Internal problem ID [10534]
Internal file name [OUTPUT/9482_Monday_June_06_2022_02_49_22_PM_50388009/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: 37.
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]
Unable to solve or complete the solution.
\[ \boxed {\left (\tan \left (\lambda x \right ) a +b \right ) y^{\prime }-y^{2}-k \tan \left (\mu x \right ) y=-d^{2}+k d \tan \left (\mu x \right )} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {y^{2}+k \tan \left (\mu x \right ) y -d^{2}+k d \tan \left (\mu x \right )}{\tan \left (\lambda x \right ) a +b} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {k d \tan \left (\mu x \right )}{\tan \left (\lambda x \right ) a +b}+\frac {k \tan \left (\mu x \right ) y}{\tan \left (\lambda x \right ) a +b}-\frac {d^{2}}{\tan \left (\lambda x \right ) a +b}+\frac {y^{2}}{\tan \left (\lambda x \right ) a +b} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {-d^{2}+k d \tan \left (\mu x \right )}{\tan \left (\lambda x \right ) a +b}\), \(f_1(x)=\frac {k \tan \left (\mu x \right )}{\tan \left (\lambda x \right ) a +b}\) and \(f_2(x)=\frac {1}{\tan \left (\lambda x \right ) a +b}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {u}{\tan \left (\lambda x \right ) a +b}} \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 {\lambda \left (1+\tan \left (\lambda x \right )^{2}\right ) a}{\left (\tan \left (\lambda x \right ) a +b \right )^{2}}\\ f_1 f_2 &=\frac {k \tan \left (\mu x \right )}{\left (\tan \left (\lambda x \right ) a +b \right )^{2}}\\ f_2^2 f_0 &=\frac {-d^{2}+k d \tan \left (\mu x \right )}{\left (\tan \left (\lambda x \right ) a +b \right )^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {u^{\prime \prime }\left (x \right )}{\tan \left (\lambda x \right ) a +b}-\left (-\frac {\lambda \left (1+\tan \left (\lambda x \right )^{2}\right ) a}{\left (\tan \left (\lambda x \right ) a +b \right )^{2}}+\frac {k \tan \left (\mu x \right )}{\left (\tan \left (\lambda x \right ) a +b \right )^{2}}\right ) u^{\prime }\left (x \right )+\frac {\left (-d^{2}+k d \tan \left (\mu x \right )\right ) u \left (x \right )}{\left (\tan \left (\lambda x \right ) a +b \right )^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives Unable to solve. Terminating.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (\tan \left (\lambda x \right ) a +b \right ) y^{\prime }-y^{2}-k \tan \left (\mu x \right ) y=-d^{2}+k d \tan \left (\mu x \right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y^{2}+k \tan \left (\mu x \right ) y-d^{2}+k d \tan \left (\mu x \right )}{\tan \left (\lambda x \right ) a +b} \end {array} \]
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: <- Riccati particular case Kamke (b) successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 351
dsolve((a*tan(lambda*x)+b)*diff(y(x),x)=y(x)^2+k*tan(mu*x)*y(x)-d^2+k*d*tan(mu*x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {-\left (\sec \left (x \lambda \right )^{2}\right )^{\frac {a d}{\lambda \left (a^{2}+b^{2}\right )}} \left (a \tan \left (x \lambda \right )+b \right )^{-\frac {2 a d}{\lambda \left (a^{2}+b^{2}\right )}} {\mathrm e}^{\frac {\lambda k \left (a^{2}+b^{2}\right ) \left (\int \frac {\tan \left (x \mu \right )}{a \tan \left (x \lambda \right )+b}d x \right )-2 \arctan \left (\tan \left (x \lambda \right )\right ) b d}{\lambda \left (a^{2}+b^{2}\right )}}-d \left (\int \left (a \tan \left (x \lambda \right )+b \right )^{\frac {\left (-a^{2}-b^{2}\right ) \lambda -2 a d}{\lambda \left (a^{2}+b^{2}\right )}} \left (\sec \left (x \lambda \right )^{2}\right )^{\frac {a d}{\lambda \left (a^{2}+b^{2}\right )}} {\mathrm e}^{\frac {\lambda k \left (a^{2}+b^{2}\right ) \left (\int \frac {\tan \left (x \mu \right )}{a \tan \left (x \lambda \right )+b}d x \right )-2 \arctan \left (\tan \left (x \lambda \right )\right ) b d}{\lambda \left (a^{2}+b^{2}\right )}}d x -c_{1} \right )}{\int \left (a \tan \left (x \lambda \right )+b \right )^{\frac {\left (-a^{2}-b^{2}\right ) \lambda -2 a d}{\lambda \left (a^{2}+b^{2}\right )}} \left (\sec \left (x \lambda \right )^{2}\right )^{\frac {a d}{\lambda \left (a^{2}+b^{2}\right )}} {\mathrm e}^{\frac {\lambda k \left (a^{2}+b^{2}\right ) \left (\int \frac {\tan \left (x \mu \right )}{a \tan \left (x \lambda \right )+b}d x \right )-2 \arctan \left (\tan \left (x \lambda \right )\right ) b d}{\lambda \left (a^{2}+b^{2}\right )}}d x -c_{1}} \]
✓ Solution by Mathematica
Time used: 130.719 (sec). Leaf size: 800
DSolve[(a*Tan[\[Lambda]*x]+b)*y'[x]==y[x]^2+k*Tan[\[Mu]*x]*y[x]-d^2+k*d*Tan[\[Mu]*x],y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\int _1^x\frac {e^{-\int _1^{K[2]}\frac {\sec (\mu K[1]) (2 d \cos (\lambda K[1]-\mu K[1])+2 d \cos (\lambda K[1]+\mu K[1])+k \sin (\lambda K[1]-\mu K[1])-k \sin (\lambda K[1]+\mu K[1]))}{2 (b \cos (\lambda K[1])+a \sin (\lambda K[1]))}dK[1]} (d \cos (\lambda K[2]-\mu K[2])-y(x) \cos (\lambda K[2]-\mu K[2])+d \cos (\lambda K[2]+\mu K[2])+k \sin (\lambda K[2]-\mu K[2])-k \sin (\lambda K[2]+\mu K[2])-\cos (\lambda K[2]+\mu K[2]) y(x))}{k \mu (b \cos (\lambda K[2]-\mu K[2])+b \cos (\lambda K[2]+\mu K[2])+a \sin (\lambda K[2]-\mu K[2])+a \sin (\lambda K[2]+\mu K[2])) (d+y(x))}dK[2]+\int _1^{y(x)}\left (\frac {e^{-\int _1^x\frac {\sec (\mu K[1]) (2 d \cos (\lambda K[1]-\mu K[1])+2 d \cos (\lambda K[1]+\mu K[1])+k \sin (\lambda K[1]-\mu K[1])-k \sin (\lambda K[1]+\mu K[1]))}{2 (b \cos (\lambda K[1])+a \sin (\lambda K[1]))}dK[1]}}{k \mu (d+K[3])^2}-\int _1^x\left (\frac {e^{-\int _1^{K[2]}\frac {\sec (\mu K[1]) (2 d \cos (\lambda K[1]-\mu K[1])+2 d \cos (\lambda K[1]+\mu K[1])+k \sin (\lambda K[1]-\mu K[1])-k \sin (\lambda K[1]+\mu K[1]))}{2 (b \cos (\lambda K[1])+a \sin (\lambda K[1]))}dK[1]} (-\cos (\lambda K[2]-\mu K[2])-\cos (\lambda K[2]+\mu K[2]))}{k \mu (d+K[3]) (b \cos (\lambda K[2]-\mu K[2])+b \cos (\lambda K[2]+\mu K[2])+a \sin (\lambda K[2]-\mu K[2])+a \sin (\lambda K[2]+\mu K[2]))}-\frac {e^{-\int _1^{K[2]}\frac {\sec (\mu K[1]) (2 d \cos (\lambda K[1]-\mu K[1])+2 d \cos (\lambda K[1]+\mu K[1])+k \sin (\lambda K[1]-\mu K[1])-k \sin (\lambda K[1]+\mu K[1]))}{2 (b \cos (\lambda K[1])+a \sin (\lambda K[1]))}dK[1]} (d \cos (\lambda K[2]-\mu K[2])-K[3] \cos (\lambda K[2]-\mu K[2])+d \cos (\lambda K[2]+\mu K[2])-\cos (\lambda K[2]+\mu K[2]) K[3]+k \sin (\lambda K[2]-\mu K[2])-k \sin (\lambda K[2]+\mu K[2]))}{k \mu (d+K[3])^2 (b \cos (\lambda K[2]-\mu K[2])+b \cos (\lambda K[2]+\mu K[2])+a \sin (\lambda K[2]-\mu K[2])+a \sin (\lambda K[2]+\mu K[2]))}\right )dK[2]\right )dK[3]=c_1,y(x)\right ] \]