problem 28

Internal problem ID [10525]
Internal ﬁle name [OUTPUT/9473_Monday_June_06_2022_02_47_13_PM_89469444/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 1.2.6-3. Equations with tangent.
Problem number: 28.
ODE order: 1.
ODE degree: 1.

\[ \boxed {y^{\prime }-y^{2}=3 \lambda a +\lambda ^{2}+a \left (-a +\lambda \right ) \tan \left (\lambda x \right )^{2}} \]

11.2.1 Solving as riccati ode

In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -a^{2} \tan \left (\lambda x \right )^{2}+a \tan \left (\lambda x \right )^{2} \lambda +3 \lambda a +\lambda ^{2}+y^{2} \end {align*}

This is a Riccati ODE. Comparing the ODE to solve \[ y' = -a^{2} \tan \left (\lambda x \right )^{2}+a \tan \left (\lambda x \right )^{2} \lambda +3 \lambda a +\lambda ^{2}+y^{2} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-a^{2} \tan \left (\lambda x \right )^{2}+a \tan \left (\lambda x \right )^{2} \lambda +3 \lambda a +\lambda ^{2}\), \(f_1(x)=0\) and \(f_2(x)=1\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{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 &=-a^{2} \tan \left (\lambda x \right )^{2}+a \tan \left (\lambda x \right )^{2} \lambda +3 \lambda a +\lambda ^{2} \end {align*}

Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (x \right )+\left (-a^{2} \tan \left (\lambda x \right )^{2}+a \tan \left (\lambda x \right )^{2} \lambda +3 \lambda a +\lambda ^{2}\right ) u \left (x \right ) &=0 \end {align*}

\[ u \left (x \right ) = \sqrt {\cos \left (\lambda x \right )}\, \left (\operatorname {LegendreP}\left (\frac {2 a +\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) c_{1} +\operatorname {LegendreQ}\left (\frac {2 a +\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) c_{2} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {-2 \operatorname {LegendreP}\left (\frac {2 a +3 \lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) c_{1} \lambda -2 \operatorname {LegendreQ}\left (\frac {2 a +3 \lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) c_{2} \lambda +\sin \left (\lambda x \right ) \left (\operatorname {LegendreP}\left (\frac {2 a +\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) c_{1} +\operatorname {LegendreQ}\left (\frac {2 a +\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) c_{2} \right ) \left (a +\lambda \right )}{\sqrt {\cos \left (\lambda x \right )}} \] Using the above in (1) gives the solution \[ y = -\frac {-2 \operatorname {LegendreP}\left (\frac {2 a +3 \lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) c_{1} \lambda -2 \operatorname {LegendreQ}\left (\frac {2 a +3 \lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) c_{2} \lambda +\sin \left (\lambda x \right ) \left (\operatorname {LegendreP}\left (\frac {2 a +\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) c_{1} +\operatorname {LegendreQ}\left (\frac {2 a +\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) c_{2} \right ) \left (a +\lambda \right )}{\cos \left (\lambda x \right ) \left (\operatorname {LegendreP}\left (\frac {2 a +\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) c_{1} +\operatorname {LegendreQ}\left (\frac {2 a +\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) c_{2} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution

\[ y = -\frac {\left (-2 \operatorname {LegendreP}\left (\frac {2 a +3 \lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) c_{3} \lambda -2 \operatorname {LegendreQ}\left (\frac {2 a +3 \lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) \lambda +\sin \left (\lambda x \right ) \left (\operatorname {LegendreP}\left (\frac {2 a +\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) c_{3} +\operatorname {LegendreQ}\left (\frac {2 a +\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right )\right ) \left (a +\lambda \right )\right ) \sec \left (\lambda x \right )}{\operatorname {LegendreP}\left (\frac {2 a +\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) c_{3} +\operatorname {LegendreQ}\left (\frac {2 a +\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right )} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\left (-2 \operatorname {LegendreP}\left (\frac {2 a +3 \lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) c_{3} \lambda -2 \operatorname {LegendreQ}\left (\frac {2 a +3 \lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) \lambda +\sin \left (\lambda x \right ) \left (\operatorname {LegendreP}\left (\frac {2 a +\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) c_{3} +\operatorname {LegendreQ}\left (\frac {2 a +\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right )\right ) \left (a +\lambda \right )\right ) \sec \left (\lambda x \right )}{\operatorname {LegendreP}\left (\frac {2 a +\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right ) c_{3} +\operatorname {LegendreQ}\left (\frac {2 a +\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (\lambda x \right )\right )} \\ \end{align*}

Veriﬁcation of solutions

11.2.2 Maple step by step solution

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

Maple trace Kovacic algorithm successful

`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) = (a^2*tan(lambda*x)^2-a*tan(lambda*x)^2*lambda-3*a*lambda-lambda^2)*y(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 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 
            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 
               -> Whittaker 
                  -> hyper3: Equivalence to 1F1 under a power @ Moebius 
               -> hypergeometric 
                  -> heuristic approach 
                  <- heuristic approach successful 
               <- hypergeometric successful 
            <- special function solution successful 
               -> Trying to convert hypergeometric functions to elementary form... 
               <- elementary form is not straightforward to achieve - returning special function solution free of uncomputed integra 
            <- Kovacics algorithm successful 
         Change of variables used: 
            [x = 1/lambda*arccos(t)] 
         Linear ODE actually solved: 
            (a^2*t^2+2*a*lambda*t^2+lambda^2*t^2-a^2+a*lambda)*u(t)-t^3*lambda^2*diff(u(t),t)+(-lambda^2*t^4+lambda^2*t^2)*diff(diff 
      <- change of variables successful 
   <- Riccati to 2nd Order successful`

\[ y \left (x \right ) = -\frac {\left (-2 \operatorname {LegendreP}\left (\frac {2 a +3 \lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (x \lambda \right )\right ) \lambda -2 \operatorname {LegendreQ}\left (\frac {2 a +3 \lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (x \lambda \right )\right ) c_{1} \lambda +\sin \left (x \lambda \right ) \left (\operatorname {LegendreQ}\left (\frac {2 a +\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (x \lambda \right )\right ) c_{1} +\operatorname {LegendreP}\left (\frac {2 a +\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (x \lambda \right )\right )\right ) \left (a +\lambda \right )\right ) \sec \left (x \lambda \right )}{\operatorname {LegendreQ}\left (\frac {2 a +\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (x \lambda \right )\right ) c_{1} +\operatorname {LegendreP}\left (\frac {2 a +\lambda }{2 \lambda }, \frac {2 a -\lambda }{2 \lambda }, \sin \left (x \lambda \right )\right )} \]

\begin{align*} y(x)\to -\frac {\sin ^{-\frac {a+\lambda }{\lambda }}(2 \lambda x) e^{-\frac {a \text {arctanh}(\cos (2 \lambda x))}{\lambda }} \left (c_1 \sin ^{\frac {a}{\lambda }}(2 \lambda x) ((a+\lambda ) \cos (2 \lambda x)-a+\lambda ) e^{\frac {a \text {arctanh}(\cos (2 \lambda x))}{\lambda }} \int _1^xe^{-\frac {(a-\lambda ) \text {arctanh}(\cos (2 \lambda K[1]))}{\lambda }} \sin ^{-\frac {a+\lambda }{\lambda }}(2 \lambda K[1])dK[1]+\sin ^{\frac {a}{\lambda }}(2 \lambda x) ((a+\lambda ) \cos (2 \lambda x)-a+\lambda ) e^{\frac {a \text {arctanh}(\cos (2 \lambda x))}{\lambda }}+c_1 e^{\text {arctanh}(\cos (2 \lambda x))}\right )}{1+c_1 \int _1^xe^{-\frac {(a-\lambda ) \text {arctanh}(\cos (2 \lambda K[1]))}{\lambda }} \sin ^{-\frac {a+\lambda }{\lambda }}(2 \lambda K[1])dK[1]} \\ y(x)\to \csc (2 \lambda x) \left (-\frac {\sin ^{-\frac {a}{\lambda }}(2 \lambda x) e^{-\frac {(a-\lambda ) \text {arctanh}(\cos (2 \lambda x))}{\lambda }}}{\int _1^xe^{-\frac {(a-\lambda ) \text {arctanh}(\cos (2 \lambda K[1]))}{\lambda }} \sin ^{-\frac {a+\lambda }{\lambda }}(2 \lambda K[1])dK[1]}-(a+\lambda ) \cos (2 \lambda x)+a-\lambda \right ) \\ \end{align*}

11.2 problem 28

11.2.1 Solving as riccati ode

11.2.2 Maple step by step solution