Internal problem ID [15118]
Internal file name [OUTPUT/15119_Sunday_April_21_2024_01_32_46_PM_81085812/index.tex]
Book: A book of problems in ordinary differential equations. M.L. KRASNOV, A.L. KISELYOV,
G.I. MARKARENKO. MIR, MOSCOW. 1983
Section: Section 9. The Riccati equation. Exercises page 75
Problem number: 233.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[[_1st_order, `_with_symmetry_[F(x),G(x)]`], _Riccati]
\[ \boxed {y^{\prime }+y^{2}-2 y \sin \left (x \right )=-\sin \left (x \right )^{2}+\cos \left (x \right )} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -y^{2}+2 y \sin \left (x \right )-\sin \left (x \right )^{2}+\cos \left (x \right ) \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -y^{2}+2 y \sin \left (x \right )-\sin \left (x \right )^{2}+\cos \left (x \right ) \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-\sin \left (x \right )^{2}+\cos \left (x \right )\), \(f_1(x)=2 \sin \left (x \right )\) and \(f_2(x)=-1\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-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 &=-2 \sin \left (x \right )\\ f_2^2 f_0 &=-\sin \left (x \right )^{2}+\cos \left (x \right ) \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} -u^{\prime \prime }\left (x \right )+2 \sin \left (x \right ) u^{\prime }\left (x \right )+\left (-\sin \left (x \right )^{2}+\cos \left (x \right )\right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = {\mathrm e}^{-\cos \left (x \right )} \left (c_{2} x +c_{1} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = {\mathrm e}^{-\cos \left (x \right )} \left (\sin \left (x \right ) \left (c_{2} x +c_{1} \right )+c_{2} \right ) \] Using the above in (1) gives the solution \[ y = \frac {\sin \left (x \right ) \left (c_{2} x +c_{1} \right )+c_{2}}{c_{2} x +c_{1}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=\textit {\_C3}\) the following solution
\[ y = \frac {\sin \left (x \right ) \left (\textit {\_C3} +x \right )+1}{\textit {\_C3} +x} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\sin \left (x \right ) \left (\textit {\_C3} +x \right )+1}{\textit {\_C3} +x} \\ \end{align*}
Verification of solutions
\[ y = \frac {\sin \left (x \right ) \left (\textit {\_C3} +x \right )+1}{\textit {\_C3} +x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }+y^{2}-2 y \sin \left (x \right )=-\sin \left (x \right )^{2}+\cos \left (x \right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-y^{2}+2 y \sin \left (x \right )-\sin \left (x \right )^{2}+\cos \left (x \right ) \end {array} \]
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: <- Riccati particular polynomial solution successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 14
dsolve(diff(y(x),x)+y(x)^2-2*y(x)*sin(x)+sin(x)^2-cos(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \sin \left (x \right )+\frac {1}{x -c_{1}} \]
✓ Solution by Mathematica
Time used: 0.257 (sec). Leaf size: 20
DSolve[y'[x]+y[x]^2-2*y[x]*Sin[x]+Sin[x]^2-Cos[x]==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \sin (x)+\frac {1}{x+c_1} \\ y(x)\to \sin (x) \\ \end{align*}