Internal problem ID [1377]
Internal file name [OUTPUT/1378_Sunday_June_05_2022_02_13_58_AM_79581135/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: 25.
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]]
\[ \boxed {x^{2} \left (1+4 x \right ) y^{\prime \prime }-x \left (1-4 x \right ) y^{\prime }+\left (x +1\right ) y=0} \] 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. \[ \left (4 x^{3}+x^{2}\right ) y^{\prime \prime }+\left (4 x^{2}-x \right ) y^{\prime }+\left (x +1\right ) y = 0 \] 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 {4 x -1}{x \left (1+4 x \right )}\\ q(x) &= \frac {x +1}{x^{2} \left (1+4 x \right )}\\ \end {align*}
| \(p(x)=\frac {4 x -1}{x \left (1+4 x \right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -{\frac {1}{4}}\) | \(\text {``regular''}\) |
| \(q(x)=\frac {x +1}{x^{2} \left (1+4 x \right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = -{\frac {1}{4}}\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \(\left [0, -{\frac {1}{4}}, \infty \right ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} \left (1+4 x \right ) y^{\prime \prime }+\left (4 x^{2}-x \right ) y^{\prime }+\left (x +1\right ) y = 0 \] 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} x^{2} \left (1+4 x \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^{2}-x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (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^{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^{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 }}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^{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 }}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} \\ \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 =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 }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\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 =1}{\sum }}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 \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )-x^{n +r} a_{n} \left (n +r \right )+a_{n} x^{n +r} = 0 \] 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 \(a_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} 4 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 -1} \left (n +r -1\right )-a_{n} \left (n +r \right )+a_{n -1}+a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -1} \left (4 n^{2}+8 n r +4 r^{2}-8 n -8 r +5\right )}{n^{2}+2 n r +r^{2}-2 n -2 r +1}\tag {4} \] Which for the root \(r = 1\) becomes \[ a_{n} = \frac {a_{n -1} \left (-4 n^{2}-1\right )}{n^{2}}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 1\) 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 \[ a_{1}=\frac {-4 r^{2}-1}{r^{2}} \] Which for the root \(r = 1\) becomes \[ a_{1}=-5 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-4 r^{2}-1}{r^{2}}\) | \(-5\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {16 r^{4}+32 r^{3}+24 r^{2}+8 r +5}{r^{2} \left (r +1\right )^{2}} \] Which for the root \(r = 1\) becomes \[ a_{2}={\frac {85}{4}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-4 r^{2}-1}{r^{2}}\) | \(-5\) |
| \(a_{2}\) | \(\frac {16 r^{4}+32 r^{3}+24 r^{2}+8 r +5}{r^{2} \left (r +1\right )^{2}}\) | \(\frac {85}{4}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-64 r^{6}-384 r^{5}-880 r^{4}-960 r^{3}-556 r^{2}-216 r -85}{r^{2} \left (r +1\right )^{2} \left (r +2\right )^{2}} \] Which for the root \(r = 1\) becomes \[ a_{3}=-{\frac {3145}{36}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-4 r^{2}-1}{r^{2}}\) | \(-5\) |
| \(a_{2}\) | \(\frac {16 r^{4}+32 r^{3}+24 r^{2}+8 r +5}{r^{2} \left (r +1\right )^{2}}\) | \(\frac {85}{4}\) |
| \(a_{3}\) | \(\frac {-64 r^{6}-384 r^{5}-880 r^{4}-960 r^{3}-556 r^{2}-216 r -85}{r^{2} \left (r +1\right )^{2} \left (r +2\right )^{2}}\) | \(-{\frac {3145}{36}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {\left (4 r^{2}+24 r +37\right ) \left (4 r^{2}+16 r +17\right ) \left (4 r^{2}+1\right ) \left (4 r^{2}+8 r +5\right )}{r^{2} \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}} \] Which for the root \(r = 1\) becomes \[ a_{4}={\frac {204425}{576}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-4 r^{2}-1}{r^{2}}\) | \(-5\) |
| \(a_{2}\) | \(\frac {16 r^{4}+32 r^{3}+24 r^{2}+8 r +5}{r^{2} \left (r +1\right )^{2}}\) | \(\frac {85}{4}\) |
| \(a_{3}\) | \(\frac {-64 r^{6}-384 r^{5}-880 r^{4}-960 r^{3}-556 r^{2}-216 r -85}{r^{2} \left (r +1\right )^{2} \left (r +2\right )^{2}}\) | \(-{\frac {3145}{36}}\) |
| \(a_{4}\) | \(\frac {\left (4 r^{2}+24 r +37\right ) \left (4 r^{2}+16 r +17\right ) \left (4 r^{2}+1\right ) \left (4 r^{2}+8 r +5\right )}{r^{2} \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\) | \(\frac {204425}{576}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {\left (4 r^{2}+24 r +37\right ) \left (4 r^{2}+16 r +17\right ) \left (4 r^{2}+1\right ) \left (4 r^{2}+8 r +5\right ) \left (4 r^{2}+32 r +65\right )}{r^{2} \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}} \] Which for the root \(r = 1\) becomes \[ a_{5}=-{\frac {825877}{576}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-4 r^{2}-1}{r^{2}}\) | \(-5\) |
| \(a_{2}\) | \(\frac {16 r^{4}+32 r^{3}+24 r^{2}+8 r +5}{r^{2} \left (r +1\right )^{2}}\) | \(\frac {85}{4}\) |
| \(a_{3}\) | \(\frac {-64 r^{6}-384 r^{5}-880 r^{4}-960 r^{3}-556 r^{2}-216 r -85}{r^{2} \left (r +1\right )^{2} \left (r +2\right )^{2}}\) | \(-{\frac {3145}{36}}\) |
| \(a_{4}\) | \(\frac {\left (4 r^{2}+24 r +37\right ) \left (4 r^{2}+16 r +17\right ) \left (4 r^{2}+1\right ) \left (4 r^{2}+8 r +5\right )}{r^{2} \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\) | \(\frac {204425}{576}\) |
| \(a_{5}\) | \(-\frac {\left (4 r^{2}+24 r +37\right ) \left (4 r^{2}+16 r +17\right ) \left (4 r^{2}+1\right ) \left (4 r^{2}+8 r +5\right ) \left (4 r^{2}+32 r +65\right )}{r^{2} \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(-{\frac {825877}{576}}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {\left (4 r^{2}+24 r +37\right ) \left (4 r^{2}+16 r +17\right ) \left (4 r^{2}+1\right ) \left (4 r^{2}+8 r +5\right ) \left (4 r^{2}+32 r +65\right ) \left (4 r^{2}+40 r +101\right )}{r^{2} \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}} \] Which for the root \(r = 1\) becomes \[ a_{6}={\frac {119752165}{20736}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-4 r^{2}-1}{r^{2}}\) | \(-5\) |
| \(a_{2}\) | \(\frac {16 r^{4}+32 r^{3}+24 r^{2}+8 r +5}{r^{2} \left (r +1\right )^{2}}\) | \(\frac {85}{4}\) |
| \(a_{3}\) | \(\frac {-64 r^{6}-384 r^{5}-880 r^{4}-960 r^{3}-556 r^{2}-216 r -85}{r^{2} \left (r +1\right )^{2} \left (r +2\right )^{2}}\) | \(-{\frac {3145}{36}}\) |
| \(a_{4}\) | \(\frac {\left (4 r^{2}+24 r +37\right ) \left (4 r^{2}+16 r +17\right ) \left (4 r^{2}+1\right ) \left (4 r^{2}+8 r +5\right )}{r^{2} \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\) | \(\frac {204425}{576}\) |
| \(a_{5}\) | \(-\frac {\left (4 r^{2}+24 r +37\right ) \left (4 r^{2}+16 r +17\right ) \left (4 r^{2}+1\right ) \left (4 r^{2}+8 r +5\right ) \left (4 r^{2}+32 r +65\right )}{r^{2} \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(-{\frac {825877}{576}}\) |
| \(a_{6}\) | \(\frac {\left (4 r^{2}+24 r +37\right ) \left (4 r^{2}+16 r +17\right ) \left (4 r^{2}+1\right ) \left (4 r^{2}+8 r +5\right ) \left (4 r^{2}+32 r +65\right ) \left (4 r^{2}+40 r +101\right )}{r^{2} \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\) | \(\frac {119752165}{20736}\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=-\frac {\left (4 r^{2}+24 r +37\right ) \left (4 r^{2}+16 r +17\right ) \left (4 r^{2}+1\right ) \left (4 r^{2}+8 r +5\right ) \left (4 r^{2}+32 r +65\right ) \left (4 r^{2}+40 r +101\right ) \left (4 r^{2}+48 r +145\right )}{r^{2} \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2} \left (r +6\right )^{2}} \] Which for the root \(r = 1\) becomes \[ a_{7}=-{\frac {23591176505}{1016064}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {-4 r^{2}-1}{r^{2}}\) | \(-5\) |
| \(a_{2}\) | \(\frac {16 r^{4}+32 r^{3}+24 r^{2}+8 r +5}{r^{2} \left (r +1\right )^{2}}\) | \(\frac {85}{4}\) |
| \(a_{3}\) | \(\frac {-64 r^{6}-384 r^{5}-880 r^{4}-960 r^{3}-556 r^{2}-216 r -85}{r^{2} \left (r +1\right )^{2} \left (r +2\right )^{2}}\) | \(-{\frac {3145}{36}}\) |
| \(a_{4}\) | \(\frac {\left (4 r^{2}+24 r +37\right ) \left (4 r^{2}+16 r +17\right ) \left (4 r^{2}+1\right ) \left (4 r^{2}+8 r +5\right )}{r^{2} \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\) | \(\frac {204425}{576}\) |
| \(a_{5}\) | \(-\frac {\left (4 r^{2}+24 r +37\right ) \left (4 r^{2}+16 r +17\right ) \left (4 r^{2}+1\right ) \left (4 r^{2}+8 r +5\right ) \left (4 r^{2}+32 r +65\right )}{r^{2} \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(-{\frac {825877}{576}}\) |
| \(a_{6}\) | \(\frac {\left (4 r^{2}+24 r +37\right ) \left (4 r^{2}+16 r +17\right ) \left (4 r^{2}+1\right ) \left (4 r^{2}+8 r +5\right ) \left (4 r^{2}+32 r +65\right ) \left (4 r^{2}+40 r +101\right )}{r^{2} \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\) | \(\frac {119752165}{20736}\) |
| \(a_{7}\) | \(-\frac {\left (4 r^{2}+24 r +37\right ) \left (4 r^{2}+16 r +17\right ) \left (4 r^{2}+1\right ) \left (4 r^{2}+8 r +5\right ) \left (4 r^{2}+32 r +65\right ) \left (4 r^{2}+40 r +101\right ) \left (4 r^{2}+48 r +145\right )}{r^{2} \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2} \left (r +6\right )^{2}}\) | \(-{\frac {23591176505}{1016064}}\) |
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 (1-5 x +\frac {85 x^{2}}{4}-\frac {3145 x^{3}}{36}+\frac {204425 x^{4}}{576}-\frac {825877 x^{5}}{576}+\frac {119752165 x^{6}}{20736}-\frac {23591176505 x^{7}}{1016064}+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 |
| \(b_{1}\) | \(\frac {-4 r^{2}-1}{r^{2}}\) | \(-5\) | \(\frac {2}{r^{3}}\) | \(2\) |
| \(b_{2}\) | \(\frac {16 r^{4}+32 r^{3}+24 r^{2}+8 r +5}{r^{2} \left (r +1\right )^{2}}\) | \(\frac {85}{4}\) | \(\frac {-16 r^{3}-24 r^{2}-28 r -10}{r^{3} \left (r +1\right )^{3}}\) | \(-{\frac {39}{4}}\) |
| \(b_{3}\) | \(\frac {-64 r^{6}-384 r^{5}-880 r^{4}-960 r^{3}-556 r^{2}-216 r -85}{r^{2} \left (r +1\right )^{2} \left (r +2\right )^{2}}\) | \(-{\frac {3145}{36}}\) | \(\frac {96 r^{6}+576 r^{5}+1584 r^{4}+2496 r^{3}+2454 r^{2}+1452 r +340}{r^{3} \left (r +1\right )^{3} \left (r +2\right )^{3}}\) | \(\frac {4499}{108}\) |
| \(b_{4}\) | \(\frac {\left (4 r^{2}+24 r +37\right ) \left (4 r^{2}+16 r +17\right ) \left (4 r^{2}+1\right ) \left (4 r^{2}+8 r +5\right )}{r^{2} \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2}}\) | \(\frac {204425}{576}\) | \(\frac {-512 r^{9}-6912 r^{8}-41856 r^{7}-149184 r^{6}-346656 r^{5}-549648 r^{4}-601864 r^{3}-444276 r^{2}-198572 r -37740}{r^{3} \left (r +1\right )^{3} \left (r +2\right )^{3} \left (r +3\right )^{3}}\) | \(-{\frac {594305}{3456}}\) |
| \(b_{5}\) | \(-\frac {\left (4 r^{2}+24 r +37\right ) \left (4 r^{2}+16 r +17\right ) \left (4 r^{2}+1\right ) \left (4 r^{2}+8 r +5\right ) \left (4 r^{2}+32 r +65\right )}{r^{2} \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2}}\) | \(-{\frac {825877}{576}}\) | \(\frac {2560 r^{12}+61440 r^{11}+670720 r^{10}+4403200 r^{9}+19377600 r^{8}+60349440 r^{7}+136874080 r^{6}+229000320 r^{5}+282485290 r^{4}+252483920 r^{3}+155837250 r^{2}+58950280 r +9812400}{r^{3} \left (r +1\right )^{3} \left (r +2\right )^{3} \left (r +3\right )^{3} \left (r +4\right )^{3}}\) | \(\frac {2420617}{3456}\) |
| \(b_{6}\) | \(\frac {\left (4 r^{2}+24 r +37\right ) \left (4 r^{2}+16 r +17\right ) \left (4 r^{2}+1\right ) \left (4 r^{2}+8 r +5\right ) \left (4 r^{2}+32 r +65\right ) \left (4 r^{2}+40 r +101\right )}{r^{2} \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2}}\) | \(\frac {119752165}{20736}\) | \(\frac {-12288 r^{15}-460800 r^{14}-7971840 r^{13}-84364800 r^{12}-610715136 r^{11}-3203946240 r^{10}-12592584320 r^{9}-37805227200 r^{8}-87591441264 r^{7}-157105182120 r^{6}-217275725940 r^{5}-228695691750 r^{4}-178174672512 r^{3}-97087162590 r^{2}-32733236200 r -4955262000}{r^{3} \left (r +1\right )^{3} \left (r +2\right )^{3} \left (r +3\right )^{3} \left (r +4\right )^{3} \left (r +5\right )^{3}}\) | \(-{\frac {117547073}{41472}}\) |
| \(b_{7}\) | \(-\frac {\left (4 r^{2}+24 r +37\right ) \left (4 r^{2}+16 r +17\right ) \left (4 r^{2}+1\right ) \left (4 r^{2}+8 r +5\right ) \left (4 r^{2}+32 r +65\right ) \left (4 r^{2}+40 r +101\right ) \left (4 r^{2}+48 r +145\right )}{r^{2} \left (r +1\right )^{2} \left (r +2\right )^{2} \left (r +3\right )^{2} \left (r +4\right )^{2} \left (r +5\right )^{2} \left (r +6\right )^{2}}\) | \(-{\frac {23591176505}{1016064}}\) | \(\frac {57344 r^{18}+3096576 r^{17}+78016512 r^{16}+1217986560 r^{15}+13199036928 r^{14}+105405576192 r^{13}+642772465664 r^{12}+3060602818560 r^{11}+11539411892512 r^{10}+34731616317888 r^{9}+83748781519536 r^{8}+161699336668800 r^{7}+248772001519566 r^{6}+301840226981244 r^{5}+283526565348838 r^{4}+199539379978680 r^{3}+99027624519600 r^{2}+30628499202000 r +4311077940000}{r^{3} \left (r +1\right )^{3} \left (r +2\right )^{3} \left (r +3\right )^{3} \left (r +4\right )^{3} \left (r +5\right )^{3} \left (r +6\right )^{3}}\) | \(\frac {162576422327}{14224896}\) |
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 (1-5 x +\frac {85 x^{2}}{4}-\frac {3145 x^{3}}{36}+\frac {204425 x^{4}}{576}-\frac {825877 x^{5}}{576}+\frac {119752165 x^{6}}{20736}-\frac {23591176505 x^{7}}{1016064}+O\left (x^{8}\right )\right ) \ln \left (x \right )+x \left (2 x -\frac {39 x^{2}}{4}+\frac {4499 x^{3}}{108}-\frac {594305 x^{4}}{3456}+\frac {2420617 x^{5}}{3456}-\frac {117547073 x^{6}}{41472}+\frac {162576422327 x^{7}}{14224896}+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 (1-5 x +\frac {85 x^{2}}{4}-\frac {3145 x^{3}}{36}+\frac {204425 x^{4}}{576}-\frac {825877 x^{5}}{576}+\frac {119752165 x^{6}}{20736}-\frac {23591176505 x^{7}}{1016064}+O\left (x^{8}\right )\right ) + c_{2} \left (x \left (1-5 x +\frac {85 x^{2}}{4}-\frac {3145 x^{3}}{36}+\frac {204425 x^{4}}{576}-\frac {825877 x^{5}}{576}+\frac {119752165 x^{6}}{20736}-\frac {23591176505 x^{7}}{1016064}+O\left (x^{8}\right )\right ) \ln \left (x \right )+x \left (2 x -\frac {39 x^{2}}{4}+\frac {4499 x^{3}}{108}-\frac {594305 x^{4}}{3456}+\frac {2420617 x^{5}}{3456}-\frac {117547073 x^{6}}{41472}+\frac {162576422327 x^{7}}{14224896}+O\left (x^{8}\right )\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x \left (1-5 x +\frac {85 x^{2}}{4}-\frac {3145 x^{3}}{36}+\frac {204425 x^{4}}{576}-\frac {825877 x^{5}}{576}+\frac {119752165 x^{6}}{20736}-\frac {23591176505 x^{7}}{1016064}+O\left (x^{8}\right )\right )+c_{2} \left (x \left (1-5 x +\frac {85 x^{2}}{4}-\frac {3145 x^{3}}{36}+\frac {204425 x^{4}}{576}-\frac {825877 x^{5}}{576}+\frac {119752165 x^{6}}{20736}-\frac {23591176505 x^{7}}{1016064}+O\left (x^{8}\right )\right ) \ln \left (x \right )+x \left (2 x -\frac {39 x^{2}}{4}+\frac {4499 x^{3}}{108}-\frac {594305 x^{4}}{3456}+\frac {2420617 x^{5}}{3456}-\frac {117547073 x^{6}}{41472}+\frac {162576422327 x^{7}}{14224896}+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 (1-5 x +\frac {85 x^{2}}{4}-\frac {3145 x^{3}}{36}+\frac {204425 x^{4}}{576}-\frac {825877 x^{5}}{576}+\frac {119752165 x^{6}}{20736}-\frac {23591176505 x^{7}}{1016064}+O\left (x^{8}\right )\right )+c_{2} \left (x \left (1-5 x +\frac {85 x^{2}}{4}-\frac {3145 x^{3}}{36}+\frac {204425 x^{4}}{576}-\frac {825877 x^{5}}{576}+\frac {119752165 x^{6}}{20736}-\frac {23591176505 x^{7}}{1016064}+O\left (x^{8}\right )\right ) \ln \left (x \right )+x \left (2 x -\frac {39 x^{2}}{4}+\frac {4499 x^{3}}{108}-\frac {594305 x^{4}}{3456}+\frac {2420617 x^{5}}{3456}-\frac {117547073 x^{6}}{41472}+\frac {162576422327 x^{7}}{14224896}+O\left (x^{8}\right )\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x \left (1-5 x +\frac {85 x^{2}}{4}-\frac {3145 x^{3}}{36}+\frac {204425 x^{4}}{576}-\frac {825877 x^{5}}{576}+\frac {119752165 x^{6}}{20736}-\frac {23591176505 x^{7}}{1016064}+O\left (x^{8}\right )\right )+c_{2} \left (x \left (1-5 x +\frac {85 x^{2}}{4}-\frac {3145 x^{3}}{36}+\frac {204425 x^{4}}{576}-\frac {825877 x^{5}}{576}+\frac {119752165 x^{6}}{20736}-\frac {23591176505 x^{7}}{1016064}+O\left (x^{8}\right )\right ) \ln \left (x \right )+x \left (2 x -\frac {39 x^{2}}{4}+\frac {4499 x^{3}}{108}-\frac {594305 x^{4}}{3456}+\frac {2420617 x^{5}}{3456}-\frac {117547073 x^{6}}{41472}+\frac {162576422327 x^{7}}{14224896}+O\left (x^{8}\right )\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (1+4 x \right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (4 x^{2}-x \right ) y^{\prime }+\left (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 (x +1\right ) y}{x^{2} \left (1+4 x \right )}-\frac {\left (4 x -1\right ) y^{\prime }}{x \left (1+4 x \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 -1\right ) y^{\prime }}{x \left (1+4 x \right )}+\frac {\left (x +1\right ) y}{x^{2} \left (1+4 x \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 -1}{x \left (1+4 x \right )}, P_{3}\left (x \right )=\frac {x +1}{x^{2} \left (1+4 x \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 (1+4 x \right ) \left (\frac {d}{d x}y^{\prime }\right )+x \left (4 x -1\right ) y^{\prime }+\left (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 =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^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..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} \left (-1+r \right )^{2} x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k} \left (k +r -1\right )^{2}+a_{k -1} \left (4 \left (k -1\right )^{2}+8 \left (k -1\right ) r +4 r^{2}+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 in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (k +r -1\right )^{2}+a_{k -1} \left (4 k^{2}+8 k r +4 r^{2}-8 k -8 r +5\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +1} \left (k +r \right )^{2}+a_{k} \left (4 \left (k +1\right )^{2}+8 \left (k +1\right ) r +4 r^{2}-8 k -3-8 r \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (4 k^{2}+8 k r +4 r^{2}+1\right )}{\left (k +r \right )^{2}} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1 \\ {} & {} & a_{k +1}=-\frac {a_{k} \left (4 k^{2}+8 k +5\right )}{\left (k +1\right )^{2}} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1}, a_{k +1}=-\frac {a_{k} \left (4 k^{2}+8 k +5\right )}{\left (k +1\right )^{2}}\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)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm <- No Liouvillian solutions exists -> 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`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 75
Order:=8; dsolve(x^2*(1+4*x)*diff(y(x),x$2)-x*(1-4*x)*diff(y(x),x)+(1+x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \left (\left (c_{2} \ln \left (x \right )+c_{1} \right ) \left (1-5 x +\frac {85}{4} x^{2}-\frac {3145}{36} x^{3}+\frac {204425}{576} x^{4}-\frac {825877}{576} x^{5}+\frac {119752165}{20736} x^{6}-\frac {23591176505}{1016064} x^{7}+\operatorname {O}\left (x^{8}\right )\right )+\left (2 x -\frac {39}{4} x^{2}+\frac {4499}{108} x^{3}-\frac {594305}{3456} x^{4}+\frac {2420617}{3456} x^{5}-\frac {117547073}{41472} x^{6}+\frac {162576422327}{14224896} x^{7}+\operatorname {O}\left (x^{8}\right )\right ) c_{2} \right ) x \]
✓ Solution by Mathematica
Time used: 0.014 (sec). Leaf size: 158
AsymptoticDSolveValue[x^2*(1+4*x)*y''[x]-x*(1-4*x)*y'[x]+(1+x)*y[x]==0,y[x],{x,0,7}]
\[ y(x)\to c_1 x \left (-\frac {23591176505 x^7}{1016064}+\frac {119752165 x^6}{20736}-\frac {825877 x^5}{576}+\frac {204425 x^4}{576}-\frac {3145 x^3}{36}+\frac {85 x^2}{4}-5 x+1\right )+c_2 \left (x \left (\frac {162576422327 x^7}{14224896}-\frac {117547073 x^6}{41472}+\frac {2420617 x^5}{3456}-\frac {594305 x^4}{3456}+\frac {4499 x^3}{108}-\frac {39 x^2}{4}+2 x\right )+x \left (-\frac {23591176505 x^7}{1016064}+\frac {119752165 x^6}{20736}-\frac {825877 x^5}{576}+\frac {204425 x^4}{576}-\frac {3145 x^3}{36}+\frac {85 x^2}{4}-5 x+1\right ) \log (x)\right ) \]