Internal problem ID [10462]
Internal file name [OUTPUT/9410_Monday_June_06_2022_02_27_12_PM_40133814/index.tex]
Book: Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev.
Second edition
Section: Chapter 1, section 1.2. Riccati Equation. subsection 1.2.4-1. Equations with hyperbolic
sine and cosine
Problem number: 15.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[_Riccati]
\[ \boxed {y^{\prime }-\cosh \left (\lambda x \right ) y^{2} a=b \cosh \left (\lambda x \right ) \sinh \left (\lambda x \right )^{n}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \cosh \left (\lambda x \right ) a \,y^{2}+b \cosh \left (\lambda x \right ) \sinh \left (\lambda x \right )^{n} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \cosh \left (\lambda x \right ) a \,y^{2}+b \cosh \left (\lambda x \right ) \sinh \left (\lambda x \right )^{n} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=b \cosh \left (\lambda x \right ) \sinh \left (\lambda x \right )^{n}\), \(f_1(x)=0\) and \(f_2(x)=a \cosh \left (\lambda x \right )\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{a \cosh \left (\lambda x \right ) u} \tag {1} \end {align*}
Using the above substitution in the given ODE results (after some simplification)in a second order ODE to solve for \(u(x)\) which is \begin {align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag {2} \end {align*}
But \begin {align*} f_2' &=a \lambda \sinh \left (\lambda x \right )\\ f_1 f_2 &=0\\ f_2^2 f_0 &=a^{2} \cosh \left (\lambda x \right )^{3} b \sinh \left (\lambda x \right )^{n} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} a \cosh \left (\lambda x \right ) u^{\prime \prime }\left (x \right )-a \lambda \sinh \left (\lambda x \right ) u^{\prime }\left (x \right )+a^{2} \cosh \left (\lambda x \right )^{3} b \sinh \left (\lambda x \right )^{n} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \frac {-\csc \left (\frac {\pi \left (n +3\right )}{2+n}\right ) c_{1} \operatorname {BesselI}\left (-\frac {1}{2+n}, 2 \sqrt {-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}}\right ) \pi \left (-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}\right )^{\frac {1}{4+2 n}}+c_{2} \sinh \left (\lambda x \right ) \operatorname {BesselI}\left (\frac {1}{2+n}, 2 \sqrt {-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}}\right ) \Gamma \left (\frac {n +3}{2+n}\right )^{2} \left (-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}\right )^{-\frac {1}{4+2 n}} \left (2+n \right )}{\left (2+n \right ) \Gamma \left (\frac {n +3}{2+n}\right )} \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\lambda \left (\Gamma \left (\frac {n +3}{2+n}\right )^{2} \left (-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}\right )^{\frac {1+n}{4+2 n}} c_{2} \cosh \left (\lambda x \right ) \left (2+n \right ) \operatorname {BesselI}\left (\frac {n +3}{2+n}, 2 \sqrt {-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}}\right )+\cosh \left (\lambda x \right ) c_{2} \operatorname {BesselI}\left (\frac {1}{2+n}, 2 \sqrt {-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}}\right ) \Gamma \left (\frac {n +3}{2+n}\right )^{2} \left (-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}\right )^{-\frac {1}{4+2 n}}-\pi c_{1} \coth \left (\lambda x \right ) \csc \left (\frac {\pi \left (n +3\right )}{2+n}\right ) \left (-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}\right )^{\frac {n +3}{4+2 n}} \operatorname {BesselI}\left (\frac {1+n}{2+n}, 2 \sqrt {-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}}\right )\right )}{\Gamma \left (\frac {n +3}{2+n}\right )} \] Using the above in (1) gives the solution \[ y = -\frac {\lambda \left (\Gamma \left (\frac {n +3}{2+n}\right )^{2} \left (-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}\right )^{\frac {1+n}{4+2 n}} c_{2} \cosh \left (\lambda x \right ) \left (2+n \right ) \operatorname {BesselI}\left (\frac {n +3}{2+n}, 2 \sqrt {-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}}\right )+\cosh \left (\lambda x \right ) c_{2} \operatorname {BesselI}\left (\frac {1}{2+n}, 2 \sqrt {-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}}\right ) \Gamma \left (\frac {n +3}{2+n}\right )^{2} \left (-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}\right )^{-\frac {1}{4+2 n}}-\pi c_{1} \coth \left (\lambda x \right ) \csc \left (\frac {\pi \left (n +3\right )}{2+n}\right ) \left (-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}\right )^{\frac {n +3}{4+2 n}} \operatorname {BesselI}\left (\frac {1+n}{2+n}, 2 \sqrt {-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}}\right )\right ) \left (2+n \right )}{a \cosh \left (\lambda x \right ) \left (-\csc \left (\frac {\pi \left (n +3\right )}{2+n}\right ) c_{1} \operatorname {BesselI}\left (-\frac {1}{2+n}, 2 \sqrt {-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}}\right ) \pi \left (-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}\right )^{\frac {1}{4+2 n}}+c_{2} \sinh \left (\lambda x \right ) \operatorname {BesselI}\left (\frac {1}{2+n}, 2 \sqrt {-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}}\right ) \Gamma \left (\frac {n +3}{2+n}\right )^{2} \left (-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}\right )^{-\frac {1}{4+2 n}} \left (2+n \right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {\left (2+n \right ) \left (\Gamma \left (\frac {n +3}{2+n}\right )^{2} \left (-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}\right )^{\frac {1+n}{4+2 n}} \left (2+n \right ) \operatorname {BesselI}\left (\frac {n +3}{2+n}, 2 \sqrt {-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}}\right )-\operatorname {csch}\left (\lambda x \right ) \pi c_{3} \csc \left (\frac {\pi \left (n +3\right )}{2+n}\right ) \left (-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}\right )^{\frac {n +3}{4+2 n}} \operatorname {BesselI}\left (\frac {1+n}{2+n}, 2 \sqrt {-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}}\right )+\operatorname {BesselI}\left (\frac {1}{2+n}, 2 \sqrt {-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}}\right ) \Gamma \left (\frac {n +3}{2+n}\right )^{2} \left (-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}\right )^{-\frac {1}{4+2 n}}\right ) \lambda }{a \left (-\csc \left (\frac {\pi \left (n +3\right )}{2+n}\right ) c_{3} \operatorname {BesselI}\left (-\frac {1}{2+n}, 2 \sqrt {-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}}\right ) \pi \left (-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}\right )^{\frac {1}{4+2 n}}+\sinh \left (\lambda x \right ) \operatorname {BesselI}\left (\frac {1}{2+n}, 2 \sqrt {-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}}\right ) \Gamma \left (\frac {n +3}{2+n}\right )^{2} \left (-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}\right )^{-\frac {1}{4+2 n}} \left (2+n \right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\left (2+n \right ) \left (\Gamma \left (\frac {n +3}{2+n}\right )^{2} \left (-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}\right )^{\frac {1+n}{4+2 n}} \left (2+n \right ) \operatorname {BesselI}\left (\frac {n +3}{2+n}, 2 \sqrt {-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}}\right )-\operatorname {csch}\left (\lambda x \right ) \pi c_{3} \csc \left (\frac {\pi \left (n +3\right )}{2+n}\right ) \left (-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}\right )^{\frac {n +3}{4+2 n}} \operatorname {BesselI}\left (\frac {1+n}{2+n}, 2 \sqrt {-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}}\right )+\operatorname {BesselI}\left (\frac {1}{2+n}, 2 \sqrt {-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}}\right ) \Gamma \left (\frac {n +3}{2+n}\right )^{2} \left (-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}\right )^{-\frac {1}{4+2 n}}\right ) \lambda }{a \left (-\csc \left (\frac {\pi \left (n +3\right )}{2+n}\right ) c_{3} \operatorname {BesselI}\left (-\frac {1}{2+n}, 2 \sqrt {-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}}\right ) \pi \left (-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}\right )^{\frac {1}{4+2 n}}+\sinh \left (\lambda x \right ) \operatorname {BesselI}\left (\frac {1}{2+n}, 2 \sqrt {-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}}\right ) \Gamma \left (\frac {n +3}{2+n}\right )^{2} \left (-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}\right )^{-\frac {1}{4+2 n}} \left (2+n \right )\right )} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\left (2+n \right ) \left (\Gamma \left (\frac {n +3}{2+n}\right )^{2} \left (-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}\right )^{\frac {1+n}{4+2 n}} \left (2+n \right ) \operatorname {BesselI}\left (\frac {n +3}{2+n}, 2 \sqrt {-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}}\right )-\operatorname {csch}\left (\lambda x \right ) \pi c_{3} \csc \left (\frac {\pi \left (n +3\right )}{2+n}\right ) \left (-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}\right )^{\frac {n +3}{4+2 n}} \operatorname {BesselI}\left (\frac {1+n}{2+n}, 2 \sqrt {-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}}\right )+\operatorname {BesselI}\left (\frac {1}{2+n}, 2 \sqrt {-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}}\right ) \Gamma \left (\frac {n +3}{2+n}\right )^{2} \left (-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}\right )^{-\frac {1}{4+2 n}}\right ) \lambda }{a \left (-\csc \left (\frac {\pi \left (n +3\right )}{2+n}\right ) c_{3} \operatorname {BesselI}\left (-\frac {1}{2+n}, 2 \sqrt {-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}}\right ) \pi \left (-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}\right )^{\frac {1}{4+2 n}}+\sinh \left (\lambda x \right ) \operatorname {BesselI}\left (\frac {1}{2+n}, 2 \sqrt {-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}}\right ) \Gamma \left (\frac {n +3}{2+n}\right )^{2} \left (-\frac {a b \sinh \left (\lambda x \right )^{2+n}}{\lambda ^{2} \left (2+n \right )^{2}}\right )^{-\frac {1}{4+2 n}} \left (2+n \right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\cosh \left (\lambda x \right ) y^{2} a =b \cosh \left (\lambda x \right ) \sinh \left (\lambda x \right )^{n} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\cosh \left (\lambda x \right ) y^{2} a +b \cosh \left (\lambda x \right ) \sinh \left (\lambda x \right )^{n} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying Chini differential order: 1; looking for linear symmetries trying exact Looking for potential symmetries trying Riccati trying Riccati sub-methods: trying Riccati_symmetries trying Riccati to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = lambda*sinh(lambda*x)*(diff(y(x), x))/cosh(lambda*x)-a*cosh(lambda*x)^ 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 -> 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 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 <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 0F1 ODE <- Kummer successful <- special function solution successful Change of variables used: [x = arccosh(t)/lambda] Linear ODE actually solved: 4*a*b*(t^2-1)^(1/2*n)*t^3*u(t)+4*lambda^2*diff(u(t),t)+(4*lambda^2*t^3-4*lambda^2*t)*diff(diff(u(t),t),t) = 0 <- change of variables successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 951
dsolve(diff(y(x),x)=a*cosh(lambda*x)*y(x)^2+b*cosh(lambda*x)*sinh(lambda*x)^n,y(x), singsol=all)
\[ \text {Expression too large to display} \]
✓ Solution by Mathematica
Time used: 1.277 (sec). Leaf size: 633
DSolve[y'[x]==a*Cosh[\[Lambda]*x]*y[x]^2+b*Cosh[\[Lambda]*x]*Sinh[\[Lambda]*x]^n,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {\text {csch}(\lambda x) \left (-\lambda \operatorname {Gamma}\left (1+\frac {1}{n+2}\right ) \operatorname {BesselJ}\left (\frac {1}{n+2},\frac {2 \sqrt {a} \sqrt {b} \sinh ^{\frac {n}{2}+1}(x \lambda )}{n \lambda +2 \lambda }\right )+\sqrt {a} \sqrt {b} \sinh ^{\frac {n}{2}+1}(\lambda x) \left (\operatorname {Gamma}\left (1+\frac {1}{n+2}\right ) \left (\operatorname {BesselJ}\left (1+\frac {1}{n+2},\frac {2 \sqrt {a} \sqrt {b} \sinh ^{\frac {n}{2}+1}(x \lambda )}{n \lambda +2 \lambda }\right )-\operatorname {BesselJ}\left (\frac {1}{n+2}-1,\frac {2 \sqrt {a} \sqrt {b} \sinh ^{\frac {n}{2}+1}(x \lambda )}{n \lambda +2 \lambda }\right )\right )+c_1 \operatorname {Gamma}\left (\frac {n+1}{n+2}\right ) \operatorname {BesselJ}\left (\frac {n+1}{n+2},\frac {2 \sqrt {a} \sqrt {b} \sinh ^{\frac {n}{2}+1}(x \lambda )}{n \lambda +2 \lambda }\right )-c_1 \operatorname {Gamma}\left (\frac {n+1}{n+2}\right ) \operatorname {BesselJ}\left (-\frac {n+3}{n+2},\frac {2 \sqrt {a} \sqrt {b} \sinh ^{\frac {n}{2}+1}(x \lambda )}{n \lambda +2 \lambda }\right )\right )-c_1 \lambda \operatorname {Gamma}\left (\frac {n+1}{n+2}\right ) \operatorname {BesselJ}\left (-\frac {1}{n+2},\frac {2 \sqrt {a} \sqrt {b} \sinh ^{\frac {n}{2}+1}(x \lambda )}{n \lambda +2 \lambda }\right )\right )}{2 a \left (\operatorname {Gamma}\left (1+\frac {1}{n+2}\right ) \operatorname {BesselJ}\left (\frac {1}{n+2},\frac {2 \sqrt {a} \sqrt {b} \sinh ^{\frac {n}{2}+1}(x \lambda )}{n \lambda +2 \lambda }\right )+c_1 \operatorname {Gamma}\left (\frac {n+1}{n+2}\right ) \operatorname {BesselJ}\left (-\frac {1}{n+2},\frac {2 \sqrt {a} \sqrt {b} \sinh ^{\frac {n}{2}+1}(x \lambda )}{n \lambda +2 \lambda }\right )\right )} \\ y(x)\to \frac {\frac {\sqrt {a} \sqrt {b} \sinh ^{\frac {n}{2}}(\lambda x) \left (\operatorname {BesselJ}\left (\frac {n+1}{n+2},\frac {2 \sqrt {a} \sqrt {b} \sinh ^{\frac {n}{2}+1}(x \lambda )}{n \lambda +2 \lambda }\right )-\operatorname {BesselJ}\left (-\frac {n+3}{n+2},\frac {2 \sqrt {a} \sqrt {b} \sinh ^{\frac {n}{2}+1}(x \lambda )}{n \lambda +2 \lambda }\right )\right )}{\operatorname {BesselJ}\left (-\frac {1}{n+2},\frac {2 \sqrt {a} \sqrt {b} \sinh ^{\frac {n}{2}+1}(x \lambda )}{n \lambda +2 \lambda }\right )}-\lambda \text {csch}(\lambda x)}{2 a} \\ \end{align*}