Internal problem ID [5216]
Internal file name [OUTPUT/4709_Sunday_June_05_2022_03_03_48_PM_60152543/index.tex]
Book: Schaums Outline Differential Equations, 4th edition. Bronson and Costa. McGraw Hill
2014
Section: Chapter 27. Power series solutions of linear DE with variable coefficients. Supplementary
Problems. page 274
Problem number: Problem 27.28.
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]]
\[ \boxed {\left (x +1\right ) y^{\prime \prime }+\frac {y^{\prime }}{x}+y x=0} \] 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. \[ \left (x +1\right ) y^{\prime \prime }+\frac {y^{\prime }}{x}+y x = 0 \] 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 {1}{x \left (x +1\right )}\\ q(x) &= \frac {x}{x +1}\\ \end {align*}
Combining everything together gives the following summary of singularities for the ode as
Regular singular points : \([-1, 0]\)
Irregular singular points : \([\infty ]\)
Since \(x = 0\) is regular singular point, then Frobenius power series is used. The ode is normalized to be \[ y^{\prime \prime } x \left (x +1\right )+y^{\prime }+y x^{2} = 0 \] 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 +1\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\right ) x^{2} = 0 \end{equation} Which simplifies to \begin{equation} \tag{2A} \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^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{2+n +r} a_{n}\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 }}x^{n +r} a_{n} \left (n +r \right ) \left (n +r -1\right ) &= \moverset {\infty }{\munderset {n =1}{\sum }}a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r -1} \\ \moverset {\infty }{\munderset {n =0}{\sum }}x^{2+n +r} a_{n} &= \moverset {\infty }{\munderset {n =3}{\sum }}a_{n -3} 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 =1}{\sum }}a_{n -1} \left (n +r -1\right ) \left (n +r -2\right ) x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )\right )+\left (\moverset {\infty }{\munderset {n =0}{\sum }}\left (n +r \right ) a_{n} x^{n +r -1}\right )+\left (\moverset {\infty }{\munderset {n =3}{\sum }}a_{n -3} x^{n +r -1}\right ) = 0 \end{equation} The indicial equation is obtained from \(n = 0\). From Eq (2B) this gives \[ x^{n +r -1} a_{n} \left (n +r \right ) \left (n +r -1\right )+\left (n +r \right ) a_{n} x^{n +r -1} = 0 \] When \(n = 0\) the above becomes \[ x^{-1+r} a_{0} r \left (-1+r \right )+r a_{0} x^{-1+r} = 0 \] Or \[ \left (x^{-1+r} r \left (-1+r \right )+r \,x^{-1+r}\right ) a_{0} = 0 \] Since \(a_{0}\neq 0\) then the above simplifies to \[ x^{-1+r} r^{2} = 0 \] Since the above is true for all \(x\) 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 \[ x^{-1+r} r^{2} = 0 \] 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 (x \right ) &= \moverset {\infty }{\munderset {n =0}{\sum }}a_{n} x^{n +r}\tag {1A} \end {align*}
Now the second solution \(y_{2}\) is found using \begin {align*} y_{2}\left (x \right ) &= y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +r}\right )\tag {1B} \end {align*}
Then the general solution will be \[ y = c_{1} y_{1}\left (x \right )+c_{2} y_{2}\left (x \right ) \] 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 \(y_{1}\left (x \right )\). 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 \[ a_{1} = -\frac {r \left (-1+r \right )}{\left (1+r \right )^{2}} \] Substituting \(n = 2\) in Eq. (2B) gives \[ a_{2} = \frac {r^{2} \left (-1+r \right )}{\left (1+r \right ) \left (2+r \right )^{2}} \] For \(3\le n\) 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} \left (n +r \right )+a_{n -3} = 0 \end{equation} Solving for \(a_{n}\) from recursive equation (4) gives \[ a_{n} = -\frac {n^{2} a_{n -1}+2 n r a_{n -1}+r^{2} a_{n -1}-3 n a_{n -1}-3 r a_{n -1}+a_{n -3}+2 a_{n -1}}{n^{2}+2 n r +r^{2}}\tag {4} \] Which for the root \(r = 0\) becomes \[ a_{n} = \frac {\left (-n^{2}+3 n -2\right ) a_{n -1}-a_{n -3}}{n^{2}}\tag {5} \] 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.
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r \left (-1+r \right )}{\left (1+r \right )^{2}}\) | \(0\) |
| \(a_{2}\) | \(\frac {r^{2} \left (-1+r \right )}{\left (1+r \right ) \left (2+r \right )^{2}}\) | \(0\) |
For \(n = 3\), using the above recursive equation gives \[ a_{3}=\frac {-r^{3}+r^{2}-r -2}{\left (r +3\right )^{2} \left (2+r \right )} \] Which for the root \(r = 0\) becomes \[ a_{3}=-{\frac {1}{9}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r \left (-1+r \right )}{\left (1+r \right )^{2}}\) | \(0\) |
| \(a_{2}\) | \(\frac {r^{2} \left (-1+r \right )}{\left (1+r \right ) \left (2+r \right )^{2}}\) | \(0\) |
| \(a_{3}\) | \(\frac {-r^{3}+r^{2}-r -2}{\left (r +3\right )^{2} \left (2+r \right )}\) | \(-{\frac {1}{9}}\) |
For \(n = 4\), using the above recursive equation gives \[ a_{4}=\frac {r^{5}+r^{4}+r^{3}+5 r^{2}+2 r +2}{\left (r +4\right )^{2} \left (1+r \right )^{2} \left (r +3\right )} \] Which for the root \(r = 0\) becomes \[ a_{4}={\frac {1}{24}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r \left (-1+r \right )}{\left (1+r \right )^{2}}\) | \(0\) |
| \(a_{2}\) | \(\frac {r^{2} \left (-1+r \right )}{\left (1+r \right ) \left (2+r \right )^{2}}\) | \(0\) |
| \(a_{3}\) | \(\frac {-r^{3}+r^{2}-r -2}{\left (r +3\right )^{2} \left (2+r \right )}\) | \(-{\frac {1}{9}}\) |
| \(a_{4}\) | \(\frac {r^{5}+r^{4}+r^{3}+5 r^{2}+2 r +2}{\left (r +4\right )^{2} \left (1+r \right )^{2} \left (r +3\right )}\) | \(\frac {1}{24}\) |
For \(n = 5\), using the above recursive equation gives \[ a_{5}=\frac {-r^{7}-5 r^{6}-10 r^{5}-17 r^{4}-25 r^{3}-26 r^{2}-16 r -8}{\left (r +5\right )^{2} \left (2+r \right )^{2} \left (1+r \right )^{2} \left (r +4\right )} \] Which for the root \(r = 0\) becomes \[ a_{5}=-{\frac {1}{50}} \] And the table now becomes
| \(n\) | \(a_{n ,r}\) | \(a_{n}\) |
| \(a_{0}\) | \(1\) | \(1\) |
| \(a_{1}\) | \(-\frac {r \left (-1+r \right )}{\left (1+r \right )^{2}}\) | \(0\) |
| \(a_{2}\) | \(\frac {r^{2} \left (-1+r \right )}{\left (1+r \right ) \left (2+r \right )^{2}}\) | \(0\) |
| \(a_{3}\) | \(\frac {-r^{3}+r^{2}-r -2}{\left (r +3\right )^{2} \left (2+r \right )}\) | \(-{\frac {1}{9}}\) |
| \(a_{4}\) | \(\frac {r^{5}+r^{4}+r^{3}+5 r^{2}+2 r +2}{\left (r +4\right )^{2} \left (1+r \right )^{2} \left (r +3\right )}\) | \(\frac {1}{24}\) |
| \(a_{5}\) | \(\frac {-r^{7}-5 r^{6}-10 r^{5}-17 r^{4}-25 r^{3}-26 r^{2}-16 r -8}{\left (r +5\right )^{2} \left (2+r \right )^{2} \left (1+r \right )^{2} \left (r +4\right )}\) | \(-{\frac {1}{50}}\) |
Using the above table, then the first solution \(y_{1}\left (x \right )\) becomes \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^{3}}{9}+\frac {x^{4}}{24}-\frac {x^{5}}{50}+O\left (x^{6}\right ) \\ \end{align*} Now the second solution is found. The second solution is given by \[ y_{2}\left (x \right ) = y_{1}\left (x \right ) \ln \left (x \right )+\left (\moverset {\infty }{\munderset {n =1}{\sum }}b_{n} x^{n +r}\right ) \] 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 |
| \(b_{1}\) | \(-\frac {r \left (-1+r \right )}{\left (1+r \right )^{2}}\) | \(0\) | \(\frac {-3 r +1}{\left (1+r \right )^{3}}\) | \(1\) |
| \(b_{2}\) | \(\frac {r^{2} \left (-1+r \right )}{\left (1+r \right ) \left (2+r \right )^{2}}\) | \(0\) | \(\frac {6 r^{3}+4 r^{2}-4 r}{\left (1+r \right )^{2} \left (2+r \right )^{3}}\) | \(0\) |
| \(b_{3}\) | \(\frac {-r^{3}+r^{2}-r -2}{\left (r +3\right )^{2} \left (2+r \right )}\) | \(-{\frac {1}{9}}\) | \(\frac {-9 r^{3}-13 r^{2}+20 r +8}{\left (r +3\right )^{3} \left (2+r \right )^{2}}\) | \(\frac {2}{27}\) |
| \(b_{4}\) | \(\frac {r^{5}+r^{4}+r^{3}+5 r^{2}+2 r +2}{\left (r +4\right )^{2} \left (1+r \right )^{2} \left (r +3\right )}\) | \(\frac {1}{24}\) | \(\frac {12 r^{6}+64 r^{5}+84 r^{4}+8 r^{3}+8 r^{2}+36 r -44}{\left (r +4\right )^{3} \left (1+r \right )^{3} \left (r +3\right )^{2}}\) | \(-{\frac {11}{144}}\) |
| \(b_{5}\) | \(\frac {-r^{7}-5 r^{6}-10 r^{5}-17 r^{4}-25 r^{3}-26 r^{2}-16 r -8}{\left (r +5\right )^{2} \left (2+r \right )^{2} \left (1+r \right )^{2} \left (r +4\right )}\) | \(-{\frac {1}{50}}\) | \(\frac {-15 r^{9}-184 r^{8}-884 r^{7}-2178 r^{6}-3099 r^{5}-2886 r^{4}-1926 r^{3}-668 r^{2}+440 r +528}{\left (r +5\right )^{3} \left (2+r \right )^{3} \left (1+r \right )^{3} \left (r +4\right )^{2}}\) | \(\frac {33}{1000}\) |
The above table gives all values of \(b_{n}\) needed. Hence the second solution is \begin{align*} y_{2}\left (x \right )&=y_{1}\left (x \right ) \ln \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 \\ &= \left (1-\frac {x^{3}}{9}+\frac {x^{4}}{24}-\frac {x^{5}}{50}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x +\frac {2 x^{3}}{27}-\frac {11 x^{4}}{144}+\frac {33 x^{5}}{1000}+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} \left (1-\frac {x^{3}}{9}+\frac {x^{4}}{24}-\frac {x^{5}}{50}+O\left (x^{6}\right )\right ) + c_{2} \left (\left (1-\frac {x^{3}}{9}+\frac {x^{4}}{24}-\frac {x^{5}}{50}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x +\frac {2 x^{3}}{27}-\frac {11 x^{4}}{144}+\frac {33 x^{5}}{1000}+O\left (x^{6}\right )\right ) \\ \end{align*} Hence the final solution is \begin{align*} y &= y_h \\ &= c_{1} \left (1-\frac {x^{3}}{9}+\frac {x^{4}}{24}-\frac {x^{5}}{50}+O\left (x^{6}\right )\right )+c_{2} \left (\left (1-\frac {x^{3}}{9}+\frac {x^{4}}{24}-\frac {x^{5}}{50}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x +\frac {2 x^{3}}{27}-\frac {11 x^{4}}{144}+\frac {33 x^{5}}{1000}+O\left (x^{6}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \left (1-\frac {x^{3}}{9}+\frac {x^{4}}{24}-\frac {x^{5}}{50}+O\left (x^{6}\right )\right )+c_{2} \left (\left (1-\frac {x^{3}}{9}+\frac {x^{4}}{24}-\frac {x^{5}}{50}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x +\frac {2 x^{3}}{27}-\frac {11 x^{4}}{144}+\frac {33 x^{5}}{1000}+O\left (x^{6}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} \left (1-\frac {x^{3}}{9}+\frac {x^{4}}{24}-\frac {x^{5}}{50}+O\left (x^{6}\right )\right )+c_{2} \left (\left (1-\frac {x^{3}}{9}+\frac {x^{4}}{24}-\frac {x^{5}}{50}+O\left (x^{6}\right )\right ) \ln \left (x \right )+x +\frac {2 x^{3}}{27}-\frac {11 x^{4}}{144}+\frac {33 x^{5}}{1000}+O\left (x^{6}\right )\right ) \] Verified OK.
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 <- Heun successful: received ODE is equivalent to the HeunC ODE, case a <> 0, e <> 0, c = 0 `
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 49
Order:=6; dsolve((x+1)*diff(y(x),x$2)+1/x*diff(y(x),x)+x*y(x)=0,y(x),type='series',x=0);
\[ y \left (x \right ) = \left (c_{2} \ln \left (x \right )+c_{1} \right ) \left (1-\frac {1}{9} x^{3}+\frac {1}{24} x^{4}-\frac {1}{50} x^{5}+\operatorname {O}\left (x^{6}\right )\right )+\left (x +\frac {2}{27} x^{3}-\frac {11}{144} x^{4}+\frac {33}{1000} x^{5}+\operatorname {O}\left (x^{6}\right )\right ) c_{2} \]
✓ Solution by Mathematica
Time used: 0.005 (sec). Leaf size: 82
AsymptoticDSolveValue[(1+x)*y''[x]+1/x*y'[x]+x*y[x]==0,y[x],{x,0,5}]
\[ y(x)\to c_1 \left (-\frac {x^5}{50}+\frac {x^4}{24}-\frac {x^3}{9}+1\right )+c_2 \left (\frac {33 x^5}{1000}-\frac {11 x^4}{144}+\frac {2 x^3}{27}+\left (-\frac {x^5}{50}+\frac {x^4}{24}-\frac {x^3}{9}+1\right ) \log (x)+x\right ) \]