Internal problem ID [10556]
Internal file name [OUTPUT/9504_Monday_June_06_2022_03_00_31_PM_30962163/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 1.2.6-5. Equations containing
combinations of trigonometric functions.
Problem number: 59.
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]
\[ \boxed {y^{\prime }-\lambda \sin \left (\lambda x \right ) y^{2}-a \sin \left (\lambda x \right ) y=-\tan \left (\lambda x \right ) a} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \lambda \sin \left (\lambda x \right ) y^{2}+a \sin \left (\lambda x \right ) y -\tan \left (\lambda x \right ) a \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \lambda \sin \left (\lambda x \right ) y^{2}+a \sin \left (\lambda x \right ) y -\tan \left (\lambda x \right ) a \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-\tan \left (\lambda x \right ) a\), \(f_1(x)=a \sin \left (\lambda x \right )\) 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 &=\sin \left (\lambda x \right )^{2} a \lambda \\ f_2^2 f_0 &=-\lambda ^{2} \sin \left (\lambda x \right )^{2} \tan \left (\lambda x \right ) a \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 )+\sin \left (\lambda x \right )^{2} a \lambda \right ) u^{\prime }\left (x \right )-\lambda ^{2} \sin \left (\lambda x \right )^{2} \tan \left (\lambda x \right ) a u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = -{\mathrm e}^{-\frac {a \cos \left (\lambda x \right )}{\lambda }} c_{2} \lambda +\cos \left (\lambda x \right ) \left (a c_{2} \operatorname {expIntegral}_{1}\left (\frac {a \cos \left (\lambda x \right )}{\lambda }\right )+c_{1} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = -\lambda \sin \left (\lambda x \right ) \left (a c_{2} \operatorname {expIntegral}_{1}\left (\frac {a \cos \left (\lambda x \right )}{\lambda }\right )+c_{1} \right ) \] Using the above in (1) gives the solution \[ y = \frac {a c_{2} \operatorname {expIntegral}_{1}\left (\frac {a \cos \left (\lambda x \right )}{\lambda }\right )+c_{1}}{-{\mathrm e}^{-\frac {a \cos \left (\lambda x \right )}{\lambda }} c_{2} \lambda +\cos \left (\lambda x \right ) \left (a c_{2} \operatorname {expIntegral}_{1}\left (\frac {a \cos \left (\lambda x \right )}{\lambda }\right )+c_{1} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {a \,\operatorname {expIntegral}_{1}\left (\frac {a \cos \left (\lambda x \right )}{\lambda }\right )+c_{3}}{-{\mathrm e}^{-\frac {a \cos \left (\lambda x \right )}{\lambda }} \lambda +\cos \left (\lambda x \right ) \left (a \,\operatorname {expIntegral}_{1}\left (\frac {a \cos \left (\lambda x \right )}{\lambda }\right )+c_{3} \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {a \,\operatorname {expIntegral}_{1}\left (\frac {a \cos \left (\lambda x \right )}{\lambda }\right )+c_{3}}{-{\mathrm e}^{-\frac {a \cos \left (\lambda x \right )}{\lambda }} \lambda +\cos \left (\lambda x \right ) \left (a \,\operatorname {expIntegral}_{1}\left (\frac {a \cos \left (\lambda x \right )}{\lambda }\right )+c_{3} \right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {a \,\operatorname {expIntegral}_{1}\left (\frac {a \cos \left (\lambda x \right )}{\lambda }\right )+c_{3}}{-{\mathrm e}^{-\frac {a \cos \left (\lambda x \right )}{\lambda }} \lambda +\cos \left (\lambda x \right ) \left (a \,\operatorname {expIntegral}_{1}\left (\frac {a \cos \left (\lambda x \right )}{\lambda }\right )+c_{3} \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\lambda \sin \left (\lambda x \right ) y^{2}-a \sin \left (\lambda x \right ) y=-\tan \left (\lambda x \right ) a \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\lambda \sin \left (\lambda x \right ) y^{2}+a \sin \left (\lambda x \right ) y-\tan \left (\lambda x \right ) a \end {array} \]
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) = (sin(lambda*x)^2*a+lambda*cos(lambda*x))*(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 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 = arccos(t)/lambda] Linear ODE actually solved: (-2*a*t^4+4*a*t^2-2*a)*u(t)+(2*a*t^5-4*a*t^3+2*a*t)*diff(u(t),t)+(2*lambda*t^5-4*lambda*t^3+2*lambda*t)*diff(diff(u(t),t <- change of variables successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 61
dsolve(diff(y(x),x)=lambda*sin(lambda*x)*y(x)^2+a*sin(lambda*x)*y(x)-a*tan(lambda*x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {\operatorname {expIntegral}_{1}\left (\frac {a \cos \left (x \lambda \right )}{\lambda }\right ) c_{1} a +1}{\cos \left (x \lambda \right ) \operatorname {expIntegral}_{1}\left (\frac {a \cos \left (x \lambda \right )}{\lambda }\right ) c_{1} a -{\mathrm e}^{-\frac {a \cos \left (x \lambda \right )}{\lambda }} c_{1} \lambda +\cos \left (x \lambda \right )} \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[y'[x]==\[Lambda]*Sin[\[Lambda]*x]*y[x]^2+a*Sin[\[Lambda]*x]*y[x]-a*Tan[\[Lambda]*x],y[x],x,IncludeSingularSolutions -> True]
Not solved