Internal problem ID [10968]
Internal file name [OUTPUT/10225_Sunday_December_31_2023_11_10_22_AM_99014734/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 2.1.2-4 Equation of form
\(x^2 y''+f(x)y'+g(x)y=0\)
Problem number: 145.
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]]
\[ \boxed {x^{2} y^{\prime \prime }+x \left (a \,x^{n}+b \right ) y^{\prime }+b \left (a \,x^{n}-1\right ) y=0} \]
In normal form the ode \begin {align*} x^{2} y^{\prime \prime }+x \left (a \,x^{n}+b \right ) y^{\prime }+b \left (a \,x^{n}-1\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^{n}+b}{x}\\ q \left (x \right )&=\frac {b \left (a \,x^{n}-1\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 \right )}{x^{2}}+\frac {b \left (a \,x^{n}-1\right )}{x^{2}}&=0 \tag {5} \end {align*}
Solving (5) for \(n\) gives \begin {align*} n&=-b \tag {6} \end {align*}
Substituting this value in (3) gives \begin {align*} v^{\prime \prime }\left (x \right )+\left (-\frac {2 b}{x}+\frac {a \,x^{n}+b}{x}\right ) v^{\prime }\left (x \right )&=0 \\ v^{\prime \prime }\left (x \right )+\frac {\left (-b +a \,x^{n}\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 (-b +a \,x^{n}\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 (-b +a \,x^{n}\right ) u}{x} \end {align*}
Where \(f(x)=-\frac {-b +a \,x^{n}}{x}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= -\frac {-b +a \,x^{n}}{x} \,d x\\ \int { \frac {1}{u} \,du} &= \int {-\frac {-b +a \,x^{n}}{x} \,d x}\\ \ln \left (u \right )&=-\frac {a \,x^{n}}{n}+\frac {b \ln \left (x^{n}\right )}{n}+c_{1}\\ u&={\mathrm e}^{-\frac {a \,x^{n}}{n}+\frac {b \ln \left (x^{n}\right )}{n}+c_{1}}\\ &=c_{1} {\mathrm e}^{-\frac {a \,x^{n}}{n}+\frac {b \ln \left (x^{n}\right )}{n}} \end {align*}
Which simplifies to \[ u \left (x \right ) = c_{1} {\mathrm e}^{-\frac {a \,x^{n}}{n}} \left (x^{n}\right )^{\frac {b}{n}} \] 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}\\ &= \frac {c_{1} \left (x^{n}\right )^{\frac {b}{n}} x^{-b} \left (\frac {a}{n}\right )^{-\frac {b}{n}-\frac {1}{n}} \left (\frac {n^{3} x^{b -n +1} \left (\frac {a}{n}\right )^{\frac {b}{n}+\frac {1}{n}} \left (a \,x^{n}+b +n +1\right ) \left (\frac {a \,x^{n}}{n}\right )^{-\frac {b +n +1}{2 n}} {\mathrm e}^{-\frac {a \,x^{n}}{2 n}} \operatorname {WhittakerM}\left (\frac {b +1}{n}-\frac {b +n +1}{2 n}, \frac {b +n +1}{2 n}+\frac {1}{2}, \frac {a \,x^{n}}{n}\right )}{\left (b +1\right ) \left (b +n +1\right ) \left (b +2 n +1\right ) a}+\frac {n^{2} x^{b -n +1} \left (\frac {a}{n}\right )^{\frac {b}{n}+\frac {1}{n}} \left (b +n +1\right ) \left (\frac {a \,x^{n}}{n}\right )^{-\frac {b +n +1}{2 n}} {\mathrm e}^{-\frac {a \,x^{n}}{2 n}} \operatorname {WhittakerM}\left (\frac {b +1}{n}-\frac {b +n +1}{2 n}+1, \frac {b +n +1}{2 n}+\frac {1}{2}, \frac {a \,x^{n}}{n}\right )}{\left (b +1\right ) a \left (b +2 n +1\right )}\right )}{n}+c_{2} \end {align*}
Hence \begin {align*} y&= v \left (x \right ) x^{n}\\ &= \left (\frac {c_{1} \left (x^{n}\right )^{\frac {b}{n}} x^{-b} \left (\frac {a}{n}\right )^{-\frac {b}{n}-\frac {1}{n}} \left (\frac {n^{3} x^{b -n +1} \left (\frac {a}{n}\right )^{\frac {b}{n}+\frac {1}{n}} \left (a \,x^{n}+b +n +1\right ) \left (\frac {a \,x^{n}}{n}\right )^{-\frac {b +n +1}{2 n}} {\mathrm e}^{-\frac {a \,x^{n}}{2 n}} \operatorname {WhittakerM}\left (\frac {b +1}{n}-\frac {b +n +1}{2 n}, \frac {b +n +1}{2 n}+\frac {1}{2}, \frac {a \,x^{n}}{n}\right )}{\left (b +1\right ) \left (b +n +1\right ) \left (b +2 n +1\right ) a}+\frac {n^{2} x^{b -n +1} \left (\frac {a}{n}\right )^{\frac {b}{n}+\frac {1}{n}} \left (b +n +1\right ) \left (\frac {a \,x^{n}}{n}\right )^{-\frac {b +n +1}{2 n}} {\mathrm e}^{-\frac {a \,x^{n}}{2 n}} \operatorname {WhittakerM}\left (\frac {b +1}{n}-\frac {b +n +1}{2 n}+1, \frac {b +n +1}{2 n}+\frac {1}{2}, \frac {a \,x^{n}}{n}\right )}{\left (b +1\right ) a \left (b +2 n +1\right )}\right )}{n}+c_{2} \right ) x^{-b}\\ &= \frac {x^{-b} \left (\left (\frac {a \,x^{n}}{n}\right )^{-\frac {b +n +1}{2 n}} {\mathrm e}^{-\frac {a \,x^{n}}{2 n}} \left (\left (b +n +1\right ) x^{-n +1}+x a \right ) n^{2} c_{1} \left (x^{n}\right )^{\frac {b}{n}} \operatorname {WhittakerM}\left (\frac {b -n +1}{2 n}, \frac {b +2 n +1}{2 n}, \frac {a \,x^{n}}{n}\right )+\left (b +n +1\right ) \left (\left (x^{n}\right )^{\frac {b}{n}} {\mathrm e}^{-\frac {a \,x^{n}}{2 n}} \left (\frac {a \,x^{n}}{n}\right )^{-\frac {b +n +1}{2 n}} n c_{1} x^{-n +1} \left (b +n +1\right ) \operatorname {WhittakerM}\left (\frac {b +n +1}{2 n}, \frac {b +2 n +1}{2 n}, \frac {a \,x^{n}}{n}\right )+a c_{2} \left (b +1\right ) \left (b +2 n +1\right )\right )\right )}{\left (b +1\right ) \left (b +n +1\right ) \left (b +2 n +1\right ) a}\\ \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \left (\frac {c_{1} \left (x^{n}\right )^{\frac {b}{n}} x^{-b} \left (\frac {a}{n}\right )^{-\frac {b}{n}-\frac {1}{n}} \left (\frac {n^{3} x^{b -n +1} \left (\frac {a}{n}\right )^{\frac {b}{n}+\frac {1}{n}} \left (a \,x^{n}+b +n +1\right ) \left (\frac {a \,x^{n}}{n}\right )^{-\frac {b +n +1}{2 n}} {\mathrm e}^{-\frac {a \,x^{n}}{2 n}} \operatorname {WhittakerM}\left (\frac {b +1}{n}-\frac {b +n +1}{2 n}, \frac {b +n +1}{2 n}+\frac {1}{2}, \frac {a \,x^{n}}{n}\right )}{\left (b +1\right ) \left (b +n +1\right ) \left (b +2 n +1\right ) a}+\frac {n^{2} x^{b -n +1} \left (\frac {a}{n}\right )^{\frac {b}{n}+\frac {1}{n}} \left (b +n +1\right ) \left (\frac {a \,x^{n}}{n}\right )^{-\frac {b +n +1}{2 n}} {\mathrm e}^{-\frac {a \,x^{n}}{2 n}} \operatorname {WhittakerM}\left (\frac {b +1}{n}-\frac {b +n +1}{2 n}+1, \frac {b +n +1}{2 n}+\frac {1}{2}, \frac {a \,x^{n}}{n}\right )}{\left (b +1\right ) a \left (b +2 n +1\right )}\right )}{n}+c_{2} \right ) x^{-b} \\ \end{align*}
Verification of solutions
\[ y = \left (\frac {c_{1} \left (x^{n}\right )^{\frac {b}{n}} x^{-b} \left (\frac {a}{n}\right )^{-\frac {b}{n}-\frac {1}{n}} \left (\frac {n^{3} x^{b -n +1} \left (\frac {a}{n}\right )^{\frac {b}{n}+\frac {1}{n}} \left (a \,x^{n}+b +n +1\right ) \left (\frac {a \,x^{n}}{n}\right )^{-\frac {b +n +1}{2 n}} {\mathrm e}^{-\frac {a \,x^{n}}{2 n}} \operatorname {WhittakerM}\left (\frac {b +1}{n}-\frac {b +n +1}{2 n}, \frac {b +n +1}{2 n}+\frac {1}{2}, \frac {a \,x^{n}}{n}\right )}{\left (b +1\right ) \left (b +n +1\right ) \left (b +2 n +1\right ) a}+\frac {n^{2} x^{b -n +1} \left (\frac {a}{n}\right )^{\frac {b}{n}+\frac {1}{n}} \left (b +n +1\right ) \left (\frac {a \,x^{n}}{n}\right )^{-\frac {b +n +1}{2 n}} {\mathrm e}^{-\frac {a \,x^{n}}{2 n}} \operatorname {WhittakerM}\left (\frac {b +1}{n}-\frac {b +n +1}{2 n}+1, \frac {b +n +1}{2 n}+\frac {1}{2}, \frac {a \,x^{n}}{n}\right )}{\left (b +1\right ) a \left (b +2 n +1\right )}\right )}{n}+c_{2} \right ) x^{-b} \] Verified OK.
Maple trace Kovacic algorithm successful
`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 Liouvillian solution using Kovacics algorithm A Liouvillian solution exists Reducible group (found an exponential solution) Group is reducible, not completely reducible <- Kovacics algorithm successful <- Equivalence, under non-integer power transformations successful`
✓ Solution by Maple
Time used: 0.11 (sec). Leaf size: 141
dsolve(x^2*diff(y(x),x$2)+x*(a*x^n+b)*diff(y(x),x)+b*(a*x^n-1)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{-\frac {a \,x^{n}}{2 n}} \left (\left (b +n +1\right ) x^{-\frac {3 n}{2}+\frac {1}{2}-\frac {b}{2}}+a \,x^{\frac {1}{2}-\frac {b}{2}-\frac {n}{2}}\right ) n c_{2} \operatorname {WhittakerM}\left (\frac {b -n +1}{2 n}, \frac {b +2 n +1}{2 n}, \frac {a \,x^{n}}{n}\right )+x^{-\frac {3 n}{2}+\frac {1}{2}-\frac {b}{2}} {\mathrm e}^{-\frac {a \,x^{n}}{2 n}} c_{2} \left (b +n +1\right )^{2} \operatorname {WhittakerM}\left (\frac {b +n +1}{2 n}, \frac {b +2 n +1}{2 n}, \frac {a \,x^{n}}{n}\right )+c_{1} x^{-b} \]
✓ Solution by Mathematica
Time used: 0.146 (sec). Leaf size: 76
DSolve[x^2*y''[x]+x*(a*x^n+b)*y'[x]+b*(a*x^n-1)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to (-1)^{-\frac {b}{n}} n^{\frac {b}{n}-1} a^{-\frac {b}{n}} \left (x^n\right )^{-\frac {b}{n}} \left ((b+1) c_1 (-1)^{b/n} \Gamma \left (\frac {b+1}{n},0,\frac {a x^n}{n}\right )+c_2 n\right ) \]