Internal problem ID [10621]
Internal file name [OUTPUT/9569_Monday_June_06_2022_03_09_51_PM_66941966/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 1.2.8-1. Equations containing
arbitrary functions (but not containing their derivatives).
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]
\[ \boxed {y^{\prime }-\lambda \sin \left (\lambda x \right ) y^{2}-f \left (x \right ) \cos \left (\lambda x \right ) y=-f \left (x \right )} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \lambda \sin \left (\lambda x \right ) y^{2}+f \left (x \right ) \cos \left (\lambda x \right ) y -f \left (x \right ) \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \lambda \sin \left (\lambda x \right ) y^{2}+f \left (x \right ) \cos \left (\lambda x \right ) y -f \left (x \right ) \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-f \left (x \right )\), \(f_1(x)=f \left (x \right ) \cos \left (\lambda x \right )\) and \(f_2(x)=\lambda \sin \left (\lambda x \right )\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\lambda \sin \left (\lambda 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' &=\lambda ^{2} \cos \left (\lambda x \right )\\ f_1 f_2 &=f \left (x \right ) \cos \left (\lambda x \right ) \lambda \sin \left (\lambda x \right )\\ f_2^2 f_0 &=-\lambda ^{2} \sin \left (\lambda x \right )^{2} f \left (x \right ) \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \lambda \sin \left (\lambda x \right ) u^{\prime \prime }\left (x \right )-\left (\lambda ^{2} \cos \left (\lambda x \right )+f \left (x \right ) \cos \left (\lambda x \right ) \lambda \sin \left (\lambda x \right )\right ) u^{\prime }\left (x \right )-\lambda ^{2} \sin \left (\lambda x \right )^{2} f \left (x \right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = -\cos \left (\lambda x \right ) \left (c_{2} \lambda \left (\int {\mathrm e}^{\int \left (f \left (x \right ) \cos \left (\lambda x \right )+2 \lambda \tan \left (\lambda x \right )\right )d x} \sin \left (\lambda x \right )d x \right )-c_{1} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = -\lambda \sin \left (\lambda x \right ) \left (\cos \left (\lambda x \right ) c_{2} {\mathrm e}^{\int \left (f \left (x \right ) \cos \left (\lambda x \right )+2 \lambda \tan \left (\lambda x \right )\right )d x}-c_{2} \lambda \left (\int {\mathrm e}^{\int \left (f \left (x \right ) \cos \left (\lambda x \right )+2 \lambda \tan \left (\lambda x \right )\right )d x} \sin \left (\lambda x \right )d x \right )+c_{1} \right ) \] Using the above in (1) gives the solution \[ y = -\frac {\cos \left (\lambda x \right ) c_{2} {\mathrm e}^{\int \left (f \left (x \right ) \cos \left (\lambda x \right )+2 \lambda \tan \left (\lambda x \right )\right )d x}-c_{2} \lambda \left (\int {\mathrm e}^{\int \left (f \left (x \right ) \cos \left (\lambda x \right )+2 \lambda \tan \left (\lambda x \right )\right )d x} \sin \left (\lambda x \right )d x \right )+c_{1}}{\cos \left (\lambda x \right ) \left (c_{2} \lambda \left (\int {\mathrm e}^{\int \left (f \left (x \right ) \cos \left (\lambda x \right )+2 \lambda \tan \left (\lambda x \right )\right )d x} \sin \left (\lambda x \right )d x \right )-c_{1} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\sec \left (\lambda x \right ) \lambda \left (\int {\mathrm e}^{\int \left (f \left (x \right ) \cos \left (\lambda x \right )+2 \lambda \tan \left (\lambda x \right )\right )d x} \sin \left (\lambda x \right )d x \right )-\sec \left (\lambda x \right ) c_{3} -{\mathrm e}^{\int \left (f \left (x \right ) \cos \left (\lambda x \right )+2 \lambda \tan \left (\lambda x \right )\right )d x}}{\lambda \left (\int {\mathrm e}^{\int \left (f \left (x \right ) \cos \left (\lambda x \right )+2 \lambda \tan \left (\lambda x \right )\right )d x} \sin \left (\lambda x \right )d x \right )-c_{3}} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\sec \left (\lambda x \right ) \lambda \left (\int {\mathrm e}^{\int \left (f \left (x \right ) \cos \left (\lambda x \right )+2 \lambda \tan \left (\lambda x \right )\right )d x} \sin \left (\lambda x \right )d x \right )-\sec \left (\lambda x \right ) c_{3} -{\mathrm e}^{\int \left (f \left (x \right ) \cos \left (\lambda x \right )+2 \lambda \tan \left (\lambda x \right )\right )d x}}{\lambda \left (\int {\mathrm e}^{\int \left (f \left (x \right ) \cos \left (\lambda x \right )+2 \lambda \tan \left (\lambda x \right )\right )d x} \sin \left (\lambda x \right )d x \right )-c_{3}} \\ \end{align*}
Verification of solutions
\[ y = \frac {\sec \left (\lambda x \right ) \lambda \left (\int {\mathrm e}^{\int \left (f \left (x \right ) \cos \left (\lambda x \right )+2 \lambda \tan \left (\lambda x \right )\right )d x} \sin \left (\lambda x \right )d x \right )-\sec \left (\lambda x \right ) c_{3} -{\mathrm e}^{\int \left (f \left (x \right ) \cos \left (\lambda x \right )+2 \lambda \tan \left (\lambda x \right )\right )d x}}{\lambda \left (\int {\mathrm e}^{\int \left (f \left (x \right ) \cos \left (\lambda x \right )+2 \lambda \tan \left (\lambda x \right )\right )d x} \sin \left (\lambda x \right )d x \right )-c_{3}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\lambda \sin \left (\lambda x \right ) y^{2}-f \left (x \right ) \cos \left (\lambda x \right ) y=-f \left (x \right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\lambda \sin \left (\lambda x \right ) y^{2}+f \left (x \right ) \cos \left (\lambda x \right ) y-f \left (x \right ) \end {array} \]
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(lambda*x)*(sin(lambda*x)*f(x)+lambda)*(diff(y(x), x))/sin(lambda*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 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 a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) <- linear symmetries successful Change of variables used: [x = arccos(t)/lambda] Linear ODE actually solved: (2*(-t^2+1)^(1/2)*f(arccos(t)/lambda)*t^2-2*(-t^2+1)^(1/2)*f(arccos(t)/lambda))*u(t)+(-2*f(arccos(t)/lambda)*(-t^2+1)^(1 <- change of variables successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 97
dsolve(diff(y(x),x)=lambda*sin(lambda*x)*y(x)^2+f(x)*cos(lambda*x)*y(x)-f(x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {\sec \left (x \lambda \right ) \lambda \left (\int {\mathrm e}^{\int \left (f \left (x \right ) \cos \left (x \lambda \right )+2 \tan \left (x \lambda \right ) \lambda \right )d x} \sin \left (x \lambda \right )d x \right ) c_{1} -c_{1} {\mathrm e}^{\int \left (f \left (x \right ) \cos \left (x \lambda \right )+2 \tan \left (x \lambda \right ) \lambda \right )d x}-\sec \left (x \lambda \right )}{\lambda \left (\int {\mathrm e}^{\int \left (f \left (x \right ) \cos \left (x \lambda \right )+2 \tan \left (x \lambda \right ) \lambda \right )d x} \sin \left (x \lambda \right )d x \right ) c_{1} -1} \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[y'[x]==\[Lambda]*Sin[\[Lambda]*x]*y[x]^2+f[x]*Cos[\[Lambda]*x]*y[x]-f[x],y[x],x,IncludeSingularSolutions -> True]
Not solved