_{2} \left (x \left (1-4 x +\frac {19 x^{2}}{2}-\frac {49 x^{3}}{3}+\frac {515 x^{4}}{24}-\frac {319 x^{5}}{15}+\frac {10093 x^{6}}{720}-\frac {647 x^{7}}{360}+O\left (x^{8}\right )\right ) \ln \left (x \right )+x \left (3 x -\frac {43 x^{2}}{4}+\frac {208 x^{3}}{9}-\frac {10379 x^{4}}{288}+\frac {76321 x^{5}}{1800}-\frac {172499 x^{6}}{4800}+\frac {39091 x^{7}}{2400}+O\left (x^{8}\right )\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x \left (1-4 x +\frac {19 x^{2}}{2}-\frac {49 x^{3}}{3}+\frac {515 x^{4}}{24}-\frac {319 x^{5}}{15}+\frac {10093 x^{6}}{720}-\frac {647 x^{7}}{360}+O\left (x^{8}\right )\right )+c_{2} \left (x \left (1-4 x +\frac {19 x^{2}}{2}-\frac {49 x^{3}}{3}+\frac {515 x^{4}}{24}-\frac {319 x^{5}}{15}+\frac {10093 x^{6}}{720}-\frac {647 x^{7}}{360}+O\left (x^{8}\right )\right ) \ln \left (x \right )+x \left (3 x -\frac {43 x^{2}}{4}+\frac {208 x^{3}}{9}-\frac {10379 x^{4}}{288}+\frac {76321 x^{5}}{1800}-\frac {172499 x^{6}}{4800}+\frac {39091 x^{7}}{2400}+O\left (x^{8}\right )\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x \left (1-4 x +\frac {19 x^{2}}{2}-\frac {49 x^{3}}{3}+\frac {515 x^{4}}{24}-\frac {319 x^{5}}{15}+\frac {10093 x^{6}}{720}-\frac {647 x^{7}}{360}+O\left (x^{8}\right )\right )+c_{2} \left (x \left (1-4 x +\frac {19 x^{2}}{2}-\frac {49 x^{3}}{3}+\frac {515 x^{4}}{24}-\frac {319 x^{5}}{15}+\frac {10093 x^{6}}{720}-\frac {647 x^{7}}{360}+O\left (x^{8}\right )\right ) \ln \left (x \right )+x \left (3 x -\frac {43 x^{2}}{4}+\frac {208 x^{3}}{9}-\frac {10379 x^{4}}{288}+\frac {76321 x^{5}}{1800}-\frac {172499 x^{6}}{4800}+\frac {39091 x^{7}}{2400}+O\left (x^{8}\right )\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x \left (1-4 x +\frac {19 x^{2}}{2}-\frac {49 x^{3}}{3}+\frac {515 x^{4}}{24}-\frac {319 x^{5}}{15}+\frac {10093 x^{6}}{720}-\frac {647 x^{7}}{360}+O\left (x^{8}\right )\right )+c_{2} \left (x \left (1-4 x +\frac {19 x^{2}}{2}-\frac {49 x^{3}}{3}+\frac {515 x^{4}}{24}-\frac {319 x^{5}}{15}+\frac {10093 x^{6}}{720}-\frac {647 x^{7}}{360}+O\left (x^{8}\right )\right ) \ln \left (x \right )+x \left (3 x -\frac {43 x^{2}}{4}+\frac {208 x^{3}}{9}-\frac {10379 x^{4}}{288}+\frac {76321 x^{5}}{1800}-\frac {172499 x^{6}}{4800}+\frac {39091 x^{7}}{2400}+O\left (x^{8}\right )\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (x^{2}+x +1\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (2 x^{3}+4 x^{2}-x \right ) y^{\prime }+y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=-\frac {y}{x^{2} \left (x^{2}+x +1\right )}-\frac {\left (2 x^{2}+4 x -1\right ) y^{\prime }}{x \left (x^{2}+x +1\right )} \\ \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 (2 x^{2}+4 x -1\right ) y^{\prime }}{x \left (x^{2}+x +1\right )}+\frac {y}{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 {2 x^{2}+4 x -1}{x \left (x^{2}+x +1\right )}, P_{3}\left (x \right )=\frac {1}{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}}}=-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}}}=1 \\ {} & \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} \\ {} & {} & x^{2} \left (x^{2}+x +1\right ) \left (\frac {d}{d x}y^{\prime }\right )+x \left (2 x^{2}+4 x -1\right ) y^{\prime }+y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..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+r \right )^{2} x^{r}+\left (a_{1} r^{2}+a_{0} r \left (3+r \right )\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (k +r -1\right )^{2}+a_{k -1} \left (k +r -1\right ) \left (k +2+r \right )+a_{k -2} \left (k -2+r \right ) \left (k +r -1\right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (-1+r \right )^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =1 \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} r^{2}+a_{0} r \left (3+r \right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=-\frac {\left (3+r \right ) a_{0}}{r} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k +r -1\right ) \left (\left (a_{k}+a_{k -2}+a_{k -1}\right ) k +\left (a_{k}+a_{k -2}+a_{k -1}\right ) r -a_{k}-2 a_{k -2}+2 a_{k -1}\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \left (k +r +1\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 -a_{k +2}-2 a_{k}+2 a_{k +1}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {k a_{k}+k a_{k +1}+r a_{k}+r a_{k +1}+4 a_{k +1}}{k +r +1} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +2}=-\frac {k a_{k}+k a_{k +1}+a_{k}+5 a_{k +1}}{k +2} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1}, a_{k +2}=-\frac {k a_{k}+k a_{k +1}+a_{k}+5 a_{k +1}}{k +2}, a_{1}=-4 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 -> Kummer -> 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 for at least one hypergeometric solution is achieved - returning with no uncomputed integrals <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 75
Order:=8; dsolve(x^2*(1+x+x^2)*diff(y(x),x$2)-x*(1-4*x-2*x^2)*diff(y(x),x)+y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \left (\left (c_{2} \ln \left (x \right )+c_{1} \right ) \left (1-4 x +\frac {19}{2} x^{2}-\frac {49}{3} x^{3}+\frac {515}{24} x^{4}-\frac {319}{15} x^{5}+\frac {10093}{720} x^{6}-\frac {647}{360} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+\left (3 x -\frac {43}{4} x^{2}+\frac {208}{9} x^{3}-\frac {10379}{288} x^{4}+\frac {76321}{1800} x^{5}-\frac {172499}{4800} x^{6}+\frac {39091}{2400} x^{7}+\operatorname {O}\left (x^{8}\right )\right ) c_{2} \right ) x \]
✓ Solution by Mathematica
Time used: 0.013 (sec). Leaf size: 158
AsymptoticDSolveValue[x^2*(1+x+x^2)*y''[x]-x*(1-4*x-2*x^2)*y'[x]+y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_1 x \left (-\frac {647 x^7}{360}+\frac {10093 x^6}{720}-\frac {319 x^5}{15}+\frac {515 x^4}{24}-\frac {49 x^3}{3}+\frac {19 x^2}{2}-4 x+1\right )+c_2 \left (x \left (\frac {39091 x^7}{2400}-\frac {172499 x^6}{4800}+\frac {76321 x^5}{1800}-\frac {10379 x^4}{288}+\frac {208 x^3}{9}-\frac {43 x^2}{4}+3 x\right )+x \left (-\frac {647 x^7}{360}+\frac {10093 x^6}{720}-\frac {319 x^5}{15}+\frac {515 x^4}{24}-\frac {49 x^3}{3}+\frac {19 x^2}{2}-4 x+1\right ) \log (x)\right ) \]