Internal problem ID [10913]
Internal file name [OUTPUT/10170_Sunday_December_31_2023_11_03_20_AM_5802521/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: 90.
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]]
\[ \boxed {x y^{\prime \prime }+\left (a \,x^{3}+b \,x^{2}+c x +d \right ) y^{\prime }+\left (d -1\right ) \left (a \,x^{2}+b x +c \right ) y=0} \]
In normal form the ode \begin {align*} x y^{\prime \prime }+\left (a \,x^{3}+b \,x^{2}+c x +d \right ) y^{\prime }+\left (d -1\right ) \left (a \,x^{2}+b x +c \right ) 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^{2}+c x +d}{x}\\ q \left (x \right )&=\frac {\left (d -1\right ) \left (a \,x^{2}+b x +c \right )}{x} \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 \,x^{2}+c x +d \right )}{x^{2}}+\frac {\left (d -1\right ) \left (a \,x^{2}+b x +c \right )}{x}&=0 \tag {5} \end {align*}
Solving (5) for \(n\) gives \begin {align*} n&=-d +1 \tag {6} \end {align*}
Substituting this value in (3) gives \begin {align*} v^{\prime \prime }\left (x \right )+\left (\frac {-2 d +2}{x}+\frac {a \,x^{3}+b \,x^{2}+c x +d}{x}\right ) v^{\prime }\left (x \right )&=0 \\ v^{\prime \prime }\left (x \right )+\frac {\left (a \,x^{3}+b \,x^{2}+c x -d +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 \,x^{2}+c x -d +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 \,x^{2}+c x -d +2\right ) u}{x} \end {align*}
Where \(f(x)=-\frac {a \,x^{3}+b \,x^{2}+c x -d +2}{x}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= -\frac {a \,x^{3}+b \,x^{2}+c x -d +2}{x} \,d x\\ \int { \frac {1}{u} \,du} &= \int {-\frac {a \,x^{3}+b \,x^{2}+c x -d +2}{x} \,d x}\\ \ln \left (u \right )&=-\frac {a \,x^{3}}{3}-\frac {b \,x^{2}}{2}-c x -\left (-d +2\right ) \ln \left (x \right )+c_{1}\\ u&={\mathrm e}^{-\frac {a \,x^{3}}{3}-\frac {b \,x^{2}}{2}-c x -\left (-d +2\right ) \ln \left (x \right )+c_{1}}\\ &=c_{1} {\mathrm e}^{-\frac {a \,x^{3}}{3}-\frac {b \,x^{2}}{2}-c x -\left (-d +2\right ) \ln \left (x \right )} \end {align*}
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}\\ &= \int c_{1} {\mathrm e}^{-\frac {a \,x^{3}}{3}-\frac {b \,x^{2}}{2}-c x -\left (-d +2\right ) \ln \left (x \right )}d x +c_{2} \end {align*}
Hence \begin {align*} y&= v \left (x \right ) x^{n}\\ &= \left (\int c_{1} {\mathrm e}^{-\frac {a \,x^{3}}{3}-\frac {b \,x^{2}}{2}-c x -\left (-d +2\right ) \ln \left (x \right )}d x +c_{2} \right ) x^{-d +1}\\ &= x^{-d +1} \left (c_{1} \left (\int x^{d -2} {\mathrm e}^{-\frac {1}{3} a \,x^{3}-\frac {1}{2} b \,x^{2}-c x}d x \right )+c_{2} \right )\\ \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (\int c_{1} {\mathrm e}^{-\frac {a \,x^{3}}{3}-\frac {b \,x^{2}}{2}-c x -\left (-d +2\right ) \ln \left (x \right )}d x +c_{2} \right ) x^{-d +1} \\ \end{align*}
Verification of solutions
\[ y = \left (\int c_{1} {\mathrm e}^{-\frac {a \,x^{3}}{3}-\frac {b \,x^{2}}{2}-c x -\left (-d +2\right ) \ln \left (x \right )}d x +c_{2} \right ) x^{-d +1} \] 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 \,x^{2}+c x +d \right ) y^{\prime }+\left (d -1\right ) \left (a \,x^{2}+b x +c \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 (d -1\right ) \left (a \,x^{2}+b x +c \right ) y}{x}-\frac {\left (a \,x^{3}+b \,x^{2}+c x +d \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^{3}+b \,x^{2}+c x +d \right ) y^{\prime }}{x}+\frac {\left (d -1\right ) \left (a \,x^{2}+b x +c \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^{3}+b \,x^{2}+c x +d}{x}, P_{3}\left (x \right )=\frac {\left (d -1\right ) \left (a \,x^{2}+b x +c \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}}}=d \\ {} & \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 \,x^{2}+c x +d \right ) y^{\prime }+\left (d -1\right ) \left (a \,x^{2}+b x +c \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 =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 +d \right ) x^{-1+r}+\left (a_{1} \left (1+r \right ) \left (r +d \right )+a_{0} c \left (-1+r +d \right )\right ) x^{r}+\left (a_{2} \left (2+r \right ) \left (1+r +d \right )+a_{1} c \left (r +d \right )+a_{0} b \left (-1+r +d \right )\right ) x^{1+r}+\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +r +d \right )+a_{k} c \left (k +r +d -1\right )+a_{k -1} b \left (k -2+r +d \right )+a_{k -2} a \left (k -3+r +d \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 +d \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -d +1\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 +d \right )+a_{0} c \left (-1+r +d \right )=0, a_{2} \left (2+r \right ) \left (1+r +d \right )+a_{1} c \left (r +d \right )+a_{0} b \left (-1+r +d \right )=0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{1}=-\frac {a_{0} c \left (-1+r +d \right )}{r d +r^{2}+d +r}, a_{2}=-\frac {a_{0} \left (b d r +b \,r^{2}-c^{2} d -c^{2} r +b d +c^{2}-b \right )}{d \,r^{2}+r^{3}+3 r d +4 r^{2}+2 d +5 r +2}\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 +d \right )+a_{k} c \left (k +r +d -1\right )+a_{k -1} b \left (k -2+r +d \right )+a_{k -2} a \left (k -3+r +d \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 +d \right )+a_{k +2} c \left (k +1+r +d \right )+a_{k +1} b \left (k +r +d \right )+a_{k} a \left (k +r +d -1\right )=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +3}=-\frac {a_{k} a d +a k a_{k}+a r a_{k}+b d a_{k +1}+b k a_{k +1}+b r a_{k +1}+c d a_{k +2}+c k a_{k +2}+c r a_{k +2}-a_{k} a +c a_{k +2}}{\left (k +3+r \right ) \left (k +2+r +d \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +3}=-\frac {a_{k} a d +a k a_{k}+b d a_{k +1}+b k a_{k +1}+c d a_{k +2}+c k a_{k +2}-a_{k} a +c a_{k +2}}{\left (k +3\right ) \left (k +2+d \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 d +a k a_{k}+b d a_{k +1}+b k a_{k +1}+c d a_{k +2}+c k a_{k +2}-a_{k} a +c a_{k +2}}{\left (k +3\right ) \left (k +2+d \right )}, a_{1}=-\frac {a_{0} c \left (d -1\right )}{d}, a_{2}=-\frac {a_{0} \left (-c^{2} d +b d +c^{2}-b \right )}{2 d +2}\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-d +1 \\ {} & {} & a_{k +3}=-\frac {a_{k} a d +a k a_{k}+a \left (-d +1\right ) a_{k}+b d a_{k +1}+b k a_{k +1}+b \left (-d +1\right ) a_{k +1}+c d a_{k +2}+c k a_{k +2}+c \left (-d +1\right ) a_{k +2}-a_{k} a +c a_{k +2}}{\left (k +4-d \right ) \left (k +3\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-d +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -d +1}, a_{k +3}=-\frac {a_{k} a d +a k a_{k}+a \left (-d +1\right ) a_{k}+b d a_{k +1}+b k a_{k +1}+b \left (-d +1\right ) a_{k +1}+c d a_{k +2}+c k a_{k +2}+c \left (-d +1\right ) a_{k +2}-a_{k} a +c a_{k +2}}{\left (k +4-d \right ) \left (k +3\right )}, a_{1}=0, a_{2}=-\frac {a_{0} \left (b d \left (-d +1\right )+b \left (-d +1\right )^{2}-c^{2} d -c^{2} \left (-d +1\right )+b d +c^{2}-b \right )}{\left (-d +1\right )^{2} d +\left (-d +1\right )^{3}+3 \left (-d +1\right ) d +4 \left (-d +1\right )^{2}-3 d +7}\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}e_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}f_{k} x^{k -d +1}\right ), e_{k +3}=-\frac {a d e_{k}+a k e_{k}+b d e_{k +1}+b k e_{k +1}+c d e_{k +2}+c k e_{k +2}-a e_{k}+c e_{k +2}}{\left (k +3\right ) \left (k +2+d \right )}, e_{1}=-\frac {e_{0} c \left (d -1\right )}{d}, e_{2}=-\frac {e_{0} \left (-c^{2} d +b d +c^{2}-b \right )}{2 d +2}, f_{k +3}=-\frac {f_{k} a d +a k f_{k}+a \left (-d +1\right ) f_{k}+b d f_{k +1}+b k f_{k +1}+b \left (-d +1\right ) f_{k +1}+c d f_{k +2}+c k f_{k +2}+c \left (-d +1\right ) f_{k +2}-f_{k} a +c f_{k +2}}{\left (k +4-d \right ) \left (k +3\right )}, f_{1}=0, f_{2}=-\frac {f_{0} \left (b d \left (-d +1\right )+b \left (-d +1\right )^{2}-c^{2} d -c^{2} \left (-d +1\right )+b d +c^{2}-b \right )}{\left (-d +1\right )^{2} d +\left (-d +1\right )^{3}+3 \left (-d +1\right ) d +4 \left (-d +1\right )^{2}-3 d +7}\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 Solution has integrals. Trying a special function solution free of integrals... -> 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 -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius No special function solution was found. <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.359 (sec). Leaf size: 42
dsolve(x*diff(y(x),x$2)+(a*x^3+b*x^2+c*x+d)*diff(y(x),x)+(d-1)*(a*x^2+b*x+c)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = x^{-d +1} \left (\left (\int x^{d -2} {\mathrm e}^{-\frac {1}{3} a \,x^{3}-\frac {1}{2} x^{2} b -c x}d x \right ) c_{2} +c_{1} \right ) \]
✓ Solution by Mathematica
Time used: 1.839 (sec). Leaf size: 57
DSolve[x*y''[x]+(a*x^3+b*x^2+c*x+d)*y'[x]+(d-1)*(a*x^2+b*x+c)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to x^{1-d} \left (c_2 \int _1^x\exp \left (-\frac {1}{6} K[1] (6 c+K[1] (3 b+2 a K[1]))\right ) K[1]^{d-2}dK[1]+c_1\right ) \]