Internal problem ID [10455]
Internal file name [OUTPUT/9403_Monday_June_06_2022_02_24_26_PM_28185864/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: 8.
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]
\[ \boxed {y^{\prime }-\alpha y^{2}=\beta +\gamma \cosh \left (x \right )} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \alpha \,y^{2}+\beta +\gamma \cosh \left (x \right ) \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \alpha \,y^{2}+\beta +\gamma \cosh \left (x \right ) \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\beta +\gamma \cosh \left (x \right )\), \(f_1(x)=0\) and \(f_2(x)=\alpha \). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\alpha 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' &=0\\ f_1 f_2 &=0\\ f_2^2 f_0 &=\alpha ^{2} \left (\beta +\gamma \cosh \left (x \right )\right ) \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \alpha u^{\prime \prime }\left (x \right )+\alpha ^{2} \left (\beta +\gamma \cosh \left (x \right )\right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} \operatorname {MathieuC}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )+c_{2} \operatorname {MathieuS}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {i \left (c_{2} \operatorname {MathieuSPrime}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )+c_{1} \operatorname {MathieuCPrime}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )\right )}{2} \] Using the above in (1) gives the solution \[ y = -\frac {i \left (c_{2} \operatorname {MathieuSPrime}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )+c_{1} \operatorname {MathieuCPrime}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )\right )}{2 \alpha \left (c_{1} \operatorname {MathieuC}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )+c_{2} \operatorname {MathieuS}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {i \left (\operatorname {MathieuSPrime}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )+c_{3} \operatorname {MathieuCPrime}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )\right )}{2 \alpha \left (c_{3} \operatorname {MathieuC}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )+\operatorname {MathieuS}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {i \left (\operatorname {MathieuSPrime}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )+c_{3} \operatorname {MathieuCPrime}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )\right )}{2 \alpha \left (c_{3} \operatorname {MathieuC}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )+\operatorname {MathieuS}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )\right )} \\ \end{align*}
Verification of solutions
\[ y = -\frac {i \left (\operatorname {MathieuSPrime}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )+c_{3} \operatorname {MathieuCPrime}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )\right )}{2 \alpha \left (c_{3} \operatorname {MathieuC}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )+\operatorname {MathieuS}\left (-4 \alpha \beta , 2 \alpha \gamma , \frac {i x}{2}\right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\alpha y^{2}=\beta +\gamma \cosh \left (x \right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\alpha y^{2}+\beta +\gamma \cosh \left (x \right ) \end {array} \]
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 Special trying Riccati sub-methods: trying Riccati_symmetries trying Riccati to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = -alpha*(beta+gamma*cosh(x))*y(x), y(x)` *** Sublevel 2 *** 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 quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> 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 -> 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 Equivalence transformation and function parameters: {z = 1/2*t+1/2}, {kappa = -16*alpha*beta+16*alpha*gamma-4, mu = 3 <- Equivalence to the rational form of Mathieu ODE successful <- Mathieu successful <- special function solution successful Change of variables used: [x = arccosh(t)] Linear ODE actually solved: (alpha*gamma*t+alpha*beta)*u(t)+t*diff(u(t),t)+(t^2-1)*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: 70
dsolve(diff(y(x),x)=alpha*y(x)^2+beta+gamma*cosh(x),y(x), singsol=all)
\[ y \left (x \right ) = -\frac {i \left (c_{1} \operatorname {MathieuSPrime}\left (-4 \alpha \beta , 2 \gamma \alpha , \frac {i x}{2}\right )+\operatorname {MathieuCPrime}\left (-4 \alpha \beta , 2 \gamma \alpha , \frac {i x}{2}\right )\right )}{2 \alpha \left (c_{1} \operatorname {MathieuS}\left (-4 \alpha \beta , 2 \gamma \alpha , \frac {i x}{2}\right )+\operatorname {MathieuC}\left (-4 \alpha \beta , 2 \gamma \alpha , \frac {i x}{2}\right )\right )} \]
✓ Solution by Mathematica
Time used: 0.543 (sec). Leaf size: 140
DSolve[y'[x]==\[Alpha]*y[x]^2+\[Beta]+\[Gamma]*Cosh[x],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {i c_1 \text {MathieuCPrime}\left [-4 \alpha \beta ,2 \alpha \gamma ,\frac {i x}{2}\right ]-i \text {MathieuSPrime}\left [-4 \alpha \beta ,2 \alpha \gamma ,\frac {i x}{2}\right ]}{2 \alpha c_1 \text {MathieuC}\left [-4 \alpha \beta ,2 \alpha \gamma ,\frac {i x}{2}\right ]-2 \alpha \text {MathieuS}\left [-4 \alpha \beta ,2 \alpha \gamma ,\frac {i x}{2}\right ]} \\ y(x)\to -\frac {i \text {MathieuCPrime}\left [-4 \alpha \beta ,2 \alpha \gamma ,\frac {i x}{2}\right ]}{2 \alpha \text {MathieuC}\left [-4 \alpha \beta ,2 \alpha \gamma ,\frac {i x}{2}\right ]} \\ \end{align*}