Internal problem ID [14808]
Internal file name [OUTPUT/14488_Monday_April_08_2024_06_25_54_AM_91222896/index.tex]
Book: INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton.
Fourth edition 2014. ElScAe. 2014
Section: Chapter 4. Higher Order Equations. Exercises 4.9, page 215
Problem number: 8.
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]]
\[ \boxed {7 x y^{\prime \prime }+10 y^{\prime }+\left (-x^{2}+1\right ) y=0} \] 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. \[ 7 x y^{\prime \prime }+10 y^{\prime }+\left (-x^{2}+1\right ) y = 0 \] 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 {10}{7 x}\\ q(x) &= -\frac {x^{2}-1}{7 x}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0, \infty , -\infty ]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ 7 x y^{\prime \prime }+10 y^{\prime }+\left (-x^{2}+1\right ) y = 0 \] 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} 7 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right ) x +10 \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (-x^{2}+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 }}7 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}10 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r +2} 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 \(n +r -1\) in each summation term. Going over each summation term above with power of \(x\) in it which is not already \(x^{n +r -1}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-x^{n +r +2} a_{n}\right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}\left (-a_{n -3} x^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r} &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r -1} \\ \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 \(n +r -1\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}7 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}10 \left (n +r \right ) a_{n} x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =3}{\sum }}\left (-a_{n -3} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 7 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+10 \left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ 7 x^{-1+r} a_{0} r \left (-1+r \right )+10 r a_{0} x^{-1+r} = 0 \] Or \[ \left (7 x^{-1+r} r \left (-1+r \right )+10 r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ r \,x^{-1+r} \left (3+7 r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 7 r^{2}+3 r = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 0\\ r_2 &= -{\frac {3}{7}} \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ r \,x^{-1+r} \left (3+7 r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \(\left [0, -{\frac {3}{7}}\right ]\).
Since \(r_1 - r_2 = {\frac {3}{7}}\) 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}\\ y_{2}\left (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}b_{n} x^{n -\frac {3}{7}} \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 \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = -\frac {1}{7 r^{2}+17 r +10} \] Substituting \(n = 2\) in Eq. (2B) gives \[ a_{2} = \frac {1}{49 r^{4}+336 r^{3}+835 r^{2}+888 r +340} \] For \(3\le n\) the recursive equation is \begin{equation} \tag{3} 7 a_{n} \left (n +r \right ) \left (n +r -1\right )+10 a_{n} \left (n +r \right )-a_{n -3}+a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {a_{n -3}-a_{n -1}}{7 n^{2}+14 n r +7 r^{2}+3 n +3 r}\tag {4} \] Which for the root \(r = 0\) becomes \[ a_{n} = \frac {a_{n -3}-a_{n -1}}{n \left (7 n +3\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 0\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{7 r^{2}+17 r +10}\) | \(-{\frac {1}{10}}\) |
| \(a_{2}\) | \(\frac {1}{49 r^{4}+336 r^{3}+835 r^{2}+888 r +340}\) | \(\frac {1}{340}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {49 r^{4}+336 r^{3}+835 r^{2}+888 r +339}{\left (49 r^{4}+336 r^{3}+835 r^{2}+888 r +340\right ) \left (7 r^{2}+45 r +72\right )} \] Which for the root \(r = 0\) becomes \[ a_{3}={\frac {113}{8160}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{7 r^{2}+17 r +10}\) | \(-{\frac {1}{10}}\) |
| \(a_{2}\) | \(\frac {1}{49 r^{4}+336 r^{3}+835 r^{2}+888 r +340}\) | \(\frac {1}{340}\) |
| \(a_{3}\) | \(\frac {49 r^{4}+336 r^{3}+835 r^{2}+888 r +339}{\left (49 r^{4}+336 r^{3}+835 r^{2}+888 r +340\right ) \left (7 r^{2}+45 r +72\right )}\) | \(\frac {113}{8160}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {-98 r^{4}-868 r^{3}-2972 r^{2}-4650 r -2787}{\left (49 r^{4}+336 r^{3}+835 r^{2}+888 r +340\right ) \left (7 r^{2}+45 r +72\right ) \left (7 r^{2}+59 r +124\right )} \] Which for the root \(r = 0\) becomes \[ a_{4}=-{\frac {929}{1011840}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{7 r^{2}+17 r +10}\) | \(-{\frac {1}{10}}\) |
| \(a_{2}\) | \(\frac {1}{49 r^{4}+336 r^{3}+835 r^{2}+888 r +340}\) | \(\frac {1}{340}\) |
| \(a_{3}\) | \(\frac {49 r^{4}+336 r^{3}+835 r^{2}+888 r +339}{\left (49 r^{4}+336 r^{3}+835 r^{2}+888 r +340\right ) \left (7 r^{2}+45 r +72\right )}\) | \(\frac {113}{8160}\) |
| \(a_{4}\) | \(\frac {-98 r^{4}-868 r^{3}-2972 r^{2}-4650 r -2787}{\left (49 r^{4}+336 r^{3}+835 r^{2}+888 r +340\right ) \left (7 r^{2}+45 r +72\right ) \left (7 r^{2}+59 r +124\right )}\) | \(-{\frac {929}{1011840}}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {147 r^{4}+1596 r^{3}+6999 r^{2}+14478 r +11715}{\left (49 r^{4}+336 r^{3}+835 r^{2}+888 r +340\right ) \left (7 r^{2}+45 r +72\right ) \left (7 r^{2}+59 r +124\right ) \left (7 r^{2}+73 r +190\right )} \] Which for the root \(r = 0\) becomes \[ a_{5}={\frac {781}{38449920}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {1}{7 r^{2}+17 r +10}\) | \(-{\frac {1}{10}}\) |
| \(a_{2}\) | \(\frac {1}{49 r^{4}+336 r^{3}+835 r^{2}+888 r +340}\) | \(\frac {1}{340}\) |
| \(a_{3}\) | \(\frac {49 r^{4}+336 r^{3}+835 r^{2}+888 r +339}{\left (49 r^{4}+336 r^{3}+835 r^{2}+888 r +340\right ) \left (7 r^{2}+45 r +72\right )}\) | \(\frac {113}{8160}\) |
| \(a_{4}\) | \(\frac {-98 r^{4}-868 r^{3}-2972 r^{2}-4650 r -2787}{\left (49 r^{4}+336 r^{3}+835 r^{2}+888 r +340\right ) \left (7 r^{2}+45 r +72\right ) \left (7 r^{2}+59 r +124\right )}\) | \(-{\frac {929}{1011840}}\) |
| \(a_{5}\) | \(\frac {147 r^{4}+1596 r^{3}+6999 r^{2}+14478 r +11715}{\left (49 r^{4}+336 r^{3}+835 r^{2}+888 r +340\right ) \left (7 r^{2}+45 r +72\right ) \left (7 r^{2}+59 r +124\right ) \left (7 r^{2}+73 r +190\right )}\) | \(\frac {781}{38449920}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= a_{0}+a_{1} x +a_{2} x^{2}+a_{3} x^{3}+a_{4} x^{4}+a_{5} x^{5}+a_{6} x^{6}\dots \\ &= 1-\frac {x}{10}+\frac {x^{2}}{340}+\frac {113 x^{3}}{8160}-\frac {929 x^{4}}{1011840}+\frac {781 x^{5}}{38449920}+O\left (x^{6}\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 \(b_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ b_{1} = -\frac {1}{7 r^{2}+17 r +10} \] Substituting \(n = 2\) in Eq. (2B) gives \[ b_{2} = \frac {1}{49 r^{4}+336 r^{3}+835 r^{2}+888 r +340} \] For \(3\le n\) the recursive equation is \begin{equation} \tag{3} 7 b_{n} \left (n +r \right ) \left (n +r -1\right )+10 \left (n +r \right ) b_{n}-b_{n -3}+b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from recursive equation (4) gives \[ b_{n} = \frac {b_{n -3}-b_{n -1}}{7 n^{2}+14 n r +7 r^{2}+3 n +3 r}\tag {4} \] Which for the root \(r = -{\frac {3}{7}}\) becomes \[ b_{n} = \frac {b_{n -3}-b_{n -1}}{n \left (7 n -3\right )}\tag {5} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = -{\frac {3}{7}}\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {1}{7 r^{2}+17 r +10}\) | \(-{\frac {1}{4}}\) |
| \(b_{2}\) | \(\frac {1}{49 r^{4}+336 r^{3}+835 r^{2}+888 r +340}\) | \(\frac {1}{88}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {49 r^{4}+336 r^{3}+835 r^{2}+888 r +339}{\left (49 r^{4}+336 r^{3}+835 r^{2}+888 r +340\right ) \left (7 r^{2}+45 r +72\right )} \] Which for the root \(r = -{\frac {3}{7}}\) becomes \[ b_{3}={\frac {29}{1584}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {1}{7 r^{2}+17 r +10}\) | \(-{\frac {1}{4}}\) |
| \(b_{2}\) | \(\frac {1}{49 r^{4}+336 r^{3}+835 r^{2}+888 r +340}\) | \(\frac {1}{88}\) |
| \(b_{3}\) | \(\frac {49 r^{4}+336 r^{3}+835 r^{2}+888 r +339}{\left (49 r^{4}+336 r^{3}+835 r^{2}+888 r +340\right ) \left (7 r^{2}+45 r +72\right )}\) | \(\frac {29}{1584}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {-98 r^{4}-868 r^{3}-2972 r^{2}-4650 r -2787}{\left (49 r^{4}+336 r^{3}+835 r^{2}+888 r +340\right ) \left (7 r^{2}+45 r +72\right ) \left (7 r^{2}+59 r +124\right )} \] Which for the root \(r = -{\frac {3}{7}}\) becomes \[ b_{4}=-{\frac {17}{6336}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {1}{7 r^{2}+17 r +10}\) | \(-{\frac {1}{4}}\) |
| \(b_{2}\) | \(\frac {1}{49 r^{4}+336 r^{3}+835 r^{2}+888 r +340}\) | \(\frac {1}{88}\) |
| \(b_{3}\) | \(\frac {49 r^{4}+336 r^{3}+835 r^{2}+888 r +339}{\left (49 r^{4}+336 r^{3}+835 r^{2}+888 r +340\right ) \left (7 r^{2}+45 r +72\right )}\) | \(\frac {29}{1584}\) |
| \(b_{4}\) | \(\frac {-98 r^{4}-868 r^{3}-2972 r^{2}-4650 r -2787}{\left (49 r^{4}+336 r^{3}+835 r^{2}+888 r +340\right ) \left (7 r^{2}+45 r +72\right ) \left (7 r^{2}+59 r +124\right )}\) | \(-{\frac {17}{6336}}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {147 r^{4}+1596 r^{3}+6999 r^{2}+14478 r +11715}{\left (49 r^{4}+336 r^{3}+835 r^{2}+888 r +340\right ) \left (7 r^{2}+45 r +72\right ) \left (7 r^{2}+59 r +124\right ) \left (7 r^{2}+73 r +190\right )} \] Which for the root \(r = -{\frac {3}{7}}\) becomes \[ b_{5}={\frac {89}{1013760}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(-\frac {1}{7 r^{2}+17 r +10}\) | \(-{\frac {1}{4}}\) |
| \(b_{2}\) | \(\frac {1}{49 r^{4}+336 r^{3}+835 r^{2}+888 r +340}\) | \(\frac {1}{88}\) |
| \(b_{3}\) | \(\frac {49 r^{4}+336 r^{3}+835 r^{2}+888 r +339}{\left (49 r^{4}+336 r^{3}+835 r^{2}+888 r +340\right ) \left (7 r^{2}+45 r +72\right )}\) | \(\frac {29}{1584}\) |
| \(b_{4}\) | \(\frac {-98 r^{4}-868 r^{3}-2972 r^{2}-4650 r -2787}{\left (49 r^{4}+336 r^{3}+835 r^{2}+888 r +340\right ) \left (7 r^{2}+45 r +72\right ) \left (7 r^{2}+59 r +124\right )}\) | \(-{\frac {17}{6336}}\) |
| \(b_{5}\) | \(\frac {147 r^{4}+1596 r^{3}+6999 r^{2}+14478 r +11715}{\left (49 r^{4}+336 r^{3}+835 r^{2}+888 r +340\right ) \left (7 r^{2}+45 r +72\right ) \left (7 r^{2}+59 r +124\right ) \left (7 r^{2}+73 r +190\right )}\) | \(\frac {89}{1013760}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= 1 \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-\frac {x}{4}+\frac {x^{2}}{88}+\frac {29 x^{3}}{1584}-\frac {17 x^{4}}{6336}+\frac {89 x^{5}}{1013760}+O\left (x^{6}\right )}{x^{\frac {3}{7}}} \end {align*}
Therefore the homogeneous solution is \begin{align*} y_h(x) &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= c_{1} \left (1-\frac {x}{10}+\frac {x^{2}}{340}+\frac {113 x^{3}}{8160}-\frac {929 x^{4}}{1011840}+\frac {781 x^{5}}{38449920}+O\left (x^{6}\right )\right ) + \frac {c_{2} \left (1-\frac {x}{4}+\frac {x^{2}}{88}+\frac {29 x^{3}}{1584}-\frac {17 x^{4}}{6336}+\frac {89 x^{5}}{1013760}+O\left (x^{6}\right )\right )}{x^{\frac {3}{7}}} \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \left (1-\frac {x}{10}+\frac {x^{2}}{340}+\frac {113 x^{3}}{8160}-\frac {929 x^{4}}{1011840}+\frac {781 x^{5}}{38449920}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-\frac {x}{4}+\frac {x^{2}}{88}+\frac {29 x^{3}}{1584}-\frac {17 x^{4}}{6336}+\frac {89 x^{5}}{1013760}+O\left (x^{6}\right )\right )}{x^{\frac {3}{7}}} \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (1-\frac {x}{10}+\frac {x^{2}}{340}+\frac {113 x^{3}}{8160}-\frac {929 x^{4}}{1011840}+\frac {781 x^{5}}{38449920}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-\frac {x}{4}+\frac {x^{2}}{88}+\frac {29 x^{3}}{1584}-\frac {17 x^{4}}{6336}+\frac {89 x^{5}}{1013760}+O\left (x^{6}\right )\right )}{x^{\frac {3}{7}}} \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (1-\frac {x}{10}+\frac {x^{2}}{340}+\frac {113 x^{3}}{8160}-\frac {929 x^{4}}{1011840}+\frac {781 x^{5}}{38449920}+O\left (x^{6}\right )\right )+\frac {c_{2} \left (1-\frac {x}{4}+\frac {x^{2}}{88}+\frac {29 x^{3}}{1584}-\frac {17 x^{4}}{6336}+\frac {89 x^{5}}{1013760}+O\left (x^{6}\right )\right )}{x^{\frac {3}{7}}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 7 y^{\prime \prime } x +10 y^{\prime }+\left (-x^{2}+1\right ) 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 {\left (x^{2}-1\right ) y}{7 x}-\frac {10 y^{\prime }}{7 x} \\ \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 {10 y^{\prime }}{7 x}-\frac {\left (x^{2}-1\right ) y}{7 x}=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=\frac {10}{7 x}, P_{3}\left (x \right )=-\frac {x^{2}-1}{7 x}\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}}}=\frac {10}{7} \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=0 \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & 7 y^{\prime \prime } x +10 y^{\prime }+\left (-x^{2}+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..2 \\ {} & {} & 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} y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & y^{\prime }=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x \cdot y^{\prime \prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & x \cdot y^{\prime \prime }=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +1+r \right ) \left (k +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (3+7 r \right ) x^{-1+r}+\left (a_{1} \left (1+r \right ) \left (10+7 r \right )+a_{0}\right ) x^{r}+\left (a_{2} \left (2+r \right ) \left (17+7 r \right )+a_{1}\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (7 k +10+7 r \right )+a_{k}-a_{k -2}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (3+7 r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -\frac {3}{7}\right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} x \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [a_{1} \left (1+r \right ) \left (10+7 r \right )+a_{0}=0, a_{2} \left (2+r \right ) \left (17+7 r \right )+a_{1}=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=-\frac {a_{0}}{7 r^{2}+17 r +10}, a_{2}=\frac {a_{0}}{49 r^{4}+336 r^{3}+835 r^{2}+888 r +340}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & 7 \left (k +\frac {10}{7}+r \right ) \left (k +1+r \right ) a_{k +1}+a_{k}-a_{k -2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & 7 \left (k +\frac {24}{7}+r \right ) \left (k +3+r \right ) a_{k +3}+a_{k +2}-a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=\frac {-a_{k +2}+a_{k}}{\left (7 k +24+7 r \right ) \left (k +3+r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +3}=\frac {-a_{k +2}+a_{k}}{\left (7 k +24\right ) \left (k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +3}=\frac {-a_{k +2}+a_{k}}{\left (7 k +24\right ) \left (k +3\right )}, a_{1}=-\frac {a_{0}}{10}, a_{2}=\frac {a_{0}}{340}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-\frac {3}{7} \\ {} & {} & a_{k +3}=\frac {-a_{k +2}+a_{k}}{\left (7 k +21\right ) \left (k +\frac {18}{7}\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-\frac {3}{7} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -\frac {3}{7}}, a_{k +3}=\frac {-a_{k +2}+a_{k}}{\left (7 k +21\right ) \left (k +\frac {18}{7}\right )}, a_{1}=-\frac {a_{0}}{4}, a_{2}=\frac {a_{0}}{88}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}b_{k} x^{k -\frac {3}{7}}\right ), a_{k +3}=\frac {-a_{k +2}+a_{k}}{\left (7 k +24\right ) \left (k +3\right )}, a_{1}=-\frac {a_{0}}{10}, a_{2}=\frac {a_{0}}{340}, b_{k +3}=\frac {-b_{k +2}+b_{k}}{\left (7 k +21\right ) \left (k +\frac {18}{7}\right )}, b_{1}=-\frac {b_{0}}{4}, b_{2}=\frac {b_{0}}{88}\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 -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius trying a solution in terms of MeijerG functions -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) trying a symmetry of the form [xi=0, eta=F(x)] trying differential order: 2; exact nonlinear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients trying 2nd order, integrating factor of the form mu(x,y) -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Whittaker -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients trying to convert to an ODE of Bessel type trying to convert to an ODE of Bessel type -> trying reduction of order to Riccati trying Riccati sub-methods: -> trying a symmetry pattern of the form [F(x)*G(y), 0] -> trying a symmetry pattern of the form [0, F(x)*G(y)] -> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] --- Trying Lie symmetry methods, 2nd order --- `, `-> Computing symmetries using: way = 3`[0, y]
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 44
Order:=6; dsolve(7*x*diff(y(x),x$2)+10*diff(y(x),x)+(1-x^2)*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \frac {c_{1} \left (1-\frac {1}{4} x +\frac {1}{88} x^{2}+\frac {29}{1584} x^{3}-\frac {17}{6336} x^{4}+\frac {89}{1013760} x^{5}+\operatorname {O}\left (x^{6}\right )\right )}{x^{\frac {3}{7}}}+c_{2} \left (1-\frac {1}{10} x +\frac {1}{340} x^{2}+\frac {113}{8160} x^{3}-\frac {929}{1011840} x^{4}+\frac {781}{38449920} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.002 (sec). Leaf size: 85
AsymptoticDSolveValue[7*x*y''[x]+10*y'[x]+(1-x^2)*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (\frac {781 x^5}{38449920}-\frac {929 x^4}{1011840}+\frac {113 x^3}{8160}+\frac {x^2}{340}-\frac {x}{10}+1\right )+\frac {c_2 \left (\frac {89 x^5}{1013760}-\frac {17 x^4}{6336}+\frac {29 x^3}{1584}+\frac {x^2}{88}-\frac {x}{4}+1\right )}{x^{3/7}} \]