Internal problem ID [10515]
Internal file name [OUTPUT/9463_Monday_June_06_2022_02_44_05_PM_18158228/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: 18.
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]
\[ \boxed {y^{\prime }-y^{2}-a \cos \left (b x \right )^{m} y=a \cos \left (b x \right )^{m}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= y^{2}+a \cos \left (b x \right )^{m} y +a \cos \left (b x \right )^{m} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = y^{2}+a \cos \left (b x \right )^{m} y +a \cos \left (b x \right )^{m} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=a \cos \left (b x \right )^{m}\), \(f_1(x)=a \cos \left (b x \right )^{m}\) 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 &=a \cos \left (b x \right )^{m}\\ f_2^2 f_0 &=a \cos \left (b x \right )^{m} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (x \right )-a \cos \left (b x \right )^{m} u^{\prime }\left (x \right )+a \cos \left (b x \right )^{m} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (x \right )+a \cos \left (b x \right )^{m} \left (-\textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y} \left (x \right )\right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (x \right )+a \cos \left (b x \right )^{m} \left (-\textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y} \left (x \right )\right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right ) \] Using the above in (1) gives the solution \[ y = -\frac {\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (x \right )+a \cos \left (b x \right )^{m} \left (-\textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y} \left (x \right )\right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )}{\operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (x \right )+a \cos \left (b x \right )^{m} \left (-\textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y} \left (x \right )\right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (x \right )+a \cos \left (b x \right )^{m} \left (-\textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y} \left (x \right )\right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )}{\operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (x \right )+a \cos \left (b x \right )^{m} \left (-\textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y} \left (x \right )\right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (x \right )+a \cos \left (b x \right )^{m} \left (-\textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y} \left (x \right )\right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )}{\operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (x \right )+a \cos \left (b x \right )^{m} \left (-\textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y} \left (x \right )\right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (x \right )+a \cos \left (b x \right )^{m} \left (-\textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y} \left (x \right )\right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )}{\operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (x \right )+a \cos \left (b x \right )^{m} \left (-\textit {\_Y}^{\prime }\left (x \right )+\textit {\_Y} \left (x \right )\right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{2}-a \cos \left (b x \right )^{m} y=a \cos \left (b x \right )^{m} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{2}+a \cos \left (b x \right )^{m} y+a \cos \left (b x \right )^{m} \end {array} \]
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) = a*cos(x*b)^m*(diff(y(x), x))-a*cos(x*b)^m*y(x), y(x)` *** Subleve 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 -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> 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) trying to convert to a linear ODE with constant coefficients -> trying with_periodic_functions in the coefficients trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Whittaker -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> 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) trying to convert to a linear ODE with constant coefficients <- unable to find a useful change of variables trying a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients trying 2nd order, integrating factor of the form mu(x,y) 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 -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> 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) trying to convert to a linear ODE with constant coefficients -> trying with_periodic_functions in the coefficients trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Whittaker -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> 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) trying to convert to a linear ODE with constant coefficients -> trying with_periodic_functions in the coefficients trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Whittaker -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> 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) trying to convert to a linear ODE with constant coefficients <- unable to find a useful change of variables trying a symmetry of the form [xi=0, eta=F(x)] trying to convert to an ODE of Bessel type -> trying with_periodic_functions in the coefficients -> Trying a change of variables to reduce to Bernoulli -> Calling odsolve with the ODE`, diff(y(x), x)-(y(x)^2+y(x)+a*cos(x*b)^m*y(x)*x+x^2*a*cos(x*b)^m)/x, y(x), explicit` *** Su 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 inverse_Riccati trying 1st order ODE linearizable_by_differentiation -> trying a symmetry pattern of the form [F(x)*G(y), 0] -> trying a symmetry pattern of the form [0, F(x)*G(y)] -> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] trying inverse_Riccati trying 1st order ODE linearizable_by_differentiation --- Trying Lie symmetry methods, 1st order --- `, `-> Computing symmetries using: way = 4 `, `-> Computing symmetries using: way = 2 `, `-> Computing symmetries using: way = 6`
✗ Solution by Maple
dsolve(diff(y(x),x)=y(x)^2+a*cos(b*x)^m*y(x)+a*cos(b*x)^m,y(x), singsol=all)
\[ \text {No solution found} \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[y'[x]==y[x]^2+a*Cos[b*x]^m*y[x]+a*Cos[b*x]^m,y[x],x,IncludeSingularSolutions -> True]
Not solved