Internal problem ID [10919]
Internal file name [OUTPUT/10176_Sunday_December_31_2023_11_03_32_AM_23395983/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: 96.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "unknown"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {x y^{\prime \prime }+\left (a \,x^{n}+b \right ) y^{\prime }+a \left (n +b -1\right ) x^{n -1} y=0} \]
In normal form the ode \begin {align*} x y^{\prime \prime }+\left (a \,x^{n}+b \right ) y^{\prime }+a \left (n +b -1\right ) x^{n -1} y = 0 \tag {1} \end {align*}
Becomes \begin {align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=r \left (x \right ) \tag {2} \end {align*}
Where \begin {align*} p \left (x \right )&=\frac {a \,x^{n}+b}{x}\\ q \left (x \right )&=a \,x^{-2+n} \left (n +b -1\right )\\ r \left (x \right )&=0 \end {align*}
The Lagrange adjoint ode is given by \begin {align*} \xi ^{''}-(\xi \, p)'+\xi q &= 0\\ \xi ^{''}-\left (\frac {\left (a \,x^{n}+b \right ) \xi \left (x \right )}{x}\right )' + \left (a \,x^{-2+n} \left (n +b -1\right ) \xi \left (x \right )\right ) &= 0\\ \xi ^{\prime \prime }\left (x \right )-\frac {\left (a \,x^{n}+b \right ) \xi ^{\prime }\left (x \right )}{x}+\left (-\frac {a \,x^{n} n}{x^{2}}+\frac {a \,x^{n}+b}{x^{2}}+a \,x^{-2+n} \left (n +b -1\right )\right ) \xi \left (x \right )&= 0 \end {align*}
Which is solved for \(\xi (x)\). In normal form the ode \begin {align*} \xi ^{\prime \prime }\left (x \right ) x^{2}+\left (-a \,x^{n}-b \right ) \xi ^{\prime }\left (x \right ) x +b \left (a \,x^{n}+1\right ) \xi \left (x \right )&=0 \tag {1} \end {align*}
Becomes \begin {align*} \xi ^{\prime \prime }\left (x \right )+p \left (x \right ) \xi ^{\prime }\left (x \right )+q \left (x \right ) \xi \left (x \right )&=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 \(\xi \left (x \right ) = v \left (x \right ) x^{n}\) to (2) gives the following ode where the dependent variables is \(v \left (x \right )\) and not \(\xi \left (x \right )\). \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 {u \left (-b +a \,x^{n}\right )}{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}\\ &= \int c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} \left (x^{n}\right )^{-\frac {b}{n}}d x +c_{2} \end {align*}
Hence \begin {align*} \xi \left (x \right )&= v \left (x \right ) x^{n}\\ &= \left (\int c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} \left (x^{n}\right )^{-\frac {b}{n}}d x +c_{2} \right ) x^{b}\\ &= \left (c_{1} \left (\int {\mathrm e}^{\frac {a \,x^{n}}{n}} \left (x^{n}\right )^{-\frac {b}{n}}d x \right )+c_{2} \right ) x^{b}\\ \end {align*}
The original ode (2) now reduces to first order ode \begin {align*} \xi \left (x \right ) y^{\prime }-y \xi ^{\prime }\left (x \right )+\xi \left (x \right ) p \left (x \right ) y&=\int \xi \left (x \right ) r \left (x \right )d x\\ y^{\prime }+y \left (p \left (x \right )-\frac {\xi ^{\prime }\left (x \right )}{\xi \left (x \right )}\right )&=\frac {\int \xi \left (x \right ) r \left (x \right )d x}{\xi \left (x \right )}\\ y^{\prime }+y \left (\frac {a \,x^{n}+b}{x}-\frac {\left (c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} \left (x^{n}\right )^{-\frac {b}{n}} x^{b}+\frac {\left (\int c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} \left (x^{n}\right )^{-\frac {b}{n}}d x +c_{2} \right ) x^{b} b}{x}\right ) x^{-b}}{\int c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} \left (x^{n}\right )^{-\frac {b}{n}}d x +c_{2}}\right )&=0 \end {align*}
Which is now a first order ode. This is now solved for \(y\). In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {y \left (-\left (\int c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} \left (x^{n}\right )^{-\frac {b}{n}}d x \right ) \left (x^{n}\right )^{\frac {b}{n}} x^{n} a -a \,x^{n} \left (x^{n}\right )^{\frac {b}{n}} c_{2} +c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} x \right ) \left (x^{n}\right )^{-\frac {b}{n}}}{x \left (\int c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} \left (x^{n}\right )^{-\frac {b}{n}}d x +c_{2} \right )} \end {align*}
Where \(f(x)=\frac {\left (-\left (\int c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} \left (x^{n}\right )^{-\frac {b}{n}}d x \right ) \left (x^{n}\right )^{\frac {b}{n}} x^{n} a -a \,x^{n} \left (x^{n}\right )^{\frac {b}{n}} c_{2} +c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} x \right ) \left (x^{n}\right )^{-\frac {b}{n}}}{x \left (\int c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} \left (x^{n}\right )^{-\frac {b}{n}}d x +c_{2} \right )}\) and \(g(y)=y\). Integrating both sides gives \begin {align*} \frac {1}{y} \,dy &= \frac {\left (-\left (\int c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} \left (x^{n}\right )^{-\frac {b}{n}}d x \right ) \left (x^{n}\right )^{\frac {b}{n}} x^{n} a -a \,x^{n} \left (x^{n}\right )^{\frac {b}{n}} c_{2} +c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} x \right ) \left (x^{n}\right )^{-\frac {b}{n}}}{x \left (\int c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} \left (x^{n}\right )^{-\frac {b}{n}}d x +c_{2} \right )} \,d x\\ \int { \frac {1}{y} \,dy} &= \int {\frac {\left (-\left (\int c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} \left (x^{n}\right )^{-\frac {b}{n}}d x \right ) \left (x^{n}\right )^{\frac {b}{n}} x^{n} a -a \,x^{n} \left (x^{n}\right )^{\frac {b}{n}} c_{2} +c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} x \right ) \left (x^{n}\right )^{-\frac {b}{n}}}{x \left (\int c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} \left (x^{n}\right )^{-\frac {b}{n}}d x +c_{2} \right )} \,d x}\\ \ln \left (y \right )&=\int \frac {\left (-\left (\int c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} \left (x^{n}\right )^{-\frac {b}{n}}d x \right ) \left (x^{n}\right )^{\frac {b}{n}} x^{n} a -a \,x^{n} \left (x^{n}\right )^{\frac {b}{n}} c_{2} +c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} x \right ) \left (x^{n}\right )^{-\frac {b}{n}}}{x \left (\int c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} \left (x^{n}\right )^{-\frac {b}{n}}d x +c_{2} \right )}d x +c_{3}\\ y&={\mathrm e}^{\int \frac {\left (-\left (\int c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} \left (x^{n}\right )^{-\frac {b}{n}}d x \right ) \left (x^{n}\right )^{\frac {b}{n}} x^{n} a -a \,x^{n} \left (x^{n}\right )^{\frac {b}{n}} c_{2} +c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} x \right ) \left (x^{n}\right )^{-\frac {b}{n}}}{x \left (\int c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} \left (x^{n}\right )^{-\frac {b}{n}}d x +c_{2} \right )}d x +c_{3}}\\ &=c_{3} {\mathrm e}^{\int \frac {\left (-\left (\int c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} \left (x^{n}\right )^{-\frac {b}{n}}d x \right ) \left (x^{n}\right )^{\frac {b}{n}} x^{n} a -a \,x^{n} \left (x^{n}\right )^{\frac {b}{n}} c_{2} +c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} x \right ) \left (x^{n}\right )^{-\frac {b}{n}}}{x \left (\int c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} \left (x^{n}\right )^{-\frac {b}{n}}d x +c_{2} \right )}d x} \end {align*}
Hence, the solution found using Lagrange adjoint equation method is \[ y = c_{3} {\mathrm e}^{\int \frac {\left (-\left (\int c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} \left (x^{n}\right )^{-\frac {b}{n}}d x \right ) \left (x^{n}\right )^{\frac {b}{n}} x^{n} a -a \,x^{n} \left (x^{n}\right )^{\frac {b}{n}} c_{2} +c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} x \right ) \left (x^{n}\right )^{-\frac {b}{n}}}{x \left (\int c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} \left (x^{n}\right )^{-\frac {b}{n}}d x +c_{2} \right )}d x} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{3} {\mathrm e}^{\int \frac {\left (-\left (\int c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} \left (x^{n}\right )^{-\frac {b}{n}}d x \right ) \left (x^{n}\right )^{\frac {b}{n}} x^{n} a -a \,x^{n} \left (x^{n}\right )^{\frac {b}{n}} c_{2} +c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} x \right ) \left (x^{n}\right )^{-\frac {b}{n}}}{x \left (\int c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} \left (x^{n}\right )^{-\frac {b}{n}}d x +c_{2} \right )}d x} \\ \end{align*}
Verification of solutions
\[ y = c_{3} {\mathrm e}^{\int \frac {\left (-\left (\int c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} \left (x^{n}\right )^{-\frac {b}{n}}d x \right ) \left (x^{n}\right )^{\frac {b}{n}} x^{n} a -a \,x^{n} \left (x^{n}\right )^{\frac {b}{n}} c_{2} +c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} x \right ) \left (x^{n}\right )^{-\frac {b}{n}}}{x \left (\int c_{1} {\mathrm e}^{\frac {a \,x^{n}}{n}} \left (x^{n}\right )^{-\frac {b}{n}}d x +c_{2} \right )}d x} \] 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 Solution has integrals. Trying a special function solution free of integrals... -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre <- Kummer successful <- special function solution successful Solution using Kummer functions still has integrals. Trying a hypergeometric solution. -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 1F1 ODE -> Trying to convert hypergeometric functions to elementary form... <- elementary form could result into a too large expression - returning special function form of solution, free of uncomput <- Kovacics algorithm successful <- Equivalence, under non-integer power transformations successful`
✓ Solution by Maple
Time used: 0.11 (sec). Leaf size: 56
dsolve(x*diff(y(x),x$2)+(a*x^n+b)*diff(y(x),x)+a*(b+n-1)*x^(n-1)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{-\frac {a \,x^{n}}{n}} \left (c_{1} +x^{-b +1} c_{2} \operatorname {hypergeom}\left (\left [\frac {-b +1}{n}\right ], \left [\frac {-b +n +1}{n}\right ], \frac {a \,x^{n}}{n}\right )\right ) \]
✓ Solution by Mathematica
Time used: 0.178 (sec). Leaf size: 93
DSolve[x*y''[x]+(a*x^n+b)*y'[x]+a*(b+n-1)*x^(n-1)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {(-1)^{-1/n} e^{-\frac {a x^n}{n}} \left ((b-1) c_2 (-1)^{b/n} \Gamma \left (\frac {1-b}{n},-\frac {a x^n}{n}\right )-(b-1) c_2 (-1)^{b/n} \operatorname {Gamma}\left (\frac {1-b}{n}\right )+c_1 (-1)^{\frac {1}{n}} n\right )}{n} \]