Internal problem ID [10906]
Internal file name [OUTPUT/10163_Sunday_December_31_2023_11_03_11_AM_95428243/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: 83.
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]]
\[ \boxed {x y^{\prime \prime }+\left (a \,x^{2}+b x +c \right ) y^{\prime }+\left (c -1\right ) \left (a x +b \right ) y=0} \]
In normal form the ode \begin {align*} x y^{\prime \prime }+\left (a \,x^{2}+b x +c \right ) y^{\prime }+\left (c -1\right ) \left (a x +b \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^{2}+b x +c}{x}\\ q \left (x \right )&=\frac {\left (c -1\right ) \left (a x +b \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^{2}+b x +c \right )}{x^{2}}+\frac {\left (c -1\right ) \left (a x +b \right )}{x}&=0 \tag {5} \end {align*}
Solving (5) for \(n\) gives \begin {align*} n&=-c +1 \tag {6} \end {align*}
Substituting this value in (3) gives \begin {align*} v^{\prime \prime }\left (x \right )+\left (\frac {-2 c +2}{x}+\frac {a \,x^{2}+b x +c}{x}\right ) v^{\prime }\left (x \right )&=0 \\ v^{\prime \prime }\left (x \right )+\frac {\left (a \,x^{2}+b x -c +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^{2}+b x -c +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^{2}+b x -c +2\right ) u}{x} \end {align*}
Where \(f(x)=-\frac {a \,x^{2}+b x -c +2}{x}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= -\frac {a \,x^{2}+b x -c +2}{x} \,d x\\ \int { \frac {1}{u} \,du} &= \int {-\frac {a \,x^{2}+b x -c +2}{x} \,d x}\\ \ln \left (u \right )&=-\frac {a \,x^{2}}{2}-b x -\left (-c +2\right ) \ln \left (x \right )+c_{1}\\ u&={\mathrm e}^{-\frac {a \,x^{2}}{2}-b x -\left (-c +2\right ) \ln \left (x \right )+c_{1}}\\ &=c_{1} {\mathrm e}^{-\frac {a \,x^{2}}{2}-b x -\left (-c +2\right ) \ln \left (x \right )} \end {align*}
Which simplifies to \[ u \left (x \right ) = \frac {c_{1} {\mathrm e}^{-\frac {a \,x^{2}}{2}} {\mathrm e}^{-b x} x^{c}}{x^{2}} \] 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}\\ &= \int \frac {c_{1} {\mathrm e}^{-\frac {a \,x^{2}}{2}} {\mathrm e}^{-b x} x^{c}}{x^{2}}d x +c_{2} \end {align*}
Hence \begin {align*} y&= v \left (x \right ) x^{n}\\ &= \left (\int \frac {c_{1} {\mathrm e}^{-\frac {a \,x^{2}}{2}} {\mathrm e}^{-b x} x^{c}}{x^{2}}d x +c_{2} \right ) x^{-c +1}\\ &= x^{-c +1} \left (c_{1} \left (\int x^{c -2} {\mathrm e}^{-\frac {1}{2} a \,x^{2}-b x}d x \right )+c_{2} \right )\\ \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (\int \frac {c_{1} {\mathrm e}^{-\frac {a \,x^{2}}{2}} {\mathrm e}^{-b x} x^{c}}{x^{2}}d x +c_{2} \right ) x^{-c +1} \\ \end{align*}
Verification of solutions
\[ y = \left (\int \frac {c_{1} {\mathrm e}^{-\frac {a \,x^{2}}{2}} {\mathrm e}^{-b x} x^{c}}{x^{2}}d x +c_{2} \right ) x^{-c +1} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}y^{\prime }\right )+\left (a \,x^{2}+b x +c \right ) y^{\prime }+\left (c -1\right ) \left (a 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 {\left (c -1\right ) \left (a x +b \right ) y}{x}-\frac {\left (a \,x^{2}+b x +c \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^{2}+b x +c \right ) y^{\prime }}{x}+\frac {\left (c -1\right ) \left (a x +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^{2}+b x +c}{x}, P_{3}\left (x \right )=\frac {\left (c -1\right ) \left (a x +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}}}=c \\ {} & \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^{2}+b x +c \right ) y^{\prime }+\left (c -1\right ) \left (a x +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..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 \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 +c \right ) x^{-1+r}+\left (a_{1} \left (1+r \right ) \left (r +c \right )+a_{0} b \left (-1+r +c \right )\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +r +c \right )+a_{k} b \left (k +r +c -1\right )+a_{k -1} a \left (k -2+r +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 +c \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -c +1\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right ) \left (r +c \right )+a_{0} b \left (-1+r +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 +c \right )+a_{k} b \left (k +r +c -1\right )+a_{k -1} a \left (k -2+r +c \right )=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +2} \left (k +2+r \right ) \left (k +1+r +c \right )+a_{k +1} b \left (k +r +c \right )+a_{k} a \left (k +r +c -1\right )=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}+a r a_{k}+b c a_{k +1}+b k a_{k +1}+b r a_{k +1}-a_{k} a}{\left (k +2+r \right ) \left (k +1+r +c \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {a_{k} a c +a k a_{k}+b c a_{k +1}+b k a_{k +1}-a_{k} a}{\left (k +2\right ) \left (k +1+c \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}+b c a_{k +1}+b k a_{k +1}-a_{k} a}{\left (k +2\right ) \left (k +1+c \right )}, a_{1} c +a_{0} b \left (c -1\right )=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-c +1 \\ {} & {} & a_{k +2}=-\frac {a_{k} a c +a k a_{k}+a \left (-c +1\right ) a_{k}+b c a_{k +1}+b k a_{k +1}+b \left (-c +1\right ) a_{k +1}-a_{k} a}{\left (k +3-c \right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-c +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -c +1}, a_{k +2}=-\frac {a_{k} a c +a k a_{k}+a \left (-c +1\right ) a_{k}+b c a_{k +1}+b k a_{k +1}+b \left (-c +1\right ) a_{k +1}-a_{k} a}{\left (k +3-c \right ) \left (k +2\right )}, a_{1} \left (-c +2\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 -c +1}\right ), d_{k +2}=-\frac {a c d_{k}+a k d_{k}+b c d_{k +1}+b k d_{k +1}-a d_{k}}{\left (k +2\right ) \left (k +1+c \right )}, d_{1} c +d_{0} b \left (c -1\right )=0, e_{k +2}=-\frac {e_{k} a c +a k e_{k}+a \left (-c +1\right ) e_{k}+b c e_{k +1}+b k e_{k +1}+b \left (-c +1\right ) e_{k +1}-e_{k} a}{\left (k +3-c \right ) \left (k +2\right )}, e_{1} \left (-c +2\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 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 <- Heun successful: received ODE is equivalent to the HeunB ODE, case c = 0 <- Kovacics algorithm successful`
✓ Solution by Maple
Time used: 0.297 (sec). Leaf size: 102
dsolve(x*diff(y(x),x$2)+(a*x^2+b*x+c)*diff(y(x),x)+(c-1)*(a*x+b)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{-\frac {x \left (a x +2 b \right )}{2}} \left (\operatorname {HeunB}\left (c -1, \frac {b \sqrt {2}}{\sqrt {a}}, c -3, -\frac {\sqrt {2}\, b \left (c -2\right )}{\sqrt {a}}, \frac {\sqrt {2}\, \sqrt {a}\, x}{2}\right ) c_{1} +\operatorname {HeunB}\left (-c +1, \frac {b \sqrt {2}}{\sqrt {a}}, c -3, -\frac {\sqrt {2}\, b \left (c -2\right )}{\sqrt {a}}, \frac {\sqrt {2}\, \sqrt {a}\, x}{2}\right ) x^{-c +1} c_{2} \right ) \]
✓ Solution by Mathematica
Time used: 1.579 (sec). Leaf size: 49
DSolve[x*y''[x]+(a*x^2+b*x+c)*y'[x]+(c-1)*(a*x+b)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to x^{1-c} \left (c_2 \int _1^xe^{-\frac {1}{2} K[1] (2 b+a K[1])} K[1]^{c-2}dK[1]+c_1\right ) \]