Internal problem ID [5471]
Internal file name [OUTPUT/4719_Sunday_June_05_2022_03_03_59_PM_3813674/index.tex]
Book: A treatise on Differential Equations by A. R. Forsyth. 6th edition. 1929. Macmillan Co. ltd.
New York, reprinted 1956
Section: Chapter VI. Note I. Integration of linear equations in series by the method of Frobenius.
page 243
Problem number: Ex. 5, page 256.
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]]
\[ \boxed {x \left (-x^{2}+2\right ) y^{\prime \prime }-\left (x^{2}+4 x +2\right ) \left (\left (1-x \right ) y^{\prime }+y\right )=0} \] 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. \[ \left (-x^{3}+2 x \right ) y^{\prime \prime }+\left (x^{3}+3 x^{2}-2 x -2\right ) y^{\prime }+\left (-x^{2}-4 x -2\right ) y = 0 \] 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 {x^{3}+3 x^{2}-2 x -2}{x \left (x^{2}-2\right )}\\ q(x) &= \frac {x^{2}+4 x +2}{\left (x^{2}-2\right ) x}\\ \end {align*}
| \(p(x)=-\frac {x^{3}+3 x^{2}-2 x -2}{x \left (x^{2}-2\right )}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = \sqrt {2}\) | \(\text {``regular''}\) |
| \(x = -\sqrt {2}\) | \(\text {``regular''}\) |
| \(q(x)=\frac {x^{2}+4 x +2}{\left (x^{2}-2\right ) x}\) | |
| singularity | type |
| \(x = 0\) | \(\text {``regular''}\) |
| \(x = \sqrt {2}\) | \(\text {``regular''}\) |
| \(x = -\sqrt {2}\) | \(\text {``regular''}\) |
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \(\left [0, \sqrt {2}, -\sqrt {2}\right ]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ -y^{\prime \prime } x \left (x^{2}-2\right )+\left (x^{3}+3 x^{2}-2 x -2\right ) y^{\prime }+\left (-x^{2}-4 x -2\right ) y = 0 \] 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} -\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} x^{n +r -2}\right ) x \left (x^{2}-2\right )+\left (x^{3}+3 x^{2}-2 x -2\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (-x^{2}-4 x -2\right ) \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 }}2 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r +2} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}3 x^{1+n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n +r} a_{n} \left (n +r \right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 \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 )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-4 x^{1+n +r} a_{n}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 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^{1+n +r} a_{n} \left (n +r \right ) \left (n +r -1\right )\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-a_{n -2} \left (n +r -2\right ) \left (n +r -3\right ) x^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r +2} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =3}{\sum }}a_{n -3} \left (n +r -3\right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}3 x^{1+n +r} a_{n} \left (n +r \right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}3 a_{n -2} \left (n +r -2\right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 x^{n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right ) \\ \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 }}\left (-4 x^{1+n +r} a_{n}\right ) &= \moverset {\infty }{\munderset {n =2}{\sum }}\left (-4 a_{n -2} x^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 a_{n} x^{n +r}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} x^{n +r -1}\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 \(n +r -1\). \begin{equation} \tag{2B} \moverset {\infty }{\munderset {n =2}{\sum }}\left (-a_{n -2} \left (n +r -2\right ) \left (n +r -3\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}2 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =3}{\sum }}a_{n -3} \left (n +r -3\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =2}{\sum }}3 a_{n -2} \left (n +r -2\right ) x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} \left (n +r -1\right ) x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-2 \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 )+\moverset {\infty }{\munderset {n =2}{\sum }}\left (-4 a_{n -2} x^{n +r -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-2 a_{n -1} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ 2 x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )-2 \left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ 2 x^{-1+r} a_{0} r \left (-1+r \right )-2 r a_{0} x^{-1+r} = 0 \] Or \[ \left (2 x^{-1+r} r \left (-1+r \right )-2 r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ 2 r \,x^{-1+r} \left (-2+r \right ) = 0 \] Since the above is true for all \(x\) then the indicial equation becomes \[ 2 r \left (-2+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 2\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ 2 r \,x^{-1+r} \left (-2+r \right ) = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([2, 0]\).
Since \(r_1 - r_2 = 2\) 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^{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 )+\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 +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}\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 \(a_{0} = 1\). Substituting \(n = 1\) in Eq. (2B) gives \[ a_{1} = \frac {1}{-1+r} \] Substituting \(n = 2\) in Eq. (2B) gives \[ a_{2} = \frac {r^{2}-5 r +10}{2 \left (-1+r \right ) \left (2+r \right )} \] For \(3\le n\) the recursive equation is \begin{equation} \tag{3} -a_{n -2} \left (n +r -2\right ) \left (n +r -3\right )+2 a_{n} \left (n +r \right ) \left (n +r -1\right )+a_{n -3} \left (n +r -3\right )+3 a_{n -2} \left (n +r -2\right )-2 a_{n -1} \left (n +r -1\right )-2 a_{n} \left (n +r \right )-a_{n -3}-4 a_{n -2}-2 a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {n^{2} a_{n -2}+2 n r a_{n -2}+r^{2} a_{n -2}-n a_{n -3}-8 n a_{n -2}+2 n a_{n -1}-r a_{n -3}-8 r a_{n -2}+2 r a_{n -1}+4 a_{n -3}+16 a_{n -2}}{2 n^{2}+4 n r +2 r^{2}-4 n -4 r}\tag {4} \] Which for the root \(r = 2\) becomes \[ a_{n} = \frac {n^{2} a_{n -2}+\left (-a_{n -3}-4 a_{n -2}+2 a_{n -1}\right ) n +2 a_{n -3}+4 a_{n -2}+4 a_{n -1}}{2 n \left (n +2\right )}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 2\) and after as more terms are found using the above recursive equation.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{-1+r}\) | \(1\) |
| \(a_{2}\) | \(\frac {r^{2}-5 r +10}{2 \left (-1+r \right ) \left (2+r \right )}\) | \(\frac {1}{2}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {r^{2}-5 r +10}{2 r^{3}+4 r^{2}-2 r -4} \] Which for the root \(r = 2\) becomes \[ a_{3}={\frac {1}{6}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{-1+r}\) | \(1\) |
| \(a_{2}\) | \(\frac {r^{2}-5 r +10}{2 \left (-1+r \right ) \left (2+r \right )}\) | \(\frac {1}{2}\) |
| \(a_{3}\) | \(\frac {r^{2}-5 r +10}{2 r^{3}+4 r^{2}-2 r -4}\) | \(\frac {1}{6}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {r^{4}-6 r^{3}+17 r^{2}-32 r +40}{4 r^{4}+24 r^{3}+28 r^{2}-24 r -32} \] Which for the root \(r = 2\) becomes \[ a_{4}={\frac {1}{24}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{-1+r}\) | \(1\) |
| \(a_{2}\) | \(\frac {r^{2}-5 r +10}{2 \left (-1+r \right ) \left (2+r \right )}\) | \(\frac {1}{2}\) |
| \(a_{3}\) | \(\frac {r^{2}-5 r +10}{2 r^{3}+4 r^{2}-2 r -4}\) | \(\frac {1}{6}\) |
| \(a_{4}\) | \(\frac {r^{4}-6 r^{3}+17 r^{2}-32 r +40}{4 r^{4}+24 r^{3}+28 r^{2}-24 r -32}\) | \(\frac {1}{24}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {r^{4}-6 r^{3}+17 r^{2}-32 r +40}{4 r^{5}+36 r^{4}+100 r^{3}+60 r^{2}-104 r -96} \] Which for the root \(r = 2\) becomes \[ a_{5}={\frac {1}{120}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{-1+r}\) | \(1\) |
| \(a_{2}\) | \(\frac {r^{2}-5 r +10}{2 \left (-1+r \right ) \left (2+r \right )}\) | \(\frac {1}{2}\) |
| \(a_{3}\) | \(\frac {r^{2}-5 r +10}{2 r^{3}+4 r^{2}-2 r -4}\) | \(\frac {1}{6}\) |
| \(a_{4}\) | \(\frac {r^{4}-6 r^{3}+17 r^{2}-32 r +40}{4 r^{4}+24 r^{3}+28 r^{2}-24 r -32}\) | \(\frac {1}{24}\) |
| \(a_{5}\) | \(\frac {r^{4}-6 r^{3}+17 r^{2}-32 r +40}{4 r^{5}+36 r^{4}+100 r^{3}+60 r^{2}-104 r -96}\) | \(\frac {1}{120}\) |
Using the above table, then the solution \(y_{1}\left (x \right )\) is \begin {align*} y_{1}\left (x \right )&= x^{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}\dots \right ) \\ &= x^{2} \left (1+x +\frac {x^{2}}{2}+\frac {x^{3}}{6}+\frac {x^{4}}{24}+\frac {x^{5}}{120}+O\left (x^{6}\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 \(N=2\). Now we need to determine if \(C\) is zero or not. This is done by finding \(\lim _{r\rightarrow r_{2}}a_{2}\left (r \right )\). 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_{2} \\ &= \frac {r^{2}-5 r +10}{2 \left (-1+r \right ) \left (2+r \right )} \end {align*}
Therefore \begin {align*} \lim _{r\rightarrow r_{2}}\frac {r^{2}-5 r +10}{2 \left (-1+r \right ) \left (2+r \right )}&= \lim _{r\rightarrow 0}\frac {r^{2}-5 r +10}{2 \left (-1+r \right ) \left (2+r \right )}\\ &= -{\frac {5}{2}} \end {align*}
The limit is \(-{\frac {5}{2}}\). 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} \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 \(b_{0} = 1\). Substituting \(n = 1\) in Eq(3) gives \[ b_{1} = \frac {1}{-1+r} \] Substituting \(n = 2\) in Eq(3) gives \[ b_{2} = \frac {r^{2}-5 r +10}{2 \left (-1+r \right ) \left (2+r \right )} \] For \(3\le n\) the recursive equation is \begin{equation} \tag{4} -b_{n -2} \left (n +r -2\right ) \left (n +r -3\right )+2 b_{n} \left (n +r \right ) \left (n +r -1\right )+b_{n -3} \left (n +r -3\right )+3 b_{n -2} \left (n +r -2\right )-2 b_{n -1} \left (n +r -1\right )-2 \left (n +r \right ) b_{n}-b_{n -3}-4 b_{n -2}-2 b_{n -1} = 0 \end{equation} Which for for the root \(r = 0\) becomes \begin{equation} \tag{4A} -b_{n -2} \left (n -2\right ) \left (n -3\right )+2 b_{n} n \left (n -1\right )+b_{n -3} \left (n -3\right )+3 b_{n -2} \left (n -2\right )-2 b_{n -1} \left (n -1\right )-2 n b_{n}-b_{n -3}-4 b_{n -2}-2 b_{n -1} = 0 \end{equation} Solving for \(b_{n}\) from the recursive equation (4) gives \[ b_{n} = \frac {n^{2} b_{n -2}+2 n r b_{n -2}+r^{2} b_{n -2}-n b_{n -3}-8 n b_{n -2}+2 n b_{n -1}-r b_{n -3}-8 r b_{n -2}+2 r b_{n -1}+4 b_{n -3}+16 b_{n -2}}{2 n^{2}+4 n r +2 r^{2}-4 n -4 r}\tag {5} \] Which for the root \(r = 0\) becomes \[ b_{n} = \frac {n^{2} b_{n -2}-n b_{n -3}-8 n b_{n -2}+2 n b_{n -1}+4 b_{n -3}+16 b_{n -2}}{2 n^{2}-4 n}\tag {6} \] At this point, it is a good idea to keep track of \(b_{n}\) in a table both before substituting \(r = 0\) and after as more terms are found using the above recursive equation.
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{-1+r}\) | \(-1\) |
| \(b_{2}\) | \(\frac {r^{2}-5 r +10}{2 \left (-1+r \right ) \left (2+r \right )}\) | \(-{\frac {5}{2}}\) |
For \(n = 3\), using the above recursive equation gives \[ b_{3}=\frac {r^{2}-5 r +10}{2 \left (1+r \right ) \left (-1+r \right ) \left (2+r \right )} \] Which for the root \(r = 0\) becomes \[ b_{3}=-{\frac {5}{2}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{-1+r}\) | \(-1\) |
| \(b_{2}\) | \(\frac {r^{2}-5 r +10}{2 \left (-1+r \right ) \left (2+r \right )}\) | \(-{\frac {5}{2}}\) |
| \(b_{3}\) | \(\frac {r^{2}-5 r +10}{2 r^{3}+4 r^{2}-2 r -4}\) | \(-{\frac {5}{2}}\) |
For \(n = 4\), using the above recursive equation gives \[ b_{4}=\frac {r^{4}-6 r^{3}+17 r^{2}-32 r +40}{4 \left (1+r \right ) \left (-1+r \right ) \left (r^{2}+6 r +8\right )} \] Which for the root \(r = 0\) becomes \[ b_{4}=-{\frac {5}{4}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{-1+r}\) | \(-1\) |
| \(b_{2}\) | \(\frac {r^{2}-5 r +10}{2 \left (-1+r \right ) \left (2+r \right )}\) | \(-{\frac {5}{2}}\) |
| \(b_{3}\) | \(\frac {r^{2}-5 r +10}{2 r^{3}+4 r^{2}-2 r -4}\) | \(-{\frac {5}{2}}\) |
| \(b_{4}\) | \(\frac {r^{4}-6 r^{3}+17 r^{2}-32 r +40}{4 r^{4}+24 r^{3}+28 r^{2}-24 r -32}\) | \(-{\frac {5}{4}}\) |
For \(n = 5\), using the above recursive equation gives \[ b_{5}=\frac {r^{4}-6 r^{3}+17 r^{2}-32 r +40}{4 \left (r +3\right ) \left (1+r \right ) \left (-1+r \right ) \left (r^{2}+6 r +8\right )} \] Which for the root \(r = 0\) becomes \[ b_{5}=-{\frac {5}{12}} \] And the table now becomes
| \(n\) | \(b_{n ,r}\) | \(b_{n}\) |
| \(b_{0}\) | \(1\) | \(1\) |
| \(b_{1}\) | \(\frac {1}{-1+r}\) | \(-1\) |
| \(b_{2}\) | \(\frac {r^{2}-5 r +10}{2 \left (-1+r \right ) \left (2+r \right )}\) | \(-{\frac {5}{2}}\) |
| \(b_{3}\) | \(\frac {r^{2}-5 r +10}{2 r^{3}+4 r^{2}-2 r -4}\) | \(-{\frac {5}{2}}\) |
| \(b_{4}\) | \(\frac {r^{4}-6 r^{3}+17 r^{2}-32 r +40}{4 r^{4}+24 r^{3}+28 r^{2}-24 r -32}\) | \(-{\frac {5}{4}}\) |
| \(b_{5}\) | \(\frac {r^{4}-6 r^{3}+17 r^{2}-32 r +40}{4 r^{5}+36 r^{4}+100 r^{3}+60 r^{2}-104 r -96}\) | \(-{\frac {5}{12}}\) |
Using the above table, then the solution \(y_{2}\left (x \right )\) is \begin {align*} y_{2}\left (x \right )&= b_{0}+b_{1} x +b_{2} x^{2}+b_{3} x^{3}+b_{4} x^{4}+b_{5} x^{5}+b_{6} x^{6}\dots \\ &= 1-x -\frac {5 x^{2}}{2}-\frac {5 x^{3}}{2}-\frac {5 x^{4}}{4}-\frac {5 x^{5}}{12}+O\left (x^{6}\right ) \end {align*}
Therefore the homogeneous solution is \begin{align*} y_h(x) &= c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \\ &= c_{1} x^{2} \left (1+x +\frac {x^{2}}{2}+\frac {x^{3}}{6}+\frac {x^{4}}{24}+\frac {x^{5}}{120}+O\left (x^{6}\right )\right ) + c_{2} \left (1-x -\frac {5 x^{2}}{2}-\frac {5 x^{3}}{2}-\frac {5 x^{4}}{4}-\frac {5 x^{5}}{12}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} x^{2} \left (1+x +\frac {x^{2}}{2}+\frac {x^{3}}{6}+\frac {x^{4}}{24}+\frac {x^{5}}{120}+O\left (x^{6}\right )\right )+c_{2} \left (1-x -\frac {5 x^{2}}{2}-\frac {5 x^{3}}{2}-\frac {5 x^{4}}{4}-\frac {5 x^{5}}{12}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} x^{2} \left (1+x +\frac {x^{2}}{2}+\frac {x^{3}}{6}+\frac {x^{4}}{24}+\frac {x^{5}}{120}+O\left (x^{6}\right )\right )+c_{2} \left (1-x -\frac {5 x^{2}}{2}-\frac {5 x^{3}}{2}-\frac {5 x^{4}}{4}-\frac {5 x^{5}}{12}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} x^{2} \left (1+x +\frac {x^{2}}{2}+\frac {x^{3}}{6}+\frac {x^{4}}{24}+\frac {x^{5}}{120}+O\left (x^{6}\right )\right )+c_{2} \left (1-x -\frac {5 x^{2}}{2}-\frac {5 x^{3}}{2}-\frac {5 x^{4}}{4}-\frac {5 x^{5}}{12}+O\left (x^{6}\right )\right ) \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -\left (\frac {d}{d x}y^{\prime }\right ) x \left (x^{2}-2\right )+\left (x^{3}+3 x^{2}-2 x -2\right ) y^{\prime }+\left (-x^{2}-4 x -2\right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=-\frac {\left (x^{2}+4 x +2\right ) y}{\left (x^{2}-2\right ) x}+\frac {\left (x^{3}+3 x^{2}-2 x -2\right ) y^{\prime }}{x \left (x^{2}-2\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 (x^{3}+3 x^{2}-2 x -2\right ) y^{\prime }}{x \left (x^{2}-2\right )}+\frac {\left (x^{2}+4 x +2\right ) y}{\left (x^{2}-2\right ) x}=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^{3}+3 x^{2}-2 x -2}{x \left (x^{2}-2\right )}, P_{3}\left (x \right )=\frac {x^{2}+4 x +2}{\left (x^{2}-2\right ) 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}}}=-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}}}=0 \\ {} & \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} \\ {} & {} & \left (\frac {d}{d x}y^{\prime }\right ) x \left (x^{2}-2\right )+\left (-x^{3}-3 x^{2}+2 x +2\right ) y^{\prime }+\left (x^{2}+4 x +2\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} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..3 \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =1..3 \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -2+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2-m \\ {} & {} & x^{m}\cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =-2+m}{\sum }}a_{k +2-m} \left (k +2-m +r \right ) \left (k +1-m +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -2 a_{0} r \left (-2+r \right ) x^{-1+r}+\left (-2 a_{1} \left (1+r \right ) \left (-1+r \right )+2 a_{0} \left (1+r \right )\right ) x^{r}+\left (-2 a_{2} \left (2+r \right ) r +2 a_{1} \left (2+r \right )+a_{0} \left (-2+r \right )^{2}\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (-2 a_{k +1} \left (k +r +1\right ) \left (k +r -1\right )+2 a_{k} \left (k +r +1\right )+a_{k -1} \left (k -3+r \right )^{2}-a_{k -2} \left (k -3+r \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -2 r \left (-2+r \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, 2\right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} x \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [-2 a_{1} \left (1+r \right ) \left (-1+r \right )+2 a_{0} \left (1+r \right )=0, -2 a_{2} \left (2+r \right ) r +2 a_{1} \left (2+r \right )+a_{0} \left (-2+r \right )^{2}=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=\frac {a_{0}}{-1+r}, a_{2}=\frac {a_{0} \left (r^{2}-5 r +10\right )}{2 \left (r^{2}+r -2\right )}\right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k -1} \left (k -3+r \right )^{2}-2 a_{k +1} \left (k +r +1\right ) \left (k +r -1\right )+\left (2 a_{k}-a_{k -2}\right ) k +\left (2 a_{k}-a_{k -2}\right ) r +2 a_{k}+3 a_{k -2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +1} \left (k +r -1\right )^{2}-2 a_{k +3} \left (k +3+r \right ) \left (k +r +1\right )+\left (2 a_{k +2}-a_{k}\right ) \left (k +2\right )+\left (2 a_{k +2}-a_{k}\right ) r +2 a_{k +2}+3 a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=\frac {k^{2} a_{k +1}+2 k r a_{k +1}+r^{2} a_{k +1}-a_{k} k -2 k a_{k +1}+2 k a_{k +2}-a_{k} r -2 r a_{k +1}+2 r a_{k +2}+a_{k}+a_{k +1}+6 a_{k +2}}{2 \left (k +3+r \right ) \left (k +r +1\right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +3}=\frac {k^{2} a_{k +1}-a_{k} k -2 k a_{k +1}+2 k a_{k +2}+a_{k}+a_{k +1}+6 a_{k +2}}{2 \left (k +3\right ) \left (k +1\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 {k^{2} a_{k +1}-a_{k} k -2 k a_{k +1}+2 k a_{k +2}+a_{k}+a_{k +1}+6 a_{k +2}}{2 \left (k +3\right ) \left (k +1\right )}, a_{1}=-a_{0}, a_{2}=-\frac {5 a_{0}}{2}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =2 \\ {} & {} & a_{k +3}=\frac {k^{2} a_{k +1}-a_{k} k +2 k a_{k +1}+2 k a_{k +2}-a_{k}+a_{k +1}+10 a_{k +2}}{2 \left (k +5\right ) \left (k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =2 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +2}, a_{k +3}=\frac {k^{2} a_{k +1}-a_{k} k +2 k a_{k +1}+2 k a_{k +2}-a_{k}+a_{k +1}+10 a_{k +2}}{2 \left (k +5\right ) \left (k +3\right )}, a_{1}=a_{0}, a_{2}=\frac {a_{0}}{2}\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 +2}\right ), a_{k +3}=\frac {k^{2} a_{k +1}-k a_{k}-2 k a_{k +1}+2 k a_{k +2}+a_{k}+a_{k +1}+6 a_{k +2}}{2 \left (k +3\right ) \left (k +1\right )}, a_{1}=-a_{0}, a_{2}=-\frac {5 a_{0}}{2}, b_{k +3}=\frac {k^{2} b_{k +1}-k b_{k}+2 k b_{k +1}+2 k b_{k +2}-b_{k}+b_{k +1}+10 b_{k +2}}{2 \left (k +5\right ) \left (k +3\right )}, b_{1}=b_{0}, b_{2}=\frac {b_{0}}{2}\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) Reducible group (found another exponential solution) <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.063 (sec). Leaf size: 44
Order:=6; dsolve(x*(2-x^2)*diff(y(x),x$2)-(x^2+4*x+2)*((1-x)*diff(y(x),x)+y(x))=0,y(x),type='series',x=0);
\[ y \left (x \right ) = c_{1} x^{2} \left (1+x +\frac {1}{2} x^{2}+\frac {1}{6} x^{3}+\frac {1}{24} x^{4}+\frac {1}{120} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+c_{2} \left (-2+2 x +4 x^{2}+4 x^{3}+2 x^{4}+\frac {2}{3} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.041 (sec). Leaf size: 64
AsymptoticDSolveValue[x*(2-x^2)*y''[x]-(x^2+4*x+2)*((1-x)*y'[x]+y[x])==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (-\frac {5 x^4}{4}-\frac {5 x^3}{2}-\frac {5 x^2}{2}-x+1\right )+c_2 \left (\frac {x^6}{24}+\frac {x^5}{6}+\frac {x^4}{2}+x^3+x^2\right ) \]