Internal problem ID [1023]
Internal file name [OUTPUT/1024_Sunday_June_05_2022_01_57_10_AM_73978969/index.tex]
Book: Elementary differential equations with boundary value problems. William F. Trench.
Brooks/Cole 2001
Section: Chapter 2, First order equations. Transformation of Nonlinear Equations into Separable
Equations. Section 2.4 Page 68
Problem number: 48.
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]
\[ \boxed {y^{\prime }-\frac {y^{2}+\tan \left (x \right ) y+\tan \left (x \right )^{2}}{\sin \left (x \right )^{2}}=0} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {y^{2}+\tan \left (x \right ) y +\tan \left (x \right )^{2}}{\sin \left (x \right )^{2}} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {\tan \left (x \right )^{2}}{\sin \left (x \right )^{2}}+\frac {\tan \left (x \right ) y}{\sin \left (x \right )^{2}}+\frac {y^{2}}{\sin \left (x \right )^{2}} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {\tan \left (x \right )^{2}}{\sin \left (x \right )^{2}}\), \(f_1(x)=\frac {\tan \left (x \right )}{\sin \left (x \right )^{2}}\) and \(f_2(x)=\frac {1}{\sin \left (x \right )^{2}}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {u}{\sin \left (x \right )^{2}}} \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 \cos \left (x \right )}{\sin \left (x \right )^{3}}\\ f_1 f_2 &=\frac {\tan \left (x \right )}{\sin \left (x \right )^{4}}\\ f_2^2 f_0 &=\frac {\tan \left (x \right )^{2}}{\sin \left (x \right )^{6}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {u^{\prime \prime }\left (x \right )}{\sin \left (x \right )^{2}}-\left (-\frac {2 \cos \left (x \right )}{\sin \left (x \right )^{3}}+\frac {\tan \left (x \right )}{\sin \left (x \right )^{4}}\right ) u^{\prime }\left (x \right )+\frac {\tan \left (x \right )^{2} u \left (x \right )}{\sin \left (x \right )^{6}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} \sin \left (\frac {\ln \left (\sin \left (x \right )+1\right )}{2}+\frac {\ln \left (\sin \left (x \right )-1\right )}{2}-\ln \left (\sin \left (x \right )\right )\right )+c_{2} \cos \left (\frac {\ln \left (\sin \left (x \right )+1\right )}{2}+\frac {\ln \left (\sin \left (x \right )-1\right )}{2}-\ln \left (\sin \left (x \right )\right )\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \sec \left (x \right ) \csc \left (x \right ) \left (c_{2} \sin \left (\frac {\ln \left (\sin \left (x \right )+1\right )}{2}+\frac {\ln \left (\sin \left (x \right )-1\right )}{2}-\ln \left (\sin \left (x \right )\right )\right )-c_{1} \cos \left (\frac {\ln \left (\sin \left (x \right )+1\right )}{2}+\frac {\ln \left (\sin \left (x \right )-1\right )}{2}-\ln \left (\sin \left (x \right )\right )\right )\right ) \] Using the above in (1) gives the solution \[ y = -\frac {\sec \left (x \right ) \csc \left (x \right ) \left (c_{2} \sin \left (\frac {\ln \left (\sin \left (x \right )+1\right )}{2}+\frac {\ln \left (\sin \left (x \right )-1\right )}{2}-\ln \left (\sin \left (x \right )\right )\right )-c_{1} \cos \left (\frac {\ln \left (\sin \left (x \right )+1\right )}{2}+\frac {\ln \left (\sin \left (x \right )-1\right )}{2}-\ln \left (\sin \left (x \right )\right )\right )\right ) \sin \left (x \right )^{2}}{c_{1} \sin \left (\frac {\ln \left (\sin \left (x \right )+1\right )}{2}+\frac {\ln \left (\sin \left (x \right )-1\right )}{2}-\ln \left (\sin \left (x \right )\right )\right )+c_{2} \cos \left (\frac {\ln \left (\sin \left (x \right )+1\right )}{2}+\frac {\ln \left (\sin \left (x \right )-1\right )}{2}-\ln \left (\sin \left (x \right )\right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (-\sin \left (\frac {\ln \left (\sin \left (x \right )+1\right )}{2}+\frac {\ln \left (\sin \left (x \right )-1\right )}{2}-\ln \left (\sin \left (x \right )\right )\right )+c_{3} \cos \left (\frac {\ln \left (\sin \left (x \right )+1\right )}{2}+\frac {\ln \left (\sin \left (x \right )-1\right )}{2}-\ln \left (\sin \left (x \right )\right )\right )\right ) \tan \left (x \right )}{c_{3} \sin \left (\frac {\ln \left (\sin \left (x \right )+1\right )}{2}+\frac {\ln \left (\sin \left (x \right )-1\right )}{2}-\ln \left (\sin \left (x \right )\right )\right )+\cos \left (\frac {\ln \left (\sin \left (x \right )+1\right )}{2}+\frac {\ln \left (\sin \left (x \right )-1\right )}{2}-\ln \left (\sin \left (x \right )\right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (-\sin \left (\frac {\ln \left (\sin \left (x \right )+1\right )}{2}+\frac {\ln \left (\sin \left (x \right )-1\right )}{2}-\ln \left (\sin \left (x \right )\right )\right )+c_{3} \cos \left (\frac {\ln \left (\sin \left (x \right )+1\right )}{2}+\frac {\ln \left (\sin \left (x \right )-1\right )}{2}-\ln \left (\sin \left (x \right )\right )\right )\right ) \tan \left (x \right )}{c_{3} \sin \left (\frac {\ln \left (\sin \left (x \right )+1\right )}{2}+\frac {\ln \left (\sin \left (x \right )-1\right )}{2}-\ln \left (\sin \left (x \right )\right )\right )+\cos \left (\frac {\ln \left (\sin \left (x \right )+1\right )}{2}+\frac {\ln \left (\sin \left (x \right )-1\right )}{2}-\ln \left (\sin \left (x \right )\right )\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (-\sin \left (\frac {\ln \left (\sin \left (x \right )+1\right )}{2}+\frac {\ln \left (\sin \left (x \right )-1\right )}{2}-\ln \left (\sin \left (x \right )\right )\right )+c_{3} \cos \left (\frac {\ln \left (\sin \left (x \right )+1\right )}{2}+\frac {\ln \left (\sin \left (x \right )-1\right )}{2}-\ln \left (\sin \left (x \right )\right )\right )\right ) \tan \left (x \right )}{c_{3} \sin \left (\frac {\ln \left (\sin \left (x \right )+1\right )}{2}+\frac {\ln \left (\sin \left (x \right )-1\right )}{2}-\ln \left (\sin \left (x \right )\right )\right )+\cos \left (\frac {\ln \left (\sin \left (x \right )+1\right )}{2}+\frac {\ln \left (\sin \left (x \right )-1\right )}{2}-\ln \left (\sin \left (x \right )\right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -y^{\prime } \sin \left (x \right )^{2}+\tan \left (x \right )^{2}+\tan \left (x \right ) y+y^{2}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y^{2}+\tan \left (x \right ) y+\tan \left (x \right )^{2}}{\sin \left (x \right )^{2}} \end {array} \]
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) = (-2*cos(x)*sin(x)+tan(x))*(diff(y(x), x))/sin(x)^2-tan(x)^2*y(x)/sin(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 trying a symmetry of the form [xi=0, eta=F(x)] <- linear_1 successful Change of variables used: [x = arcsin(t)] Linear ODE actually solved: -8*t^2*u(t)+(-24*t^7+32*t^5-8*t^3)*diff(u(t),t)+(-8*t^8+16*t^6-8*t^4)*diff(diff(u(t),t),t) = 0 <- change of variables successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 102
dsolve(diff(y(x),x)=(y(x)^2+y(x)*tan(x)+tan(x)^2)/sin(x)^2,y(x), singsol=all)
\[ y \left (x \right ) = -\frac {\tan \left (x \right ) \left (c_{1} \sin \left (\frac {\ln \left (\sin \left (x \right )-1\right )}{2}+\frac {\ln \left (\sin \left (x \right )+1\right )}{2}-\ln \left (\sin \left (x \right )\right )\right )-\cos \left (\frac {\ln \left (\sin \left (x \right )-1\right )}{2}+\frac {\ln \left (\sin \left (x \right )+1\right )}{2}-\ln \left (\sin \left (x \right )\right )\right )\right )}{c_{1} \cos \left (\frac {\ln \left (\sin \left (x \right )-1\right )}{2}+\frac {\ln \left (\sin \left (x \right )+1\right )}{2}-\ln \left (\sin \left (x \right )\right )\right )+\sin \left (\frac {\ln \left (\sin \left (x \right )-1\right )}{2}+\frac {\ln \left (\sin \left (x \right )+1\right )}{2}-\ln \left (\sin \left (x \right )\right )\right )} \]
✓ Solution by Mathematica
Time used: 0.678 (sec). Leaf size: 20
DSolve[y'[x]==(y[x]^2+y[x]*Tan[x]+Tan[x]^2)/Sin[x]^2,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \tan (x) \tan (\log (\sin (x))-\log (\cos (x))+c_1) \]