Internal problem ID [10915]
Internal file name [OUTPUT/10172_Sunday_December_31_2023_11_03_22_AM_74756653/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: 92.
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]]
\[ \boxed {x y^{\prime \prime }+\left (a \,x^{n}+2\right ) y^{\prime }+x^{n -1} y a=0} \]
In normal form the ode \begin {align*} x y^{\prime \prime }+\left (a \,x^{n}+2\right ) y^{\prime }+x^{n -1} y a&=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}+2}{x}\\ q \left (x \right )&=a \,x^{-2+n} \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}+2\right )}{x^{2}}+a \,x^{-2+n}&=0 \tag {5} \end {align*}
Solving (5) for \(n\) gives \begin {align*} n&=-1 \tag {6} \end {align*}
Substituting this value in (3) gives \begin {align*} v^{\prime \prime }\left (x \right )+\left (-\frac {2}{x}+\frac {a \,x^{n}+2}{x}\right ) v^{\prime }\left (x \right )&=0 \\ v^{\prime \prime }\left (x \right )+a \,x^{n -1} v^{\prime }\left (x \right )&=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 )+a \,x^{n -1} u \left (x \right ) = 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)\\ &= -a \,x^{n -1} u \end {align*}
Where \(f(x)=-a \,x^{n -1}\) and \(g(u)=u\). Integrating both sides gives \begin {align*} \frac {1}{u} \,du &= -a \,x^{n -1} \,d x\\ \int { \frac {1}{u} \,du} &= \int {-a \,x^{n -1} \,d x}\\ \ln \left (u \right )&=-\frac {a \,x^{n}}{n}+c_{1}\\ u&={\mathrm e}^{-\frac {a \,x^{n}}{n}+c_{1}}\\ &=c_{1} {\mathrm e}^{-\frac {a \,x^{n}}{n}} \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}\\ &= \frac {c_{1} \left (\frac {a}{n}\right )^{-\frac {1}{n}} \left (\frac {n^{3} x^{1-n} \left (\frac {a}{n}\right )^{\frac {1}{n}} \left (a \,x^{n}+n +1\right ) \left (\frac {a \,x^{n}}{n}\right )^{-\frac {1+n}{2 n}} {\mathrm e}^{-\frac {a \,x^{n}}{2 n}} \operatorname {WhittakerM}\left (\frac {1}{n}-\frac {1+n}{2 n}, \frac {1+n}{2 n}+\frac {1}{2}, \frac {a \,x^{n}}{n}\right )}{\left (1+n \right ) \left (1+2 n \right ) a}+\frac {n^{2} x^{1-n} \left (\frac {a}{n}\right )^{\frac {1}{n}} \left (1+n \right ) \left (\frac {a \,x^{n}}{n}\right )^{-\frac {1+n}{2 n}} {\mathrm e}^{-\frac {a \,x^{n}}{2 n}} \operatorname {WhittakerM}\left (\frac {1}{n}-\frac {1+n}{2 n}+1, \frac {1+n}{2 n}+\frac {1}{2}, \frac {a \,x^{n}}{n}\right )}{a \left (1+2 n \right )}\right )}{n}+c_{2} \end {align*}
Hence \begin {align*} y&= v \left (x \right ) x^{n}\\ &= \frac {\frac {c_{1} \left (\frac {a}{n}\right )^{-\frac {1}{n}} \left (\frac {n^{3} x^{1-n} \left (\frac {a}{n}\right )^{\frac {1}{n}} \left (a \,x^{n}+n +1\right ) \left (\frac {a \,x^{n}}{n}\right )^{-\frac {1+n}{2 n}} {\mathrm e}^{-\frac {a \,x^{n}}{2 n}} \operatorname {WhittakerM}\left (\frac {1}{n}-\frac {1+n}{2 n}, \frac {1+n}{2 n}+\frac {1}{2}, \frac {a \,x^{n}}{n}\right )}{\left (1+n \right ) \left (1+2 n \right ) a}+\frac {n^{2} x^{1-n} \left (\frac {a}{n}\right )^{\frac {1}{n}} \left (1+n \right ) \left (\frac {a \,x^{n}}{n}\right )^{-\frac {1+n}{2 n}} {\mathrm e}^{-\frac {a \,x^{n}}{2 n}} \operatorname {WhittakerM}\left (\frac {1}{n}-\frac {1+n}{2 n}+1, \frac {1+n}{2 n}+\frac {1}{2}, \frac {a \,x^{n}}{n}\right )}{a \left (1+2 n \right )}\right )}{n}+c_{2}}{x}\\ &= \frac {n c_{1} x^{1-n} {\mathrm e}^{-\frac {a \,x^{n}}{2 n}} \left (\frac {a \,x^{n}}{n}\right )^{-\frac {1+n}{2 n}} \left (1+n \right )^{2} \operatorname {WhittakerM}\left (\frac {1+n}{2 n}, \frac {1+2 n}{2 n}, \frac {a \,x^{n}}{n}\right )+{\mathrm e}^{-\frac {a \,x^{n}}{2 n}} c_{1} n^{2} \left (\frac {a \,x^{n}}{n}\right )^{-\frac {1+n}{2 n}} \left (\left (1+n \right ) x^{1-n}+x a \right ) \operatorname {WhittakerM}\left (-\frac {n -1}{2 n}, \frac {1+2 n}{2 n}, \frac {a \,x^{n}}{n}\right )+2 \left (1+n \right ) a \left (n +\frac {1}{2}\right ) c_{2}}{\left (1+n \right ) \left (1+2 n \right ) a x}\\ \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\frac {c_{1} \left (\frac {a}{n}\right )^{-\frac {1}{n}} \left (\frac {n^{3} x^{1-n} \left (\frac {a}{n}\right )^{\frac {1}{n}} \left (a \,x^{n}+n +1\right ) \left (\frac {a \,x^{n}}{n}\right )^{-\frac {1+n}{2 n}} {\mathrm e}^{-\frac {a \,x^{n}}{2 n}} \operatorname {WhittakerM}\left (\frac {1}{n}-\frac {1+n}{2 n}, \frac {1+n}{2 n}+\frac {1}{2}, \frac {a \,x^{n}}{n}\right )}{\left (1+n \right ) \left (1+2 n \right ) a}+\frac {n^{2} x^{1-n} \left (\frac {a}{n}\right )^{\frac {1}{n}} \left (1+n \right ) \left (\frac {a \,x^{n}}{n}\right )^{-\frac {1+n}{2 n}} {\mathrm e}^{-\frac {a \,x^{n}}{2 n}} \operatorname {WhittakerM}\left (\frac {1}{n}-\frac {1+n}{2 n}+1, \frac {1+n}{2 n}+\frac {1}{2}, \frac {a \,x^{n}}{n}\right )}{a \left (1+2 n \right )}\right )}{n}+c_{2}}{x} \\ \end{align*}
Verification of solutions
\[ y = \frac {\frac {c_{1} \left (\frac {a}{n}\right )^{-\frac {1}{n}} \left (\frac {n^{3} x^{1-n} \left (\frac {a}{n}\right )^{\frac {1}{n}} \left (a \,x^{n}+n +1\right ) \left (\frac {a \,x^{n}}{n}\right )^{-\frac {1+n}{2 n}} {\mathrm e}^{-\frac {a \,x^{n}}{2 n}} \operatorname {WhittakerM}\left (\frac {1}{n}-\frac {1+n}{2 n}, \frac {1+n}{2 n}+\frac {1}{2}, \frac {a \,x^{n}}{n}\right )}{\left (1+n \right ) \left (1+2 n \right ) a}+\frac {n^{2} x^{1-n} \left (\frac {a}{n}\right )^{\frac {1}{n}} \left (1+n \right ) \left (\frac {a \,x^{n}}{n}\right )^{-\frac {1+n}{2 n}} {\mathrm e}^{-\frac {a \,x^{n}}{2 n}} \operatorname {WhittakerM}\left (\frac {1}{n}-\frac {1+n}{2 n}+1, \frac {1+n}{2 n}+\frac {1}{2}, \frac {a \,x^{n}}{n}\right )}{a \left (1+2 n \right )}\right )}{n}+c_{2}}{x} \] 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: 0.203 (sec). Leaf size: 122
dsolve(x*diff(y(x),x$2)+(a*x^n+2)*diff(y(x),x)+a*x^(n-1)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {n c_{2} {\mathrm e}^{-\frac {a \,x^{n}}{2 n}} \left (\left (n +1\right ) x^{-\frac {3 n}{2}+\frac {1}{2}}+x^{-\frac {n}{2}+\frac {1}{2}} a \right ) \operatorname {WhittakerM}\left (-\frac {n -1}{2 n}, \frac {2 n +1}{2 n}, \frac {a \,x^{n}}{n}\right )+c_{2} x^{-\frac {3 n}{2}+\frac {1}{2}} {\mathrm e}^{-\frac {a \,x^{n}}{2 n}} \left (n +1\right )^{2} \operatorname {WhittakerM}\left (\frac {n +1}{2 n}, \frac {2 n +1}{2 n}, \frac {a \,x^{n}}{n}\right )+c_{1}}{x} \]
✓ Solution by Mathematica
Time used: 0.099 (sec). Leaf size: 62
DSolve[x*y''[x]+(a*x^n+2)*y'[x]+a*x^(n-1)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to (-1)^{-1/n} n^{\frac {1}{n}-1} a^{-1/n} \left (x^n\right )^{-1/n} \left (c_1 (-1)^{\frac {1}{n}} \Gamma \left (\frac {1}{n},0,\frac {a x^n}{n}\right )+c_2 n\right ) \]