Internal problem ID [10909]
Internal file name [OUTPUT/10166_Sunday_December_31_2023_11_03_14_AM_61480679/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: 86.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_change_of_variable_on_y_method_2"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x y^{\prime \prime }+\left (a \,x^{3}+b \right ) y^{\prime }+a \left (-1+b \right ) x^{2} y=0} \]
In normal form the ode \begin {align*} x y^{\prime \prime }+\left (a \,x^{3}+b \right ) y^{\prime }+a \left (-1+b \right ) x^{2} y&=0 \tag {1} \end {align*}
Becomes \begin {align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=0 \tag {2} \end {align*}
Where \begin {align*} p \left (x \right )&=\frac {a \,x^{3}+b}{x}\\ q \left (x \right )&=a x \left (-1+b \right ) \end {align*}
Applying change of variables on the depndent variable \(y = v \left (x \right ) x^{n}\) to (2) gives the following ode where the dependent variables is \(v \left (x \right )\) and not \(y\). \begin {align*} v^{\prime \prime }\left (x \right )+\left (\frac {2 n}{x}+p \right ) v^{\prime }\left (x \right )+\left (\frac {n \left (n -1\right )}{x^{2}}+\frac {n p}{x}+q \right ) v \left (x \right )&=0 \tag {3} \end {align*}
Let the coefficient of \(v \left (x \right )\) above be zero. Hence \begin {align*} \frac {n \left (n -1\right )}{x^{2}}+\frac {n p}{x}+q&=0 \tag {4} \end {align*}
Substituting the earlier values found for \(p \left (x \right )\) and \(q \left (x \right )\) into (4) gives \begin {align*} \frac {n \left (n -1\right )}{x^{2}}+\frac {n \left (a \,x^{3}+b \right )}{x^{2}}+a x \left (-1+b \right )&=0 \tag {5} \end {align*}
Solving (5) for \(n\) gives \begin {align*} n&=1-b \tag {6} \end {align*}
Substituting this value in (3) gives \begin {align*} v^{\prime \prime }\left (x \right )+\left (\frac {2-2 b}{x}+\frac {a \,x^{3}+b}{x}\right ) v^{\prime }\left (x \right )&=0 \\ v^{\prime \prime }\left (x \right )+\frac {\left (a \,x^{3}-b +2\right ) v^{\prime }\left (x \right )}{x}&=0 \tag {7} \\ \end {align*}
Using the substitution \begin {align*} u \left (x \right ) = v^{\prime }\left (x \right ) \end {align*}
Then (7) becomes \begin {align*} u^{\prime }\left (x \right )+\frac {\left (a \,x^{3}-b +2\right ) u \left (x \right )}{x} = 0 \tag {8} \\ \end {align*}
The above is now solved for \(u \left (x \right )\). In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= -\frac {\left (a \,x^{3}-b +2\right ) u}{x} \end {align*}
Where \(f(x)=-\frac {a \,x^{3}-b +2}{x}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= -\frac {a \,x^{3}-b +2}{x} \,d x\\ \int { \frac {1}{u} \,du} &= \int {-\frac {a \,x^{3}-b +2}{x} \,d x}\\ \ln \left (u \right )&=-\frac {a \,x^{3}}{3}-\left (-b +2\right ) \ln \left (x \right )+c_{1}\\ u&={\mathrm e}^{-\frac {a \,x^{3}}{3}-\left (-b +2\right ) \ln \left (x \right )+c_{1}}\\ &=c_{1} {\mathrm e}^{-\frac {a \,x^{3}}{3}-\left (-b +2\right ) \ln \left (x \right )} \end {align*}
Which simplifies to \[ u \left (x \right ) = \frac {c_{1} {\mathrm e}^{-\frac {a \,x^{3}}{3}} x^{b}}{x^{2}} \] Now that \(u \left (x \right )\) is known, then \begin {align*} v^{\prime }\left (x \right )&= u \left (x \right )\\ v \left (x \right )&= \int u \left (x \right )d x +c_{2}\\ &= 3^{-\frac {4}{3}+\frac {b}{3}} a^{-\frac {b}{3}+\frac {1}{3}} c_{1} \left (\frac {3^{-\frac {b}{6}+\frac {8}{3}} x^{b +2} a^{\frac {2}{3}+\frac {b}{3}} \left (a \,x^{3}\right )^{-\frac {1}{3}-\frac {b}{6}} {\mathrm e}^{-\frac {a \,x^{3}}{6}} \operatorname {WhittakerM}\left (\frac {1}{3}+\frac {b}{6}, \frac {b}{6}+\frac {5}{6}, \frac {a \,x^{3}}{3}\right )}{\left (-1+b \right ) \left (b +2\right ) \left (5+b \right )}+\frac {3^{-\frac {b}{6}+\frac {8}{3}} x^{-4+b} a^{-\frac {4}{3}+\frac {b}{3}} \left (a \,x^{3}+b +2\right ) \left (a \,x^{3}\right )^{-\frac {1}{3}-\frac {b}{6}} {\mathrm e}^{-\frac {a \,x^{3}}{6}} \operatorname {WhittakerM}\left (\frac {4}{3}+\frac {b}{6}, \frac {b}{6}+\frac {5}{6}, \frac {a \,x^{3}}{3}\right )}{\left (-1+b \right ) \left (b +2\right )}\right )+c_{2} \end {align*}
Hence \begin {align*} y&= v \left (x \right ) x^{n}\\ &= \left (3^{-\frac {4}{3}+\frac {b}{3}} a^{-\frac {b}{3}+\frac {1}{3}} c_{1} \left (\frac {3^{-\frac {b}{6}+\frac {8}{3}} x^{b +2} a^{\frac {2}{3}+\frac {b}{3}} \left (a \,x^{3}\right )^{-\frac {1}{3}-\frac {b}{6}} {\mathrm e}^{-\frac {a \,x^{3}}{6}} \operatorname {WhittakerM}\left (\frac {1}{3}+\frac {b}{6}, \frac {b}{6}+\frac {5}{6}, \frac {a \,x^{3}}{3}\right )}{\left (-1+b \right ) \left (b +2\right ) \left (5+b \right )}+\frac {3^{-\frac {b}{6}+\frac {8}{3}} x^{-4+b} a^{-\frac {4}{3}+\frac {b}{3}} \left (a \,x^{3}+b +2\right ) \left (a \,x^{3}\right )^{-\frac {1}{3}-\frac {b}{6}} {\mathrm e}^{-\frac {a \,x^{3}}{6}} \operatorname {WhittakerM}\left (\frac {4}{3}+\frac {b}{6}, \frac {b}{6}+\frac {5}{6}, \frac {a \,x^{3}}{3}\right )}{\left (-1+b \right ) \left (b +2\right )}\right )+c_{2} \right ) x^{1-b}\\ &= \frac {3^{\frac {4}{3}+\frac {b}{6}} a c_{1} x^{3} \left (a \,x^{3}\right )^{-\frac {1}{3}-\frac {b}{6}} {\mathrm e}^{-\frac {a \,x^{3}}{6}} \operatorname {WhittakerM}\left (\frac {1}{3}+\frac {b}{6}, \frac {b}{6}+\frac {5}{6}, \frac {a \,x^{3}}{3}\right )+\left (c_{1} \left (a \,x^{3}+b +2\right ) {\mathrm e}^{-\frac {a \,x^{3}}{3}}+c_{2} x^{1-b} \left (b +2\right ) \left (-1+b \right )\right ) \left (5+b \right )}{\left (-1+b \right ) \left (5+b \right ) \left (b +2\right )}\\ \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (3^{-\frac {4}{3}+\frac {b}{3}} a^{-\frac {b}{3}+\frac {1}{3}} c_{1} \left (\frac {3^{-\frac {b}{6}+\frac {8}{3}} x^{b +2} a^{\frac {2}{3}+\frac {b}{3}} \left (a \,x^{3}\right )^{-\frac {1}{3}-\frac {b}{6}} {\mathrm e}^{-\frac {a \,x^{3}}{6}} \operatorname {WhittakerM}\left (\frac {1}{3}+\frac {b}{6}, \frac {b}{6}+\frac {5}{6}, \frac {a \,x^{3}}{3}\right )}{\left (-1+b \right ) \left (b +2\right ) \left (5+b \right )}+\frac {3^{-\frac {b}{6}+\frac {8}{3}} x^{-4+b} a^{-\frac {4}{3}+\frac {b}{3}} \left (a \,x^{3}+b +2\right ) \left (a \,x^{3}\right )^{-\frac {1}{3}-\frac {b}{6}} {\mathrm e}^{-\frac {a \,x^{3}}{6}} \operatorname {WhittakerM}\left (\frac {4}{3}+\frac {b}{6}, \frac {b}{6}+\frac {5}{6}, \frac {a \,x^{3}}{3}\right )}{\left (-1+b \right ) \left (b +2\right )}\right )+c_{2} \right ) x^{1-b} \\ \end{align*}
Verification of solutions
\[ y = \left (3^{-\frac {4}{3}+\frac {b}{3}} a^{-\frac {b}{3}+\frac {1}{3}} c_{1} \left (\frac {3^{-\frac {b}{6}+\frac {8}{3}} x^{b +2} a^{\frac {2}{3}+\frac {b}{3}} \left (a \,x^{3}\right )^{-\frac {1}{3}-\frac {b}{6}} {\mathrm e}^{-\frac {a \,x^{3}}{6}} \operatorname {WhittakerM}\left (\frac {1}{3}+\frac {b}{6}, \frac {b}{6}+\frac {5}{6}, \frac {a \,x^{3}}{3}\right )}{\left (-1+b \right ) \left (b +2\right ) \left (5+b \right )}+\frac {3^{-\frac {b}{6}+\frac {8}{3}} x^{-4+b} a^{-\frac {4}{3}+\frac {b}{3}} \left (a \,x^{3}+b +2\right ) \left (a \,x^{3}\right )^{-\frac {1}{3}-\frac {b}{6}} {\mathrm e}^{-\frac {a \,x^{3}}{6}} \operatorname {WhittakerM}\left (\frac {4}{3}+\frac {b}{6}, \frac {b}{6}+\frac {5}{6}, \frac {a \,x^{3}}{3}\right )}{\left (-1+b \right ) \left (b +2\right )}\right )+c_{2} \right ) x^{1-b} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}y^{\prime }\right )+\left (a \,x^{3}+b \right ) y^{\prime }+a \left (-1+b \right ) x^{2} 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 (a \,x^{3}+b \right ) y^{\prime }}{x}-a x \left (-1+b \right ) y \\ \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^{3}+b \right ) y^{\prime }}{x}+a x \left (-1+b \right ) y=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^{3}+b}{x}, P_{3}\left (x \right )=a x \left (-1+b \right )\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^{3}+b \right ) y^{\prime }+a \left (-1+b \right ) x^{2} 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^{2}\cdot y\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x^{2}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +2} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -2 \\ {} & {} & x^{2}\cdot y=\moverset {\infty }{\munderset {k =2}{\sum }}a_{k -2} x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..3 \\ {} & {} & 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}+a_{1} \left (1+r \right ) \left (r +b \right ) x^{r}+a_{2} \left (2+r \right ) \left (1+r +b \right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +r +b \right )+a_{k -2} a \left (k -3+r +b \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, 1-b \right \} \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} x \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [a_{1} \left (1+r \right ) \left (r +b \right )=0, a_{2} \left (2+r \right ) \left (1+r +b \right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=0, a_{2}=0\right \} \\ \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 -2} a \left (k -3+r +b \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & a_{k +3} \left (k +3+r \right ) \left (k +2+r +b \right )+a_{k} a \left (k +r +b -1\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=-\frac {a_{k} a \left (k +r +b -1\right )}{\left (k +3+r \right ) \left (k +2+r +b \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +3}=-\frac {a_{k} a \left (k -1+b \right )}{\left (k +3\right ) \left (k +2+b \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +3}=-\frac {a_{k} a \left (k -1+b \right )}{\left (k +3\right ) \left (k +2+b \right )}, a_{1}=0, a_{2}=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =1-b \\ {} & {} & a_{k +3}=-\frac {a_{k} a k}{\left (k +4-b \right ) \left (k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =1-b \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +1-b}, a_{k +3}=-\frac {a_{k} a k}{\left (k +4-b \right ) \left (k +3\right )}, a_{1}=0, a_{2}=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k +1-b}\right ), c_{k +3}=-\frac {c_{k} a \left (k -1+b \right )}{\left (k +3\right ) \left (k +2+b \right )}, c_{1}=0, c_{2}=0, d_{k +3}=-\frac {d_{k} a k}{\left (k +4-b \right ) \left (k +3\right )}, d_{1}=0, d_{2}=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: 107
dsolve(x*diff(y(x),x$2)+(a*x^3+b)*diff(y(x),x)+a*(b-1)*x^2*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {9 c_{2} a^{2} x^{-\frac {b}{2}+3} {\mathrm e}^{-\frac {a \,x^{3}}{6}} \operatorname {WhittakerM}\left (\frac {1}{3}+\frac {b}{6}, \frac {b}{6}+\frac {5}{6}, \frac {a \,x^{3}}{3}\right )+\left (a \,x^{-\frac {b}{2}+3}+x^{-\frac {b}{2}} \left (b +2\right )\right ) c_{2} {\mathrm e}^{-\frac {a \,x^{3}}{3}} a 3^{-\frac {b}{6}+\frac {2}{3}} \left (b +5\right ) \left (a \,x^{3}\right )^{\frac {1}{3}+\frac {b}{6}}+9 c_{1} x^{-b +2}}{9 x} \]
✓ Solution by Mathematica
Time used: 0.424 (sec). Leaf size: 60
DSolve[x*y''[x]+(a*x^3+b)*y'[x]+a*(b-1)*x^2*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_1 x^{1-b}-3^{\frac {b-4}{3}} c_2 \left (a x^3\right )^{\frac {1}{3}-\frac {b}{3}} \Gamma \left (\frac {b-1}{3},\frac {a x^3}{3}\right ) \]