Internal problem ID [10575]
Internal file name [OUTPUT/9523_Monday_June_06_2022_03_04_01_PM_31783764/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: 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]
\[ \boxed {y^{\prime }-y^{2}-\lambda \arctan \left (x \right )^{n} y=-a^{2}+a \lambda \arctan \left (x \right )^{n}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= y^{2}+\lambda \arctan \left (x \right )^{n} y -a^{2}+a \lambda \arctan \left (x \right )^{n} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = y^{2}+\lambda \arctan \left (x \right )^{n} y -a^{2}+a \lambda \arctan \left (x \right )^{n} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-a^{2}+a \lambda \arctan \left (x \right )^{n}\), \(f_1(x)=\arctan \left (x \right )^{n} \lambda \) 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 &=\arctan \left (x \right )^{n} \lambda \\ f_2^2 f_0 &=-a^{2}+a \lambda \arctan \left (x \right )^{n} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (x \right )-\arctan \left (x \right )^{n} \lambda u^{\prime }\left (x \right )+\left (-a^{2}+a \lambda \arctan \left (x \right )^{n}\right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = {\mathrm e}^{\int \frac {a \left (\int {\mathrm e}^{-\left (\int \left (-\arctan \left (x \right )^{n} \lambda +2 a \right )d x \right )}d x \right )-c_{1} a +{\mathrm e}^{-\left (\int \left (-\arctan \left (x \right )^{n} \lambda +2 a \right )d x \right )}}{-c_{1} +\int {\mathrm e}^{-\left (\int \left (-\arctan \left (x \right )^{n} \lambda +2 a \right )d x \right )}d x}d x} c_{2} \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\left (a \left (\int {\mathrm e}^{-\left (\int \left (-\arctan \left (x \right )^{n} \lambda +2 a \right )d x \right )}d x \right )-c_{1} a +{\mathrm e}^{-\left (\int \left (-\arctan \left (x \right )^{n} \lambda +2 a \right )d x \right )}\right ) {\mathrm e}^{\int \frac {a \left (\int {\mathrm e}^{-\left (\int \left (-\arctan \left (x \right )^{n} \lambda +2 a \right )d x \right )}d x \right )-c_{1} a +{\mathrm e}^{-\left (\int \left (-\arctan \left (x \right )^{n} \lambda +2 a \right )d x \right )}}{-c_{1} +\int {\mathrm e}^{-\left (\int \left (-\arctan \left (x \right )^{n} \lambda +2 a \right )d x \right )}d x}d x} c_{2}}{-c_{1} +\int {\mathrm e}^{-\left (\int \left (-\arctan \left (x \right )^{n} \lambda +2 a \right )d x \right )}d x} \] Using the above in (1) gives the solution \[ y = -\frac {a \left (\int {\mathrm e}^{-\left (\int \left (-\arctan \left (x \right )^{n} \lambda +2 a \right )d x \right )}d x \right )-c_{1} a +{\mathrm e}^{-\left (\int \left (-\arctan \left (x \right )^{n} \lambda +2 a \right )d x \right )}}{-c_{1} +\int {\mathrm e}^{-\left (\int \left (-\arctan \left (x \right )^{n} \lambda +2 a \right )d x \right )}d x} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {-a \left (\int {\mathrm e}^{-\left (\int \left (-\arctan \left (x \right )^{n} \lambda +2 a \right )d x \right )}d x \right )+c_{3} a -{\mathrm e}^{-\left (\int \left (-\arctan \left (x \right )^{n} \lambda +2 a \right )d x \right )}}{-c_{3} +\int {\mathrm e}^{-\left (\int \left (-\arctan \left (x \right )^{n} \lambda +2 a \right )d x \right )}d x} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-a \left (\int {\mathrm e}^{-\left (\int \left (-\arctan \left (x \right )^{n} \lambda +2 a \right )d x \right )}d x \right )+c_{3} a -{\mathrm e}^{-\left (\int \left (-\arctan \left (x \right )^{n} \lambda +2 a \right )d x \right )}}{-c_{3} +\int {\mathrm e}^{-\left (\int \left (-\arctan \left (x \right )^{n} \lambda +2 a \right )d x \right )}d x} \\ \end{align*}
Verification of solutions
\[ y = \frac {-a \left (\int {\mathrm e}^{-\left (\int \left (-\arctan \left (x \right )^{n} \lambda +2 a \right )d x \right )}d x \right )+c_{3} a -{\mathrm e}^{-\left (\int \left (-\arctan \left (x \right )^{n} \lambda +2 a \right )d x \right )}}{-c_{3} +\int {\mathrm e}^{-\left (\int \left (-\arctan \left (x \right )^{n} \lambda +2 a \right )d x \right )}d x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{2}-\lambda \arctan \left (x \right )^{n} y=-a^{2}+a \lambda \arctan \left (x \right )^{n} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{2}+\lambda \arctan \left (x \right )^{n} y-a^{2}+a \lambda \arctan \left (x \right )^{n} \end {array} \]
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) = arctan(x)^n*lambda*(diff(y(x), x))+(a^2-a*lambda*arctan(x)^n)*y(x), y( 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)-(y(x)^2+y(x)+arctan(x)^n*lambda*y(x)*x+x^2*(-a^2+a*lambda*arctan(x)^n))/x, y(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)] <- symmetry pattern of the form [0, F(x)*G(y)] successful <- Riccati with symmetry pattern of the form [0,F(x)*G(y)] successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 71
dsolve(diff(y(x),x)=y(x)^2+lambda*arctan(x)^n*y(x)-a^2+a*lambda*arctan(x)^n,y(x), singsol=all)
\[ y \left (x \right ) = \frac {-c_{1} a -a \left (\int {\mathrm e}^{-\left (\int \left (-\arctan \left (x \right )^{n} \lambda +2 a \right )d x \right )}d x \right )-{\mathrm e}^{-\left (\int \left (-\arctan \left (x \right )^{n} \lambda +2 a \right )d x \right )}}{c_{1} +\int {\mathrm e}^{-\left (\int \left (-\arctan \left (x \right )^{n} \lambda +2 a \right )d x \right )}d x} \]
✓ Solution by Mathematica
Time used: 7.862 (sec). Leaf size: 210
DSolve[y'[x]==y[x]^2+\[Lambda]*ArcTan[x]^n*y[x]-a^2+a*\[Lambda]*ArcTan[x]^n,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\int _1^x\frac {\exp \left (-\int _1^{K[2]}\left (2 a-\lambda \arctan (K[1])^n\right )dK[1]\right ) \left (-\lambda \arctan (K[2])^n+a-y(x)\right )}{n \lambda (a+y(x))}dK[2]+\int _1^{y(x)}\left (\frac {\exp \left (-\int _1^x\left (2 a-\lambda \arctan (K[1])^n\right )dK[1]\right )}{n \lambda (a+K[3])^2}-\int _1^x\left (-\frac {\exp \left (-\int _1^{K[2]}\left (2 a-\lambda \arctan (K[1])^n\right )dK[1]\right ) \left (-\lambda \arctan (K[2])^n+a-K[3]\right )}{n \lambda (a+K[3])^2}-\frac {\exp \left (-\int _1^{K[2]}\left (2 a-\lambda \arctan (K[1])^n\right )dK[1]\right )}{n \lambda (a+K[3])}\right )dK[2]\right )dK[3]=c_1,y(x)\right ] \]