Internal problem ID [4519]
Internal file name [OUTPUT/4012_Sunday_June_05_2022_12_08_08_PM_29117838/index.tex]
Book: Ordinary Differential Equations, By Tenenbaum and Pollard. Dover, NY 1963
Section: Chapter 2. Special types of differential equations of the first kind. Lesson 11, Bernoulli
Equations
Problem number: Exercise 11.27, page 97.
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]
\[ \boxed {y^{\prime }+y^{2} \sin \left (x \right )=2 \sec \left (x \right ) \tan \left (x \right )} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= 2 \sec \left (x \right ) \tan \left (x \right )-\sin \left (x \right ) y^{2} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = 2 \sec \left (x \right ) \tan \left (x \right )-\sin \left (x \right ) y^{2} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=2 \sec \left (x \right ) \tan \left (x \right )\), \(f_1(x)=0\) and \(f_2(x)=-\sin \left (x \right )\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-\sin \left (x \right ) 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' &=-\cos \left (x \right )\\ f_1 f_2 &=0\\ f_2^2 f_0 &=2 \sin \left (x \right )^{2} \sec \left (x \right ) \tan \left (x \right ) \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} -\sin \left (x \right ) u^{\prime \prime }\left (x \right )+\cos \left (x \right ) u^{\prime }\left (x \right )+2 \sin \left (x \right )^{2} \sec \left (x \right ) \tan \left (x \right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} \sec \left (x \right )+c_{2} \cos \left (x \right )^{2} \] The above shows that \[ u^{\prime }\left (x \right ) = \sec \left (x \right ) \tan \left (x \right ) \left (-2 c_{2} \cos \left (x \right )^{3}+c_{1} \right ) \] Using the above in (1) gives the solution \[ y = \frac {\sec \left (x \right ) \tan \left (x \right ) \left (-2 c_{2} \cos \left (x \right )^{3}+c_{1} \right )}{\sin \left (x \right ) \left (c_{1} \sec \left (x \right )+c_{2} \cos \left (x \right )^{2}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {-2 \cos \left (x \right )^{2}+c_{3} \sec \left (x \right )}{\cos \left (x \right )^{3}+c_{3}} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-2 \cos \left (x \right )^{2}+c_{3} \sec \left (x \right )}{\cos \left (x \right )^{3}+c_{3}} \\ \end{align*}
Verification of solutions
\[ y = \frac {-2 \cos \left (x \right )^{2}+c_{3} \sec \left (x \right )}{\cos \left (x \right )^{3}+c_{3}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }+y^{2} \sin \left (x \right )=2 \sec \left (x \right ) \tan \left (x \right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=2 \sec \left (x \right ) \tan \left (x \right )-y^{2} \sin \left (x \right ) \end {array} \]
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) = cos(x)*(diff(y(x), x))/sin(x)+2*tan(x)*sec(x)*sin(x)*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 = arccos(t)] Linear ODE actually solved: (2*t^2-2)*u(t)+(-t^4+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: 25
dsolve(diff(y(x),x)=2*tan(x)*sec(x)-y(x)^2*sin(x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {-2 c_{1} \cos \left (x \right )^{2}+\sec \left (x \right )}{c_{1} \cos \left (x \right )^{3}+1} \]
✓ Solution by Mathematica
Time used: 0.88 (sec). Leaf size: 32
DSolve[y'[x]==2*Tan[x]*Sec[x]-y[x]^2*Sin[x],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {\sec (x) \left (-2 \cos ^3(x)+c_1\right )}{\cos ^3(x)+c_1} \\ y(x)\to \sec (x) \\ \end{align*}