Internal problem ID [10926]
Internal file name [OUTPUT/10183_Sunday_December_31_2023_11_03_40_AM_86763740/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: 103.
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]]
\[ \boxed {x y^{\prime \prime }+\left (a \,x^{n}+b \,x^{m}+c \right ) y^{\prime }+\left (c -1\right ) \left (a \,x^{n -1}+b \,x^{m -1}\right ) y=0} \]
In normal form the ode \begin {align*} x y^{\prime \prime }+\left (a \,x^{n}+b \,x^{m}+c \right ) y^{\prime }+\frac {\left (c -1\right ) \left (a \,x^{n}+b \,x^{m}\right ) y}{x}&=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^{n}+b \,x^{m}+c}{x}\\ q \left (x \right )&=\frac {\left (c -1\right ) \left (a \,x^{n}+b \,x^{m}\right )}{x^{2}} \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^{n}+b \,x^{m}+c \right )}{x^{2}}+\frac {\left (c -1\right ) \left (a \,x^{n}+b \,x^{m}\right )}{x^{2}}&=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^{n}+b \,x^{m}+c}{x}\right ) v^{\prime }\left (x \right )&=0 \\ v^{\prime \prime }\left (x \right )+\frac {\left (-c +2+a \,x^{n}+b \,x^{m}\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 (-c +2+a \,x^{n}+b \,x^{m}\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 (-c +2+a \,x^{n}+b \,x^{m}\right ) u}{x} \end {align*}
Where \(f(x)=-\frac {-c +2+a \,x^{n}+b \,x^{m}}{x}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= -\frac {-c +2+a \,x^{n}+b \,x^{m}}{x} \,d x\\ \int { \frac {1}{u} \,du} &= \int {-\frac {-c +2+a \,x^{n}+b \,x^{m}}{x} \,d x}\\ \ln \left (u \right )&=\left (-2+c \right ) \ln \left (x \right )-\frac {a \,{\mathrm e}^{n \ln \left (x \right )}}{n}-\frac {b \,{\mathrm e}^{m \ln \left (x \right )}}{m}+c_{1}\\ u&={\mathrm e}^{\left (-2+c \right ) \ln \left (x \right )-\frac {a \,{\mathrm e}^{n \ln \left (x \right )}}{n}-\frac {b \,{\mathrm e}^{m \ln \left (x \right )}}{m}+c_{1}}\\ &=c_{1} {\mathrm e}^{\left (-2+c \right ) \ln \left (x \right )-\frac {a \,{\mathrm e}^{n \ln \left (x \right )}}{n}-\frac {b \,{\mathrm e}^{m \ln \left (x \right )}}{m}} \end {align*}
Which simplifies to \[ u \left (x \right ) = \frac {c_{1} x^{c} {\mathrm e}^{-\frac {a \,x^{n}}{n}} {\mathrm e}^{-\frac {b \,x^{m}}{m}}}{x^{2}} \] 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} x^{c} {\mathrm e}^{-\frac {a \,x^{n}}{n}} {\mathrm e}^{-\frac {b \,x^{m}}{m}}}{x^{2}}d x +c_{2} \end {align*}
Hence \begin {align*} y&= v \left (x \right ) x^{n}\\ &= \left (\int \frac {c_{1} x^{c} {\mathrm e}^{-\frac {a \,x^{n}}{n}} {\mathrm e}^{-\frac {b \,x^{m}}{m}}}{x^{2}}d x +c_{2} \right ) x^{-c +1}\\ &= x^{-c +1} \left (c_{1} \left (\int x^{-2+c} {\mathrm e}^{-\frac {a \,x^{n}}{n}-\frac {b \,x^{m}}{m}}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} x^{c} {\mathrm e}^{-\frac {a \,x^{n}}{n}} {\mathrm e}^{-\frac {b \,x^{m}}{m}}}{x^{2}}d x +c_{2} \right ) x^{-c +1} \\ \end{align*}
Verification of solutions
\[ y = \left (\int \frac {c_{1} x^{c} {\mathrm e}^{-\frac {a \,x^{n}}{n}} {\mathrm e}^{-\frac {b \,x^{m}}{m}}}{x^{2}}d x +c_{2} \right ) x^{-c +1} \] Verified OK.
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying an equivalence, under non-integer power transformations, to LODEs admitting Liouvillian solutions. -> 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: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) -> Trying changes of variables to rationalize or make the ODE simpler <- unable to find a useful change of variables trying a symmetry of the form [xi=0, eta=F(x)] trying differential order: 2; exact nonlinear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients trying 2nd order, integrating factor of the form mu(x,y) trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying an equivalence, under non-integer power transformations, to LODEs admitting Liouvillian solutions. -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Whittaker -> 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: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) -> Trying changes of variables to rationalize or make the ODE simpler <- unable to find a useful change of variables trying a symmetry of the form [xi=0, eta=F(x)] trying to convert to an ODE of Bessel type -> trying reduction of order to Riccati trying Riccati sub-methods: trying Riccati_symmetries -> trying a symmetry pattern of the form [F(x)*G(y), 0] -> trying a symmetry pattern of the form [0, F(x)*G(y)] -> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] --- Trying Lie symmetry methods, 2nd order --- `, `-> Computing symmetries using: way = 3`[0, y]
✗ Solution by Maple
dsolve(x*diff(y(x),x$2)+(a*x^n+b*x^m+c)*diff(y(x),x)+(c-1)*(a*x^(n-1)+b*x^(m-1))*y(x)=0,y(x), singsol=all)
\[ \text {No solution found} \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[x*y''[x]+(a*x^n+b*x^m+c)*y'[x]+(c-1)*(a*x^(n-1)+b*x^(m-1))*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
Not solved