Internal problem ID [10577]
Internal file name [OUTPUT/9525_Monday_June_06_2022_03_04_13_PM_78552728/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 1.2.7-3. Equations containing
arctangent.
Problem number: 21.
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]
\[ \boxed {y^{\prime }+\left (k +1\right ) x^{k} y^{2}-\lambda \arctan \left (x \right )^{n} \left (x^{k +1} y-1\right )=0} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -x^{k} y^{2} k +\arctan \left (x \right )^{n} x^{k +1} \lambda y -x^{k} y^{2}-\arctan \left (x \right )^{n} \lambda \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -x^{k} y^{2} k +\arctan \left (x \right )^{n} x^{k} x \lambda y -x^{k} y^{2}-\arctan \left (x \right )^{n} \lambda \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-\arctan \left (x \right )^{n} \lambda \), \(f_1(x)=\lambda \arctan \left (x \right )^{n} x^{k +1}\) and \(f_2(x)=-x^{k} k -x^{k}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\left (-x^{k} k -x^{k}\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' &=-\frac {x^{k} k^{2}}{x}-\frac {x^{k} k}{x}\\ f_1 f_2 &=\lambda \arctan \left (x \right )^{n} x^{k +1} \left (-x^{k} k -x^{k}\right )\\ f_2^2 f_0 &=-\left (-x^{k} k -x^{k}\right )^{2} \arctan \left (x \right )^{n} \lambda \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \left (-x^{k} k -x^{k}\right ) u^{\prime \prime }\left (x \right )-\left (-\frac {x^{k} k^{2}}{x}-\frac {x^{k} k}{x}+\lambda \arctan \left (x \right )^{n} x^{k +1} \left (-x^{k} k -x^{k}\right )\right ) u^{\prime }\left (x \right )-\left (-x^{k} k -x^{k}\right )^{2} \arctan \left (x \right )^{n} \lambda u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = x^{k +1} \left (\left (\int x^{-2 k -2} {\mathrm e}^{\int \left (\lambda \arctan \left (x \right )^{n} x^{k +1}+\frac {k}{x}\right )d x}d x \right ) c_{2} +c_{1} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \left (k +1\right ) \left (\left (\int {\mathrm e}^{\int \frac {x^{2+k} \lambda \arctan \left (x \right )^{n}+k}{x}d x} x^{-2 k -2}d x \right ) c_{2} +c_{1} \right ) x^{k}+c_{2} x^{-k -1} {\mathrm e}^{\int \frac {x^{2+k} \lambda \arctan \left (x \right )^{n}+k}{x}d x} \] Using the above in (1) gives the solution \[ y = -\frac {\left (\left (k +1\right ) \left (\left (\int {\mathrm e}^{\int \frac {x^{2+k} \lambda \arctan \left (x \right )^{n}+k}{x}d x} x^{-2 k -2}d x \right ) c_{2} +c_{1} \right ) x^{k}+c_{2} x^{-k -1} {\mathrm e}^{\int \frac {x^{2+k} \lambda \arctan \left (x \right )^{n}+k}{x}d x}\right ) x^{-k -1}}{\left (-x^{k} k -x^{k}\right ) \left (\left (\int x^{-2 k -2} {\mathrm e}^{\int \left (\lambda \arctan \left (x \right )^{n} x^{k +1}+\frac {k}{x}\right )d x}d x \right ) c_{2} +c_{1} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (k +1\right ) \left (\int x^{-2 k -2} {\mathrm e}^{\int \left (\lambda \arctan \left (x \right )^{n} x^{k +1}+\frac {k}{x}\right )d x}d x +c_{3} \right ) x^{-k -1}+x^{-2-3 k} {\mathrm e}^{\int \left (\lambda \arctan \left (x \right )^{n} x^{k +1}+\frac {k}{x}\right )d x}}{\left (k +1\right ) \left (\int {\mathrm e}^{\int \frac {x^{2+k} \lambda \arctan \left (x \right )^{n}+k}{x}d x} x^{-2 k -2}d x +c_{3} \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (k +1\right ) \left (\int x^{-2 k -2} {\mathrm e}^{\int \left (\lambda \arctan \left (x \right )^{n} x^{k +1}+\frac {k}{x}\right )d x}d x +c_{3} \right ) x^{-k -1}+x^{-2-3 k} {\mathrm e}^{\int \left (\lambda \arctan \left (x \right )^{n} x^{k +1}+\frac {k}{x}\right )d x}}{\left (k +1\right ) \left (\int {\mathrm e}^{\int \frac {x^{2+k} \lambda \arctan \left (x \right )^{n}+k}{x}d x} x^{-2 k -2}d x +c_{3} \right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (k +1\right ) \left (\int x^{-2 k -2} {\mathrm e}^{\int \left (\lambda \arctan \left (x \right )^{n} x^{k +1}+\frac {k}{x}\right )d x}d x +c_{3} \right ) x^{-k -1}+x^{-2-3 k} {\mathrm e}^{\int \left (\lambda \arctan \left (x \right )^{n} x^{k +1}+\frac {k}{x}\right )d x}}{\left (k +1\right ) \left (\int {\mathrm e}^{\int \frac {x^{2+k} \lambda \arctan \left (x \right )^{n}+k}{x}d x} x^{-2 k -2}d x +c_{3} \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }+\left (k +1\right ) x^{k} y^{2}-\lambda \arctan \left (x \right )^{n} \left (x^{k +1} y-1\right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\left (k +1\right ) x^{k} y^{2}+\lambda \arctan \left (x \right )^{n} \left (x^{k +1} y-1\right ) \end {array} \]
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) = (x^(1+k)*arctan(x)^n*x*lambda+k)*(diff(y(x), x))/x-x^k*(1+k)*arctan(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) trying to convert to a linear ODE with constant coefficients -> trying with_periodic_functions in the 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 -> 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 <- 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 a change of variables to reduce to Bernoulli -> Calling odsolve with the ODE`, diff(y(x), x)-((-x^k*k-x^k)*y(x)^2+y(x)+arctan(x)^n*x^(1+k)*lambda*y(x)*x-arctan(x)^n*lambda*x^ 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
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 180
dsolve(diff(y(x),x)=-(k+1)*x^k*y(x)^2+lambda*arctan(x)^n*(x^(k+1)*y(x)-1),y(x), singsol=all)
\[ y \left (x \right ) = \frac {x^{-1-k} \left (\left (\int x^{k} {\mathrm e}^{\lambda \left (\int \arctan \left (x \right )^{n} x^{1+k}d x \right )-2 \left (\int \frac {1}{x}d x \right ) \left (1+k \right )}d x \right ) k +x^{1+k} {\mathrm e}^{\int \frac {x^{1+k} \arctan \left (x \right )^{n} x \lambda -2 k -2}{x}d x}+\int x^{k} {\mathrm e}^{\lambda \left (\int \arctan \left (x \right )^{n} x^{1+k}d x \right )-2 \left (\int \frac {1}{x}d x \right ) \left (1+k \right )}d x +c_{1} \right )}{\left (\int x^{k} {\mathrm e}^{\lambda \left (\int \arctan \left (x \right )^{n} x^{1+k}d x \right )-2 \left (\int \frac {1}{x}d x \right ) \left (1+k \right )}d x \right ) k +\int x^{k} {\mathrm e}^{\lambda \left (\int \arctan \left (x \right )^{n} x^{1+k}d x \right )-2 \left (\int \frac {1}{x}d x \right ) \left (1+k \right )}d x +c_{1}} \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[y'[x]==-(k+1)*x^k*y[x]^2+\[Lambda]*ArcTan[x]^n*(x^(k+1)*y[x]-1),y[x],x,IncludeSingularSolutions -> True]
Not solved