problem 1

Internal problem ID [10498]
Internal ﬁle name [OUTPUT/9446_Monday_June_06_2022_02_32_55_PM_8580992/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 1.2.6-1. Equations with sine
Problem number: 1.
ODE order: 1.
ODE degree: 1.

\[ \boxed {y^{\prime }-\alpha y^{2}=\beta +\gamma \sin \left (\lambda x \right )} \]

9.1.1 Solving as riccati ode

In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \alpha \,y^{2}+\beta +\gamma \sin \left (\lambda x \right ) \end {align*}

This is a Riccati ODE. Comparing the ODE to solve \[ y' = \alpha \,y^{2}+\beta +\gamma \sin \left (\lambda x \right ) \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\beta +\gamma \sin \left (\lambda x \right )\), \(f_1(x)=0\) 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 \sin \left (\lambda 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 \sin \left (\lambda x \right )\right ) u \left (x \right ) &=0 \end {align*}

\[ u \left (x \right ) = c_{1} \operatorname {MathieuC}\left (\frac {4 \alpha \beta }{\lambda ^{2}}, -\frac {2 \alpha \gamma }{\lambda ^{2}}, -\frac {\pi }{4}+\frac {\lambda x}{2}\right )+c_{2} \operatorname {MathieuS}\left (\frac {4 \alpha \beta }{\lambda ^{2}}, -\frac {2 \alpha \gamma }{\lambda ^{2}}, -\frac {\pi }{4}+\frac {\lambda x}{2}\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\lambda \left (c_{1} \operatorname {MathieuCPrime}\left (\frac {4 \alpha \beta }{\lambda ^{2}}, -\frac {2 \alpha \gamma }{\lambda ^{2}}, -\frac {\pi }{4}+\frac {\lambda x}{2}\right )+c_{2} \operatorname {MathieuSPrime}\left (\frac {4 \alpha \beta }{\lambda ^{2}}, -\frac {2 \alpha \gamma }{\lambda ^{2}}, -\frac {\pi }{4}+\frac {\lambda x}{2}\right )\right )}{2} \] Using the above in (1) gives the solution \[ y = -\frac {\lambda \left (c_{1} \operatorname {MathieuCPrime}\left (\frac {4 \alpha \beta }{\lambda ^{2}}, -\frac {2 \alpha \gamma }{\lambda ^{2}}, -\frac {\pi }{4}+\frac {\lambda x}{2}\right )+c_{2} \operatorname {MathieuSPrime}\left (\frac {4 \alpha \beta }{\lambda ^{2}}, -\frac {2 \alpha \gamma }{\lambda ^{2}}, -\frac {\pi }{4}+\frac {\lambda x}{2}\right )\right )}{2 \alpha \left (c_{1} \operatorname {MathieuC}\left (\frac {4 \alpha \beta }{\lambda ^{2}}, -\frac {2 \alpha \gamma }{\lambda ^{2}}, -\frac {\pi }{4}+\frac {\lambda x}{2}\right )+c_{2} \operatorname {MathieuS}\left (\frac {4 \alpha \beta }{\lambda ^{2}}, -\frac {2 \alpha \gamma }{\lambda ^{2}}, -\frac {\pi }{4}+\frac {\lambda x}{2}\right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution

\[ y = -\frac {\lambda \left (c_{3} \operatorname {MathieuCPrime}\left (\frac {4 \alpha \beta }{\lambda ^{2}}, -\frac {2 \alpha \gamma }{\lambda ^{2}}, -\frac {\pi }{4}+\frac {\lambda x}{2}\right )+\operatorname {MathieuSPrime}\left (\frac {4 \alpha \beta }{\lambda ^{2}}, -\frac {2 \alpha \gamma }{\lambda ^{2}}, -\frac {\pi }{4}+\frac {\lambda x}{2}\right )\right )}{2 \alpha \left (c_{3} \operatorname {MathieuC}\left (\frac {4 \alpha \beta }{\lambda ^{2}}, -\frac {2 \alpha \gamma }{\lambda ^{2}}, -\frac {\pi }{4}+\frac {\lambda x}{2}\right )+\operatorname {MathieuS}\left (\frac {4 \alpha \beta }{\lambda ^{2}}, -\frac {2 \alpha \gamma }{\lambda ^{2}}, -\frac {\pi }{4}+\frac {\lambda x}{2}\right )\right )} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\lambda \left (c_{3} \operatorname {MathieuCPrime}\left (\frac {4 \alpha \beta }{\lambda ^{2}}, -\frac {2 \alpha \gamma }{\lambda ^{2}}, -\frac {\pi }{4}+\frac {\lambda x}{2}\right )+\operatorname {MathieuSPrime}\left (\frac {4 \alpha \beta }{\lambda ^{2}}, -\frac {2 \alpha \gamma }{\lambda ^{2}}, -\frac {\pi }{4}+\frac {\lambda x}{2}\right )\right )}{2 \alpha \left (c_{3} \operatorname {MathieuC}\left (\frac {4 \alpha \beta }{\lambda ^{2}}, -\frac {2 \alpha \gamma }{\lambda ^{2}}, -\frac {\pi }{4}+\frac {\lambda x}{2}\right )+\operatorname {MathieuS}\left (\frac {4 \alpha \beta }{\lambda ^{2}}, -\frac {2 \alpha \gamma }{\lambda ^{2}}, -\frac {\pi }{4}+\frac {\lambda x}{2}\right )\right )} \\ \end{align*}

Veriﬁcation of solutions

9.1.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\alpha y^{2}=\beta +\gamma \sin \left (\lambda x \right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\alpha y^{2}+\beta +\gamma \sin \left (\lambda x \right ) \end {array} \]

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*sin(lambda*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 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 
            -> 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: {t = 1/2*t+1/2}, {kappa = 4*(4*alpha*beta-4*alpha*gamma-lambda^2) 
               <- Equivalence to the rational form of Mathieu ODE successful 
            <- Mathieu successful 
         <- special function solution successful 
         Change of variables used: 
            [x = 1/lambda*arccos(t)] 
         Linear ODE actually solved: 
            alpha*(beta+gamma*(-t^2+1)^(1/2))*u(t)-lambda^2*t*diff(u(t),t)+(-lambda^2*t^2+lambda^2)*diff(diff(u(t),t),t) = 0 
      <- change of variables successful 
   <- Riccati to 2nd Order successful`

\[ y \left (x \right ) = -\frac {\lambda \left (c_{1} \operatorname {MathieuSPrime}\left (\frac {4 \alpha \beta }{\lambda ^{2}}, -\frac {2 \gamma \alpha }{\lambda ^{2}}, -\frac {\pi }{4}+\frac {x \lambda }{2}\right )+\operatorname {MathieuCPrime}\left (\frac {4 \alpha \beta }{\lambda ^{2}}, -\frac {2 \gamma \alpha }{\lambda ^{2}}, -\frac {\pi }{4}+\frac {x \lambda }{2}\right )\right )}{2 \alpha \left (c_{1} \operatorname {MathieuS}\left (\frac {4 \alpha \beta }{\lambda ^{2}}, -\frac {2 \gamma \alpha }{\lambda ^{2}}, -\frac {\pi }{4}+\frac {x \lambda }{2}\right )+\operatorname {MathieuC}\left (\frac {4 \alpha \beta }{\lambda ^{2}}, -\frac {2 \gamma \alpha }{\lambda ^{2}}, -\frac {\pi }{4}+\frac {x \lambda }{2}\right )\right )} \]

\begin{align*} y(x)\to -\frac {\lambda \left (\text {MathieuSPrime}\left [\frac {4 \alpha \beta }{\lambda ^2},-\frac {2 \alpha \gamma }{\lambda ^2},\frac {1}{4} (\pi -2 \lambda x)\right ]+c_1 \text {MathieuCPrime}\left [\frac {4 \alpha \beta }{\lambda ^2},-\frac {2 \alpha \gamma }{\lambda ^2},\frac {1}{4} (2 \lambda x-\pi )\right ]\right )}{2 \alpha \left (\text {MathieuS}\left [\frac {4 \alpha \beta }{\lambda ^2},-\frac {2 \alpha \gamma }{\lambda ^2},\frac {1}{4} (2 \lambda x-\pi )\right ]+c_1 \text {MathieuC}\left [\frac {4 \alpha \beta }{\lambda ^2},-\frac {2 \alpha \gamma }{\lambda ^2},\frac {1}{4} (\pi -2 \lambda x)\right ]\right )} \\ y(x)\to \frac {\lambda \text {MathieuCPrime}\left [\frac {4 \alpha \beta }{\lambda ^2},-\frac {2 \alpha \gamma }{\lambda ^2},\frac {1}{4} (\pi -2 \lambda x)\right ]}{2 \alpha \text {MathieuC}\left [\frac {4 \alpha \beta }{\lambda ^2},-\frac {2 \alpha \gamma }{\lambda ^2},\frac {1}{4} (\pi -2 \lambda x)\right ]} \\ \end{align*}

9.1 problem 1

9.1.1 Solving as riccati ode

9.1.2 Maple step by step solution