Internal problem ID [2382]
Internal file name [OUTPUT/2382_Tuesday_February_27_2024_08_36_39_AM_61879799/index.tex]
Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath.
Boston. 1964
Section: Exercise 41, page 195
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. Difference not integer"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x^{2} \left (4+x \right ) y^{\prime \prime }+7 y^{\prime } x -y=0} \] 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. \[ \left (x^{3}+4 x^{2}\right ) y^{\prime \prime }+7 y^{\prime } x -y = 0 \] 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 {7}{x \left (4+x \right )}\\ q(x) &= -\frac {1}{x^{2} \left (4+x \right )}\\ \end {align*}
| \(p(x)=\frac {7}{x \left (4+x \right )}\) | |
| singularity | type |
| \(x = -4\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
| \(q(x)=-\frac {1}{x^{2} \left (4+x \right )}\) | |
| singularity | type |
| \(x = -4\) | \(\text {``regular''}\) |
| \(x = 0\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([-4, 0, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ x^{2} \left (4+x \right ) y^{\prime \prime }+7 y^{\prime } x -y = 0 \] 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 (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 )+7 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right ) x -\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^{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 }}7 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-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^{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} \\ \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 }}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 =0}{\sum }}7 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 4 x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+7 x^{n +r} a_{n} \left (n +r \right )-a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ 4 x^{r} a_{0} r \left (-1+r \right )+7 x^{r} a_{0} r -a_{0} x^{r} = 0 \] Or \[ \left (4 x^{r} r \left (-1+r \right )+7 x^{r} r -x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (4 r^{2}+3 r -1\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 4 r^{2}+3 r -1 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= {\frac {1}{4}}\\ r_2 &= -1 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (4 r^{2}+3 r -1\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [{\frac {1}{4}}, -1\right ]\).
Since \(r_1 - r_2 = {\frac {5}{4}}\) is not 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 ) &= 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 ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +\frac {1}{4}}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -1} \end {align*}
We start by finding \(y_{1}\left (x \right )\). 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} 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 )+7 a_{n} \left (n +r \right )-a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )}{4 n^{2}+8 n r +4 r^{2}+3 n +3 r -1}\tag {4} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ a_{n} = -\frac {a_{n -1} \left (16 n^{2}-40 n +21\right )}{64 n^{2}+80 n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = {\frac {1}{4}}\) 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 \[ a_{1}=-\frac {r \left (-1+r \right )}{4 r^{2}+11 r +6} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ a_{1}={\frac {1}{48}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r \left (-1+r \right )}{4 r^{2}+11 r +6}\) | \(\frac {1}{48}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {r^{2} \left (r^{2}-1\right )}{\left (4 r^{2}+11 r +6\right ) \left (4 r^{2}+19 r +21\right )} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ a_{2}=-{\frac {5}{19968}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r \left (-1+r \right )}{4 r^{2}+11 r +6}\) | \(\frac {1}{48}\) |
| \(a_{2}\) | \(\frac {r^{2} \left (r^{2}-1\right )}{\left (4 r^{2}+11 r +6\right ) \left (4 r^{2}+19 r +21\right )}\) | \(-{\frac {5}{19968}}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=-\frac {\left (1+r \right )^{2} r^{2} \left (-1+r \right )}{64 r^{5}+784 r^{4}+3644 r^{3}+7931 r^{2}+7905 r +2772} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ a_{3}={\frac {25}{1810432}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r \left (-1+r \right )}{4 r^{2}+11 r +6}\) | \(\frac {1}{48}\) |
| \(a_{2}\) | \(\frac {r^{2} \left (r^{2}-1\right )}{\left (4 r^{2}+11 r +6\right ) \left (4 r^{2}+19 r +21\right )}\) | \(-{\frac {5}{19968}}\) |
| \(a_{3}\) | \(-\frac {\left (1+r \right )^{2} r^{2} \left (-1+r \right )}{64 r^{5}+784 r^{4}+3644 r^{3}+7931 r^{2}+7905 r +2772}\) | \(\frac {25}{1810432}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {\left (2+r \right ) \left (1+r \right )^{2} r^{2} \left (-1+r \right )}{256 r^{6}+4608 r^{5}+32992 r^{4}+119088 r^{3}+225241 r^{2}+206865 r +69300} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ a_{4}=-{\frac {75}{62390272}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r \left (-1+r \right )}{4 r^{2}+11 r +6}\) | \(\frac {1}{48}\) |
| \(a_{2}\) | \(\frac {r^{2} \left (r^{2}-1\right )}{\left (4 r^{2}+11 r +6\right ) \left (4 r^{2}+19 r +21\right )}\) | \(-{\frac {5}{19968}}\) |
| \(a_{3}\) | \(-\frac {\left (1+r \right )^{2} r^{2} \left (-1+r \right )}{64 r^{5}+784 r^{4}+3644 r^{3}+7931 r^{2}+7905 r +2772}\) | \(\frac {25}{1810432}\) |
| \(a_{4}\) | \(\frac {\left (2+r \right ) \left (1+r \right )^{2} r^{2} \left (-1+r \right )}{256 r^{6}+4608 r^{5}+32992 r^{4}+119088 r^{3}+225241 r^{2}+206865 r +69300}\) | \(-{\frac {75}{62390272}}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=-\frac {\left (3+r \right ) \left (2+r \right ) \left (1+r \right )^{2} r^{2} \left (-1+r \right )}{1024 r^{7}+25344 r^{6}+257920 r^{5}+1388640 r^{4}+4228276 r^{3}+7175751 r^{2}+6146865 r +1975050} \] Which for the root \(r = {\frac {1}{4}}\) becomes \[ a_{5}={\frac {39}{293601280}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r \left (-1+r \right )}{4 r^{2}+11 r +6}\) | \(\frac {1}{48}\) |
| \(a_{2}\) | \(\frac {r^{2} \left (r^{2}-1\right )}{\left (4 r^{2}+11 r +6\right ) \left (4 r^{2}+19 r +21\right )}\) | \(-{\frac {5}{19968}}\) |
| \(a_{3}\) | \(-\frac {\left (1+r \right )^{2} r^{2} \left (-1+r \right )}{64 r^{5}+784 r^{4}+3644 r^{3}+7931 r^{2}+7905 r +2772}\) | \(\frac {25}{1810432}\) |
| \(a_{4}\) | \(\frac {\left (2+r \right ) \left (1+r \right )^{2} r^{2} \left (-1+r \right )}{256 r^{6}+4608 r^{5}+32992 r^{4}+119088 r^{3}+225241 r^{2}+206865 r +69300}\) | \(-{\frac {75}{62390272}}\) |
| \(a_{5}\) | \(-\frac {\left (3+r \right ) \left (2+r \right ) \left (1+r \right )^{2} r^{2} \left (-1+r \right )}{1024 r^{7}+25344 r^{6}+257920 r^{5}+1388640 r^{4}+4228276 r^{3}+7175751 r^{2}+6146865 r +1975050}\) | \(\frac {39}{293601280}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{\frac {1}{4}} \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}\dots \right ) \\ &= x^{\frac {1}{4}} \left (1+\frac {x}{48}-\frac {5 x^{2}}{19968}+\frac {25 x^{3}}{1810432}-\frac {75 x^{4}}{62390272}+\frac {39 x^{5}}{293601280}+O\left (x^{6}\right )\right ) \end {align*}
Now the second solution \(y_{2}\left (x \right )\) is found. Eq (2B) derived above is now 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{3} 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 )+7 b_{n} \left (n +r \right )-b_{n} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = -\frac {b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )}{4 n^{2}+8 n r +4 r^{2}+3 n +3 r -1}\tag {4} \] Which for the root \(r = -1\) becomes \[ b_{n} = -\frac {b_{n -1} \left (n -2\right ) \left (n -3\right )}{n \left (4 n -5\right )}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -1\) 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 \[ b_{1}=-\frac {r \left (-1+r \right )}{4 r^{2}+11 r +6} \] Which for the root \(r = -1\) becomes \[ b_{1}=2 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {r \left (-1+r \right )}{4 r^{2}+11 r +6}\) | \(2\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {r^{2} \left (r^{2}-1\right )}{\left (4 r^{2}+11 r +6\right ) \left (4 r^{2}+19 r +21\right )} \] Which for the root \(r = -1\) becomes \[ b_{2}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {r \left (-1+r \right )}{4 r^{2}+11 r +6}\) | \(2\) |
| \(b_{2}\) | \(\frac {r^{2} \left (r^{2}-1\right )}{\left (4 r^{2}+11 r +6\right ) \left (4 r^{2}+19 r +21\right )}\) | \(0\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=-\frac {\left (1+r \right )^{2} r^{2} \left (-1+r \right )}{64 r^{5}+784 r^{4}+3644 r^{3}+7931 r^{2}+7905 r +2772} \] Which for the root \(r = -1\) becomes \[ b_{3}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {r \left (-1+r \right )}{4 r^{2}+11 r +6}\) | \(2\) |
| \(b_{2}\) | \(\frac {r^{2} \left (r^{2}-1\right )}{\left (4 r^{2}+11 r +6\right ) \left (4 r^{2}+19 r +21\right )}\) | \(0\) |
| \(b_{3}\) | \(-\frac {\left (1+r \right )^{2} r^{2} \left (-1+r \right )}{64 r^{5}+784 r^{4}+3644 r^{3}+7931 r^{2}+7905 r +2772}\) | \(0\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {\left (2+r \right ) \left (1+r \right )^{2} r^{2} \left (-1+r \right )}{256 r^{6}+4608 r^{5}+32992 r^{4}+119088 r^{3}+225241 r^{2}+206865 r +69300} \] Which for the root \(r = -1\) becomes \[ b_{4}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {r \left (-1+r \right )}{4 r^{2}+11 r +6}\) | \(2\) |
| \(b_{2}\) | \(\frac {r^{2} \left (r^{2}-1\right )}{\left (4 r^{2}+11 r +6\right ) \left (4 r^{2}+19 r +21\right )}\) | \(0\) |
| \(b_{3}\) | \(-\frac {\left (1+r \right )^{2} r^{2} \left (-1+r \right )}{64 r^{5}+784 r^{4}+3644 r^{3}+7931 r^{2}+7905 r +2772}\) | \(0\) |
| \(b_{4}\) | \(\frac {\left (2+r \right ) \left (1+r \right )^{2} r^{2} \left (-1+r \right )}{256 r^{6}+4608 r^{5}+32992 r^{4}+119088 r^{3}+225241 r^{2}+206865 r +69300}\) | \(0\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=-\frac {\left (3+r \right ) \left (2+r \right ) \left (1+r \right )^{2} r^{2} \left (-1+r \right )}{1024 r^{7}+25344 r^{6}+257920 r^{5}+1388640 r^{4}+4228276 r^{3}+7175751 r^{2}+6146865 r +1975050} \] Which for the root \(r = -1\) becomes \[ b_{5}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {r \left (-1+r \right )}{4 r^{2}+11 r +6}\) | \(2\) |
| \(b_{2}\) | \(\frac {r^{2} \left (r^{2}-1\right )}{\left (4 r^{2}+11 r +6\right ) \left (4 r^{2}+19 r +21\right )}\) | \(0\) |
| \(b_{3}\) | \(-\frac {\left (1+r \right )^{2} r^{2} \left (-1+r \right )}{64 r^{5}+784 r^{4}+3644 r^{3}+7931 r^{2}+7905 r +2772}\) | \(0\) |
| \(b_{4}\) | \(\frac {\left (2+r \right ) \left (1+r \right )^{2} r^{2} \left (-1+r \right )}{256 r^{6}+4608 r^{5}+32992 r^{4}+119088 r^{3}+225241 r^{2}+206865 r +69300}\) | \(0\) |
| \(b_{5}\) | \(-\frac {\left (3+r \right ) \left (2+r \right ) \left (1+r \right )^{2} r^{2} \left (-1+r \right )}{1024 r^{7}+25344 r^{6}+257920 r^{5}+1388640 r^{4}+4228276 r^{3}+7175751 r^{2}+6146865 r +1975050}\) | \(0\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x^{\frac {1}{4}} \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}\dots \right ) \\ &= \frac {1+2 x +O\left (x^{6}\right )}{x} \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 {1}{4}} \left (1+\frac {x}{48}-\frac {5 x^{2}}{19968}+\frac {25 x^{3}}{1810432}-\frac {75 x^{4}}{62390272}+\frac {39 x^{5}}{293601280}+O\left (x^{6}\right )\right ) + \frac {c_{2} \left (1+2 x +O\left (x^{6}\right )\right )}{x} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{\frac {1}{4}} \left (1+\frac {x}{48}-\frac {5 x^{2}}{19968}+\frac {25 x^{3}}{1810432}-\frac {75 x^{4}}{62390272}+\frac {39 x^{5}}{293601280}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+2 x +O\left (x^{6}\right )\right )}{x} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{\frac {1}{4}} \left (1+\frac {x}{48}-\frac {5 x^{2}}{19968}+\frac {25 x^{3}}{1810432}-\frac {75 x^{4}}{62390272}+\frac {39 x^{5}}{293601280}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+2 x +O\left (x^{6}\right )\right )}{x} \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{\frac {1}{4}} \left (1+\frac {x}{48}-\frac {5 x^{2}}{19968}+\frac {25 x^{3}}{1810432}-\frac {75 x^{4}}{62390272}+\frac {39 x^{5}}{293601280}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1+2 x +O\left (x^{6}\right )\right )}{x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (4+x \right ) y^{\prime \prime }+7 y^{\prime } x -y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\frac {y}{x^{2} \left (4+x \right )}-\frac {7 y^{\prime }}{x \left (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} \\ {} & {} & y^{\prime \prime }+\frac {7 y^{\prime }}{x \left (4+x \right )}-\frac {y}{x^{2} \left (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 {7}{x \left (4+x \right )}, P_{3}\left (x \right )=-\frac {1}{x^{2} \left (4+x \right )}\right ] \\ {} & \circ & \left (4+x \right )\cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-4 \\ {} & {} & \left (\left (4+x \right )\cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-4}}}=-\frac {7}{4} \\ {} & \circ & \left (4+x \right )^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =-4 \\ {} & {} & \left (\left (4+x \right )^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}-4}}}=0 \\ {} & \circ & x =-4\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}=-4 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x^{2} \left (4+x \right ) y^{\prime \prime }+7 y^{\prime } x -y=0 \\ \bullet & {} & \textrm {Change variables using}\hspace {3pt} x =u -4\hspace {3pt}\textrm {so that the regular singular point is at}\hspace {3pt} u =0 \\ {} & {} & \left (u^{3}-8 u^{2}+16 u \right ) \left (\frac {d^{2}}{d u^{2}}y \left (u \right )\right )+\left (7 u -28\right ) \left (\frac {d}{d u}y \left (u \right )\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 \left (\frac {d}{d u}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & 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^{2}}{d u^{2}}y \left (u \right )\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & u^{m}\cdot \left (\frac {d^{2}}{d u^{2}}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^{2}}{d u^{2}}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 (-11+4 r \right ) u^{-1+r}+\left (4 a_{1} \left (1+r \right ) \left (-7+4 r \right )-a_{0} \left (8 r^{2}-15 r +1\right )\right ) u^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (4 a_{k +1} \left (k +1+r \right ) \left (4 k -7+4 r \right )-a_{k} \left (8 k^{2}+16 k r +8 r^{2}-15 k -15 r +1\right )+a_{k -1} \left (k +r -1\right ) \left (k -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 (-11+4 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, \frac {11}{4}\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & 4 a_{1} \left (1+r \right ) \left (-7+4 r \right )-a_{0} \left (8 r^{2}-15 r +1\right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & \left (-8 a_{k}+a_{k -1}+16 a_{k +1}\right ) k^{2}+\left (2 \left (-8 a_{k}+a_{k -1}+16 a_{k +1}\right ) r +15 a_{k}-3 a_{k -1}-12 a_{k +1}\right ) k +\left (-8 a_{k}+a_{k -1}+16 a_{k +1}\right ) r^{2}+3 \left (5 a_{k}-a_{k -1}-4 a_{k +1}\right ) r -a_{k}+2 a_{k -1}-28 a_{k +1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & \left (-8 a_{k +1}+a_{k}+16 a_{k +2}\right ) \left (k +1\right )^{2}+\left (2 \left (-8 a_{k +1}+a_{k}+16 a_{k +2}\right ) r +15 a_{k +1}-3 a_{k}-12 a_{k +2}\right ) \left (k +1\right )+\left (-8 a_{k +1}+a_{k}+16 a_{k +2}\right ) r^{2}+3 \left (5 a_{k +1}-a_{k}-4 a_{k +2}\right ) r -a_{k +1}+2 a_{k}-28 a_{k +2}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-8 k^{2} a_{k +1}+2 k r a_{k}-16 k r a_{k +1}+r^{2} a_{k}-8 r^{2} a_{k +1}-k a_{k}-k a_{k +1}-r a_{k}-r a_{k +1}+6 a_{k +1}}{4 \left (4 k^{2}+8 k r +4 r^{2}+5 k +5 r -6\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-8 k^{2} a_{k +1}-k a_{k}-k a_{k +1}+6 a_{k +1}}{4 \left (4 k^{2}+5 k -6\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 {k^{2} a_{k}-8 k^{2} a_{k +1}-k a_{k}-k a_{k +1}+6 a_{k +1}}{4 \left (4 k^{2}+5 k -6\right )}, -28 a_{1}-a_{0}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =4+x \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (4+x \right )^{k}, a_{k +2}=-\frac {k^{2} a_{k}-8 k^{2} a_{k +1}-k a_{k}-k a_{k +1}+6 a_{k +1}}{4 \left (4 k^{2}+5 k -6\right )}, -28 a_{1}-a_{0}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =\frac {11}{4} \\ {} & {} & a_{k +2}=-\frac {k^{2} a_{k}-8 k^{2} a_{k +1}+\frac {9}{2} k a_{k}-45 k a_{k +1}+\frac {77}{16} a_{k}-\frac {229}{4} a_{k +1}}{4 \left (4 k^{2}+27 k +38\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =\frac {11}{4} \\ {} & {} & \left [y \left (u \right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} u^{k +\frac {11}{4}}, a_{k +2}=-\frac {k^{2} a_{k}-8 k^{2} a_{k +1}+\frac {9}{2} k a_{k}-45 k a_{k +1}+\frac {77}{16} a_{k}-\frac {229}{4} a_{k +1}}{4 \left (4 k^{2}+27 k +38\right )}, 60 a_{1}-\frac {81 a_{0}}{4}=0\right ] \\ \bullet & {} & \textrm {Revert the change of variables}\hspace {3pt} u =4+x \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (4+x \right )^{k +\frac {11}{4}}, a_{k +2}=-\frac {k^{2} a_{k}-8 k^{2} a_{k +1}+\frac {9}{2} k a_{k}-45 k a_{k +1}+\frac {77}{16} a_{k}-\frac {229}{4} a_{k +1}}{4 \left (4 k^{2}+27 k +38\right )}, 60 a_{1}-\frac {81 a_{0}}{4}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (4+x \right )^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} \left (4+x \right )^{k +\frac {11}{4}}\right ), a_{k +2}=-\frac {k^{2} a_{k}-8 k^{2} a_{k +1}-k a_{k}-k a_{k +1}+6 a_{k +1}}{4 \left (4 k^{2}+5 k -6\right )}, -28 a_{1}-a_{0}=0, b_{k +2}=-\frac {k^{2} b_{k}-8 k^{2} b_{k +1}+\frac {9}{2} k b_{k}-45 k b_{k +1}+\frac {77}{16} b_{k}-\frac {229}{4} b_{k +1}}{4 \left (4 k^{2}+27 k +38\right )}, 60 b_{1}-\frac {81 b_{0}}{4}=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: 0.016 (sec). Leaf size: 39
Order:=6; dsolve(x^2*(4+x)*diff(y(x),x$2)+7*x*diff(y(x),x)-y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{2} x^{\frac {5}{4}} \left (1+\frac {1}{48} x -\frac {5}{19968} x^{2}+\frac {25}{1810432} x^{3}-\frac {75}{62390272} x^{4}+\frac {39}{293601280} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{1} \left (1+2 x +\operatorname {O}\left (x^{6}\right )\right )}{x} \]
✓ Solution by Mathematica
Time used: 0.005 (sec). Leaf size: 58
AsymptoticDSolveValue[x^2*(4+x)*y''[x]+7*x*y'[x]-y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \sqrt [4]{x} \left (\frac {39 x^5}{293601280}-\frac {75 x^4}{62390272}+\frac {25 x^3}{1810432}-\frac {5 x^2}{19968}+\frac {x}{48}+1\right )+\frac {c_2 (2 x+1)}{x} \]