Internal problem ID [10582]
Internal file name [OUTPUT/9530_Monday_June_06_2022_03_05_10_PM_55650788/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: 26.
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]
\[ \boxed {x y^{\prime }-\lambda \arctan \left (x \right )^{n} y^{2}-y k=\lambda \,b^{2} x^{2 k} \arctan \left (x \right )^{n}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {\lambda \arctan \left (x \right )^{n} y^{2}+y k +\lambda \,b^{2} x^{2 k} \arctan \left (x \right )^{n}}{x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {\lambda \,b^{2} x^{2 k} \arctan \left (x \right )^{n}}{x}+\frac {\lambda \arctan \left (x \right )^{n} y^{2}}{x}+\frac {y k}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {\lambda \,b^{2} x^{2 k} \arctan \left (x \right )^{n}}{x}\), \(f_1(x)=\frac {k}{x}\) and \(f_2(x)=\frac {\lambda \arctan \left (x \right )^{n}}{x}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {\lambda \arctan \left (x \right )^{n} u}{x}} \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 {\lambda \arctan \left (x \right )^{n} n}{\left (x^{2}+1\right ) \arctan \left (x \right ) x}-\frac {\lambda \arctan \left (x \right )^{n}}{x^{2}}\\ f_1 f_2 &=\frac {k \lambda \arctan \left (x \right )^{n}}{x^{2}}\\ f_2^2 f_0 &=\frac {\lambda ^{3} \arctan \left (x \right )^{3 n} b^{2} x^{2 k}}{x^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {\lambda \arctan \left (x \right )^{n} u^{\prime \prime }\left (x \right )}{x}-\left (\frac {\lambda \arctan \left (x \right )^{n} n}{\left (x^{2}+1\right ) \arctan \left (x \right ) x}-\frac {\lambda \arctan \left (x \right )^{n}}{x^{2}}+\frac {k \lambda \arctan \left (x \right )^{n}}{x^{2}}\right ) u^{\prime }\left (x \right )+\frac {\lambda ^{3} \arctan \left (x \right )^{3 n} b^{2} x^{2 k} u \left (x \right )}{x^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} {\mathrm e}^{i b \lambda \left (\int x^{k -1} \arctan \left (x \right )^{n}d x \right )}+c_{2} {\mathrm e}^{-i b \lambda \left (\int x^{k -1} \arctan \left (x \right )^{n}d x \right )} \] The above shows that \[ u^{\prime }\left (x \right ) = i b \,x^{k -1} \lambda \arctan \left (x \right )^{n} {\mathrm e}^{-i b \lambda \left (\int x^{k -1} \arctan \left (x \right )^{n}d x \right )} \left (c_{1} {\mathrm e}^{2 i b \lambda \left (\int x^{k -1} \arctan \left (x \right )^{n}d x \right )}-c_{2} \right ) \] Using the above in (1) gives the solution \[ y = -\frac {i b \,x^{k -1} {\mathrm e}^{-i b \lambda \left (\int x^{k -1} \arctan \left (x \right )^{n}d x \right )} \left (c_{1} {\mathrm e}^{2 i b \lambda \left (\int x^{k -1} \arctan \left (x \right )^{n}d x \right )}-c_{2} \right ) x}{c_{1} {\mathrm e}^{i b \lambda \left (\int x^{k -1} \arctan \left (x \right )^{n}d x \right )}+c_{2} {\mathrm e}^{-i b \lambda \left (\int x^{k -1} \arctan \left (x \right )^{n}d x \right )}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {i b \,x^{k} \left (c_{3} {\mathrm e}^{2 i b \lambda \left (\int x^{k -1} \arctan \left (x \right )^{n}d x \right )}-1\right )}{c_{3} {\mathrm e}^{2 i b \lambda \left (\int x^{k -1} \arctan \left (x \right )^{n}d x \right )}+1} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {i b \,x^{k} \left (c_{3} {\mathrm e}^{2 i b \lambda \left (\int x^{k -1} \arctan \left (x \right )^{n}d x \right )}-1\right )}{c_{3} {\mathrm e}^{2 i b \lambda \left (\int x^{k -1} \arctan \left (x \right )^{n}d x \right )}+1} \\ \end{align*}
Verification of solutions
\[ y = -\frac {i b \,x^{k} \left (c_{3} {\mathrm e}^{2 i b \lambda \left (\int x^{k -1} \arctan \left (x \right )^{n}d x \right )}-1\right )}{c_{3} {\mathrm e}^{2 i b \lambda \left (\int x^{k -1} \arctan \left (x \right )^{n}d x \right )}+1} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x y^{\prime }-\lambda \arctan \left (x \right )^{n} y^{2}-y k =\lambda \,b^{2} x^{2 k} \arctan \left (x \right )^{n} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\lambda \arctan \left (x \right )^{n} y^{2}+y k +\lambda \,b^{2} x^{2 k} \arctan \left (x \right )^{n}}{x} \end {array} \]
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 <- Chini successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 29
dsolve(x*diff(y(x),x)=lambda*arctan(x)^n*y(x)^2+k*y(x)+lambda*b^2*x^(2*k)*arctan(x)^n,y(x), singsol=all)
\[ y \left (x \right ) = -\tan \left (-\lambda b \left (\int x^{-1+k} \arctan \left (x \right )^{n}d x \right )+c_{1} \right ) b \,x^{k} \]
✓ Solution by Mathematica
Time used: 1.992 (sec). Leaf size: 48
DSolve[x*y'[x]==\[Lambda]*ArcTan[x]^n*y[x]^2+k*y[x]+\[Lambda]*b^2*x^(2*k)*ArcTan[x]^n,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \sqrt {b^2} x^k \tan \left (\sqrt {b^2} \int _1^x\lambda \arctan (K[1])^n K[1]^{k-1}dK[1]+c_1\right ) \]