Internal problem ID [8369]
Internal file name [OUTPUT/7302_Sunday_June_05_2022_05_43_46_PM_3280305/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 32.
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]
\[ \boxed {y^{\prime }+y^{2} \sin \left (x \right )=\frac {2 \sin \left (x \right )}{\cos \left (x \right )^{2}}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -\frac {\sin \left (x \right ) \left (\cos \left (x \right )^{2} y^{2}-2\right )}{\cos \left (x \right )^{2}} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -y^{2} \sin \left (x \right )+\frac {2 \sin \left (x \right )}{\cos \left (x \right )^{2}} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {2 \sin \left (x \right )}{\cos \left (x \right )^{2}}\), \(f_1(x)=0\) 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 &=\frac {2 \sin \left (x \right )^{3}}{\cos \left (x \right )^{2}} \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 )+\frac {2 \sin \left (x \right )^{3} u \left (x \right )}{\cos \left (x \right )^{2}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \sec \left (x \right ) c_{1} +c_{2} \cos \left (x \right )^{2} \] The above shows that \[ u^{\prime }\left (x \right ) = \left (-2 c_{2} \cos \left (x \right )^{3}+c_{1} \right ) \tan \left (x \right ) \sec \left (x \right ) \] Using the above in (1) gives the solution \[ y = \frac {\left (-2 c_{2} \cos \left (x \right )^{3}+c_{1} \right ) \tan \left (x \right ) \sec \left (x \right )}{\sin \left (x \right ) \left (\sec \left (x \right ) c_{1} +c_{2} \cos \left (x \right )^{2}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {-2 \cos \left (x \right )^{2}+\sec \left (x \right ) c_{3}}{\cos \left (x \right )^{3}+c_{3}} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-2 \cos \left (x \right )^{2}+\sec \left (x \right ) c_{3}}{\cos \left (x \right )^{3}+c_{3}} \\ \end{align*}
Verification of solutions
\[ y = \frac {-2 \cos \left (x \right )^{2}+\sec \left (x \right ) c_{3}}{\cos \left (x \right )^{3}+c_{3}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{2} \sin \left (x \right ) \cos \left (x \right )^{2}+y^{\prime } \cos \left (x \right )^{2}-2 \sin \left (x \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {-y^{2} \sin \left (x \right ) \cos \left (x \right )^{2}+2 \sin \left (x \right )}{\cos \left (x \right )^{2}} \end {array} \]
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 <- Chini successful`
✓ Solution by Maple
Time used: 0.25 (sec). Leaf size: 27
dsolve(diff(y(x),x) + y(x)^2*sin(x) - 2*sin(x)/cos(x)^2=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {-2 \cos \left (x \right )^{2} c_{1} -2 \sec \left (x \right )}{\cos \left (x \right )^{3} c_{1} -2} \]
✓ Solution by Mathematica
Time used: 0.926 (sec). Leaf size: 32
DSolve[y'[x] + y[x]^2*Sin[x] - 2*Sin[x]/Cos[x]^2==0,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*}