Internal problem ID [10576]
Internal file name [OUTPUT/9524_Monday_June_06_2022_03_04_08_PM_65810070/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: 20.
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]
\[ \boxed {y^{\prime }-y^{2}-\lambda x \arctan \left (x \right )^{n} y=\arctan \left (x \right )^{n} \lambda } \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= y^{2}+\arctan \left (x \right )^{n} \lambda x y +\arctan \left (x \right )^{n} \lambda \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = y^{2}+\arctan \left (x \right )^{n} \lambda x y +\arctan \left (x \right )^{n} \lambda \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\arctan \left (x \right )^{n} \lambda \), \(f_1(x)=\arctan \left (x \right )^{n} \lambda x\) 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 x\\ f_2^2 f_0 &=\arctan \left (x \right )^{n} \lambda \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 x u^{\prime }\left (x \right )+\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
\[ u \left (x \right ) = x \left (c_{1} +c_{2} \left (\int \frac {{\mathrm e}^{\int \frac {x \left (2+\left (x^{2}+1\right ) \arctan \left (x \right )^{n} \lambda \right )}{x^{2}+1}d x}}{x^{2} \left (x^{2}+1\right )}d x \right )\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {x c_{2} \left (x^{2}+1\right ) \left (\int \frac {{\mathrm e}^{\int \frac {x \left (2+\left (x^{2}+1\right ) \arctan \left (x \right )^{n} \lambda \right )}{x^{2}+1}d x}}{x^{2} \left (x^{2}+1\right )}d x \right )+c_{2} {\mathrm e}^{\int \frac {x \left (2+\left (x^{2}+1\right ) \arctan \left (x \right )^{n} \lambda \right )}{x^{2}+1}d x}+c_{1} x \left (x^{2}+1\right )}{x \left (x^{2}+1\right )} \] Using the above in (1) gives the solution \[ y = -\frac {x c_{2} \left (x^{2}+1\right ) \left (\int \frac {{\mathrm e}^{\int \frac {x \left (2+\left (x^{2}+1\right ) \arctan \left (x \right )^{n} \lambda \right )}{x^{2}+1}d x}}{x^{2} \left (x^{2}+1\right )}d x \right )+c_{2} {\mathrm e}^{\int \frac {x \left (2+\left (x^{2}+1\right ) \arctan \left (x \right )^{n} \lambda \right )}{x^{2}+1}d x}+c_{1} x \left (x^{2}+1\right )}{x^{2} \left (x^{2}+1\right ) \left (c_{1} +c_{2} \left (\int \frac {{\mathrm e}^{\int \frac {x \left (2+\left (x^{2}+1\right ) \arctan \left (x \right )^{n} \lambda \right )}{x^{2}+1}d x}}{x^{2} \left (x^{2}+1\right )}d x \right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (-x^{3}-x \right ) \left (\int \frac {{\mathrm e}^{\int \frac {x \left (2+\left (x^{2}+1\right ) \arctan \left (x \right )^{n} \lambda \right )}{x^{2}+1}d x}}{x^{2} \left (x^{2}+1\right )}d x \right )-c_{3} x^{3}-c_{3} x -{\mathrm e}^{\int \frac {x \left (2+\left (x^{2}+1\right ) \arctan \left (x \right )^{n} \lambda \right )}{x^{2}+1}d x}}{x^{2} \left (x^{2}+1\right ) \left (c_{3} +\int \frac {{\mathrm e}^{\int \frac {x \left (2+\left (x^{2}+1\right ) \arctan \left (x \right )^{n} \lambda \right )}{x^{2}+1}d x}}{x^{2} \left (x^{2}+1\right )}d x \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (-x^{3}-x \right ) \left (\int \frac {{\mathrm e}^{\int \frac {x \left (2+\left (x^{2}+1\right ) \arctan \left (x \right )^{n} \lambda \right )}{x^{2}+1}d x}}{x^{2} \left (x^{2}+1\right )}d x \right )-c_{3} x^{3}-c_{3} x -{\mathrm e}^{\int \frac {x \left (2+\left (x^{2}+1\right ) \arctan \left (x \right )^{n} \lambda \right )}{x^{2}+1}d x}}{x^{2} \left (x^{2}+1\right ) \left (c_{3} +\int \frac {{\mathrm e}^{\int \frac {x \left (2+\left (x^{2}+1\right ) \arctan \left (x \right )^{n} \lambda \right )}{x^{2}+1}d x}}{x^{2} \left (x^{2}+1\right )}d x \right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (-x^{3}-x \right ) \left (\int \frac {{\mathrm e}^{\int \frac {x \left (2+\left (x^{2}+1\right ) \arctan \left (x \right )^{n} \lambda \right )}{x^{2}+1}d x}}{x^{2} \left (x^{2}+1\right )}d x \right )-c_{3} x^{3}-c_{3} x -{\mathrm e}^{\int \frac {x \left (2+\left (x^{2}+1\right ) \arctan \left (x \right )^{n} \lambda \right )}{x^{2}+1}d x}}{x^{2} \left (x^{2}+1\right ) \left (c_{3} +\int \frac {{\mathrm e}^{\int \frac {x \left (2+\left (x^{2}+1\right ) \arctan \left (x \right )^{n} \lambda \right )}{x^{2}+1}d x}}{x^{2} \left (x^{2}+1\right )}d x \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{2}-\lambda x \arctan \left (x \right )^{n} y=\arctan \left (x \right )^{n} \lambda \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{2}+\lambda x \arctan \left (x \right )^{n} y+\arctan \left (x \right )^{n} \lambda \end {array} \]
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 found: 2 potential symmetries. Proceeding with integration step`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 78
dsolve(diff(y(x),x)=y(x)^2+lambda*x*arctan(x)^n*y(x)+lambda*arctan(x)^n,y(x), singsol=all)
\[ y \left (x \right ) = \frac {{\mathrm e}^{\int \frac {\arctan \left (x \right )^{n} \lambda \,x^{2}-2}{x}d x} x +\int {\mathrm e}^{\int \frac {\arctan \left (x \right )^{n} \lambda \,x^{2}-2}{x}d x}d x -c_{1}}{\left (c_{1} -\left (\int {\mathrm e}^{\int \frac {\arctan \left (x \right )^{n} \lambda \,x^{2}-2}{x}d x}d x \right )\right ) x} \]
✓ Solution by Mathematica
Time used: 7.063 (sec). Leaf size: 120
DSolve[y'[x]==y[x]^2+\[Lambda]*x*ArcTan[x]^n*y[x]+\[Lambda]*ArcTan[x]^n,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {\exp \left (-\int _1^x-\lambda \arctan (K[1])^n K[1]dK[1]\right )+x \int _1^x\frac {\exp \left (-\int _1^{K[2]}-\lambda \arctan (K[1])^n K[1]dK[1]\right )}{K[2]^2}dK[2]+c_1 x}{x^2 \left (\int _1^x\frac {\exp \left (-\int _1^{K[2]}-\lambda \arctan (K[1])^n K[1]dK[1]\right )}{K[2]^2}dK[2]+c_1\right )} \\ y(x)\to -\frac {1}{x} \\ \end{align*}