Internal problem ID [10550]
Internal file name [OUTPUT/9498_Monday_June_06_2022_02_58_18_PM_50440844/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 1.2.6-5. Equations containing
combinations of trigonometric functions.
Problem number: 53.
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]
\[ \boxed {y^{\prime }-y^{2}+y \tan \left (x \right )=a \left (1-a \right ) \cot \left (x \right )^{2}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -a^{2} \cot \left (x \right )^{2}+a \cot \left (x \right )^{2}-y \tan \left (x \right )+y^{2} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -a^{2} \cot \left (x \right )^{2}+a \cot \left (x \right )^{2}-y \tan \left (x \right )+y^{2} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-a^{2} \cot \left (x \right )^{2}+a \cot \left (x \right )^{2}\), \(f_1(x)=-\tan \left (x \right )\) 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 &=-\tan \left (x \right )\\ f_2^2 f_0 &=-a^{2} \cot \left (x \right )^{2}+a \cot \left (x \right )^{2} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (x \right )+\tan \left (x \right ) u^{\prime }\left (x \right )+\left (-a^{2} \cot \left (x \right )^{2}+a \cot \left (x \right )^{2}\right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} \sin \left (x \right )^{a}+c_{2} \sin \left (x \right )^{1-a} \] The above shows that \[ u^{\prime }\left (x \right ) = \cot \left (x \right ) \left (-c_{2} \left (a -1\right ) \sin \left (x \right )^{1-a}+c_{1} \sin \left (x \right )^{a} a \right ) \] Using the above in (1) gives the solution \[ y = -\frac {\cot \left (x \right ) \left (-c_{2} \left (a -1\right ) \sin \left (x \right )^{1-a}+c_{1} \sin \left (x \right )^{a} a \right )}{c_{1} \sin \left (x \right )^{a}+c_{2} \sin \left (x \right )^{1-a}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {-\cot \left (x \right ) c_{3} \sin \left (x \right )^{2 a} a +\cos \left (x \right ) \left (a -1\right )}{c_{3} \sin \left (x \right )^{2 a}+\sin \left (x \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-\cot \left (x \right ) c_{3} \sin \left (x \right )^{2 a} a +\cos \left (x \right ) \left (a -1\right )}{c_{3} \sin \left (x \right )^{2 a}+\sin \left (x \right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {-\cot \left (x \right ) c_{3} \sin \left (x \right )^{2 a} a +\cos \left (x \right ) \left (a -1\right )}{c_{3} \sin \left (x \right )^{2 a}+\sin \left (x \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{2}+y \tan \left (x \right )=a \left (1-a \right ) \cot \left (x \right )^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{2}-y \tan \left (x \right )+a \left (1-a \right ) \cot \left (x \right )^{2} \end {array} \]
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) = -tan(x)*(diff(y(x), x))+(a^2*cot(x)^2-a*cot(x)^2)*y(x), 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 <- LODE of Euler type successful Change of variables used: [x = arcsin(t)] Linear ODE actually solved: (a^2-a)*u(t)-t^2*diff(diff(u(t),t),t) = 0 <- change of variables successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 37
dsolve(diff(y(x),x)=y(x)^2-y(x)*tan(x)+a*(1-a)*cot(x)^2,y(x), singsol=all)
\[ y \left (x \right ) = \frac {-\cot \left (x \right ) \sin \left (x \right )^{2 a} a +c_{1} \cos \left (x \right ) \left (a -1\right )}{c_{1} \sin \left (x \right )+\sin \left (x \right )^{2 a}} \]
✓ Solution by Mathematica
Time used: 7.444 (sec). Leaf size: 230
DSolve[y'[x]==y[x]^2-y[x]*Tan[x]+a*(1-a)*Cot[x]^2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {i \cot (x) \left (\left (\sqrt {a-1} \sqrt {a} \sqrt {-\frac {(2 a-1)^2}{(a-1) a}}-i\right ) \left (-\sin ^2(x)\right )^{\frac {1}{2} i \sqrt {a-1} \sqrt {a} \sqrt {-\frac {(2 a-1)^2}{(a-1) a}}}-\left (\sqrt {a-1} \sqrt {a} \sqrt {-\frac {(2 a-1)^2}{(a-1) a}}+i\right ) c_1\right )}{2 \left (\left (-\sin ^2(x)\right )^{\frac {1}{2} i \sqrt {a-1} \sqrt {a} \sqrt {\frac {1}{a-a^2}-4}}+c_1\right )} \\ y(x)\to \frac {1}{2} i \left (\sqrt {a-1} \sqrt {a} \sqrt {\frac {1}{a-a^2}-4}+i\right ) \cot (x) \\ \end{align*}