Internal problem ID [10894]
Internal file name [OUTPUT/10151_Sunday_December_24_2023_05_15_36_PM_79692819/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 2.1.2-3 Equation of form
\((a x + b)y''+f(x)y'+g(x)y=0\)
Problem number: 71.
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.
\[ \boxed {x y^{\prime \prime }+\left (a x +b \right ) y^{\prime }+c \left (\left (a -c \right ) x +b \right ) y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}y^{\prime }\right )+\left (a x +b \right ) y^{\prime }+c \left (\left (a -c \right ) x +b \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 {c \left (a x -x c +b \right ) y}{x}-\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 {c \left (a x -x c +b \right ) y}{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 {a x +b}{x}, P_{3}\left (x \right )=\frac {c \left (a x -x c +b \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}}}=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}}}=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} \\ {} & {} & x \left (\frac {d}{d x}y^{\prime }\right )+\left (a x +b \right ) y^{\prime }+c \left (a x -x c +b \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..1 \\ {} & {} & 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..1 \\ {} & {} & 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 \cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \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 -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & x \cdot \left (\frac {d}{d x}y^{\prime }\right )=\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 (-1+r +b \right ) x^{-1+r}+\left (a_{1} \left (1+r \right ) \left (r +b \right )+a_{0} \left (a r +b c \right )\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +r +b \right )+a_{k} \left (a k +a r +b c \right )+c a_{k -1} \left (a -c \right )\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & r \left (-1+r +b \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -b +1\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right ) \left (r +b \right )+a_{0} \left (a r +b c \right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +1} \left (k +1+r \right ) \left (k +r +b \right )+a k a_{k}+a r a_{k}+\left (a_{k} b +a_{k -1} \left (a -c \right )\right ) c =0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +2} \left (k +2+r \right ) \left (k +1+r +b \right )+a \left (k +1\right ) a_{k +1}+a r a_{k +1}+\left (b a_{k +1}+a_{k} \left (a -c \right )\right ) c =0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=-\frac {a_{k} a c +a k a_{k +1}+a r a_{k +1}+b c a_{k +1}-a_{k} c^{2}+a a_{k +1}}{\left (k +2+r \right ) \left (k +1+r +b \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {a_{k} a c +a k a_{k +1}+b c a_{k +1}-a_{k} c^{2}+a a_{k +1}}{\left (k +2\right ) \left (k +1+b \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +2}=-\frac {a_{k} a c +a k a_{k +1}+b c a_{k +1}-a_{k} c^{2}+a a_{k +1}}{\left (k +2\right ) \left (k +1+b \right )}, a_{0} b c +a_{1} b =0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-b +1 \\ {} & {} & a_{k +2}=-\frac {a_{k} a c +a k a_{k +1}+a \left (-b +1\right ) a_{k +1}+b c a_{k +1}-a_{k} c^{2}+a a_{k +1}}{\left (k +3-b \right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-b +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -b +1}, a_{k +2}=-\frac {a_{k} a c +a k a_{k +1}+a \left (-b +1\right ) a_{k +1}+b c a_{k +1}-a_{k} c^{2}+a a_{k +1}}{\left (k +3-b \right ) \left (k +2\right )}, a_{1} \left (2-b \right )+a_{0} \left (a \left (-b +1\right )+b c \right )=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}e_{k} x^{k -b +1}\right ), d_{k +2}=-\frac {a c d_{k}+a k d_{k +1}+b c d_{k +1}-c^{2} d_{k}+a d_{k +1}}{\left (k +2\right ) \left (k +1+b \right )}, b c d_{0}+b d_{1}=0, e_{k +2}=-\frac {e_{k} a c +a k e_{k +1}+a \left (-b +1\right ) e_{k +1}+b c e_{k +1}-e_{k} c^{2}+a e_{k +1}}{\left (k +3-b \right ) \left (k +2\right )}, e_{1} \left (2-b \right )+e_{0} \left (a \left (-b +1\right )+b c \right )=0\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: 0.0 (sec). Leaf size: 40
dsolve(x*diff(y(x),x$2)+(a*x+b)*diff(y(x),x)+c*((a-c)*x+b)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{1} {\mathrm e}^{-c x}+c_{2} x^{-\frac {b}{2}} \operatorname {WhittakerM}\left (-\frac {b}{2}, \frac {1}{2}-\frac {b}{2}, \left (-2 c +a \right ) x \right ) {\mathrm e}^{-\frac {a x}{2}} \]
✓ Solution by Mathematica
Time used: 0.333 (sec). Leaf size: 50
DSolve[x*y''[x]+(a*x+b)*y'[x]+c*((a-c)*x+b)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{-c x} \left (c_1-c_2 x^{1-b} (x (a-2 c))^{b-1} \Gamma (1-b,(a-2 c) x)\right ) \]