Internal problem ID [10526]
Internal file name [OUTPUT/9474_Monday_June_06_2022_02_47_15_PM_7792648/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: 29.
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]
\[ \boxed {y^{\prime }-y^{2} a -b \tan \left (x \right ) y=c} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= a \,y^{2}+b \tan \left (x \right ) y +c \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = a \,y^{2}+b \tan \left (x \right ) y +c \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=c\), \(f_1(x)=b \tan \left (x \right )\) and \(f_2(x)=a\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{a 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 &=b \tan \left (x \right ) a\\ f_2^2 f_0 &=a^{2} c \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} a u^{\prime \prime }\left (x \right )-b \tan \left (x \right ) a u^{\prime }\left (x \right )+a^{2} c u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \cos \left (x \right )^{\frac {1}{2}-\frac {b}{2}} \left (c_{1} \operatorname {LegendreP}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )+c_{2} \operatorname {LegendreQ}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = -\frac {\left (\left (\cos \left (x \right ) \sin \left (x \right ) \sqrt {4 a c +b^{2}}+\cos \left (x \right ) \sin \left (x \right )-\tan \left (x \right ) \left (\sin \left (x \right )-1\right ) \left (\sin \left (x \right )+1\right ) \left (b -1\right )\right ) c_{1} \operatorname {LegendreP}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )+c_{2} \left (\cos \left (x \right ) \sin \left (x \right ) \sqrt {4 a c +b^{2}}+\cos \left (x \right ) \sin \left (x \right )-\tan \left (x \right ) \left (\sin \left (x \right )-1\right ) \left (\sin \left (x \right )+1\right ) \left (b -1\right )\right ) \operatorname {LegendreQ}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )-\cos \left (x \right ) \left (c_{1} \operatorname {LegendreP}\left (\frac {\sqrt {4 a c +b^{2}}}{2}+\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )+c_{2} \operatorname {LegendreQ}\left (\frac {\sqrt {4 a c +b^{2}}}{2}+\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )\right ) \left (\sqrt {4 a c +b^{2}}-b +2\right )\right ) \cos \left (x \right )^{\frac {1}{2}-\frac {b}{2}}}{2 \sin \left (x \right )^{2}-2} \] Using the above in (1) gives the solution \[ y = \frac {\left (\cos \left (x \right ) \sin \left (x \right ) \sqrt {4 a c +b^{2}}+\cos \left (x \right ) \sin \left (x \right )-\tan \left (x \right ) \left (\sin \left (x \right )-1\right ) \left (\sin \left (x \right )+1\right ) \left (b -1\right )\right ) c_{1} \operatorname {LegendreP}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )+c_{2} \left (\cos \left (x \right ) \sin \left (x \right ) \sqrt {4 a c +b^{2}}+\cos \left (x \right ) \sin \left (x \right )-\tan \left (x \right ) \left (\sin \left (x \right )-1\right ) \left (\sin \left (x \right )+1\right ) \left (b -1\right )\right ) \operatorname {LegendreQ}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )-\cos \left (x \right ) \left (c_{1} \operatorname {LegendreP}\left (\frac {\sqrt {4 a c +b^{2}}}{2}+\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )+c_{2} \operatorname {LegendreQ}\left (\frac {\sqrt {4 a c +b^{2}}}{2}+\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )\right ) \left (\sqrt {4 a c +b^{2}}-b +2\right )}{\left (2 \sin \left (x \right )^{2}-2\right ) a \left (c_{1} \operatorname {LegendreP}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )+c_{2} \operatorname {LegendreQ}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {\left (\left (\sqrt {4 a c +b^{2}}+b \right ) \left (c_{3} \operatorname {LegendreP}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )+\operatorname {LegendreQ}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )\right ) \sin \left (x \right )-\left (\sqrt {4 a c +b^{2}}-b +2\right ) \left (c_{3} \operatorname {LegendreP}\left (\frac {\sqrt {4 a c +b^{2}}}{2}+\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )+\operatorname {LegendreQ}\left (\frac {\sqrt {4 a c +b^{2}}}{2}+\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )\right )\right ) \sec \left (x \right )}{2 \left (c_{3} \operatorname {LegendreP}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )+\operatorname {LegendreQ}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )\right ) a} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\left (\left (\sqrt {4 a c +b^{2}}+b \right ) \left (c_{3} \operatorname {LegendreP}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )+\operatorname {LegendreQ}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )\right ) \sin \left (x \right )-\left (\sqrt {4 a c +b^{2}}-b +2\right ) \left (c_{3} \operatorname {LegendreP}\left (\frac {\sqrt {4 a c +b^{2}}}{2}+\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )+\operatorname {LegendreQ}\left (\frac {\sqrt {4 a c +b^{2}}}{2}+\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )\right )\right ) \sec \left (x \right )}{2 \left (c_{3} \operatorname {LegendreP}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )+\operatorname {LegendreQ}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )\right ) a} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\left (\left (\sqrt {4 a c +b^{2}}+b \right ) \left (c_{3} \operatorname {LegendreP}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )+\operatorname {LegendreQ}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )\right ) \sin \left (x \right )-\left (\sqrt {4 a c +b^{2}}-b +2\right ) \left (c_{3} \operatorname {LegendreP}\left (\frac {\sqrt {4 a c +b^{2}}}{2}+\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )+\operatorname {LegendreQ}\left (\frac {\sqrt {4 a c +b^{2}}}{2}+\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )\right )\right ) \sec \left (x \right )}{2 \left (c_{3} \operatorname {LegendreP}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )+\operatorname {LegendreQ}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )\right ) a} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{2} a -b \tan \left (x \right ) y=c \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{2} a +b \tan \left (x \right ) y+c \end {array} \]
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) = b*tan(x)*(diff(y(x), x))-a*c*y(x), y(x)` *** Sublevel 2 *** 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 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 <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre <- Legendre successful <- special function solution successful Change of variables used: [x = arcsin(t)] Linear ODE actually solved: a*c*u(t)+(-b*t-t)*diff(u(t),t)+(-t^2+1)*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: 187
dsolve(diff(y(x),x)=a*y(x)^2+b*tan(x)*y(x)+c,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\sec \left (x \right ) \left (-\left (\operatorname {LegendreQ}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right ) c_{1} +\operatorname {LegendreP}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )\right ) \left (b +\sqrt {4 a c +b^{2}}\right ) \sin \left (x \right )+\left (\sqrt {4 a c +b^{2}}-b +2\right ) \left (\operatorname {LegendreQ}\left (\frac {\sqrt {4 a c +b^{2}}}{2}+\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right ) c_{1} +\operatorname {LegendreP}\left (\frac {\sqrt {4 a c +b^{2}}}{2}+\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )\right )\right )}{2 \left (\operatorname {LegendreQ}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right ) c_{1} +\operatorname {LegendreP}\left (\frac {\sqrt {4 a c +b^{2}}}{2}-\frac {1}{2}, -\frac {1}{2}+\frac {b}{2}, \sin \left (x \right )\right )\right ) a} \]
✓ Solution by Mathematica
Time used: 2.211 (sec). Leaf size: 608
DSolve[y'[x]==a*y[x]^2+b*Tan[x]*y[x]+c,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {\sin (x) \left (\left (-b^3+3 b^2+b-3\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{4} \left (-b-\sqrt {b^2+4 a c}+2\right ),\frac {1}{4} \left (-b+\sqrt {b^2+4 a c}+2\right ),\frac {3-b}{2},\cos ^2(x)\right )+\cos (x) \left ((b+1) \cos (x) (a c+b-1) \operatorname {Hypergeometric2F1}\left (\frac {1}{4} \left (-b-\sqrt {b^2+4 a c}+6\right ),\frac {1}{4} \left (-b+\sqrt {b^2+4 a c}+6\right ),\frac {5-b}{2},\cos ^2(x)\right )+a i^{b+1} (b-3) c c_1 \cos ^b(x) \operatorname {Hypergeometric2F1}\left (\frac {1}{4} \left (b-\sqrt {b^2+4 a c}+4\right ),\frac {1}{4} \left (b+\sqrt {b^2+4 a c}+4\right ),\frac {b+3}{2},\cos ^2(x)\right )\right )\right )}{a (b-3) (b+1) \left (\cos (x) \operatorname {Hypergeometric2F1}\left (\frac {1}{4} \left (-b-\sqrt {b^2+4 a c}+2\right ),\frac {1}{4} \left (-b+\sqrt {b^2+4 a c}+2\right ),\frac {3-b}{2},\cos ^2(x)\right )-i i^b c_1 \cos ^b(x) \operatorname {Hypergeometric2F1}\left (\frac {1}{4} \left (b-\sqrt {b^2+4 a c}\right ),\frac {1}{4} \left (b+\sqrt {b^2+4 a c}\right ),\frac {b+1}{2},\cos ^2(x)\right )\right )} \\ y(x)\to -\frac {c \sin (x) \cos (x) \operatorname {Hypergeometric2F1}\left (\frac {1}{4} \left (b-\sqrt {b^2+4 a c}+4\right ),\frac {1}{4} \left (b+\sqrt {b^2+4 a c}+4\right ),\frac {b+3}{2},\cos ^2(x)\right )}{(b+1) \operatorname {Hypergeometric2F1}\left (\frac {1}{4} \left (b-\sqrt {b^2+4 a c}\right ),\frac {1}{4} \left (b+\sqrt {b^2+4 a c}\right ),\frac {b+1}{2},\cos ^2(x)\right )} \\ y(x)\to -\frac {c \sin (x) \cos (x) \operatorname {Hypergeometric2F1}\left (\frac {1}{4} \left (b-\sqrt {b^2+4 a c}+4\right ),\frac {1}{4} \left (b+\sqrt {b^2+4 a c}+4\right ),\frac {b+3}{2},\cos ^2(x)\right )}{(b+1) \operatorname {Hypergeometric2F1}\left (\frac {1}{4} \left (b-\sqrt {b^2+4 a c}\right ),\frac {1}{4} \left (b+\sqrt {b^2+4 a c}\right ),\frac {b+1}{2},\cos ^2(x)\right )} \\ \end{align*}