Internal problem ID [10852]
Internal file name [OUTPUT/10109_Sunday_December_24_2023_05_12_23_PM_15034228/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 2.1.2-2 Equation of form
\(y''+f(x)y'+g(x)y=0\)
Problem number: 29.
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.
\[ \boxed {y^{\prime \prime }+\left (a x +b \right ) y^{\prime }+c \left (\left (a -c \right ) x^{2}+b x +1\right ) y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }+\left (a x +b \right ) y^{\prime }+c \left (\left (a -c \right ) x^{2}+b x +1\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 }=-c \left (a \,x^{2}-x^{2} c +b x +1\right ) y-\left (a x +b \right ) y^{\prime } \\ \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 }+\left (a x +b \right ) y^{\prime }+c \left (a \,x^{2}-x^{2} c +b x +1\right ) y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k} \\ \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 =\max \left (0, -m \right )}{\sum }}a_{k} x^{k +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =\max \left (0, -m \right )+m}{\sum }}a_{k -m} x^{k} \\ {} & \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 =\max \left (0, 1-m \right )}{\sum }}a_{k} k \,x^{k -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =\max \left (0, 1-m \right )+m -1}{\sum }}a_{k +1-m} \left (k +1-m \right ) x^{k} \\ {} & \circ & \textrm {Convert}\hspace {3pt} \frac {d}{d x}y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=\moverset {\infty }{\munderset {k =2}{\sum }}a_{k} k \left (k -1\right ) x^{k -2} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \frac {d}{d x}y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k +2} \left (k +2\right ) \left (k +1\right ) x^{k} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{1} b +a_{0} c +2 a_{2}+\left (6 a_{3}+2 a_{2} b +a_{1} \left (a +c \right )+a_{0} b c \right ) x +\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k +2} \left (k +2\right ) \left (k +1\right )+a_{k +1} \left (k +1\right ) b +a_{k} \left (a k +c \right )+a_{k -1} b c +c a_{k -2} \left (a -c \right )\right ) x^{k}\right )=0 \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} x \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [2 a_{2}+a_{1} b +a_{0} c =0, 6 a_{3}+2 a_{2} b +a_{1} \left (a +c \right )+a_{0} b c =0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{2}=-\frac {a_{1} b}{2}-\frac {a_{0} c}{2}, a_{3}=\frac {1}{6} a_{1} b^{2}-\frac {1}{6} a_{1} a -\frac {1}{6} a_{1} c \right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & -c^{2} a_{k -2}+\left (a a_{k -2}+b a_{k -1}+a_{k}\right ) c +k^{2} a_{k +2}+\left (a a_{k}+a_{k +1} b +3 a_{k +2}\right ) k +a_{k +1} b +2 a_{k +2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & -a_{k} c^{2}+\left (a a_{k}+a_{k +1} b +a_{k +2}\right ) c +\left (k +2\right )^{2} a_{k +4}+\left (a a_{k +2}+a_{k +3} b +3 a_{k +4}\right ) \left (k +2\right )+a_{k +3} b +2 a_{k +4}=0 \\ \bullet & {} & \textrm {Recursion relation that defines the series solution to the ODE}\hspace {3pt} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +4}=-\frac {a_{k} a c +a k a_{k +2}+b c a_{k +1}+b k a_{k +3}-a_{k} c^{2}+2 a a_{k +2}+3 a_{k +3} b +c a_{k +2}}{k^{2}+7 k +12}, a_{2}=-\frac {a_{1} b}{2}-\frac {a_{0} c}{2}, a_{3}=\frac {1}{6} a_{1} b^{2}-\frac {1}{6} a_{1} a -\frac {1}{6} a_{1} c \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: 36
dsolve(diff(y(x),x$2)+(a*x+b)*diff(y(x),x)+c*((a-c)*x^2+b*x+1)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{-\frac {c \,x^{2}}{2}} \left (c_{1} +\operatorname {erf}\left (\frac {\left (-2 c +a \right ) x +b}{\sqrt {2 a -4 c}}\right ) c_{2} \right ) \]
✓ Solution by Mathematica
Time used: 0.135 (sec). Leaf size: 81
DSolve[y''[x]+(a*x+b)*y'[x]+c*((a-c)*x^2+b*x+1)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{-\frac {1}{2} x (x (a-c)+2 b)} \left (c_1 \operatorname {HermiteH}\left (-1,\frac {b+(a-2 c) x}{\sqrt {2} \sqrt {a-2 c}}\right )+c_2 e^{\frac {(x (a-2 c)+b)^2}{2 (a-2 c)}}\right ) \]