Internal problem ID [5469]
Internal file name [OUTPUT/4960_Wednesday_February_14_2024_02_06_01_AM_85208353/index.tex]
Book: Schaums Outline. Theory and problems of Differential Equations, 1st edition. Frank Ayres.
McGraw Hill 1952
Section: Chapter 26. Integration in series (singular points). Supplemetary problems. Page
218
Problem number: 21.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second order series method. Regular singular point. Repeated root"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x^{3} y^{\prime \prime }+\left (x^{2}+x \right ) y^{\prime }-y=0} \] With the expansion point for the power series method at \(x = \infty \).
Since expansion is around \(\infty \), then the independent variable \(x\) is replaced by \(\frac {1}{t}\) and the expansion is made around \(t = 0\) and after solving, the solution is changed back to \(x\) using \(x = \frac {1}{t}\). Changing variables results in the new ode \begin {align*} -\frac {-\left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right ) t^{2}-2 t \left (\frac {d}{d t}y \left (t \right )\right )}{t}-\left (\frac {1}{t^{2}}+\frac {1}{t}\right ) \left (\frac {d}{d t}y \left (t \right )\right ) t^{2}-y \left (t \right ) = 0 \end {align*}
The transformed ODE is now solved. The type of the expansion point is first determined. This is done on the homogeneous part of the ODE. \[ \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right ) t +\left (-t +1\right ) \left (\frac {d}{d t}y \left (t \right )\right )-y \left (t \right ) = 0 \] The following is summary of singularities for the above ode. Writing the ode as \begin {align*} \frac {d^{2}}{d t^{2}}y \left (t \right )+p(t) \frac {d}{d t}y \left (t \right ) + q(t) y \left (t \right ) &=0 \end {align*}
Where \begin {align*} p(t) &= -\frac {t -1}{t}\\ q(t) &= -\frac {1}{t}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([0]\)
Irregular singular points : \([\infty ]\)
Since \(t = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right ) t +\left (-t +1\right ) \left (\frac {d}{d t}y \left (t \right )\right )-y \left (t \right ) = 0 \] Let the solution be represented as Frobenius power series of the form \[ y \left (t \right ) = \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r} \] Then \begin{align*} \frac {d}{d t}y \left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1} \\ \frac {d^{2}}{d t^{2}}y \left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) \left (n +r -1\right ) a_{n} t^{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} t^{n +r -2}\right ) t +\left (-t +1\right ) \left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1}\right )-\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r}\right ) = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-t^{n +r} a_{n} \left (n +r \right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1}\right )+\moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} t^{n +r}\right ) = 0 \end{equation} The next step is to make all powers of \(t\) be \(n +r -1\) in each summation term. Going over each summation term above with power of \(t\) in it which is not already \(t^{n +r -1}\) and adjusting the power and the corresponding index gives \begin{align*} \moverset {\infty }{\munderset {n =0}{\sum }}\left (-t^{n +r} a_{n} \left (n +r \right )\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} \left (n +r -1\right ) t^{n +r -1}\right ) \\ \moverset {\infty }{\munderset {n =0}{\sum }}\left (-a_{n} t^{n +r}\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} t^{n +r -1}\right ) \\ \end{align*} Substituting all the above in Eq (2A) gives the following equation where now all powers of \(t\) are the same and equal to \(n +r -1\). \begin{equation} \tag{2B} \left (\moverset {\infty }{\munderset {n =0}{\sum }}t^{n +r -1} 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 ) t^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} t^{n +r -1}\right )+\moverset {\infty }{\munderset {n =1}{\sum }}\left (-a_{n -1} t^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ t^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+\left (n +r \right ) a_{n} t^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ t^{-1+r} a_{0} r \left (-1+r \right )+r a_{0} t^{-1+r} = 0 \] Or \[ \left (t^{-1+r} r \left (-1+r \right )+r \,t^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ t^{-1+r} r^{2} = 0 \] Since the above is true for all \(t\) then the indicial equation becomes \[ r^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \begin {align*} r_1 &= 0\\ r_2 &= 0 \end {align*}
Since \(a_{0}\neq 0\) then the indicial equation becomes \[ t^{-1+r} r^{2} = 0 \] Solving for \(r\) gives the roots of the indicial equation as \([0, 0]\).
Since the root of the indicial equation is repeated, then we can construct two linearly independent solutions. The first solution has the form \begin {align*} y_{1}\left (t \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} t^{n +r}\tag {1A} \end {align*}
Now the second solution \(y_{2}\) is found using \begin {align*} y_{2}\left (t \right ) &= y_{1}\left (t \right ) \ln \left (t \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} t^{n +r}\right )\tag {1B} \end {align*}
Then the general solution will be \[ y \left (t \right ) = c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right ) \] In Eq (1B) the sum starts from 1 and not zero. In Eq (1A), \(a_{0}\) is never zero, and is arbitrary and is typically taken as \(a_{0} = 1\), and \(\{c_{1}, c_{2}\}\) are two arbitray constants of integration which can be found from initial conditions. We start by finding the first solution \(y_{1}\left (t \right )\). 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} \left (n +r \right ) \left (n +r -1\right )-a_{n -1} \left (n +r -1\right )+a_{n} \left (n +r \right )-a_{n -1} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = \frac {a_{n -1}}{n +r}\tag {4} \] Which for the root \(r = 0\) becomes \[ a_{n} = \frac {a_{n -1}}{n}\tag {5} \] At this point, it is a good idea to keep track of \(a_{n}\) in a table both before substituting \(r = 0\) 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 \[ a_{1}=\frac {1}{1+r} \] Which for the root \(r = 0\) becomes \[ a_{1}=1 \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{1+r}\) | \(1\) |
For \(n = 2\), using the above recursive equation gives \[ a_{2}=\frac {1}{\left (1+r \right ) \left (2+r \right )} \] Which for the root \(r = 0\) becomes \[ a_{2}={\frac {1}{2}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{1+r}\) | \(1\) |
| \(a_{2}\) | \(\frac {1}{\left (1+r \right ) \left (2+r \right )}\) | \(\frac {1}{2}\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {1}{\left (3+r \right ) \left (1+r \right ) \left (2+r \right )} \] Which for the root \(r = 0\) becomes \[ a_{3}={\frac {1}{6}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{1+r}\) | \(1\) |
| \(a_{2}\) | \(\frac {1}{\left (1+r \right ) \left (2+r \right )}\) | \(\frac {1}{2}\) |
| \(a_{3}\) | \(\frac {1}{\left (3+r \right ) \left (1+r \right ) \left (2+r \right )}\) | \(\frac {1}{6}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {1}{\left (3+r \right ) \left (1+r \right ) \left (2+r \right ) \left (4+r \right )} \] Which for the root \(r = 0\) becomes \[ a_{4}={\frac {1}{24}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{1+r}\) | \(1\) |
| \(a_{2}\) | \(\frac {1}{\left (1+r \right ) \left (2+r \right )}\) | \(\frac {1}{2}\) |
| \(a_{3}\) | \(\frac {1}{\left (3+r \right ) \left (1+r \right ) \left (2+r \right )}\) | \(\frac {1}{6}\) |
| \(a_{4}\) | \(\frac {1}{\left (3+r \right ) \left (1+r \right ) \left (2+r \right ) \left (4+r \right )}\) | \(\frac {1}{24}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {1}{\left (5+r \right ) \left (3+r \right ) \left (1+r \right ) \left (2+r \right ) \left (4+r \right )} \] Which for the root \(r = 0\) becomes \[ a_{5}={\frac {1}{120}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(\frac {1}{1+r}\) | \(1\) |
| \(a_{2}\) | \(\frac {1}{\left (1+r \right ) \left (2+r \right )}\) | \(\frac {1}{2}\) |
| \(a_{3}\) | \(\frac {1}{\left (3+r \right ) \left (1+r \right ) \left (2+r \right )}\) | \(\frac {1}{6}\) |
| \(a_{4}\) | \(\frac {1}{\left (3+r \right ) \left (1+r \right ) \left (2+r \right ) \left (4+r \right )}\) | \(\frac {1}{24}\) |
| \(a_{5}\) | \(\frac {1}{\left (5+r \right ) \left (3+r \right ) \left (1+r \right ) \left (2+r \right ) \left (4+r \right )}\) | \(\frac {1}{120}\) |
Using the above table, then the first solution \(y_{1}\left (t \right )\) becomes \begin{align*} y_{1}\left (t \right )&= a_{0}+a_{1} t +a_{2} t^{2}+a_{3} t^{3}+a_{4} t^{4}+a_{5} t^{5}+a_{6} t^{6}\dots \\ &= t +1+\frac {t^{2}}{2}+\frac {t^{3}}{6}+\frac {t^{4}}{24}+\frac {t^{5}}{120}+O\left (t^{6}\right ) \\ \end{align*} Now the second solution is found. The second solution is given by \[ y_{2}\left (t \right ) = y_{1}\left (t \right ) \ln \left (t \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} t^{n +r}\right ) \] Where \(b_{n}\) is found using \[ b_{n} = \frac {d}{d r}a_{n ,r} \] And the above is then evaluated at \(r = 0\). The above table for \(a_{n ,r}\) is used for this purpose. Computing the derivatives gives the following table
| \(n\) | \(b_{n ,r}\) | \(a_{n}\) | \(b_{n ,r} = \frac {d}{d r}a_{n ,r}\) | \(b_{n}\left (r =0\right )\) |
| \(b_{0}\) | \(1\) | \(1\) | N/A since \(b_{n}\) starts from 1 | N/A |
| \(b_{1}\) | \(\frac {1}{1+r}\) | \(1\) | \(-\frac {1}{\left (1+r \right )^{2}}\) | \(-1\) |
| \(b_{2}\) | \(\frac {1}{\left (1+r \right ) \left (2+r \right )}\) | \(\frac {1}{2}\) | \(\frac {-3-2 r}{\left (1+r \right )^{2} \left (2+r \right )^{2}}\) | \(-{\frac {3}{4}}\) |
| \(b_{3}\) | \(\frac {1}{\left (3+r \right ) \left (1+r \right ) \left (2+r \right )}\) | \(\frac {1}{6}\) | \(\frac {-3 r^{2}-12 r -11}{\left (3+r \right )^{2} \left (1+r \right )^{2} \left (2+r \right )^{2}}\) | \(-{\frac {11}{36}}\) |
| \(b_{4}\) | \(\frac {1}{\left (3+r \right ) \left (1+r \right ) \left (2+r \right ) \left (4+r \right )}\) | \(\frac {1}{24}\) | \(\frac {-4 r^{3}-30 r^{2}-70 r -50}{\left (3+r \right )^{2} \left (1+r \right )^{2} \left (2+r \right )^{2} \left (4+r \right )^{2}}\) | \(-{\frac {25}{288}}\) |
| \(b_{5}\) | \(\frac {1}{\left (5+r \right ) \left (3+r \right ) \left (1+r \right ) \left (2+r \right ) \left (4+r \right )}\) | \(\frac {1}{120}\) | \(\frac {-5 r^{4}-60 r^{3}-255 r^{2}-450 r -274}{\left (5+r \right )^{2} \left (3+r \right )^{2} \left (1+r \right )^{2} \left (2+r \right )^{2} \left (4+r \right )^{2}}\) | \(-{\frac {137}{7200}}\) |
The above table gives all values of \(b_{n}\) needed. Hence the second solution is \begin{align*} y_{2}\left (t \right )&=y_{1}\left (t \right ) \ln \left (t \right )+b_{0}+b_{1} t +b_{2} t^{2}+b_{3} t^{3}+b_{4} t^{4}+b_{5} t^{5}+b_{6} t^{6}\dots \\ &= \left (t +1+\frac {t^{2}}{2}+\frac {t^{3}}{6}+\frac {t^{4}}{24}+\frac {t^{5}}{120}+O\left (t^{6}\right )\right ) \ln \left (t \right )-t -\frac {3 t^{2}}{4}-\frac {11 t^{3}}{36}-\frac {25 t^{4}}{288}-\frac {137 t^{5}}{7200}+O\left (t^{6}\right ) \\ \end{align*} Therefore the homogeneous solution is \begin{align*} y_h(t) &= c_{1} y_{1}\left (t \right )+c_{2} y_{2}\left (t \right ) \\ &= c_{1} \left (t +1+\frac {t^{2}}{2}+\frac {t^{3}}{6}+\frac {t^{4}}{24}+\frac {t^{5}}{120}+O\left (t^{6}\right )\right ) + c_{2} \left (\left (t +1+\frac {t^{2}}{2}+\frac {t^{3}}{6}+\frac {t^{4}}{24}+\frac {t^{5}}{120}+O\left (t^{6}\right )\right ) \ln \left (t \right )-t -\frac {3 t^{2}}{4}-\frac {11 t^{3}}{36}-\frac {25 t^{4}}{288}-\frac {137 t^{5}}{7200}+O\left (t^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y \left (t \right ) &= y_h \\ &= c_{1} \left (t +1+\frac {t^{2}}{2}+\frac {t^{3}}{6}+\frac {t^{4}}{24}+\frac {t^{5}}{120}+O\left (t^{6}\right )\right )+c_{2} \left (\left (t +1+\frac {t^{2}}{2}+\frac {t^{3}}{6}+\frac {t^{4}}{24}+\frac {t^{5}}{120}+O\left (t^{6}\right )\right ) \ln \left (t \right )-t -\frac {3 t^{2}}{4}-\frac {11 t^{3}}{36}-\frac {25 t^{4}}{288}-\frac {137 t^{5}}{7200}+O\left (t^{6}\right )\right ) \\ \end{align*} Replacing \(t\) by \(\frac {1}{x}\) gives \begin {align*} y = c_{1} \left (\frac {1}{x}+1+\frac {1}{2 x^{2}}+\frac {1}{6 x^{3}}+\frac {1}{24 x^{4}}+\frac {1}{120 x^{5}}+O\left (\frac {1}{x^{6}}\right )\right )+c_{2} \left (\left (\frac {1}{x}+1+\frac {1}{2 x^{2}}+\frac {1}{6 x^{3}}+\frac {1}{24 x^{4}}+\frac {1}{120 x^{5}}+O\left (\frac {1}{x^{6}}\right )\right ) \ln \left (\frac {1}{x}\right )-\frac {1}{x}-\frac {3}{4 x^{2}}-\frac {11}{36 x^{3}}-\frac {25}{288 x^{4}}-\frac {137}{7200 x^{5}}+O\left (\frac {1}{x^{6}}\right )\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (\frac {1}{x}+1+\frac {1}{2 x^{2}}+\frac {1}{6 x^{3}}+\frac {1}{24 x^{4}}+\frac {1}{120 x^{5}}+O\left (\frac {1}{x^{6}}\right )\right )+c_{2} \left (\left (\frac {1}{x}+1+\frac {1}{2 x^{2}}+\frac {1}{6 x^{3}}+\frac {1}{24 x^{4}}+\frac {1}{120 x^{5}}+O\left (\frac {1}{x^{6}}\right )\right ) \ln \left (\frac {1}{x}\right )-\frac {1}{x}-\frac {3}{4 x^{2}}-\frac {11}{36 x^{3}}-\frac {25}{288 x^{4}}-\frac {137}{7200 x^{5}}+O\left (\frac {1}{x^{6}}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (\frac {1}{x}+1+\frac {1}{2 x^{2}}+\frac {1}{6 x^{3}}+\frac {1}{24 x^{4}}+\frac {1}{120 x^{5}}+O\left (\frac {1}{x^{6}}\right )\right )+c_{2} \left (\left (\frac {1}{x}+1+\frac {1}{2 x^{2}}+\frac {1}{6 x^{3}}+\frac {1}{24 x^{4}}+\frac {1}{120 x^{5}}+O\left (\frac {1}{x^{6}}\right )\right ) \ln \left (\frac {1}{x}\right )-\frac {1}{x}-\frac {3}{4 x^{2}}-\frac {11}{36 x^{3}}-\frac {25}{288 x^{4}}-\frac {137}{7200 x^{5}}+O\left (\frac {1}{x^{6}}\right )\right ) \] Verified OK.
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: 0.0 (sec). Leaf size: 179
Order:=6; dsolve(x^3*diff(y(x),x$2)+(x^2+x)*diff(y(x),x)-y(x)=0,y(x),type='series',x=infinity);
\[ y \left (x \right ) = \left (1+\frac {\left (x -\operatorname {Infinity} \right )^{2}}{2 \operatorname {Infinity}^{3}}+\frac {\left (-4 \operatorname {Infinity} -1\right ) \left (x -\operatorname {Infinity} \right )^{3}}{6 \operatorname {Infinity}^{5}}+\frac {\left (18 \operatorname {Infinity}^{2}+10 \operatorname {Infinity} +1\right ) \left (x -\operatorname {Infinity} \right )^{4}}{24 \operatorname {Infinity}^{7}}+\frac {\left (-96 \operatorname {Infinity}^{3}-86 \operatorname {Infinity}^{2}-18 \operatorname {Infinity} -1\right ) \left (x -\operatorname {Infinity} \right )^{5}}{120 \operatorname {Infinity}^{9}}\right ) y \left (\operatorname {Infinity} \right )+\left (x -\operatorname {Infinity} +\frac {\left (-\operatorname {Infinity}^{2}-\operatorname {Infinity} \right ) \left (x -\operatorname {Infinity} \right )^{2}}{2 \operatorname {Infinity}^{3}}+\frac {\left (2 \operatorname {Infinity}^{3}+5 \operatorname {Infinity}^{2}+\operatorname {Infinity} \right ) \left (x -\operatorname {Infinity} \right )^{3}}{6 \operatorname {Infinity}^{5}}+\frac {\left (-6 \operatorname {Infinity}^{4}-26 \operatorname {Infinity}^{3}-11 \operatorname {Infinity}^{2}-\operatorname {Infinity} \right ) \left (x -\operatorname {Infinity} \right )^{4}}{24 \operatorname {Infinity}^{7}}+\frac {\left (24 \operatorname {Infinity}^{5}+154 \operatorname {Infinity}^{4}+102 \operatorname {Infinity}^{3}+19 \operatorname {Infinity}^{2}+\operatorname {Infinity} \right ) \left (x -\operatorname {Infinity} \right )^{5}}{120 \operatorname {Infinity}^{9}}\right ) D\left (y \right )\left (\operatorname {Infinity} \right )+O\left (x^{6}\right ) \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 124
AsymptoticDSolveValue[x^3*y''[x]+(x^2+x)*y'[x]-y[x]==0,y[x],{x,infinity,5}]
\[ y(x)\to c_1 \left (\frac {1}{120 x^5}+\frac {1}{24 x^4}+\frac {1}{6 x^3}+\frac {1}{2 x^2}+\frac {1}{x}+1\right )+c_2 \left (-\frac {137}{7200 x^5}-\frac {\log (x)}{120 x^5}-\frac {25}{288 x^4}-\frac {\log (x)}{24 x^4}-\frac {11}{36 x^3}-\frac {\log (x)}{6 x^3}-\frac {3}{4 x^2}-\frac {\log (x)}{2 x^2}-\frac {1}{x}-\frac {\log (x)}{x}-\log (x)\right ) \]