Internal problem ID [1437]
Internal file name [OUTPUT/1438_Sunday_June_05_2022_02_17_23_AM_43414470/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 III. Exercises 7.7. Page 389
Problem number: 21.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Difference is integer"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {4 x^{2} \left (1+x \right ) y^{\prime \prime }+4 x \left (1+4 x \right ) y^{\prime }-\left (49+27 x \right ) y=0} \] 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. \[ \left (4 x^{3}+4 x^{2}\right ) y^{\prime \prime }+\left (16 x^{2}+4 x \right ) y^{\prime }+\left (-27 x -49\right ) y = 0 \] 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 {1+4 x}{x \left (1+x \right )}\\ q(x) &= -\frac {49+27 x}{4 x^{2} \left (1+x \right )}\\ \end {align*}
| \(p(x)=\frac {1+4 x}{x \left (1+x \right )}\) | |
| singularity | type |
| \(x = -1\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
| \(q(x)=-\frac {49+27 x}{4 x^{2} \left (1+x \right )}\) | |
| 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 : \([-1, 0, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ 4 x^{2} \left (1+x \right ) y^{\prime \prime }+\left (16 x^{2}+4 x \right ) y^{\prime }+\left (-27 x -49\right ) y = 0 \] 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 (1+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 (16 x^{2}+4 x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (-27 x -49\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 }}4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}16 x^{1+n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-27 x^{1+n +r} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-49 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 }}16 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}16 a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-27 x^{1+n +r} a_{n}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-27 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 =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 =1}{\sum }}16 a_{n -1} \left (n +r -1\right ) x^{n +r}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}4 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-27 a_{n -1} x^{n +r}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-49 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+4 x^{n +r} a_{n} \left (n +r \right )-49 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 4 x^{r} a_{0} r \left (-1+r \right )+4 x^{r} a_{0} r -49 a_{0} x^{r} = 0 \] Or \[ \left (4 x^{r} r \left (-1+r \right )+4 x^{r} r -49 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (4 r^{2}-49\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 4 r^{2}-49 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {7}{2}}\\ r_2 &= -{\frac {7}{2}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (4 r^{2}-49\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {7}{2}}, -{\frac {7}{2}}\right ]\).
Since \(r_1 - r_2 = 7\) is an integer, then we can construct two linearly independent solutions \begin {align*} y_{1}\left (x \right ) &= x^{r_{1}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+x^{r_{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}\right ) \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= x^{\frac {7}{2}} \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n}\right )\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\frac {\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n}}{x^{\frac {7}{2}}} \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +\frac {7}{2}}\\ y_{2}\left (x \right ) &= C y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -\frac {7}{2}}\right ) \end {align*}
Where \(C\) above can be zero. We start by finding \(y_{1}\). Eq (2B) derived above is now used to find all \(a_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(a_{0}\) is arbitrary and taken as \(a_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{3} 4 a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+4 a_{n} \left (n +r \right ) \left (n +r -1\right )+16 a_{n -1} \left (n +r -1\right )+4 a_{n} \left (n +r \right )-27 a_{n -1}-49 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {\left (2 n +2 r -5\right ) a_{n -1}}{2 n +2 r -7}\tag {4} \] Which for the root \(r = {\frac {7}{2}}\) becomes \[ a_{n} = -\frac {\left (1+n \right ) a_{n -1}}{n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {7}{2}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ a_{1}=\frac {3-2 r}{2 r -5} \] Which for the root \(r = {\frac {7}{2}}\) becomes \[ a_{1}=-2 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {3-2 r}{2 r -5}\) | \(-2\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {2 r -1}{2 r -5} \] Which for the root \(r = {\frac {7}{2}}\) becomes \[ a_{2}=3 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {3-2 r}{2 r -5}\) | \(-2\) |
| \(a_{2}\) | \(\frac {2 r -1}{2 r -5}\) | \(3\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-1-2 r}{2 r -5} \] Which for the root \(r = {\frac {7}{2}}\) becomes \[ a_{3}=-4 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {3-2 r}{2 r -5}\) | \(-2\) |
| \(a_{2}\) | \(\frac {2 r -1}{2 r -5}\) | \(3\) |
| \(a_{3}\) | \(\frac {-1-2 r}{2 r -5}\) | \(-4\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {3+2 r}{2 r -5} \] Which for the root \(r = {\frac {7}{2}}\) becomes \[ a_{4}=5 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {3-2 r}{2 r -5}\) | \(-2\) |
| \(a_{2}\) | \(\frac {2 r -1}{2 r -5}\) | \(3\) |
| \(a_{3}\) | \(\frac {-1-2 r}{2 r -5}\) | \(-4\) |
| \(a_{4}\) | \(\frac {3+2 r}{2 r -5}\) | \(5\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-5-2 r}{2 r -5} \] Which for the root \(r = {\frac {7}{2}}\) becomes \[ a_{5}=-6 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {3-2 r}{2 r -5}\) | \(-2\) |
| \(a_{2}\) | \(\frac {2 r -1}{2 r -5}\) | \(3\) |
| \(a_{3}\) | \(\frac {-1-2 r}{2 r -5}\) | \(-4\) |
| \(a_{4}\) | \(\frac {3+2 r}{2 r -5}\) | \(5\) |
| \(a_{5}\) | \(\frac {-5-2 r}{2 r -5}\) | \(-6\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {7+2 r}{2 r -5} \] Which for the root \(r = {\frac {7}{2}}\) becomes \[ a_{6}=7 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {3-2 r}{2 r -5}\) | \(-2\) |
| \(a_{2}\) | \(\frac {2 r -1}{2 r -5}\) | \(3\) |
| \(a_{3}\) | \(\frac {-1-2 r}{2 r -5}\) | \(-4\) |
| \(a_{4}\) | \(\frac {3+2 r}{2 r -5}\) | \(5\) |
| \(a_{5}\) | \(\frac {-5-2 r}{2 r -5}\) | \(-6\) |
| \(a_{6}\) | \(\frac {7+2 r}{2 r -5}\) | \(7\) |
For \(n = 7\), using the above recursive equation gives \[ a_{7}=\frac {-9-2 r}{2 r -5} \] Which for the root \(r = {\frac {7}{2}}\) becomes \[ a_{7}=-8 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {3-2 r}{2 r -5}\) | \(-2\) |
| \(a_{2}\) | \(\frac {2 r -1}{2 r -5}\) | \(3\) |
| \(a_{3}\) | \(\frac {-1-2 r}{2 r -5}\) | \(-4\) |
| \(a_{4}\) | \(\frac {3+2 r}{2 r -5}\) | \(5\) |
| \(a_{5}\) | \(\frac {-5-2 r}{2 r -5}\) | \(-6\) |
| \(a_{6}\) | \(\frac {7+2 r}{2 r -5}\) | \(7\) |
| \(a_{7}\) | \(\frac {-9-2 r}{2 r -5}\) | \(-8\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{\frac {7}{2}} \left (a_{0}+a_{1} x +a_{2} x^{2}+a_{3} x^{3}+a_{4} x^{4}+a_{5} x^{5}+a_{6} x^{6}+a_{7} x^{7}+a_{8} x^{8}\dots \right ) \\ &= x^{\frac {7}{2}} \left (1-2 x +3 x^{2}-4 x^{3}+5 x^{4}-6 x^{5}+7 x^{6}-8 x^{7}+O\left (x^{8}\right )\right ) \end {align*}
Now the second solution \(y_{2}\left (x \right )\) is found. Let \[ r_{1}-r_{2} = N \] Where \(N\) is positive integer which is the difference between the two roots. \(r_{1}\) is taken as the larger root. Hence for this problem we have \(N=7\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{7}\left (r \right )\). If this limit exists, then \(C = 0\), else we need to keep the log term and \(C \neq 0\). The above table shows that \begin {align*} a_N &= a_{7} \\ &= \frac {-9-2 r}{2 r -5} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {-9-2 r}{2 r -5}&= \lim _{r\rightarrow -{\frac {7}{2}}}\frac {-9-2 r}{2 r -5}\\ &= {\frac {1}{6}} \end {align*}
The limit is \(\frac {1}{6}\). Since the limit exists then the log term is not needed and we can set \(C = 0\). Therefore the second solution has the form \begin {align*} y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n +r}\\ &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -\frac {7}{2}} \end {align*}
Eq (3) derived above is used to find all \(b_{n}\) coefficients. The case \(n = 0\) is skipped since it was used to find the roots of the indicial equation. \(b_{0}\) is arbitrary and taken as \(b_{0} = 1\). For \(1\le n\) the recursive equation is \begin{equation} \tag{4} 4 b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+4 b_{n} \left (n +r \right ) \left (n +r -1\right )+16 b_{n -1} \left (n +r -1\right )+4 b_{n} \left (n +r \right )-27 b_{n -1}-49 b_{n} = 0 \end{equation} Which for for the root \(r = -{\frac {7}{2}}\) becomes \begin{equation} \tag{4A} 4 b_{n -1} \left (n -\frac {9}{2}\right ) \left (n -\frac {11}{2}\right )+4 b_{n} \left (n -\frac {7}{2}\right ) \left (n -\frac {9}{2}\right )+16 b_{n -1} \left (n -\frac {9}{2}\right )+4 b_{n} \left (n -\frac {7}{2}\right )-27 b_{n -1}-49 b_{n} = 0 \end{equation} Solving for \(b_{n}\) from the recursive equation (4) gives \[ b_{n} = -\frac {\left (2 n +2 r -5\right ) b_{n -1}}{2 n +2 r -7}\tag {5} \] Which for the root \(r = -{\frac {7}{2}}\) becomes \[ b_{n} = -\frac {\left (2 n -12\right ) b_{n -1}}{2 n -14}\tag {6} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -{\frac {7}{2}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
For \(n = 1\), using the above recursive equation gives \[ b_{1}=-\frac {-3+2 r}{2 r -5} \] Which for the root \(r = -{\frac {7}{2}}\) becomes \[ b_{1}=-{\frac {5}{6}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {3-2 r}{2 r -5}\) | \(-{\frac {5}{6}}\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {2 r -1}{2 r -5} \] Which for the root \(r = -{\frac {7}{2}}\) becomes \[ b_{2}={\frac {2}{3}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {3-2 r}{2 r -5}\) | \(-{\frac {5}{6}}\) |
| \(b_{2}\) | \(\frac {2 r -1}{2 r -5}\) | \(\frac {2}{3}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=-\frac {1+2 r}{2 r -5} \] Which for the root \(r = -{\frac {7}{2}}\) becomes \[ b_{3}=-{\frac {1}{2}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {3-2 r}{2 r -5}\) | \(-{\frac {5}{6}}\) |
| \(b_{2}\) | \(\frac {2 r -1}{2 r -5}\) | \(\frac {2}{3}\) |
| \(b_{3}\) | \(\frac {-1-2 r}{2 r -5}\) | \(-{\frac {1}{2}}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {3+2 r}{2 r -5} \] Which for the root \(r = -{\frac {7}{2}}\) becomes \[ b_{4}={\frac {1}{3}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {3-2 r}{2 r -5}\) | \(-{\frac {5}{6}}\) |
| \(b_{2}\) | \(\frac {2 r -1}{2 r -5}\) | \(\frac {2}{3}\) |
| \(b_{3}\) | \(\frac {-1-2 r}{2 r -5}\) | \(-{\frac {1}{2}}\) |
| \(b_{4}\) | \(\frac {3+2 r}{2 r -5}\) | \(\frac {1}{3}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-\frac {5+2 r}{2 r -5} \] Which for the root \(r = -{\frac {7}{2}}\) becomes \[ b_{5}=-{\frac {1}{6}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {3-2 r}{2 r -5}\) | \(-{\frac {5}{6}}\) |
| \(b_{2}\) | \(\frac {2 r -1}{2 r -5}\) | \(\frac {2}{3}\) |
| \(b_{3}\) | \(\frac {-1-2 r}{2 r -5}\) | \(-{\frac {1}{2}}\) |
| \(b_{4}\) | \(\frac {3+2 r}{2 r -5}\) | \(\frac {1}{3}\) |
| \(b_{5}\) | \(\frac {-5-2 r}{2 r -5}\) | \(-{\frac {1}{6}}\) |
For \(n = 6\), using the above recursive equation gives \[ b_{6}=\frac {7+2 r}{2 r -5} \] Which for the root \(r = -{\frac {7}{2}}\) becomes \[ b_{6}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {3-2 r}{2 r -5}\) | \(-{\frac {5}{6}}\) |
| \(b_{2}\) | \(\frac {2 r -1}{2 r -5}\) | \(\frac {2}{3}\) |
| \(b_{3}\) | \(\frac {-1-2 r}{2 r -5}\) | \(-{\frac {1}{2}}\) |
| \(b_{4}\) | \(\frac {3+2 r}{2 r -5}\) | \(\frac {1}{3}\) |
| \(b_{5}\) | \(\frac {-5-2 r}{2 r -5}\) | \(-{\frac {1}{6}}\) |
| \(b_{6}\) | \(\frac {7+2 r}{2 r -5}\) | \(0\) |
For \(n = 7\), using the above recursive equation gives \[ b_{7}=-\frac {9+2 r}{2 r -5} \] Which for the root \(r = -{\frac {7}{2}}\) becomes \[ b_{7}={\frac {1}{6}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {3-2 r}{2 r -5}\) | \(-{\frac {5}{6}}\) |
| \(b_{2}\) | \(\frac {2 r -1}{2 r -5}\) | \(\frac {2}{3}\) |
| \(b_{3}\) | \(\frac {-1-2 r}{2 r -5}\) | \(-{\frac {1}{2}}\) |
| \(b_{4}\) | \(\frac {3+2 r}{2 r -5}\) | \(\frac {1}{3}\) |
| \(b_{5}\) | \(\frac {-5-2 r}{2 r -5}\) | \(-{\frac {1}{6}}\) |
| \(b_{6}\) | \(\frac {7+2 r}{2 r -5}\) | \(0\) |
| \(b_{7}\) | \(\frac {-9-2 r}{2 r -5}\) | \(\frac {1}{6}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x^{\frac {7}{2}} \left (b_{0}+b_{1} x +b_{2} x^{2}+b_{3} x^{3}+b_{4} x^{4}+b_{5} x^{5}+b_{6} x^{6}+b_{7} x^{7}+b_{8} x^{8}\dots \right ) \\ &= \frac {1-\frac {5 x}{6}+\frac {2 x^{2}}{3}-\frac {x^{3}}{2}+\frac {x^{4}}{3}-\frac {x^{5}}{6}+\frac {x^{7}}{6}+O\left (x^{8}\right )}{x^{\frac {7}{2}}} \end {align*}
Therefore the homogeneous solution is \begin{align*} y_h(x) &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= c_{1} x^{\frac {7}{2}} \left (1-2 x +3 x^{2}-4 x^{3}+5 x^{4}-6 x^{5}+7 x^{6}-8 x^{7}+O\left (x^{8}\right )\right ) + \frac {c_{2} \left (1-\frac {5 x}{6}+\frac {2 x^{2}}{3}-\frac {x^{3}}{2}+\frac {x^{4}}{3}-\frac {x^{5}}{6}+\frac {x^{7}}{6}+O\left (x^{8}\right )\right )}{x^{\frac {7}{2}}} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {7}{2}} \left (1-2 x +3 x^{2}-4 x^{3}+5 x^{4}-6 x^{5}+7 x^{6}-8 x^{7}+O\left (x^{8}\right )\right )+\frac {c_{2} \left (1-\frac {5 x}{6}+\frac {2 x^{2}}{3}-\frac {x^{3}}{2}+\frac {x^{4}}{3}-\frac {x^{5}}{6}+\frac {x^{7}}{6}+O\left (x^{8}\right )\right )}{x^{\frac {7}{2}}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {7}{2}} \left (1-2 x +3 x^{2}-4 x^{3}+5 x^{4}-6 x^{5}+7 x^{6}-8 x^{7}+O\left (x^{8}\right )\right )+\frac {c_{2} \left (1-\frac {5 x}{6}+\frac {2 x^{2}}{3}-\frac {x^{3}}{2}+\frac {x^{4}}{3}-\frac {x^{5}}{6}+\frac {x^{7}}{6}+O\left (x^{8}\right )\right )}{x^{\frac {7}{2}}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {7}{2}} \left (1-2 x +3 x^{2}-4 x^{3}+5 x^{4}-6 x^{5}+7 x^{6}-8 x^{7}+O\left (x^{8}\right )\right )+\frac {c_{2} \left (1-\frac {5 x}{6}+\frac {2 x^{2}}{3}-\frac {x^{3}}{2}+\frac {x^{4}}{3}-\frac {x^{5}}{6}+\frac {x^{7}}{6}+O\left (x^{8}\right )\right )}{x^{\frac {7}{2}}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 4 x^{2} \left (1+x \right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (16 x^{2}+4 x \right ) y^{\prime }+\left (-27 x -49\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 (49+27 x \right ) y}{4 x^{2} \left (1+x \right )}-\frac {\left (1+4 x \right ) y^{\prime }}{x \left (1+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 (1+4 x \right ) y^{\prime }}{x \left (1+x \right )}-\frac {\left (49+27 x \right ) y}{4 x^{2} \left (1+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 {1+4 x}{x \left (1+x \right )}, P_{3}\left (x \right )=-\frac {49+27 x}{4 x^{2} \left (1+x \right )}\right ] \\ {} & \circ & \left (1+x \right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (1+x \right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=3 \\ {} & \circ & \left (1+x \right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-1 \\ {} & {} & \left (\left (1+x \right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-1}}}=0 \\ {} & \circ & x =-1\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}=-1 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 4 x^{2} \left (1+x \right ) \left (\frac {d}{d x}y^{\prime }\right )+4 x \left (1+4 x \right ) y^{\prime }+\left (-27 x -49\right ) y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -1\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (4 u^{3}-8 u^{2}+4 u \right ) \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )+\left (16 u^{2}-28 u +12\right ) \left (\frac {d}{d u}y \left (u \right )\right )+\left (-27 u -22\right ) y \left (u \right )=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \left (u \right ) \\ {} & {} & y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot y \left (u \right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & u^{m}\cdot y \left (u \right )=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) u^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) u^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) u^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & u^{m}\cdot \left (\frac {d}{d u}\frac {d}{d u}y \left (u \right )\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) u^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & 4 a_{0} r \left (2+r \right ) u^{-1+r}+\left (4 a_{1} \left (1+r \right ) \left (3+r \right )-2 a_{0} \left (4 r^{2}+10 r +11\right )\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (4 a_{k +1} \left (k +1+r \right ) \left (k +3+r \right )-2 a_{k} \left (4 k^{2}+8 k r +4 r^{2}+10 k +10 r +11\right )+a_{k -1} \left (2 k +7+2 r \right ) \left (2 k -5+2 r \right )\right ) u^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & 4 r \left (2+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-2, 0\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & 4 a_{1} \left (1+r \right ) \left (3+r \right )-2 a_{0} \left (4 r^{2}+10 r +11\right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 4 \left (-2 a_{k}+a_{k -1}+a_{k +1}\right ) k^{2}+4 \left (2 \left (-2 a_{k}+a_{k -1}+a_{k +1}\right ) r -5 a_{k}+a_{k -1}+4 a_{k +1}\right ) k +4 \left (-2 a_{k}+a_{k -1}+a_{k +1}\right ) r^{2}+4 \left (-5 a_{k}+a_{k -1}+4 a_{k +1}\right ) r -22 a_{k}-35 a_{k -1}+12 a_{k +1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & 4 \left (-2 a_{k +1}+a_{k}+a_{k +2}\right ) \left (k +1\right )^{2}+4 \left (2 \left (-2 a_{k +1}+a_{k}+a_{k +2}\right ) r -5 a_{k +1}+a_{k}+4 a_{k +2}\right ) \left (k +1\right )+4 \left (-2 a_{k +1}+a_{k}+a_{k +2}\right ) r^{2}+4 \left (-5 a_{k +1}+a_{k}+4 a_{k +2}\right ) r -22 a_{k +1}-35 a_{k}+12 a_{k +2}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {4 k^{2} a_{k}-8 k^{2} a_{k +1}+8 k r a_{k}-16 k r a_{k +1}+4 r^{2} a_{k}-8 r^{2} a_{k +1}+12 k a_{k}-36 k a_{k +1}+12 r a_{k}-36 r a_{k +1}-27 a_{k}-50 a_{k +1}}{4 \left (k^{2}+2 k r +r^{2}+6 k +6 r +8\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-2 \\ {} & {} & a_{k +2}=-\frac {4 k^{2} a_{k}-8 k^{2} a_{k +1}-4 k a_{k}-4 k a_{k +1}-35 a_{k}-10 a_{k +1}}{4 \left (k^{2}+2 k \right )} \\ \bullet & {} & \textrm {Series not valid for}\hspace {3pt} r =-2\hspace {3pt}\textrm {, division by}\hspace {3pt} 0\hspace {3pt}\textrm {in the recursion relation at}\hspace {3pt} k =0 \\ {} & {} & a_{k +2}=-\frac {4 k^{2} a_{k}-8 k^{2} a_{k +1}-4 k a_{k}-4 k a_{k +1}-35 a_{k}-10 a_{k +1}}{4 \left (k^{2}+2 k \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {4 k^{2} a_{k}-8 k^{2} a_{k +1}+12 k a_{k}-36 k a_{k +1}-27 a_{k}-50 a_{k +1}}{4 \left (k^{2}+6 k +8\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k}, a_{k +2}=-\frac {4 k^{2} a_{k}-8 k^{2} a_{k +1}+12 k a_{k}-36 k a_{k +1}-27 a_{k}-50 a_{k +1}}{4 \left (k^{2}+6 k +8\right )}, 12 a_{1}-22 a_{0}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =1+x \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (1+x \right )^{k}, a_{k +2}=-\frac {4 k^{2} a_{k}-8 k^{2} a_{k +1}+12 k a_{k}-36 k a_{k +1}-27 a_{k}-50 a_{k +1}}{4 \left (k^{2}+6 k +8\right )}, 12 a_{1}-22 a_{0}=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) <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 47
Order:=6; dsolve(4*x^2*(1+x)*diff(y(x),x$2)+4*x*(1+4*x)*diff(y(x),x)-(49+27*x)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{1} x^{7} \left (1-2 x +3 x^{2}-4 x^{3}+5 x^{4}-6 x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (3628800-3024000 x +2419200 x^{2}-1814400 x^{3}+1209600 x^{4}-604800 x^{5}+\operatorname {O}\left (x^{6}\right )\right )}{x^{\frac {7}{2}}} \]
✓ Solution by Mathematica
Time used: 0.044 (sec). Leaf size: 86
AsymptoticDSolveValue[4*x^2*(1+x)*y''[x]+4*x*(1+4*x)*y'[x]-(49+27*x)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (\frac {2}{3 x^{3/2}}-\frac {5}{6 x^{5/2}}+\frac {1}{x^{7/2}}+\frac {\sqrt {x}}{3}-\frac {1}{2 \sqrt {x}}\right )+c_2 \left (5 x^{15/2}-4 x^{13/2}+3 x^{11/2}-2 x^{9/2}+x^{7/2}\right ) \]