Internal problem ID [10516]
Internal file name [OUTPUT/9464_Monday_June_06_2022_02_44_10_PM_6467390/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 1.2.6-2. Equations with
cosine.
Problem number: 19.
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]
\[ \boxed {y^{\prime }-\cos \left (\lambda x \right ) y^{2} \lambda =\lambda \cos \left (\lambda x \right )^{3}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \cos \left (\lambda x \right ) \lambda \,y^{2}+\lambda \cos \left (\lambda x \right )^{3} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \cos \left (\lambda x \right ) \lambda \,y^{2}+\lambda \cos \left (\lambda x \right )^{3} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\lambda \cos \left (\lambda x \right )^{3}\), \(f_1(x)=0\) and \(f_2(x)=\lambda \cos \left (\lambda x \right )\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\lambda \cos \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' &=-\sin \left (\lambda x \right ) \lambda ^{2}\\ f_1 f_2 &=0\\ f_2^2 f_0 &=\lambda ^{3} \cos \left (\lambda x \right )^{5} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \lambda \cos \left (\lambda x \right ) u^{\prime \prime }\left (x \right )+\sin \left (\lambda x \right ) \lambda ^{2} u^{\prime }\left (x \right )+\lambda ^{3} \cos \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
\[ u \left (x \right ) = \frac {\sqrt {\pi }\, \left (\left (c_{1} -2 c_{2} \right ) \operatorname {erf}\left (\sqrt {-\sin \left (\lambda x \right )^{2}}\right )+2 c_{2} \right ) {\mathrm e}^{-\sin \left (\lambda x \right )^{2}-\frac {\cos \left (\lambda x \right )^{2}}{2}} \sin \left (\lambda x \right )}{2 \sqrt {-\sin \left (\lambda x \right )^{2}}} \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\cos \left (\lambda x \right ) \left (-\frac {\sqrt {\pi }\, \sin \left (\lambda x \right )^{2} \left (c_{1} -2 c_{2} \right ) \operatorname {erf}\left (\sqrt {-\sin \left (\lambda x \right )^{2}}\right )}{2}+{\mathrm e}^{\sin \left (\lambda x \right )^{2}} \left (c_{1} -2 c_{2} \right ) \sqrt {-\sin \left (\lambda x \right )^{2}}-\sin \left (\lambda x \right )^{2} \sqrt {\pi }\, c_{2} \right ) \lambda \,{\mathrm e}^{\frac {\cos \left (\lambda x \right )^{2}}{2}-1}}{\sqrt {-\sin \left (\lambda x \right )^{2}}} \] Using the above in (1) gives the solution \[ y = -\frac {2 \left (-\frac {\sqrt {\pi }\, \sin \left (\lambda x \right )^{2} \left (c_{1} -2 c_{2} \right ) \operatorname {erf}\left (\sqrt {-\sin \left (\lambda x \right )^{2}}\right )}{2}+{\mathrm e}^{\sin \left (\lambda x \right )^{2}} \left (c_{1} -2 c_{2} \right ) \sqrt {-\sin \left (\lambda x \right )^{2}}-\sin \left (\lambda x \right )^{2} \sqrt {\pi }\, c_{2} \right ) {\mathrm e}^{\frac {\cos \left (\lambda x \right )^{2}}{2}-1} {\mathrm e}^{\frac {3}{4}-\frac {\cos \left (2 \lambda x \right )}{4}}}{\sqrt {\pi }\, \left (\left (c_{1} -2 c_{2} \right ) \operatorname {erf}\left (\sqrt {-\sin \left (\lambda x \right )^{2}}\right )+2 c_{2} \right ) \sin \left (\lambda x \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {2 \left (-\frac {\sin \left (\lambda x \right ) \sqrt {\pi }\, \left (c_{3} -2\right ) \operatorname {erf}\left (\sqrt {-\sin \left (\lambda x \right )^{2}}\right )}{2}+\csc \left (\lambda x \right ) {\mathrm e}^{\sin \left (\lambda x \right )^{2}} \left (c_{3} -2\right ) \sqrt {-\sin \left (\lambda x \right )^{2}}-\sin \left (\lambda x \right ) \sqrt {\pi }\right )}{\left (\left (c_{3} -2\right ) \operatorname {erf}\left (\sqrt {-\sin \left (\lambda x \right )^{2}}\right )+2\right ) \sqrt {\pi }} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {2 \left (-\frac {\sin \left (\lambda x \right ) \sqrt {\pi }\, \left (c_{3} -2\right ) \operatorname {erf}\left (\sqrt {-\sin \left (\lambda x \right )^{2}}\right )}{2}+\csc \left (\lambda x \right ) {\mathrm e}^{\sin \left (\lambda x \right )^{2}} \left (c_{3} -2\right ) \sqrt {-\sin \left (\lambda x \right )^{2}}-\sin \left (\lambda x \right ) \sqrt {\pi }\right )}{\left (\left (c_{3} -2\right ) \operatorname {erf}\left (\sqrt {-\sin \left (\lambda x \right )^{2}}\right )+2\right ) \sqrt {\pi }} \\ \end{align*}
Verification of solutions
\[ y = -\frac {2 \left (-\frac {\sin \left (\lambda x \right ) \sqrt {\pi }\, \left (c_{3} -2\right ) \operatorname {erf}\left (\sqrt {-\sin \left (\lambda x \right )^{2}}\right )}{2}+\csc \left (\lambda x \right ) {\mathrm e}^{\sin \left (\lambda x \right )^{2}} \left (c_{3} -2\right ) \sqrt {-\sin \left (\lambda x \right )^{2}}-\sin \left (\lambda x \right ) \sqrt {\pi }\right )}{\left (\left (c_{3} -2\right ) \operatorname {erf}\left (\sqrt {-\sin \left (\lambda x \right )^{2}}\right )+2\right ) \sqrt {\pi }} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-\cos \left (\lambda x \right ) y^{2} \lambda =\lambda \cos \left (\lambda x \right )^{3} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\cos \left (\lambda x \right ) y^{2} \lambda +\lambda \cos \left (\lambda x \right )^{3} \end {array} \]
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*sin(lambda*x)*(diff(y(x), x))/cos(lambda*x)-cos(lambda*x)^4*la 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 Solution has integrals. Trying a special function solution free of integrals... -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 1F1 ODE <- Kummer successful <- special function solution successful -> Trying to convert hypergeometric functions to elementary form... <- elementary form is not straightforward to achieve - returning special function solution free of uncomputed integra <- Kovacics algorithm successful Change of variables used: [x = arccos(t)/lambda] Linear ODE actually solved: 16*t^5*u(t)-16*diff(u(t),t)+(-16*t^3+16*t)*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: 95
dsolve(diff(y(x),x)=lambda*cos(lambda*x)*y(x)^2+lambda*cos(lambda*x)^3,y(x), singsol=all)
\[ y \left (x \right ) = -\frac {4 \csc \left (x \lambda \right ) \left (-\frac {\sqrt {\pi }\, \sin \left (x \lambda \right )^{2} \left (c_{1} -\frac {1}{2}\right ) \operatorname {erf}\left (\sqrt {-\sin \left (x \lambda \right )^{2}}\right )}{2}+\left (c_{1} -\frac {1}{2}\right ) {\mathrm e}^{\sin \left (x \lambda \right )^{2}} \sqrt {-\sin \left (x \lambda \right )^{2}}+\frac {\sin \left (x \lambda \right )^{2} \sqrt {\pi }\, c_{1}}{2}\right )}{\sqrt {\pi }\, \left (\operatorname {erf}\left (\sqrt {-\sin \left (x \lambda \right )^{2}}\right ) \left (2 c_{1} -1\right )-2 c_{1} \right )} \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[y'[x]==\[Lambda]*Cos[\[Lambda]*x]*y[x]^2+\[Lambda]*Cos[\[Lambda]*x]^3,y[x],x,IncludeSingularSolutions -> True]
Not solved