Internal problem ID [10518]
Internal file name [OUTPUT/9466_Monday_June_06_2022_02_44_15_PM_26441355/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 1.2.6-2. Equations with
cosine.
Problem number: 21.
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]
\[ \boxed {y^{\prime }-\left (\lambda +a \cos \left (\lambda x \right )^{2}\right ) y^{2}=\lambda -a +a \cos \left (\lambda x \right )^{2}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \cos \left (\lambda x \right )^{2} a \,y^{2}+a \cos \left (\lambda x \right )^{2}+\lambda \,y^{2}-a +\lambda \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \cos \left (\lambda x \right )^{2} a \,y^{2}+a \cos \left (\lambda x \right )^{2}+\lambda \,y^{2}-a +\lambda \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\lambda -a +a \cos \left (\lambda x \right )^{2}\), \(f_1(x)=0\) and \(f_2(x)=\lambda +a \cos \left (\lambda x \right )^{2}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\left (\lambda +a \cos \left (\lambda x \right )^{2}\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' &=-2 \sin \left (\lambda x \right ) a \lambda \cos \left (\lambda x \right )\\ f_1 f_2 &=0\\ f_2^2 f_0 &=\left (\lambda +a \cos \left (\lambda x \right )^{2}\right )^{2} \left (\lambda -a +a \cos \left (\lambda x \right )^{2}\right ) \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \left (\lambda +a \cos \left (\lambda x \right )^{2}\right ) u^{\prime \prime }\left (x \right )+2 \sin \left (\lambda x \right ) a \lambda \cos \left (\lambda x \right ) u^{\prime }\left (x \right )+\left (\lambda +a \cos \left (\lambda x \right )^{2}\right )^{2} \left (\lambda -a +a \cos \left (\lambda x \right )^{2}\right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \cos \left (\lambda x \right ) {\mathrm e}^{\frac {\cos \left (2 \lambda x \right ) a}{4 \lambda }} \left (c_{1} +2 i c_{2} \lambda \left (\int {\mathrm e}^{-\frac {\cos \left (2 \lambda x \right ) a}{2 \lambda }} \left (\sec \left (\lambda x \right )^{2} \lambda +a \right )d x \right )\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = -\frac {\sec \left (\lambda x \right ) \left (\lambda +a \cos \left (\lambda x \right )^{2}\right ) \left (2 i \sin \left (2 \lambda x \right ) \left (\int {\mathrm e}^{-\frac {\cos \left (2 \lambda x \right ) a}{2 \lambda }} \left (\sec \left (\lambda x \right )^{2} \lambda +a \right )d x \right ) c_{2} \lambda \,{\mathrm e}^{\frac {\cos \left (2 \lambda x \right ) a}{4 \lambda }}+\sin \left (2 \lambda x \right ) c_{1} {\mathrm e}^{\frac {\cos \left (2 \lambda x \right ) a}{4 \lambda }}-4 i \lambda c_{2} {\mathrm e}^{-\frac {\cos \left (2 \lambda x \right ) a}{4 \lambda }}\right )}{2} \] Using the above in (1) gives the solution \[ y = \frac {\sec \left (\lambda x \right ) \left (2 i \sin \left (2 \lambda x \right ) \left (\int {\mathrm e}^{-\frac {\cos \left (2 \lambda x \right ) a}{2 \lambda }} \left (\sec \left (\lambda x \right )^{2} \lambda +a \right )d x \right ) c_{2} \lambda \,{\mathrm e}^{\frac {\cos \left (2 \lambda x \right ) a}{4 \lambda }}+\sin \left (2 \lambda x \right ) c_{1} {\mathrm e}^{\frac {\cos \left (2 \lambda x \right ) a}{4 \lambda }}-4 i \lambda c_{2} {\mathrm e}^{-\frac {\cos \left (2 \lambda x \right ) a}{4 \lambda }}\right ) {\mathrm e}^{-\frac {\cos \left (2 \lambda x \right ) a}{4 \lambda }}}{2 \cos \left (\lambda x \right ) \left (c_{1} +2 i c_{2} \lambda \left (\int {\mathrm e}^{-\frac {\cos \left (2 \lambda x \right ) a}{2 \lambda }} \left (\sec \left (\lambda x \right )^{2} \lambda +a \right )d x \right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {i c_{3} \tan \left (\lambda x \right )-2 \lambda \left (\int {\mathrm e}^{-\frac {\cos \left (2 \lambda x \right ) a}{2 \lambda }} \left (\sec \left (\lambda x \right )^{2} \lambda +a \right )d x \right ) \tan \left (\lambda x \right )+2 \sec \left (\lambda x \right )^{2} {\mathrm e}^{-\frac {\cos \left (2 \lambda x \right ) a}{2 \lambda }} \lambda }{i c_{3} -2 \lambda \left (\int {\mathrm e}^{-\frac {\cos \left (2 \lambda x \right ) a}{2 \lambda }} \left (\sec \left (\lambda x \right )^{2} \lambda +a \right )d x \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {i c_{3} \tan \left (\lambda x \right )-2 \lambda \left (\int {\mathrm e}^{-\frac {\cos \left (2 \lambda x \right ) a}{2 \lambda }} \left (\sec \left (\lambda x \right )^{2} \lambda +a \right )d x \right ) \tan \left (\lambda x \right )+2 \sec \left (\lambda x \right )^{2} {\mathrm e}^{-\frac {\cos \left (2 \lambda x \right ) a}{2 \lambda }} \lambda }{i c_{3} -2 \lambda \left (\int {\mathrm e}^{-\frac {\cos \left (2 \lambda x \right ) a}{2 \lambda }} \left (\sec \left (\lambda x \right )^{2} \lambda +a \right )d x \right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {i c_{3} \tan \left (\lambda x \right )-2 \lambda \left (\int {\mathrm e}^{-\frac {\cos \left (2 \lambda x \right ) a}{2 \lambda }} \left (\sec \left (\lambda x \right )^{2} \lambda +a \right )d x \right ) \tan \left (\lambda x \right )+2 \sec \left (\lambda x \right )^{2} {\mathrm e}^{-\frac {\cos \left (2 \lambda x \right ) a}{2 \lambda }} \lambda }{i c_{3} -2 \lambda \left (\int {\mathrm e}^{-\frac {\cos \left (2 \lambda x \right ) a}{2 \lambda }} \left (\sec \left (\lambda x \right )^{2} \lambda +a \right )d x \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\left (\lambda +a \cos \left (\lambda x \right )^{2}\right ) y^{2}=\lambda -a +a \cos \left (\lambda x \right )^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\left (\lambda +a \cos \left (\lambda x \right )^{2}\right ) y^{2}+\lambda -a +a \cos \left (\lambda x \right )^{2} \end {array} \]
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 sub-methods: trying Riccati_symmetries trying Riccati to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = -2*a*cos(lambda*x)*lambda*sin(lambda*x)*(diff(y(x), x))/(lambda+cos(la 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 -> 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 -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius No special function solution was found. <- Kovacics algorithm successful Change of variables used: [x = 1/2*arccos(t)/lambda] Linear ODE actually solved: (4*a^3*t^3+4*a^3*t^2+24*a^2*lambda*t^2-4*a^3*t+16*a^2*lambda*t+48*a*lambda^2*t-4*a^3-8*a^2*lambda+16*a*lambda^2+32*lambd <- change of variables successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 102
dsolve(diff(y(x),x)=(lambda+a*cos(lambda*x)^2)*y(x)^2+lambda-a+a*cos(lambda*x)^2,y(x), singsol=all)
\[ y \left (x \right ) = \frac {2 \sec \left (x \lambda \right )^{2} {\mathrm e}^{-\frac {a \cos \left (2 x \lambda \right )}{2 \lambda }} c_{1} \lambda -2 \tan \left (x \lambda \right ) \lambda \left (\int {\mathrm e}^{-\frac {a \cos \left (2 x \lambda \right )}{2 \lambda }} \left (\sec \left (x \lambda \right )^{2} \lambda +a \right )d x \right ) c_{1} +i \tan \left (x \lambda \right )}{-2 \lambda \left (\int {\mathrm e}^{-\frac {a \cos \left (2 x \lambda \right )}{2 \lambda }} \left (\sec \left (x \lambda \right )^{2} \lambda +a \right )d x \right ) c_{1} +i} \]
✓ Solution by Mathematica
Time used: 36.333 (sec). Leaf size: 263
DSolve[y'[x]==(\[Lambda]+a*Cos[\[Lambda]*x]^2)*y[x]^2+\[Lambda]-a+a*Cos[\[Lambda]*x]^2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {2 \left (c_1 \tan (\lambda x) \int _1^xe^{-\frac {a \cos ^2(\lambda K[1])}{\lambda }} \left (\lambda \sec ^2(\lambda K[1])+a\right )dK[1]+c_1 \sec ^2(\lambda x) \left (-e^{-\frac {a \cos ^2(\lambda x)}{\lambda }}\right )+\tan (\lambda x)\right )}{2+2 c_1 \int _1^xe^{-\frac {a \cos ^2(\lambda K[1])}{\lambda }} \left (\lambda \sec ^2(\lambda K[1])+a\right )dK[1]} \\ y(x)\to \frac {1}{2} \sec ^2(\lambda x) \left (\sin (2 \lambda x)-\frac {2 e^{-\frac {a \cos ^2(\lambda x)}{\lambda }}}{\int _1^xe^{-\frac {a \cos ^2(\lambda K[1])}{\lambda }} \left (\lambda \sec ^2(\lambda K[1])+a\right )dK[1]}\right ) \\ y(x)\to \frac {1}{2} \sec ^2(\lambda x) \left (\sin (2 \lambda x)-\frac {2 e^{-\frac {a \cos ^2(\lambda x)}{\lambda }}}{\int _1^xe^{-\frac {a \cos ^2(\lambda K[1])}{\lambda }} \left (\lambda \sec ^2(\lambda K[1])+a\right )dK[1]}\right ) \\ \end{align*}