Internal problem ID [10961]
Internal file name [OUTPUT/10218_Sunday_December_31_2023_11_10_11_AM_65170388/index.tex]
Book: Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev.
Second edition
Section: Chapter 2, Second-Order Differential Equations. section 2.1.2-4 Equation of form
\(x^2 y''+f(x)y'+g(x)y=0\)
Problem number: 138.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "unknown"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
Unable to solve or complete the solution.
\[ \boxed {x^{2} y^{\prime \prime }+\left (a \,x^{2}+b x \right ) y^{\prime }+\left (k \left (a -k \right ) x^{2}+\left (a n +b k -2 k n \right ) x +n \left (b -n -1\right )\right ) y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (\frac {d}{d x}y^{\prime }\right )+\left (a x +b \right ) y^{\prime } x +\left (-n^{2}+\left (\left (a -2 k \right ) x +b -1\right ) n +k \left (\left (a -k \right ) x +b \right ) x \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 (k \,x^{2} a -k^{2} x^{2}+a n x +b k x -2 k n x +n b -n^{2}-n \right ) y}{x^{2}}-\frac {\left (a x +b \right ) y^{\prime }}{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} \\ {} & {} & \frac {d}{d x}y^{\prime }+\frac {\left (a x +b \right ) y^{\prime }}{x}+\frac {\left (k \,x^{2} a -k^{2} x^{2}+a n x +b k x -2 k n x +n b -n^{2}-n \right ) y}{x^{2}}=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 {a x +b}{x}, P_{3}\left (x \right )=\frac {k \,x^{2} a -k^{2} x^{2}+a n x +b k x -2 k n x +n b -n^{2}-n}{x^{2}}\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}}}=b \\ {} & \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}}}=n b -n^{2}-n \\ {} & \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} \\ {} & {} & x^{2} \left (\frac {d}{d x}y^{\prime }\right )+\left (a x +b \right ) y^{\prime } x +\left (k \,x^{2} a -k^{2} x^{2}+a n x +b k x -2 k n x +n b -n^{2}-n \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 =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^{2}\cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\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} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} \left (n +r \right ) \left (b +r -1-n \right ) x^{r}+\left (a_{1} \left (1+n +r \right ) \left (b +r -n \right )+a_{0} \left (a n +a r +b k -2 k n \right )\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k} \left (k +n +r \right ) \left (k +b +r -1-n \right )+a_{k -1} \left (a \left (k -1\right )+a n +a r +b k -2 k n \right )+k a_{k -2} \left (a -k \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & \left (n +r \right ) \left (b +r -1-n \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{-n , -b +n +1\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+n +r \right ) \left (b +r -n \right )+a_{0} \left (a n +a r +b k -2 k n \right )=0 \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & a_{1}=-\frac {a_{0} \left (a n +a r +b k -2 k n \right )}{n b +b r -n^{2}+r^{2}+b -n +r} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k} \left (k +n +r \right ) \left (k +b +r -1-n \right )+\left (\left (k +n +r -1\right ) a +k \left (b -2 n \right )\right ) a_{k -1}+k a_{k -2} \left (a -k \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +2} \left (k +2+n +r \right ) \left (k +1+b +r -n \right )+\left (\left (k +1+n +r \right ) a +k \left (b -2 n \right )\right ) a_{k +1}+k a_{k} \left (a -k \right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a_{k} a k +a k a_{k +1}+a n a_{k +1}+a r a_{k +1}+b k a_{k +1}-a_{k} k^{2}-2 k n a_{k +1}+a a_{k +1}}{\left (k +2+n +r \right ) \left (k +1+b +r -n \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-n \\ {} & {} & a_{k +2}=-\frac {a_{k} a k +a k a_{k +1}+b k a_{k +1}-a_{k} k^{2}-2 k n a_{k +1}+a a_{k +1}}{\left (k +2\right ) \left (k +1+b -2 n \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-n \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -n}, a_{k +2}=-\frac {a_{k} a k +a k a_{k +1}+b k a_{k +1}-a_{k} k^{2}-2 k n a_{k +1}+a a_{k +1}}{\left (k +2\right ) \left (k +1+b -2 n \right )}, a_{1}=-\frac {a_{0} \left (b k -2 k n \right )}{b -2 n}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-b +n +1 \\ {} & {} & a_{k +2}=-\frac {a_{k} a k +a k a_{k +1}+a n a_{k +1}+a \left (-b +n +1\right ) a_{k +1}+b k a_{k +1}-a_{k} k^{2}-2 k n a_{k +1}+a a_{k +1}}{\left (k +3+2 n -b \right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-b +n +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -b +n +1}, a_{k +2}=-\frac {a_{k} a k +a k a_{k +1}+a n a_{k +1}+a \left (-b +n +1\right ) a_{k +1}+b k a_{k +1}-a_{k} k^{2}-2 k n a_{k +1}+a a_{k +1}}{\left (k +3+2 n -b \right ) \left (k +2\right )}, a_{1}=-\frac {a_{0} \left (a n +a \left (-b +n +1\right )+b k -2 k n \right )}{n b +b \left (-b +n +1\right )-n^{2}+\left (-b +n +1\right )^{2}+1}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {m =0}{\sum }}c_{m} x^{m -n}\right )+\left (\moverset {\infty }{\munderset {m =0}{\sum }}d_{m} x^{m -b +n +1}\right ), c_{m +2}=-\frac {a k c_{m}+a m c_{m +1}+b k c_{m +1}-k^{2} c_{m}-2 k n c_{m +1}+a c_{m +1}}{\left (m +2\right ) \left (m +1+b -2 n \right )}, c_{1}=-\frac {c_{0} \left (b k -2 k n \right )}{b -2 n}, d_{m +2}=-\frac {d_{m} a k +a m d_{m +1}+a n d_{m +1}+a \left (-b +n +1\right ) d_{m +1}+b k d_{m +1}-d_{m} k^{2}-2 k n d_{m +1}+a d_{m +1}}{\left (m +3+2 n -b \right ) \left (m +2\right )}, d_{1}=-\frac {d_{0} \left (a n +a \left (-b +n +1\right )+b k -2 k n \right )}{n b +b \left (-b +n +1\right )-n^{2}+\left (-b +n +1\right )^{2}+1}\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: 0.016 (sec). Leaf size: 48
dsolve(x^2*diff(y(x),x$2)+(a*x^2+b*x)*diff(y(x),x)+(k*(a-k)*x^2+(a*n+b*k-2*k*n)*x+n*(b-n-1))*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} x^{-n} {\mathrm e}^{-k x}+c_{2} x^{-\frac {b}{2}} \operatorname {WhittakerM}\left (-\frac {b}{2}+n , -\frac {b}{2}+n +\frac {1}{2}, \left (-2 k +a \right ) x \right ) {\mathrm e}^{-\frac {a x}{2}} \]
✓ Solution by Mathematica
Time used: 0.504 (sec). Leaf size: 64
DSolve[x^2*y''[x]+(a*x^2+b*x)*y'[x]+(k*(a-k)*x^2+(a*n+b*k-2*k*n)*x+n*(b-n-1))*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{-k x} x^{-n} \left (c_1-c_2 x^{-b+2 n+1} (x (a-2 k))^{b-2 n-1} \Gamma (-b+2 n+1,(a-2 k) x)\right ) \]