Internal problem ID [10510]
Internal file name [OUTPUT/9458_Monday_June_06_2022_02_39_36_PM_69357783/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: 13.
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]
\[ \boxed {\left (a \sin \left (\lambda x \right )+b \right ) \left (y^{\prime }-y^{2}\right )=\sin \left (\lambda x \right ) a \,\lambda ^{2}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {\sin \left (\lambda x \right ) a \,y^{2}+\sin \left (\lambda x \right ) a \,\lambda ^{2}+b \,y^{2}}{a \sin \left (\lambda x \right )+b} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {a \,\lambda ^{2} \sin \left (\lambda x \right )}{a \sin \left (\lambda x \right )+b}+\frac {\sin \left (\lambda x \right ) a \,y^{2}}{a \sin \left (\lambda x \right )+b}+\frac {b \,y^{2}}{a \sin \left (\lambda x \right )+b} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {a \,\lambda ^{2} \sin \left (\lambda x \right )}{a \sin \left (\lambda x \right )+b}\), \(f_1(x)=0\) 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 &=\frac {a \,\lambda ^{2} \sin \left (\lambda x \right )}{a \sin \left (\lambda x \right )+b} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (x \right )+\frac {a \,\lambda ^{2} \sin \left (\lambda x \right ) u \left (x \right )}{a \sin \left (\lambda x \right )+b} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = 2 b^{2} \left (\sin \left (\frac {\lambda x}{2}\right ) a \cos \left (\frac {\lambda x}{2}\right )+\frac {b}{2}\right ) c_{1} \left (a +b \right ) \left (a -b \right ) \arctan \left (\frac {b \tan \left (\frac {\lambda x}{2}\right )+a}{\sqrt {-a^{2}+b^{2}}}\right )+a c_{1} \cos \left (\frac {\lambda x}{2}\right ) \left (a -b \right ) \left (a +b \right ) \left (a \sin \left (\frac {\lambda x}{2}\right )+\cos \left (\frac {\lambda x}{2}\right ) b \right ) \sqrt {-a^{2}+b^{2}}+2 \left (\sin \left (\frac {\lambda x}{2}\right ) a \cos \left (\frac {\lambda x}{2}\right )+\frac {b}{2}\right ) c_{2} \] The above shows that \[ u^{\prime }\left (x \right ) = -\frac {\lambda \left (-2 b^{2} c_{1} \left (a +b \right ) \left (a -b \right ) a \sqrt {-a^{2}+b^{2}}\, \left (\cos \left (\frac {\lambda x}{2}\right )^{2}-\frac {1}{2}\right ) \arctan \left (\frac {b \tan \left (\frac {\lambda x}{2}\right )+a}{\sqrt {-a^{2}+b^{2}}}\right )-2 a \left (\cos \left (\frac {\lambda x}{2}\right )^{2}-\frac {1}{2}\right ) c_{2} \sqrt {-a^{2}+b^{2}}+\left (a^{2} \cos \left (\frac {\lambda x}{2}\right )^{2}-\sin \left (\frac {\lambda x}{2}\right ) a b \cos \left (\frac {\lambda x}{2}\right )-\frac {a^{2}}{2}+\frac {b^{2}}{2}\right ) c_{1} \left (a +b \right )^{2} \left (a -b \right )^{2}\right )}{\sqrt {-a^{2}+b^{2}}} \] Using the above in (1) gives the solution \[ y = \frac {\lambda \left (-2 b^{2} c_{1} \left (a +b \right ) \left (a -b \right ) a \sqrt {-a^{2}+b^{2}}\, \left (\cos \left (\frac {\lambda x}{2}\right )^{2}-\frac {1}{2}\right ) \arctan \left (\frac {b \tan \left (\frac {\lambda x}{2}\right )+a}{\sqrt {-a^{2}+b^{2}}}\right )-2 a \left (\cos \left (\frac {\lambda x}{2}\right )^{2}-\frac {1}{2}\right ) c_{2} \sqrt {-a^{2}+b^{2}}+\left (a^{2} \cos \left (\frac {\lambda x}{2}\right )^{2}-\sin \left (\frac {\lambda x}{2}\right ) a b \cos \left (\frac {\lambda x}{2}\right )-\frac {a^{2}}{2}+\frac {b^{2}}{2}\right ) c_{1} \left (a +b \right )^{2} \left (a -b \right )^{2}\right )}{\sqrt {-a^{2}+b^{2}}\, \left (2 b^{2} \left (\sin \left (\frac {\lambda x}{2}\right ) a \cos \left (\frac {\lambda x}{2}\right )+\frac {b}{2}\right ) c_{1} \left (a +b \right ) \left (a -b \right ) \arctan \left (\frac {b \tan \left (\frac {\lambda x}{2}\right )+a}{\sqrt {-a^{2}+b^{2}}}\right )+a c_{1} \cos \left (\frac {\lambda x}{2}\right ) \left (a -b \right ) \left (a +b \right ) \left (a \sin \left (\frac {\lambda x}{2}\right )+\cos \left (\frac {\lambda x}{2}\right ) b \right ) \sqrt {-a^{2}+b^{2}}+2 \left (\sin \left (\frac {\lambda x}{2}\right ) a \cos \left (\frac {\lambda x}{2}\right )+\frac {b}{2}\right ) c_{2} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (-2 b^{2} c_{3} \left (a +b \right ) \left (a -b \right ) a \sqrt {-a^{2}+b^{2}}\, \left (\cos \left (\frac {\lambda x}{2}\right )^{2}-\frac {1}{2}\right ) \arctan \left (\frac {b \tan \left (\frac {\lambda x}{2}\right )+a}{\sqrt {-a^{2}+b^{2}}}\right )+a \left (-2 \cos \left (\frac {\lambda x}{2}\right )^{2}+1\right ) \sqrt {-a^{2}+b^{2}}+\left (a^{2} \cos \left (\frac {\lambda x}{2}\right )^{2}-\sin \left (\frac {\lambda x}{2}\right ) a b \cos \left (\frac {\lambda x}{2}\right )-\frac {a^{2}}{2}+\frac {b^{2}}{2}\right ) c_{3} \left (a +b \right )^{2} \left (a -b \right )^{2}\right ) \lambda }{\sqrt {-a^{2}+b^{2}}\, \left (2 b^{2} \left (\sin \left (\frac {\lambda x}{2}\right ) a \cos \left (\frac {\lambda x}{2}\right )+\frac {b}{2}\right ) c_{3} \left (a +b \right ) \left (a -b \right ) \arctan \left (\frac {b \tan \left (\frac {\lambda x}{2}\right )+a}{\sqrt {-a^{2}+b^{2}}}\right )+a c_{3} \cos \left (\frac {\lambda x}{2}\right ) \left (a -b \right ) \left (a +b \right ) \left (a \sin \left (\frac {\lambda x}{2}\right )+\cos \left (\frac {\lambda x}{2}\right ) b \right ) \sqrt {-a^{2}+b^{2}}+2 \sin \left (\frac {\lambda x}{2}\right ) a \cos \left (\frac {\lambda x}{2}\right )+b \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (-2 b^{2} c_{3} \left (a +b \right ) \left (a -b \right ) a \sqrt {-a^{2}+b^{2}}\, \left (\cos \left (\frac {\lambda x}{2}\right )^{2}-\frac {1}{2}\right ) \arctan \left (\frac {b \tan \left (\frac {\lambda x}{2}\right )+a}{\sqrt {-a^{2}+b^{2}}}\right )+a \left (-2 \cos \left (\frac {\lambda x}{2}\right )^{2}+1\right ) \sqrt {-a^{2}+b^{2}}+\left (a^{2} \cos \left (\frac {\lambda x}{2}\right )^{2}-\sin \left (\frac {\lambda x}{2}\right ) a b \cos \left (\frac {\lambda x}{2}\right )-\frac {a^{2}}{2}+\frac {b^{2}}{2}\right ) c_{3} \left (a +b \right )^{2} \left (a -b \right )^{2}\right ) \lambda }{\sqrt {-a^{2}+b^{2}}\, \left (2 b^{2} \left (\sin \left (\frac {\lambda x}{2}\right ) a \cos \left (\frac {\lambda x}{2}\right )+\frac {b}{2}\right ) c_{3} \left (a +b \right ) \left (a -b \right ) \arctan \left (\frac {b \tan \left (\frac {\lambda x}{2}\right )+a}{\sqrt {-a^{2}+b^{2}}}\right )+a c_{3} \cos \left (\frac {\lambda x}{2}\right ) \left (a -b \right ) \left (a +b \right ) \left (a \sin \left (\frac {\lambda x}{2}\right )+\cos \left (\frac {\lambda x}{2}\right ) b \right ) \sqrt {-a^{2}+b^{2}}+2 \sin \left (\frac {\lambda x}{2}\right ) a \cos \left (\frac {\lambda x}{2}\right )+b \right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (-2 b^{2} c_{3} \left (a +b \right ) \left (a -b \right ) a \sqrt {-a^{2}+b^{2}}\, \left (\cos \left (\frac {\lambda x}{2}\right )^{2}-\frac {1}{2}\right ) \arctan \left (\frac {b \tan \left (\frac {\lambda x}{2}\right )+a}{\sqrt {-a^{2}+b^{2}}}\right )+a \left (-2 \cos \left (\frac {\lambda x}{2}\right )^{2}+1\right ) \sqrt {-a^{2}+b^{2}}+\left (a^{2} \cos \left (\frac {\lambda x}{2}\right )^{2}-\sin \left (\frac {\lambda x}{2}\right ) a b \cos \left (\frac {\lambda x}{2}\right )-\frac {a^{2}}{2}+\frac {b^{2}}{2}\right ) c_{3} \left (a +b \right )^{2} \left (a -b \right )^{2}\right ) \lambda }{\sqrt {-a^{2}+b^{2}}\, \left (2 b^{2} \left (\sin \left (\frac {\lambda x}{2}\right ) a \cos \left (\frac {\lambda x}{2}\right )+\frac {b}{2}\right ) c_{3} \left (a +b \right ) \left (a -b \right ) \arctan \left (\frac {b \tan \left (\frac {\lambda x}{2}\right )+a}{\sqrt {-a^{2}+b^{2}}}\right )+a c_{3} \cos \left (\frac {\lambda x}{2}\right ) \left (a -b \right ) \left (a +b \right ) \left (a \sin \left (\frac {\lambda x}{2}\right )+\cos \left (\frac {\lambda x}{2}\right ) b \right ) \sqrt {-a^{2}+b^{2}}+2 \sin \left (\frac {\lambda x}{2}\right ) a \cos \left (\frac {\lambda x}{2}\right )+b \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (a \sin \left (\lambda x \right )+b \right ) \left (y^{\prime }-y^{2}\right )=\sin \left (\lambda x \right ) a \,\lambda ^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\sin \left (\lambda x \right ) y^{2} a +\sin \left (\lambda x \right ) a \,\lambda ^{2}+y^{2} b}{a \sin \left (\lambda x \right )+b} \end {array} \]
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: trying Riccati_symmetries trying Riccati to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = -a*lambda^2*sin(lambda*x)*y(x)/(a*sin(lambda*x)+b), y(x)` *** Sub Methods for second order ODEs: --- Trying classification methods --- trying a symmetry of the form [xi=0, eta=F(x)] <- linear_1 successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 261
dsolve((a*sin(lambda*x)+b)*(diff(y(x),x)-y(x)^2)-a*lambda^2*sin(lambda*x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (-2 \left (a -b \right ) b^{2} \left (a +b \right ) \left (\cos \left (\frac {x \lambda }{2}\right )^{2}-\frac {1}{2}\right ) a \sqrt {-a^{2}+b^{2}}\, \arctan \left (\frac {\tan \left (\frac {x \lambda }{2}\right ) b +a}{\sqrt {-a^{2}+b^{2}}}\right )+2 c_{1} \left (\cos \left (\frac {x \lambda }{2}\right )^{2}-\frac {1}{2}\right ) a \sqrt {-a^{2}+b^{2}}+\left (a -b \right )^{2} \left (a^{2} \cos \left (\frac {x \lambda }{2}\right )^{2}-\sin \left (\frac {x \lambda }{2}\right ) a b \cos \left (\frac {x \lambda }{2}\right )-\frac {a^{2}}{2}+\frac {b^{2}}{2}\right ) \left (a +b \right )^{2}\right ) \lambda }{\sqrt {-a^{2}+b^{2}}\, \left (2 \left (a -b \right ) b^{2} \left (a +b \right ) \left (\sin \left (\frac {x \lambda }{2}\right ) a \cos \left (\frac {x \lambda }{2}\right )+\frac {b}{2}\right ) \arctan \left (\frac {\tan \left (\frac {x \lambda }{2}\right ) b +a}{\sqrt {-a^{2}+b^{2}}}\right )+a \cos \left (\frac {x \lambda }{2}\right ) \left (a -b \right ) \left (a +b \right ) \left (a \sin \left (\frac {x \lambda }{2}\right )+b \cos \left (\frac {x \lambda }{2}\right )\right ) \sqrt {-a^{2}+b^{2}}-2 c_{1} \left (\sin \left (\frac {x \lambda }{2}\right ) a \cos \left (\frac {x \lambda }{2}\right )+\frac {b}{2}\right )\right )} \]
✓ Solution by Mathematica
Time used: 24.795 (sec). Leaf size: 189
DSolve[(a*Sin[\[Lambda]*x]+b)*(y'[x]-y[x]^2)-a*\[Lambda]^2*Sin[\[Lambda]*x]==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {\lambda \left (2 a b \cos (\lambda x) \arctan \left (\frac {a+b \tan \left (\frac {\lambda x}{2}\right )}{\sqrt {b^2-a^2}}\right )+\sqrt {b^2-a^2} \left (-a c_1 \lambda \left (a^2-b^2\right ) \cos (\lambda x)-a \sin (\lambda x)+b\right )\right )}{-2 b (a \sin (\lambda x)+b) \arctan \left (\frac {a+b \tan \left (\frac {\lambda x}{2}\right )}{\sqrt {b^2-a^2}}\right )+\sqrt {b^2-a^2} \left (-a \cos (\lambda x)+c_1 \lambda \left (a^2-b^2\right ) (a \sin (\lambda x)+b)\right )} \\ y(x)\to -\frac {a \lambda \cos (\lambda x)}{a \sin (\lambda x)+b} \\ \end{align*}