Internal problem ID [1355]
Internal file name [OUTPUT/1356_Sunday_June_05_2022_02_12_41_AM_68323968/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: 3.
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], [_2nd_order, _linear, `_with_symmetry_[0,F(x)]`]]
\[ \boxed {x^{2} \left (x^{2}+2 x +1\right ) y^{\prime \prime }+x \left (4 x^{2}+3 x +1\right ) y^{\prime }-x \left (1-2 x \right ) y=0} \] With the expansion point for the power series method at \(x = 0\).
The ODE is \[ y^{\prime \prime } x^{2} \left (x +1\right )^{2}+\left (4 x^{3}+3 x^{2}+x \right ) y^{\prime }+\left (2 x^{2}-x \right ) y = 0 \] Or \[ x \left (y^{\prime \prime } x^{3}+2 x^{2} y^{\prime \prime }+4 y^{\prime } x^{2}+x y^{\prime \prime }+3 x y^{\prime }+2 y x +y^{\prime }-y\right ) = 0 \] For \(x \neq 0\) the above simplifies to \[ \left (2 x -1\right ) y+\left (4 x^{2}+3 x +1\right ) y^{\prime }+x y^{\prime \prime } \left (x +1\right )^{2} = 0 \] The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ \left (x^{4}+2 x^{3}+x^{2}\right ) y^{\prime \prime }+\left (4 x^{3}+3 x^{2}+x \right ) y^{\prime }+\left (2 x^{2}-x \right ) y = 0 \] 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 {4 x^{2}+3 x +1}{x \left (x +1\right )^{2}}\\ q(x) &= \frac {2 x -1}{x \left (x +1\right )^{2}}\\ \end {align*}
| \(p(x)=\frac {4 x^{2}+3 x +1}{x \left (x +1\right )^{2}}\) | |
| singularity | type |
| \(x = -1\) | \(\text {``irregular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
| \(q(x)=\frac {2 x -1}{x \left (x +1\right )^{2}}\) | |
| singularity | type |
| \(x = -1\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, \infty ]\)
Irregular singular points : \([-1]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} \left (x^{2}+2 x +1\right ) y^{\prime \prime }+\left (4 x^{3}+3 x^{2}+x \right ) y^{\prime }+\left (2 x^{2}-x \right ) y = 0 \] 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} x^{2} \left (x^{2}+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 (4 x^{3}+3 x^{2}+x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (2 x^{2}-x \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 }}2 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 }}4 x^{n +r +2} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r +2} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{1+n +r} a_{n}\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 }}2 x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r +2} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}4 a_{n -2} \left (n +r -2\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}3 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}3 a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r +2} a_{n} &= \moverset {\infty }{\munderset {n =2}{\sum }}2 a_{n -2} x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{1+n +r} a_{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} x^{n +r}\right ) \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =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 }}2 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 }}4 a_{n -2} \left (n +r -2\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}3 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}2 a_{n -2} x^{n +r}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+x^{n +r} a_{n} \left (n +r \right ) = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )+x^{r} a_{0} r = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )+x^{r} r \right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ x^{r} r^{2} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 0\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ x^{r} r^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([0, 0]\).
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. 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 \[ a_{1} = \frac {-2 r +1}{1+r} \] 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 )+2 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+a_{n} \left (n +r \right ) \left (n +r -1\right )+4 a_{n -2} \left (n +r -2\right )+3 a_{n -1} \left (n +r -1\right )+a_{n} \left (n +r \right )+2 a_{n -2}-a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {n a_{n -2}+2 n a_{n -1}+r a_{n -2}+2 r a_{n -1}-a_{n -2}-3 a_{n -1}}{n +r}\tag {4} \] Which for the root \(r = 0\) becomes \[ a_{n} = \frac {\left (-a_{n -2}-2 a_{n -1}\right ) n +a_{n -2}+3 a_{n -1}}{n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 0\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2 r +1}{1+r}\) | \(1\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {3 r^{2}-2 r -2}{\left (2+r \right ) \left (1+r \right )} \] Which for the root \(r = 0\) becomes \[ a_{2}=-1 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2 r +1}{1+r}\) | \(1\) |
| \(a_{2}\) | \(\frac {3 r^{2}-2 r -2}{\left (2+r \right ) \left (1+r \right )}\) | \(-1\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-4 r^{3}+2 r^{2}+14 r +2}{\left (1+r \right ) \left (2+r \right ) \left (3+r \right )} \] Which for the root \(r = 0\) becomes \[ a_{3}={\frac {1}{3}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2 r +1}{1+r}\) | \(1\) |
| \(a_{2}\) | \(\frac {3 r^{2}-2 r -2}{\left (2+r \right ) \left (1+r \right )}\) | \(-1\) |
| \(a_{3}\) | \(\frac {-4 r^{3}+2 r^{2}+14 r +2}{\left (1+r \right ) \left (2+r \right ) \left (3+r \right )}\) | \(\frac {1}{3}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {5 r^{4}-51 r^{2}-44 r +8}{\left (4+r \right ) \left (1+r \right ) \left (2+r \right ) \left (3+r \right )} \] Which for the root \(r = 0\) becomes \[ a_{4}={\frac {1}{3}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2 r +1}{1+r}\) | \(1\) |
| \(a_{2}\) | \(\frac {3 r^{2}-2 r -2}{\left (2+r \right ) \left (1+r \right )}\) | \(-1\) |
| \(a_{3}\) | \(\frac {-4 r^{3}+2 r^{2}+14 r +2}{\left (1+r \right ) \left (2+r \right ) \left (3+r \right )}\) | \(\frac {1}{3}\) |
| \(a_{4}\) | \(\frac {5 r^{4}-51 r^{2}-44 r +8}{\left (4+r \right ) \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}\) | \(\frac {1}{3}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-6 r^{5}-5 r^{4}+136 r^{3}+299 r^{2}+52 r -88}{\left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right )} \] Which for the root \(r = 0\) becomes \[ a_{5}=-{\frac {11}{15}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2 r +1}{1+r}\) | \(1\) |
| \(a_{2}\) | \(\frac {3 r^{2}-2 r -2}{\left (2+r \right ) \left (1+r \right )}\) | \(-1\) |
| \(a_{3}\) | \(\frac {-4 r^{3}+2 r^{2}+14 r +2}{\left (1+r \right ) \left (2+r \right ) \left (3+r \right )}\) | \(\frac {1}{3}\) |
| \(a_{4}\) | \(\frac {5 r^{4}-51 r^{2}-44 r +8}{\left (4+r \right ) \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}\) | \(\frac {1}{3}\) |
| \(a_{5}\) | \(\frac {-6 r^{5}-5 r^{4}+136 r^{3}+299 r^{2}+52 r -88}{\left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right )}\) | \(-{\frac {11}{15}}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {7 r^{6}+14 r^{5}-301 r^{4}-1268 r^{3}-1088 r^{2}+728 r +592}{\left (6+r \right ) \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right )} \] Which for the root \(r = 0\) becomes \[ a_{6}={\frac {37}{45}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2 r +1}{1+r}\) | \(1\) |
| \(a_{2}\) | \(\frac {3 r^{2}-2 r -2}{\left (2+r \right ) \left (1+r \right )}\) | \(-1\) |
| \(a_{3}\) | \(\frac {-4 r^{3}+2 r^{2}+14 r +2}{\left (1+r \right ) \left (2+r \right ) \left (3+r \right )}\) | \(\frac {1}{3}\) |
| \(a_{4}\) | \(\frac {5 r^{4}-51 r^{2}-44 r +8}{\left (4+r \right ) \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}\) | \(\frac {1}{3}\) |
| \(a_{5}\) | \(\frac {-6 r^{5}-5 r^{4}+136 r^{3}+299 r^{2}+52 r -88}{\left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right )}\) | \(-{\frac {11}{15}}\) |
| \(a_{6}\) | \(\frac {7 r^{6}+14 r^{5}-301 r^{4}-1268 r^{3}-1088 r^{2}+728 r +592}{\left (6+r \right ) \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right )}\) | \(\frac {37}{45}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=\frac {-8 r^{7}-28 r^{6}+588 r^{5}+4096 r^{4}+7588 r^{3}-788 r^{2}-10008 r -3344}{\left (7+r \right ) \left (6+r \right ) \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right )} \] Which for the root \(r = 0\) becomes \[ a_{7}=-{\frac {209}{315}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-2 r +1}{1+r}\) | \(1\) |
| \(a_{2}\) | \(\frac {3 r^{2}-2 r -2}{\left (2+r \right ) \left (1+r \right )}\) | \(-1\) |
| \(a_{3}\) | \(\frac {-4 r^{3}+2 r^{2}+14 r +2}{\left (1+r \right ) \left (2+r \right ) \left (3+r \right )}\) | \(\frac {1}{3}\) |
| \(a_{4}\) | \(\frac {5 r^{4}-51 r^{2}-44 r +8}{\left (4+r \right ) \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}\) | \(\frac {1}{3}\) |
| \(a_{5}\) | \(\frac {-6 r^{5}-5 r^{4}+136 r^{3}+299 r^{2}+52 r -88}{\left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right )}\) | \(-{\frac {11}{15}}\) |
| \(a_{6}\) | \(\frac {7 r^{6}+14 r^{5}-301 r^{4}-1268 r^{3}-1088 r^{2}+728 r +592}{\left (6+r \right ) \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right )}\) | \(\frac {37}{45}\) |
| \(a_{7}\) | \(\frac {-8 r^{7}-28 r^{6}+588 r^{5}+4096 r^{4}+7588 r^{3}-788 r^{2}-10008 r -3344}{\left (7+r \right ) \left (6+r \right ) \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right )}\) | \(-{\frac {209}{315}}\) |
Using the above table, then the first solution \(y_{1}\left (x \right )\) becomes \begin{align*} y_{1}\left (x \right )&= 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 \\ &= -x^{2}+x +1+\frac {x^{3}}{3}+\frac {x^{4}}{3}-\frac {11 x^{5}}{15}+\frac {37 x^{6}}{45}-\frac {209 x^{7}}{315}+O\left (x^{8}\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 \(r = 0\). The above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table
| \(n\) | \(b_{n ,r}\) | \(a_{n}\) | \(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) | \(b_{n}\left (r =0\right )\) |
| \(b_{0}\) | \(1\) | \(1\) | N/A since \(b_{n}\) starts from 1 | N/A |
| \(b_{1}\) | \(\frac {-2 r +1}{1+r}\) | \(1\) | \(-\frac {3}{\left (1+r \right )^{2}}\) | \(-3\) |
| \(b_{2}\) | \(\frac {3 r^{2}-2 r -2}{\left (2+r \right ) \left (1+r \right )}\) | \(-1\) | \(\frac {11 r^{2}+16 r +2}{\left (2+r \right )^{2} \left (1+r \right )^{2}}\) | \(\frac {1}{2}\) |
| \(b_{3}\) | \(\frac {-4 r^{3}+2 r^{2}+14 r +2}{\left (1+r \right ) \left (2+r \right ) \left (3+r \right )}\) | \(\frac {1}{3}\) | \(\frac {-26 r^{4}-116 r^{3}-140 r^{2}+62}{\left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2}}\) | \(\frac {31}{18}\) |
| \(b_{4}\) | \(\frac {5 r^{4}-51 r^{2}-44 r +8}{\left (4+r \right ) \left (1+r \right ) \left (2+r \right ) \left (3+r \right )}\) | \(\frac {1}{3}\) | \(\frac {50 r^{6}+452 r^{5}+1392 r^{4}+1328 r^{3}-1250 r^{2}-3008 r -1456}{\left (4+r \right )^{2} \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2}}\) | \(-{\frac {91}{36}}\) |
| \(b_{5}\) | \(\frac {-6 r^{5}-5 r^{4}+136 r^{3}+299 r^{2}+52 r -88}{\left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right )}\) | \(-{\frac {11}{15}}\) | \(\frac {-85 r^{8}-1292 r^{7}-7412 r^{6}-18004 r^{5}-4425 r^{4}+68568 r^{3}+141626 r^{2}+111360 r +30352}{\left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2}}\) | \(\frac {1897}{900}\) |
| \(b_{6}\) | \(\frac {7 r^{6}+14 r^{5}-301 r^{4}-1268 r^{3}-1088 r^{2}+728 r +592}{\left (6+r \right ) \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right )}\) | \(\frac {37}{45}\) | \(\frac {133 r^{10}+3052 r^{9}+28010 r^{8}+123660 r^{7}+195517 r^{6}-532528 r^{5}-3246404 r^{4}-6824944 r^{3}-7145744 r^{2}-3489536 r -520128}{\left (6+r \right )^{2} \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2}}\) | \(-{\frac {301}{300}}\) |
| \(b_{7}\) | \(\frac {-8 r^{7}-28 r^{6}+588 r^{5}+4096 r^{4}+7588 r^{3}-788 r^{2}-10008 r -3344}{\left (7+r \right ) \left (6+r \right ) \left (1+r \right ) \left (2+r \right ) \left (3+r \right ) \left (4+r \right ) \left (5+r \right )}\) | \(-{\frac {209}{315}}\) | \(-\frac {4 \left (49 r^{12}+1582 r^{11}+21202 r^{10}+146524 r^{9}+473440 r^{8}-280982 r^{7}-9023042 r^{6}-38501904 r^{5}-86151913 r^{4}-110650148 r^{3}-75941160 r^{2}-19970944 r +1685232\right )}{\left (7+r \right )^{2} \left (6+r \right )^{2} \left (1+r \right )^{2} \left (2+r \right )^{2} \left (3+r \right )^{2} \left (4+r \right )^{2} \left (5+r \right )^{2}}\) | \(-{\frac {3901}{14700}}\) |
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 \\ &= \left (-x^{2}+x +1+\frac {x^{3}}{3}+\frac {x^{4}}{3}-\frac {11 x^{5}}{15}+\frac {37 x^{6}}{45}-\frac {209 x^{7}}{315}+O\left (x^{8}\right )\right ) \ln \left (x \right )-3 x +\frac {x^{2}}{2}+\frac {31 x^{3}}{18}-\frac {91 x^{4}}{36}+\frac {1897 x^{5}}{900}-\frac {301 x^{6}}{300}-\frac {3901 x^{7}}{14700}+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} \left (-x^{2}+x +1+\frac {x^{3}}{3}+\frac {x^{4}}{3}-\frac {11 x^{5}}{15}+\frac {37 x^{6}}{45}-\frac {209 x^{7}}{315}+O\left (x^{8}\right )\right ) + c_{2} \left (\left (-x^{2}+x +1+\frac {x^{3}}{3}+\frac {x^{4}}{3}-\frac {11 x^{5}}{15}+\frac {37 x^{6}}{45}-\frac {209 x^{7}}{315}+O\left (x^{8}\right )\right ) \ln \left (x \right )-3 x +\frac {x^{2}}{2}+\frac {31 x^{3}}{18}-\frac {91 x^{4}}{36}+\frac {1897 x^{5}}{900}-\frac {301 x^{6}}{300}-\frac {3901 x^{7}}{14700}+O\left (x^{8}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \left (-x^{2}+x +1+\frac {x^{3}}{3}+\frac {x^{4}}{3}-\frac {11 x^{5}}{15}+\frac {37 x^{6}}{45}-\frac {209 x^{7}}{315}+O\left (x^{8}\right )\right )+c_{2} \left (\left (-x^{2}+x +1+\frac {x^{3}}{3}+\frac {x^{4}}{3}-\frac {11 x^{5}}{15}+\frac {37 x^{6}}{45}-\frac {209 x^{7}}{315}+O\left (x^{8}\right )\right ) \ln \left (x \right )-3 x +\frac {x^{2}}{2}+\frac {31 x^{3}}{18}-\frac {91 x^{4}}{36}+\frac {1897 x^{5}}{900}-\frac {301 x^{6}}{300}-\frac {3901 x^{7}}{14700}+O\left (x^{8}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (-x^{2}+x +1+\frac {x^{3}}{3}+\frac {x^{4}}{3}-\frac {11 x^{5}}{15}+\frac {37 x^{6}}{45}-\frac {209 x^{7}}{315}+O\left (x^{8}\right )\right )+c_{2} \left (\left (-x^{2}+x +1+\frac {x^{3}}{3}+\frac {x^{4}}{3}-\frac {11 x^{5}}{15}+\frac {37 x^{6}}{45}-\frac {209 x^{7}}{315}+O\left (x^{8}\right )\right ) \ln \left (x \right )-3 x +\frac {x^{2}}{2}+\frac {31 x^{3}}{18}-\frac {91 x^{4}}{36}+\frac {1897 x^{5}}{900}-\frac {301 x^{6}}{300}-\frac {3901 x^{7}}{14700}+O\left (x^{8}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (-x^{2}+x +1+\frac {x^{3}}{3}+\frac {x^{4}}{3}-\frac {11 x^{5}}{15}+\frac {37 x^{6}}{45}-\frac {209 x^{7}}{315}+O\left (x^{8}\right )\right )+c_{2} \left (\left (-x^{2}+x +1+\frac {x^{3}}{3}+\frac {x^{4}}{3}-\frac {11 x^{5}}{15}+\frac {37 x^{6}}{45}-\frac {209 x^{7}}{315}+O\left (x^{8}\right )\right ) \ln \left (x \right )-3 x +\frac {x^{2}}{2}+\frac {31 x^{3}}{18}-\frac {91 x^{4}}{36}+\frac {1897 x^{5}}{900}-\frac {301 x^{6}}{300}-\frac {3901 x^{7}}{14700}+O\left (x^{8}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (x^{2}+2 x +1\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (4 x^{3}+3 x^{2}+x \right ) y^{\prime }+\left (2 x^{2}-x \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 (2 x -1\right ) y}{x \left (x^{2}+2 x +1\right )}-\frac {\left (4 x^{2}+3 x +1\right ) y^{\prime }}{x \left (x^{2}+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 (4 x^{2}+3 x +1\right ) y^{\prime }}{x \left (x^{2}+2 x +1\right )}+\frac {\left (2 x -1\right ) y}{x \left (x^{2}+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 {4 x^{2}+3 x +1}{x \left (x^{2}+2 x +1\right )}, P_{3}\left (x \right )=\frac {2 x -1}{x \left (x^{2}+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}}}=0 \\ {} & \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} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) \left (x^{2}+2 x +1\right ) x +\left (4 x^{2}+3 x +1\right ) y^{\prime }+\left (2 x -1\right ) y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..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^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & 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} r^{2} x^{-1+r}+\left (a_{1} \left (1+r \right )^{2}+a_{0} \left (1+r \right ) \left (-1+2 r \right )\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +r +1\right )^{2}+a_{k} \left (k +r +1\right ) \left (2 k +2 r -1\right )+a_{k -1} \left (k +r +1\right ) \left (k +r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r^{2}=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r =0 \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right )^{2}+a_{0} \left (1+r \right ) \left (-1+2 r \right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (k +1\right ) \left (2 k a_{k}+a_{k -1} k +k a_{k +1}-a_{k}+a_{k +1}\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (k +2\right ) \left (2 a_{k +1} \left (k +1\right )+a_{k} \left (k +1\right )+\left (k +1\right ) a_{k +2}-a_{k +1}+a_{k +2}\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {k a_{k}+2 k a_{k +1}+a_{k}+a_{k +1}}{k +2} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {k a_{k}+2 k a_{k +1}+a_{k}+a_{k +1}}{k +2} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +2}=-\frac {k a_{k}+2 k a_{k +1}+a_{k}+a_{k +1}}{k +2}, a_{1}-a_{0}=0\right ] \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] <- linear_1 successful`
✓ Solution by Maple
Time used: 0.032 (sec). Leaf size: 71
Order:=8; dsolve(x^2*(1+2*x+x^2)*diff(y(x),x$2)+x*(1+3*x+4*x^2)*diff(y(x),x)-x*(1-2*x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \left (c_{2} \ln \left (x \right )+c_{1} \right ) \left (1+x -x^{2}+\frac {1}{3} x^{3}+\frac {1}{3} x^{4}-\frac {11}{15} x^{5}+\frac {37}{45} x^{6}-\frac {209}{315} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+\left (\left (-3\right ) x +\frac {1}{2} x^{2}+\frac {31}{18} x^{3}-\frac {91}{36} x^{4}+\frac {1897}{900} x^{5}-\frac {301}{300} x^{6}-\frac {3901}{14700} x^{7}+\operatorname {O}\left (x^{8}\right )\right ) c_{2} \]
✓ Solution by Mathematica
Time used: 0.013 (sec). Leaf size: 145
AsymptoticDSolveValue[x^2*(1+2*x+x^2)*y''[x]+x*(1+3*x+4*x^2)*y'[x]-x*(1-2*x)*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_1 \left (-\frac {209 x^7}{315}+\frac {37 x^6}{45}-\frac {11 x^5}{15}+\frac {x^4}{3}+\frac {x^3}{3}-x^2+x+1\right )+c_2 \left (-\frac {3901 x^7}{14700}-\frac {301 x^6}{300}+\frac {1897 x^5}{900}-\frac {91 x^4}{36}+\frac {31 x^3}{18}+\frac {x^2}{2}+\left (-\frac {209 x^7}{315}+\frac {37 x^6}{45}-\frac {11 x^5}{15}+\frac {x^4}{3}+\frac {x^3}{3}-x^2+x+1\right ) \log (x)-3 x\right ) \]