Internal problem ID [10531]
Internal file name [OUTPUT/9479_Monday_June_06_2022_02_48_09_PM_7980074/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: 34.
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]
\[ \boxed {y^{\prime }-a \tan \left (\lambda x \right )^{n} y^{2}=-a \,b^{2} \tan \left (\lambda x \right )^{n +2}+b \lambda \tan \left (\lambda x \right )^{2}+b \lambda } \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= a \tan \left (\lambda x \right )^{n} y^{2}-a \,b^{2} \tan \left (\lambda x \right )^{n +2}+b \lambda \tan \left (\lambda x \right )^{2}+b \lambda \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = a \tan \left (\lambda x \right )^{n} y^{2}-a \,b^{2} \tan \left (\lambda x \right )^{n} \tan \left (\lambda x \right )^{2}+b \lambda \tan \left (\lambda x \right )^{2}+b \lambda \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-a \,b^{2} \tan \left (\lambda x \right )^{n +2}+b \lambda \tan \left (\lambda x \right )^{2}+b \lambda \), \(f_1(x)=0\) and \(f_2(x)=a \tan \left (\lambda x \right )^{n}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{a \tan \left (\lambda x \right )^{n} 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' &=\frac {a \tan \left (\lambda x \right )^{n} n \lambda \left (1+\tan \left (\lambda x \right )^{2}\right )}{\tan \left (\lambda x \right )}\\ f_1 f_2 &=0\\ f_2^2 f_0 &=a^{2} \tan \left (\lambda x \right )^{2 n} \left (-a \,b^{2} \tan \left (\lambda x \right )^{n +2}+b \lambda \tan \left (\lambda x \right )^{2}+b \lambda \right ) \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} a \tan \left (\lambda x \right )^{n} u^{\prime \prime }\left (x \right )-\frac {a \tan \left (\lambda x \right )^{n} n \lambda \left (1+\tan \left (\lambda x \right )^{2}\right ) u^{\prime }\left (x \right )}{\tan \left (\lambda x \right )}+a^{2} \tan \left (\lambda x \right )^{2 n} \left (-a \,b^{2} \tan \left (\lambda x \right )^{n +2}+b \lambda \tan \left (\lambda x \right )^{2}+b \lambda \right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = {\mathrm e}^{-a \left (\int \frac {\left (a \left (\int \tan \left (\lambda x \right )^{1+n} {\mathrm e}^{-\left (\int \left (-2 \cot \left (\lambda x \right ) \tan \left (\lambda x \right )^{n +2} a b +\cot \left (\lambda x \right ) \lambda \left (1+\tan \left (\lambda x \right )^{2}\right )\right )d x \right )}d x \right ) b -c_{1} b -{\mathrm e}^{-\left (\int \left (-2 \cot \left (\lambda x \right ) \tan \left (\lambda x \right )^{n +2} a b +\cot \left (\lambda x \right ) \lambda \left (1+\tan \left (\lambda x \right )^{2}\right )\right )d x \right )}\right ) \tan \left (\lambda x \right )^{1+n}}{a \left (\int \tan \left (\lambda x \right )^{1+n} {\mathrm e}^{-\left (\int \left (-2 \cot \left (\lambda x \right ) \tan \left (\lambda x \right )^{n +2} a b +\cot \left (\lambda x \right ) \lambda \left (1+\tan \left (\lambda x \right )^{2}\right )\right )d x \right )}d x \right )-c_{1}}d x \right )} c_{2} \] The above shows that \[ u^{\prime }\left (x \right ) = -\frac {a c_{2} \tan \left (\lambda x \right )^{1+n} \left (a \left (\int \tan \left (\lambda x \right )^{1+n} {\mathrm e}^{-\left (\int \left (-2 a b \tan \left (\lambda x \right )^{1+n}+\sec \left (\lambda x \right ) \csc \left (\lambda x \right ) \lambda \right )d x \right )}d x \right ) b -c_{1} b -{\mathrm e}^{-\left (\int \left (-2 a b \tan \left (\lambda x \right )^{1+n}+\sec \left (\lambda x \right ) \csc \left (\lambda x \right ) \lambda \right )d x \right )}\right ) {\mathrm e}^{-a \left (\int \frac {\tan \left (\lambda x \right )^{1+n} \left (a \left (\int \tan \left (\lambda x \right )^{1+n} {\mathrm e}^{-\left (\int \left (-2 a b \tan \left (\lambda x \right )^{1+n}+\sec \left (\lambda x \right ) \csc \left (\lambda x \right ) \lambda \right )d x \right )}d x \right ) b -c_{1} b -{\mathrm e}^{-\left (\int \left (-2 a b \tan \left (\lambda x \right )^{1+n}+\sec \left (\lambda x \right ) \csc \left (\lambda x \right ) \lambda \right )d x \right )}\right )}{a \left (\int \tan \left (\lambda x \right )^{1+n} {\mathrm e}^{-\left (\int \left (-2 a b \tan \left (\lambda x \right )^{1+n}+\sec \left (\lambda x \right ) \csc \left (\lambda x \right ) \lambda \right )d x \right )}d x \right )-c_{1}}d x \right )}}{a \left (\int \tan \left (\lambda x \right )^{1+n} {\mathrm e}^{-\left (\int \left (-2 \cot \left (\lambda x \right ) \tan \left (\lambda x \right )^{n +2} a b +\cot \left (\lambda x \right ) \lambda \left (1+\tan \left (\lambda x \right )^{2}\right )\right )d x \right )}d x \right )-c_{1}} \] Using the above in (1) gives the solution \[ y = \frac {\tan \left (\lambda x \right )^{1+n} \left (a \left (\int \tan \left (\lambda x \right )^{1+n} {\mathrm e}^{-\left (\int \left (-2 a b \tan \left (\lambda x \right )^{1+n}+\sec \left (\lambda x \right ) \csc \left (\lambda x \right ) \lambda \right )d x \right )}d x \right ) b -c_{1} b -{\mathrm e}^{-\left (\int \left (-2 a b \tan \left (\lambda x \right )^{1+n}+\sec \left (\lambda x \right ) \csc \left (\lambda x \right ) \lambda \right )d x \right )}\right ) {\mathrm e}^{-a \left (\int \frac {\tan \left (\lambda x \right )^{1+n} \left (a \left (\int \tan \left (\lambda x \right )^{1+n} {\mathrm e}^{-\left (\int \left (-2 a b \tan \left (\lambda x \right )^{1+n}+\sec \left (\lambda x \right ) \csc \left (\lambda x \right ) \lambda \right )d x \right )}d x \right ) b -c_{1} b -{\mathrm e}^{-\left (\int \left (-2 a b \tan \left (\lambda x \right )^{1+n}+\sec \left (\lambda x \right ) \csc \left (\lambda x \right ) \lambda \right )d x \right )}\right )}{a \left (\int \tan \left (\lambda x \right )^{1+n} {\mathrm e}^{-\left (\int \left (-2 a b \tan \left (\lambda x \right )^{1+n}+\sec \left (\lambda x \right ) \csc \left (\lambda x \right ) \lambda \right )d x \right )}d x \right )-c_{1}}d x \right )} \tan \left (\lambda x \right )^{-n} {\mathrm e}^{\int \frac {\left (-\left (\int {\mathrm e}^{\int \cot \left (\lambda x \right ) \left (2 \tan \left (\lambda x \right )^{n +2} a b -\tan \left (\lambda x \right )^{2} \lambda -\lambda \right )d x} \tan \left (\lambda x \right )^{1+n} a b d x \right )+{\mathrm e}^{\int \cot \left (\lambda x \right ) \left (2 \tan \left (\lambda x \right )^{n +2} a b -\tan \left (\lambda x \right )^{2} \lambda -\lambda \right )d x}+c_{1} b \right ) \tan \left (\lambda x \right )^{1+n} a}{-\left (\int {\mathrm e}^{\int \cot \left (\lambda x \right ) \left (2 \tan \left (\lambda x \right )^{n +2} a b -\tan \left (\lambda x \right )^{2} \lambda -\lambda \right )d x} \tan \left (\lambda x \right )^{1+n} a d x \right )+c_{1}}d x}}{a \left (\int \tan \left (\lambda x \right )^{1+n} {\mathrm e}^{-\left (\int \left (-2 \cot \left (\lambda x \right ) \tan \left (\lambda x \right )^{n +2} a b +\cot \left (\lambda x \right ) \lambda \left (1+\tan \left (\lambda x \right )^{2}\right )\right )d x \right )}d x \right )-c_{1}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\tan \left (\lambda x \right ) \left (a \left (\int \tan \left (\lambda x \right )^{1+n} {\mathrm e}^{-\left (\int \left (-2 a b \tan \left (\lambda x \right )^{1+n}+\sec \left (\lambda x \right ) \csc \left (\lambda x \right ) \lambda \right )d x \right )}d x \right ) b -c_{3} b -{\mathrm e}^{-\left (\int \left (-2 a b \tan \left (\lambda x \right )^{1+n}+\sec \left (\lambda x \right ) \csc \left (\lambda x \right ) \lambda \right )d x \right )}\right )}{a \left (\int \tan \left (\lambda x \right )^{1+n} {\mathrm e}^{-\left (\int \left (-2 \cot \left (\lambda x \right ) \tan \left (\lambda x \right )^{n +2} a b +\cot \left (\lambda x \right ) \lambda \left (1+\tan \left (\lambda x \right )^{2}\right )\right )d x \right )}d x \right )-c_{3}} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\tan \left (\lambda x \right ) \left (a \left (\int \tan \left (\lambda x \right )^{1+n} {\mathrm e}^{-\left (\int \left (-2 a b \tan \left (\lambda x \right )^{1+n}+\sec \left (\lambda x \right ) \csc \left (\lambda x \right ) \lambda \right )d x \right )}d x \right ) b -c_{3} b -{\mathrm e}^{-\left (\int \left (-2 a b \tan \left (\lambda x \right )^{1+n}+\sec \left (\lambda x \right ) \csc \left (\lambda x \right ) \lambda \right )d x \right )}\right )}{a \left (\int \tan \left (\lambda x \right )^{1+n} {\mathrm e}^{-\left (\int \left (-2 \cot \left (\lambda x \right ) \tan \left (\lambda x \right )^{n +2} a b +\cot \left (\lambda x \right ) \lambda \left (1+\tan \left (\lambda x \right )^{2}\right )\right )d x \right )}d x \right )-c_{3}} \\ \end{align*}
Verification of solutions
\[ y = \frac {\tan \left (\lambda x \right ) \left (a \left (\int \tan \left (\lambda x \right )^{1+n} {\mathrm e}^{-\left (\int \left (-2 a b \tan \left (\lambda x \right )^{1+n}+\sec \left (\lambda x \right ) \csc \left (\lambda x \right ) \lambda \right )d x \right )}d x \right ) b -c_{3} b -{\mathrm e}^{-\left (\int \left (-2 a b \tan \left (\lambda x \right )^{1+n}+\sec \left (\lambda x \right ) \csc \left (\lambda x \right ) \lambda \right )d x \right )}\right )}{a \left (\int \tan \left (\lambda x \right )^{1+n} {\mathrm e}^{-\left (\int \left (-2 \cot \left (\lambda x \right ) \tan \left (\lambda x \right )^{n +2} a b +\cot \left (\lambda x \right ) \lambda \left (1+\tan \left (\lambda x \right )^{2}\right )\right )d x \right )}d x \right )-c_{3}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-a \tan \left (\lambda x \right )^{n} y^{2}=-a \,b^{2} \tan \left (\lambda x \right )^{n +2}+b \lambda \tan \left (\lambda x \right )^{2}+b \lambda \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=a \tan \left (\lambda x \right )^{n} y^{2}-a \,b^{2} \tan \left (\lambda x \right )^{n +2}+b \lambda \tan \left (\lambda x \right )^{2}+b \lambda \end {array} \]
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) = n*lambda*(1+tan(lambda*x)^2)*(diff(y(x), x))/tan(lambda*x)-a*tan(lambd 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 a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Whittaker -> 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 -> 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 a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients 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 solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Whittaker -> 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 -> 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 a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients <- unable to find a useful change of variables trying a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients trying 2nd order, integrating factor of the form mu(x,y) 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 a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Whittaker -> 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 -> 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 a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients 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 solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Whittaker -> 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 -> 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 a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients <- unable to find a useful change of variables trying a symmetry of the form [xi=0, eta=F(x)] trying to convert to an ODE of Bessel type -> trying with_periodic_functions in the coefficients -> Trying a change of variables to reduce to Bernoulli -> Calling odsolve with the ODE`, diff(y(x), x)-(a*tan(lambda*x)^n*y(x)^2+y(x)+x^2*(-a*b^2*tan(lambda*x)^(n+2)+b*lambda*tan(lambd 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 inverse_Riccati trying 1st order ODE linearizable_by_differentiation -> trying a symmetry pattern of the form [F(x)*G(y), 0] -> trying a symmetry pattern of the form [0, F(x)*G(y)] -> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] trying inverse_Riccati trying 1st order ODE linearizable_by_differentiation --- Trying Lie symmetry methods, 1st order --- `, `-> Computing symmetries using: way = 4 `, `-> Computing symmetries using: way = 2 `, `-> Computing symmetries using: way = 6`
✗ Solution by Maple
dsolve(diff(y(x),x)=a*tan(lambda*x)^n*y(x)^2-a*b^2*tan(lambda*x)^(n+2)+b*lambda*tan(lambda*x)^2+b*lambda,y(x), singsol=all)
\[ \text {No solution found} \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[y'[x]==a*Tan[\[Lambda]*x]^n*y[x]^2-a*b^2*Tan[\[Lambda]*x]^(n+2)+b*\[Lambda]*Tan[\[Lambda]*x]^2+b*\[Lambda],y[x],x,IncludeSingularSolutions -> True]
Not solved