Internal problem ID [10549]
Internal file name [OUTPUT/9497_Monday_June_06_2022_02_57_18_PM_24345774/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: 52.
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]
\[ \boxed {\sin \left (2 x \right )^{n +1} y^{\prime }-a y^{2} \sin \left (x \right )^{2 n}=b \cos \left (x \right )^{2 n}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \left (a \,y^{2} \sin \left (x \right )^{2 n}+b \cos \left (x \right )^{2 n}\right ) \sin \left (2 x \right )^{-n -1} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {a \,y^{2} \sin \left (x \right )^{2 n} \left (\frac {\sin \left (2 x \right )}{2}\right )^{-n} 2^{-n}}{2 \sin \left (x \right ) \cos \left (x \right )}+\frac {b \cos \left (x \right )^{2 n} \left (\frac {\sin \left (2 x \right )}{2}\right )^{-n} 2^{-n}}{2 \sin \left (x \right ) \cos \left (x \right )} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\sin \left (2 x \right )^{-n -1} b \cos \left (x \right )^{2 n}\), \(f_1(x)=0\) and \(f_2(x)=a \sin \left (x \right )^{2 n} \sin \left (2 x \right )^{-n -1}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{a \sin \left (x \right )^{2 n} \sin \left (2 x \right )^{-n -1} 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 {2 a \sin \left (x \right )^{2 n} n \cos \left (x \right ) \sin \left (2 x \right )^{-n -1}}{\sin \left (x \right )}-\frac {2 a \sin \left (x \right )^{2 n} \sin \left (2 x \right )^{-n -1} \left (n +1\right ) \cos \left (2 x \right )}{\sin \left (2 x \right )}\\ f_1 f_2 &=0\\ f_2^2 f_0 &=a^{2} \sin \left (x \right )^{4 n} \sin \left (2 x \right )^{-3 n -3} b \cos \left (x \right )^{2 n} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} a \sin \left (x \right )^{2 n} \sin \left (2 x \right )^{-n -1} u^{\prime \prime }\left (x \right )-\left (\frac {2 a \sin \left (x \right )^{2 n} n \cos \left (x \right ) \sin \left (2 x \right )^{-n -1}}{\sin \left (x \right )}-\frac {2 a \sin \left (x \right )^{2 n} \sin \left (2 x \right )^{-n -1} \left (n +1\right ) \cos \left (2 x \right )}{\sin \left (2 x \right )}\right ) u^{\prime }\left (x \right )+a^{2} \sin \left (x \right )^{4 n} \sin \left (2 x \right )^{-3 n -3} b \cos \left (x \right )^{2 n} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \cot \left (x \right )^{-\frac {n}{2}} \left (c_{1} \cot \left (x \right )^{\frac {\sqrt {n^{2}-4^{-n} b a}}{2}}+c_{2} \cot \left (x \right )^{-\frac {\sqrt {n^{2}-4^{-n} b a}}{2}}\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\cot \left (x \right )^{-\frac {n}{2}} \left (\cot \left (x \right )+\tan \left (x \right )\right ) \left (c_{2} \left (n +\sqrt {n^{2}-4^{-n} b a}\right ) \cot \left (x \right )^{-\frac {\sqrt {n^{2}-4^{-n} b a}}{2}}-\cot \left (x \right )^{\frac {\sqrt {n^{2}-4^{-n} b a}}{2}} c_{1} \left (-n +\sqrt {n^{2}-4^{-n} b a}\right )\right )}{2} \] Using the above in (1) gives the solution \[ y = -\frac {\left (\cot \left (x \right )+\tan \left (x \right )\right ) \left (c_{2} \left (n +\sqrt {n^{2}-4^{-n} b a}\right ) \cot \left (x \right )^{-\frac {\sqrt {n^{2}-4^{-n} b a}}{2}}-\cot \left (x \right )^{\frac {\sqrt {n^{2}-4^{-n} b a}}{2}} c_{1} \left (-n +\sqrt {n^{2}-4^{-n} b a}\right )\right ) \sin \left (x \right )^{-2 n} \sin \left (2 x \right )^{n +1}}{2 a \left (c_{1} \cot \left (x \right )^{\frac {\sqrt {n^{2}-4^{-n} b a}}{2}}+c_{2} \cot \left (x \right )^{-\frac {\sqrt {n^{2}-4^{-n} b a}}{2}}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {\csc \left (x \right ) \sin \left (x \right )^{-2 n} \sin \left (2 x \right )^{n +1} \left (\left (n +\sqrt {n^{2}-4^{-n} b a}\right ) \cot \left (x \right )^{-\frac {\sqrt {n^{2}-4^{-n} b a}}{2}}-\cot \left (x \right )^{\frac {\sqrt {n^{2}-4^{-n} b a}}{2}} c_{3} \left (-n +\sqrt {n^{2}-4^{-n} b a}\right )\right ) \sec \left (x \right )}{2 \left (c_{3} \cot \left (x \right )^{\frac {\sqrt {n^{2}-4^{-n} b a}}{2}}+\cot \left (x \right )^{-\frac {\sqrt {n^{2}-4^{-n} b a}}{2}}\right ) a} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\csc \left (x \right ) \sin \left (x \right )^{-2 n} \sin \left (2 x \right )^{n +1} \left (\left (n +\sqrt {n^{2}-4^{-n} b a}\right ) \cot \left (x \right )^{-\frac {\sqrt {n^{2}-4^{-n} b a}}{2}}-\cot \left (x \right )^{\frac {\sqrt {n^{2}-4^{-n} b a}}{2}} c_{3} \left (-n +\sqrt {n^{2}-4^{-n} b a}\right )\right ) \sec \left (x \right )}{2 \left (c_{3} \cot \left (x \right )^{\frac {\sqrt {n^{2}-4^{-n} b a}}{2}}+\cot \left (x \right )^{-\frac {\sqrt {n^{2}-4^{-n} b a}}{2}}\right ) a} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\csc \left (x \right ) \sin \left (x \right )^{-2 n} \sin \left (2 x \right )^{n +1} \left (\left (n +\sqrt {n^{2}-4^{-n} b a}\right ) \cot \left (x \right )^{-\frac {\sqrt {n^{2}-4^{-n} b a}}{2}}-\cot \left (x \right )^{\frac {\sqrt {n^{2}-4^{-n} b a}}{2}} c_{3} \left (-n +\sqrt {n^{2}-4^{-n} b a}\right )\right ) \sec \left (x \right )}{2 \left (c_{3} \cot \left (x \right )^{\frac {\sqrt {n^{2}-4^{-n} b a}}{2}}+\cot \left (x \right )^{-\frac {\sqrt {n^{2}-4^{-n} b a}}{2}}\right ) a} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \sin \left (2 x \right )^{n +1} y^{\prime }-a y^{2} \sin \left (x \right )^{2 n}=b \cos \left (x \right )^{2 n} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {a y^{2} \sin \left (x \right )^{2 n}+b \cos \left (x \right )^{2 n}}{\sin \left (2 x \right )^{n +1}} \end {array} \]
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) = (sin(x)^2*n+cos(x)^2*n+sin(x)^2-cos(x)^2)*(diff(y(x), x))/(cos(x)*sin( 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 Liouvillian solution using Kovacics algorithm A Liouvillian solution exists Group is reducible or imprimitive <- Kovacics algorithm successful Change of variables used: [x = arccos(t)] Linear ODE actually solved: b*a*(-t^2+1)^n*t^(2*n)*u(t)+2^(2*n+2)*t^(2*n+1)*(-t^2+1)^(n+1)*(-3*t^2+n+1)*diff(u(t),t)+2^(2*n+2)*t^(2*n+2)*(-t^2+1)^(n <- change of variables successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.172 (sec). Leaf size: 232
dsolve(sin(2*x)^(n+1)*diff(y(x),x)=a*y(x)^2*sin(x)^(2*n)+b*cos(x)^(2*n),y(x), singsol=all)
\[ y \left (x \right ) = -\frac {\csc \left (x \right ) \sin \left (2 x \right )^{n} \left (\sin \left (x \right )^{\frac {\sqrt {n^{2}-4^{-n} a b}}{2}} \left (n +\sqrt {n^{2}-4^{-n} a b}\right ) \cos \left (x \right )^{-\frac {\sqrt {n^{2}-4^{-n} a b}}{2}}-\cos \left (x \right )^{\frac {\sqrt {n^{2}-4^{-n} a b}}{2}} \sin \left (x \right )^{-\frac {\sqrt {n^{2}-4^{-n} a b}}{2}} c_{1} \left (-n +\sqrt {n^{2}-4^{-n} a b}\right )\right ) \sin \left (x \right )^{-2 n +1}}{a \left (\cos \left (x \right )^{\frac {\sqrt {n^{2}-4^{-n} a b}}{2}} \sin \left (x \right )^{-\frac {\sqrt {n^{2}-4^{-n} a b}}{2}} c_{1} +\cos \left (x \right )^{-\frac {\sqrt {n^{2}-4^{-n} a b}}{2}} \sin \left (x \right )^{\frac {\sqrt {n^{2}-4^{-n} a b}}{2}}\right )} \]
✓ Solution by Mathematica
Time used: 33.745 (sec). Leaf size: 132
DSolve[Sin[2*x]^(n+1)*y'[x]==a*y[x]^2*Sin[x]^(2*n)+b*Cos[x]^(2*n),y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\int _1^{\sqrt {\frac {a \cos ^{-2 n}(x) \sin ^{2 n}(x)}{b}} y(x)}\frac {1}{K[1]^2-\sqrt {\frac {2^{2 n+2} n^2}{a b}} K[1]+1}dK[1]=\frac {1}{2} b \sin ^{-n}(2 x) \cos ^{2 n}(x) \left (\log \left (\tan \left (\frac {x}{2}\right )\right )-\log \left (\cos (x) \sec ^2\left (\frac {x}{2}\right )\right )\right ) \sqrt {\frac {a \sin ^{2 n}(x) \cos ^{-2 n}(x)}{b}}+c_1,y(x)\right ] \]