Internal problem ID [2412]
Internal file name [OUTPUT/2412_Tuesday_February_27_2024_08_37_03_AM_26828799/index.tex]
Book: Differential Equations by Alfred L. Nelson, Karl W. Folley, Max Coral. 3rd ed. DC heath.
Boston. 1964
Section: Exercise 42, page 206
Problem number: 11.
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]]
\[ \boxed {x^{2} \left (1-x \right ) y^{\prime \prime }+x \left (x +1\right ) y^{\prime }-9 y=0} \] 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. \[ \left (-x^{3}+x^{2}\right ) y^{\prime \prime }+\left (x^{2}+x \right ) y^{\prime }-9 y = 0 \] 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 {x +1}{x \left (x -1\right )}\\ q(x) &= \frac {9}{x^{2} \left (x -1\right )}\\ \end {align*}
| \(p(x)=-\frac {x +1}{x \left (x -1\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = 1\) | \(\text {``regular''}\) |
| \(q(x)=\frac {9}{x^{2} \left (x -1\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = 1\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, 1, \infty ]\)
Irregular singular points : \([]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ -y^{\prime \prime } x^{2} \left (x -1\right )+\left (x^{2}+x \right ) y^{\prime }-9 y = 0 \] 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} -\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right ) x^{2} \left (x -1\right )+\left (x^{2}+x \right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )-9 \left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-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 }}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 )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-9 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 }}\left (-x^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) x^{n +r} \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(x\) are the same and equal to \(n +r\). \begin{equation} \tag{2B} \moverset {\infty }{\munderset {n =1}{\sum }}\left (-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 }}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 )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-9 a_{n} x^{n +r}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )+x^{n +r} a_{n} \left (n +r \right )-9 a_{n} x^{n +r} = 0 \] When \(n = 0\) the above becomes \[ x^{r} a_{0} r \left (-1+r \right )+x^{r} a_{0} r -9 a_{0} x^{r} = 0 \] Or \[ \left (x^{r} r \left (-1+r \right )+x^{r} r -9 x^{r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ \left (r^{2}-9\right ) x^{r} = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ r^{2}-9 = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 3\\ r_2 &= -3 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ \left (r^{2}-9\right ) x^{r} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([3, -3]\).
Since \(r_1 - r_2 = 6\) 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^{3} \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^{3}} \end {align*}
Or \begin {align*} y_{1}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +3}\\ 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 -3}\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} -a_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n -1} \left (n +r -1\right )+a_{n} \left (n +r \right )-9 a_{n} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {a_{n -1} \left (n +r -1\right )}{n +r +3}\tag {4} \] Which for the root \(r = 3\) becomes \[ a_{n} = \frac {a_{n -1} \left (n +2\right )}{n +6}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 3\) 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 \[ a_{1}=\frac {r}{4+r} \] Which for the root \(r = 3\) becomes \[ a_{1}={\frac {3}{7}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{4+r}\) | \(\frac {3}{7}\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {\left (1+r \right ) r}{\left (4+r \right ) \left (5+r \right )} \] Which for the root \(r = 3\) becomes \[ a_{2}={\frac {3}{14}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{4+r}\) | \(\frac {3}{7}\) |
| \(a_{2}\) | \(\frac {\left (1+r \right ) r}{\left (4+r \right ) \left (5+r \right )}\) | \(\frac {3}{14}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {r \left (1+r \right ) \left (2+r \right )}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right )} \] Which for the root \(r = 3\) becomes \[ a_{3}={\frac {5}{42}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{4+r}\) | \(\frac {3}{7}\) |
| \(a_{2}\) | \(\frac {\left (1+r \right ) r}{\left (4+r \right ) \left (5+r \right )}\) | \(\frac {3}{14}\) |
| \(a_{3}\) | \(\frac {r \left (1+r \right ) \left (2+r \right )}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right )}\) | \(\frac {5}{42}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {\left (1+r \right ) r \left (2+r \right ) \left (r +3\right )}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right ) \left (7+r \right )} \] Which for the root \(r = 3\) becomes \[ a_{4}={\frac {1}{14}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{4+r}\) | \(\frac {3}{7}\) |
| \(a_{2}\) | \(\frac {\left (1+r \right ) r}{\left (4+r \right ) \left (5+r \right )}\) | \(\frac {3}{14}\) |
| \(a_{3}\) | \(\frac {r \left (1+r \right ) \left (2+r \right )}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right )}\) | \(\frac {5}{42}\) |
| \(a_{4}\) | \(\frac {\left (1+r \right ) r \left (2+r \right ) \left (r +3\right )}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right ) \left (7+r \right )}\) | \(\frac {1}{14}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {\left (r +3\right ) \left (2+r \right ) \left (1+r \right ) r}{\left (7+r \right ) \left (6+r \right ) \left (5+r \right ) \left (8+r \right )} \] Which for the root \(r = 3\) becomes \[ a_{5}={\frac {1}{22}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{4+r}\) | \(\frac {3}{7}\) |
| \(a_{2}\) | \(\frac {\left (1+r \right ) r}{\left (4+r \right ) \left (5+r \right )}\) | \(\frac {3}{14}\) |
| \(a_{3}\) | \(\frac {r \left (1+r \right ) \left (2+r \right )}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right )}\) | \(\frac {5}{42}\) |
| \(a_{4}\) | \(\frac {\left (1+r \right ) r \left (2+r \right ) \left (r +3\right )}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right ) \left (7+r \right )}\) | \(\frac {1}{14}\) |
| \(a_{5}\) | \(\frac {\left (r +3\right ) \left (2+r \right ) \left (1+r \right ) r}{\left (7+r \right ) \left (6+r \right ) \left (5+r \right ) \left (8+r \right )}\) | \(\frac {1}{22}\) |
For \(n = 6\), using the above recursive equation gives \[ a_{6}=\frac {\left (2+r \right ) r \left (1+r \right ) \left (r +3\right )}{\left (8+r \right ) \left (7+r \right ) \left (6+r \right ) \left (9+r \right )} \] Which for the root \(r = 3\) becomes \[ a_{6}={\frac {1}{33}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {r}{4+r}\) | \(\frac {3}{7}\) |
| \(a_{2}\) | \(\frac {\left (1+r \right ) r}{\left (4+r \right ) \left (5+r \right )}\) | \(\frac {3}{14}\) |
| \(a_{3}\) | \(\frac {r \left (1+r \right ) \left (2+r \right )}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right )}\) | \(\frac {5}{42}\) |
| \(a_{4}\) | \(\frac {\left (1+r \right ) r \left (2+r \right ) \left (r +3\right )}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right ) \left (7+r \right )}\) | \(\frac {1}{14}\) |
| \(a_{5}\) | \(\frac {\left (r +3\right ) \left (2+r \right ) \left (1+r \right ) r}{\left (7+r \right ) \left (6+r \right ) \left (5+r \right ) \left (8+r \right )}\) | \(\frac {1}{22}\) |
| \(a_{6}\) | \(\frac {\left (2+r \right ) r \left (1+r \right ) \left (r +3\right )}{\left (8+r \right ) \left (7+r \right ) \left (6+r \right ) \left (9+r \right )}\) | \(\frac {1}{33}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{3} \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}\dots \right ) \\ &= x^{3} \left (1+\frac {3 x}{7}+\frac {3 x^{2}}{14}+\frac {5 x^{3}}{42}+\frac {x^{4}}{14}+\frac {x^{5}}{22}+\frac {x^{6}}{33}+O\left (x^{7}\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 \(N=6\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{6}\left (r \right )\). 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_{6} \\ &= \frac {\left (2+r \right ) r \left (1+r \right ) \left (r +3\right )}{\left (8+r \right ) \left (7+r \right ) \left (6+r \right ) \left (9+r \right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {\left (2+r \right ) r \left (1+r \right ) \left (r +3\right )}{\left (8+r \right ) \left (7+r \right ) \left (6+r \right ) \left (9+r \right )}&= \lim _{r\rightarrow -3}\frac {\left (2+r \right ) r \left (1+r \right ) \left (r +3\right )}{\left (8+r \right ) \left (7+r \right ) \left (6+r \right ) \left (9+r \right )}\\ &= 0 \end {align*}
The limit is \(0\). 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 -3} \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} -b_{n -1} \left (n +r -1\right ) \left (n +r -2\right )+b_{n} \left (n +r \right ) \left (n +r -1\right )+b_{n -1} \left (n +r -1\right )+b_{n} \left (n +r \right )-9 b_{n} = 0 \end{equation} Which for for the root \(r = -3\) becomes \begin{equation} \tag{4A} -b_{n -1} \left (n -4\right ) \left (n -5\right )+b_{n} \left (n -3\right ) \left (n -4\right )+b_{n -1} \left (n -4\right )+b_{n} \left (n -3\right )-9 b_{n} = 0 \end{equation} Solving for \(b_{n}\) from the recursive equation (4) gives \[ b_{n} = \frac {b_{n -1} \left (n +r -1\right )}{n +r +3}\tag {5} \] Which for the root \(r = -3\) becomes \[ b_{n} = \frac {b_{n -1} \left (n -4\right )}{n}\tag {6} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -3\) 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 \[ b_{1}=\frac {r}{4+r} \] Which for the root \(r = -3\) becomes \[ b_{1}=-3 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {r}{4+r}\) | \(-3\) |
For \(n = 2\), using the above recursive equation gives \[ b_{2}=\frac {\left (1+r \right ) r}{\left (4+r \right ) \left (5+r \right )} \] Which for the root \(r = -3\) becomes \[ b_{2}=3 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {r}{4+r}\) | \(-3\) |
| \(b_{2}\) | \(\frac {\left (1+r \right ) r}{\left (4+r \right ) \left (5+r \right )}\) | \(3\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {r \left (1+r \right ) \left (2+r \right )}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right )} \] Which for the root \(r = -3\) becomes \[ b_{3}=-1 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {r}{4+r}\) | \(-3\) |
| \(b_{2}\) | \(\frac {\left (1+r \right ) r}{\left (4+r \right ) \left (5+r \right )}\) | \(3\) |
| \(b_{3}\) | \(\frac {r \left (1+r \right ) \left (2+r \right )}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right )}\) | \(-1\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {\left (1+r \right ) r \left (2+r \right ) \left (r +3\right )}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right ) \left (7+r \right )} \] Which for the root \(r = -3\) becomes \[ b_{4}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {r}{4+r}\) | \(-3\) |
| \(b_{2}\) | \(\frac {\left (1+r \right ) r}{\left (4+r \right ) \left (5+r \right )}\) | \(3\) |
| \(b_{3}\) | \(\frac {r \left (1+r \right ) \left (2+r \right )}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right )}\) | \(-1\) |
| \(b_{4}\) | \(\frac {\left (1+r \right ) r \left (2+r \right ) \left (r +3\right )}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right ) \left (7+r \right )}\) | \(0\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {\left (r +3\right ) \left (2+r \right ) \left (1+r \right ) r}{\left (7+r \right ) \left (6+r \right ) \left (5+r \right ) \left (8+r \right )} \] Which for the root \(r = -3\) becomes \[ b_{5}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {r}{4+r}\) | \(-3\) |
| \(b_{2}\) | \(\frac {\left (1+r \right ) r}{\left (4+r \right ) \left (5+r \right )}\) | \(3\) |
| \(b_{3}\) | \(\frac {r \left (1+r \right ) \left (2+r \right )}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right )}\) | \(-1\) |
| \(b_{4}\) | \(\frac {\left (1+r \right ) r \left (2+r \right ) \left (r +3\right )}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right ) \left (7+r \right )}\) | \(0\) |
| \(b_{5}\) | \(\frac {\left (r +3\right ) \left (2+r \right ) \left (1+r \right ) r}{\left (7+r \right ) \left (6+r \right ) \left (5+r \right ) \left (8+r \right )}\) | \(0\) |
For \(n = 6\), using the above recursive equation gives \[ b_{6}=\frac {\left (2+r \right ) r \left (1+r \right ) \left (r +3\right )}{\left (8+r \right ) \left (7+r \right ) \left (6+r \right ) \left (9+r \right )} \] Which for the root \(r = -3\) becomes \[ b_{6}=0 \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {r}{4+r}\) | \(-3\) |
| \(b_{2}\) | \(\frac {\left (1+r \right ) r}{\left (4+r \right ) \left (5+r \right )}\) | \(3\) |
| \(b_{3}\) | \(\frac {r \left (1+r \right ) \left (2+r \right )}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right )}\) | \(-1\) |
| \(b_{4}\) | \(\frac {\left (1+r \right ) r \left (2+r \right ) \left (r +3\right )}{\left (4+r \right ) \left (5+r \right ) \left (6+r \right ) \left (7+r \right )}\) | \(0\) |
| \(b_{5}\) | \(\frac {\left (r +3\right ) \left (2+r \right ) \left (1+r \right ) r}{\left (7+r \right ) \left (6+r \right ) \left (5+r \right ) \left (8+r \right )}\) | \(0\) |
| \(b_{6}\) | \(\frac {\left (2+r \right ) r \left (1+r \right ) \left (r +3\right )}{\left (8+r \right ) \left (7+r \right ) \left (6+r \right ) \left (9+r \right )}\) | \(0\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= x^{3} \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}\dots \right ) \\ &= \frac {1-3 x +3 x^{2}-x^{3}+O\left (x^{7}\right )}{x^{3}} \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^{3} \left (1+\frac {3 x}{7}+\frac {3 x^{2}}{14}+\frac {5 x^{3}}{42}+\frac {x^{4}}{14}+\frac {x^{5}}{22}+\frac {x^{6}}{33}+O\left (x^{7}\right )\right ) + \frac {c_{2} \left (1-3 x +3 x^{2}-x^{3}+O\left (x^{7}\right )\right )}{x^{3}} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{3} \left (1+\frac {3 x}{7}+\frac {3 x^{2}}{14}+\frac {5 x^{3}}{42}+\frac {x^{4}}{14}+\frac {x^{5}}{22}+\frac {x^{6}}{33}+O\left (x^{7}\right )\right )+\frac {c_{2} \left (1-3 x +3 x^{2}-x^{3}+O\left (x^{7}\right )\right )}{x^{3}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{3} \left (1+\frac {3 x}{7}+\frac {3 x^{2}}{14}+\frac {5 x^{3}}{42}+\frac {x^{4}}{14}+\frac {x^{5}}{22}+\frac {x^{6}}{33}+O\left (x^{7}\right )\right )+\frac {c_{2} \left (1-3 x +3 x^{2}-x^{3}+O\left (x^{7}\right )\right )}{x^{3}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{3} \left (1+\frac {3 x}{7}+\frac {3 x^{2}}{14}+\frac {5 x^{3}}{42}+\frac {x^{4}}{14}+\frac {x^{5}}{22}+\frac {x^{6}}{33}+O\left (x^{7}\right )\right )+\frac {c_{2} \left (1-3 x +3 x^{2}-x^{3}+O\left (x^{7}\right )\right )}{x^{3}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -y^{\prime \prime } x^{2} \left (x -1\right )+\left (x^{2}+x \right ) y^{\prime }-9 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 {9 y}{x^{2} \left (x -1\right )}+\frac {\left (x +1\right ) y^{\prime }}{x \left (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} \\ {} & {} & y^{\prime \prime }-\frac {\left (x +1\right ) y^{\prime }}{x \left (x -1\right )}+\frac {9 y}{x^{2} \left (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 {x +1}{x \left (x -1\right )}, P_{3}\left (x \right )=\frac {9}{x^{2} \left (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}}}=-9 \\ {} & \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} \\ {} & {} & y^{\prime \prime } x^{2} \left (x -1\right )-x \left (x +1\right ) y^{\prime }+9 y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..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 y^{\prime \prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =2..3 \\ {} & {} & x^{m}\cdot y^{\prime \prime }=\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 y^{\prime \prime }=\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 (3+r \right ) \left (-3+r \right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (-a_{k} \left (k +r +3\right ) \left (k +r -3\right )+a_{k -1} \left (k +r -1\right ) \left (k +r -3\right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -\left (3+r \right ) \left (-3+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-3, 3\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -\left (k +r -3\right ) \left (\left (-k -r +1\right ) a_{k -1}+a_{k} \left (k +r +3\right )\right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & -\left (k +r -2\right ) \left (\left (-k -r \right ) a_{k}+a_{k +1} \left (k +4+r \right )\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +1}=\frac {\left (k +r \right ) a_{k}}{k +4+r} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-3\hspace {3pt}\textrm {; series terminates at}\hspace {3pt} k =3 \\ {} & {} & a_{k +1}=\frac {\left (k -3\right ) a_{k}}{k +1} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =0 \\ {} & {} & a_{1}=-3 a_{0} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =1 \\ {} & {} & a_{2}=-a_{1} \\ \bullet & {} & \textrm {Express in terms of}\hspace {3pt} a_{0} \\ {} & {} & a_{2}=3 a_{0} \\ \bullet & {} & \textrm {Apply recursion relation for}\hspace {3pt} k =2 \\ {} & {} & a_{3}=-\frac {a_{2}}{3} \\ \bullet & {} & \textrm {Express in terms of}\hspace {3pt} a_{0} \\ {} & {} & a_{3}=-a_{0} \\ \bullet & {} & \textrm {Terminating series solution of the ODE for}\hspace {3pt} r =-3\hspace {3pt}\textrm {. Use reduction of order to find the second linearly independent solution}\hspace {3pt} \\ {} & {} & y=a_{0}\cdot \left (1-3 x +3 x^{2}-x^{3}\right ) \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =3 \\ {} & {} & a_{k +1}=\frac {\left (k +3\right ) a_{k}}{k +7} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =3 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +3}, a_{k +1}=\frac {\left (k +3\right ) a_{k}}{k +7}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=a_{0}\cdot \left (1-3 x +3 x^{2}-x^{3}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k +3}\right ), b_{k +1}=\frac {\left (k +3\right ) b_{k}}{k +7}\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 <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 43
Order:=6; dsolve(x^2*(1-x)*diff(y(x),x$2)+x*(1+x)*diff(y(x),x)-9*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{3} \left (1+\frac {3}{7} x +\frac {3}{14} x^{2}+\frac {5}{42} x^{3}+\frac {1}{14} x^{4}+\frac {1}{22} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\frac {c_{2} \left (-86400+259200 x -259200 x^{2}+86400 x^{3}+\operatorname {O}\left (x^{6}\right )\right )}{x^{3}} \]
✓ Solution by Mathematica
Time used: 0.032 (sec). Leaf size: 57
AsymptoticDSolveValue[x^2*(1-x)*y''[x]+x*(1+x)*y'[x]-9*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (\frac {1}{x^3}-\frac {3}{x^2}+\frac {3}{x}-1\right )+c_2 \left (\frac {x^7}{14}+\frac {5 x^6}{42}+\frac {3 x^5}{14}+\frac {3 x^4}{7}+x^3\right ) \]