Internal problem ID [10503]
Internal file name [OUTPUT/9451_Monday_June_06_2022_02_36_31_PM_98522936/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 1.2.6-1. Equations with
sine
Problem number: 6.
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]
\[ \boxed {y^{\prime }-\lambda \sin \left (\lambda x \right ) y^{2}=\lambda \sin \left (\lambda x \right )^{3}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= y^{2} \sin \left (\lambda x \right ) \lambda +\lambda \sin \left (\lambda x \right )^{3} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = y^{2} \sin \left (\lambda x \right ) \lambda +\lambda \sin \left (\lambda x \right )^{3} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\lambda \sin \left (\lambda x \right )^{3}\), \(f_1(x)=0\) 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' &=\cos \left (\lambda x \right ) \lambda ^{2}\\ f_1 f_2 &=0\\ f_2^2 f_0 &=\lambda ^{3} \sin \left (\lambda x \right )^{5} \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 )-\cos \left (\lambda x \right ) \lambda ^{2} u^{\prime }\left (x \right )+\lambda ^{3} \sin \left (\lambda x \right )^{5} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = {\mathrm e}^{-\frac {\cos \left (\lambda x \right )^{2}}{2}} \left (c_{2} \operatorname {erfi}\left (\cos \left (\lambda x \right )\right )+c_{1} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\left (\sqrt {\pi }\, \cos \left (\lambda x \right ) \left (c_{2} \operatorname {erfi}\left (\cos \left (\lambda x \right )\right )+c_{1} \right ) {\mathrm e}^{-\frac {\cos \left (\lambda x \right )^{2}}{2}}-2 c_{2} {\mathrm e}^{\frac {\cos \left (\lambda x \right )^{2}}{2}}\right ) \lambda \sin \left (\lambda x \right )}{\sqrt {\pi }} \] Using the above in (1) gives the solution \[ y = -\frac {\left (\sqrt {\pi }\, \cos \left (\lambda x \right ) \left (c_{2} \operatorname {erfi}\left (\cos \left (\lambda x \right )\right )+c_{1} \right ) {\mathrm e}^{-\frac {\cos \left (\lambda x \right )^{2}}{2}}-2 c_{2} {\mathrm e}^{\frac {\cos \left (\lambda x \right )^{2}}{2}}\right ) {\mathrm e}^{\frac {\cos \left (2 \lambda x \right )}{4}+\frac {1}{4}}}{\sqrt {\pi }\, \left (c_{2} \operatorname {erfi}\left (\cos \left (\lambda x \right )\right )+c_{1} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {-2 \,{\mathrm e}^{\cos \left (\lambda x \right )^{2}}+\cos \left (\lambda x \right ) \sqrt {\pi }\, \left (\operatorname {erfi}\left (\cos \left (\lambda x \right )\right )+c_{3} \right )}{\sqrt {\pi }\, \left (\operatorname {erfi}\left (\cos \left (\lambda x \right )\right )+c_{3} \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {-2 \,{\mathrm e}^{\cos \left (\lambda x \right )^{2}}+\cos \left (\lambda x \right ) \sqrt {\pi }\, \left (\operatorname {erfi}\left (\cos \left (\lambda x \right )\right )+c_{3} \right )}{\sqrt {\pi }\, \left (\operatorname {erfi}\left (\cos \left (\lambda x \right )\right )+c_{3} \right )} \\ \end{align*}
Verification of solutions
\[ y = -\frac {-2 \,{\mathrm e}^{\cos \left (\lambda x \right )^{2}}+\cos \left (\lambda x \right ) \sqrt {\pi }\, \left (\operatorname {erfi}\left (\cos \left (\lambda x \right )\right )+c_{3} \right )}{\sqrt {\pi }\, \left (\operatorname {erfi}\left (\cos \left (\lambda x \right )\right )+c_{3} \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\lambda \sin \left (\lambda x \right ) y^{2}=\lambda \sin \left (\lambda x \right )^{3} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\lambda \sin \left (\lambda x \right ) y^{2}+\lambda \sin \left (\lambda x \right )^{3} \end {array} \]
Maple trace Kovacic algorithm successful
`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) = lambda*cos(lambda*x)*(diff(y(x), x))/sin(lambda*x)-lambda^2*sin(lambda 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)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm A Liouvillian solution exists Reducible group (found an exponential solution) Group is reducible, not completely reducible <- Kovacics algorithm successful Change of variables used: [x = arccos(t)/lambda] Linear ODE actually solved: 16*(-t^2+1)^(1/2)*(t^4-2*t^2+1)*u(t)+16*(-t^2+1)^(3/2)*diff(diff(u(t),t),t) = 0 <- change of variables successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 51
dsolve(diff(y(x),x)=lambda*sin(lambda*x)*y(x)^2+lambda*sin(lambda*x)^3,y(x), singsol=all)
\[ y \left (x \right ) = \frac {2 \,{\mathrm e}^{\frac {\cos \left (2 x \lambda \right )}{2}+\frac {1}{2}} c_{1} -\cos \left (x \lambda \right ) \sqrt {\pi }\, \left (\operatorname {erfi}\left (\cos \left (x \lambda \right )\right ) c_{1} +1\right )}{\sqrt {\pi }\, \left (\operatorname {erfi}\left (\cos \left (x \lambda \right )\right ) c_{1} +1\right )} \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[y'[x]==\[Lambda]*Sin[\[Lambda]*x]*y[x]^2+\[Lambda]*Sin[\[Lambda]*x]^3,y[x],x,IncludeSingularSolutions -> True]
Not solved