61.11.7 problem 33

Internal problem ID [12207]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 1, section 1.2. Riccati Equation. subsection 1.2.6-3. Equations with tangent.
Problem number : 33
Date solved : Tuesday, January 28, 2025 at 01:07:44 AM
CAS classification : [_Riccati]

\begin{align*} y^{\prime }&=-\left (k +1\right ) x^{k} y^{2}+a \,x^{k +1} \tan \left (x \right )^{m} y-a \tan \left (x \right )^{m} \end{align*}

Solution by Maple

Time used: 0.037 (sec). Leaf size: 174

dsolve(diff(y(x),x)=-(k+1)*x^k*y(x)^2+a*x^(k+1)*tan(x)^m*y(x)-a*tan(x)^m,y(x), singsol=all)
 
\[ y = \frac {x^{-k -1} \left (x^{k +1} {\mathrm e}^{\int \frac {x^{k +1} \tan \left (x \right )^{m} a x -2 k -2}{x}d x}+\left (\int x^{k} {\mathrm e}^{\int \frac {x^{k +1} \tan \left (x \right )^{m} a x -2 k -2}{x}d x}d x \right ) k +\int x^{k} {\mathrm e}^{\int \frac {x^{k +1} \tan \left (x \right )^{m} a x -2 k -2}{x}d x}d x -c_{1} \right )}{\left (\int x^{k} {\mathrm e}^{\int \frac {a \,x^{k +2} \tan \left (x \right )^{m}-2 k -2}{x}d x}d x \right ) k +\int x^{k} {\mathrm e}^{\int \frac {a \,x^{k +2} \tan \left (x \right )^{m}-2 k -2}{x}d x}d x -c_{1}} \]

Solution by Mathematica

Time used: 3.568 (sec). Leaf size: 248

DSolve[D[y[x],x]==-(k+1)*x^k*y[x]^2+a*x^(k+1)*Tan[x]^m*y[x]-a*Tan[x]^m,y[x],x,IncludeSingularSolutions -> True]
 
\begin{align*} y(x)\to \frac {x^{-k-1} \left (c_1 x \exp \left (\int _1^x-\frac {-a \tan ^m(K[1]) K[1]^{k+2}+k+2}{K[1]}dK[1]\right )+c_1 (k+1) \int _1^x\exp \left (\int _1^{K[2]}-\frac {-a \tan ^m(K[1]) K[1]^{k+2}+k+2}{K[1]}dK[1]\right )dK[2]+k+1\right )}{(k+1) \left (1+c_1 \int _1^x\exp \left (\int _1^{K[2]}-\frac {-a \tan ^m(K[1]) K[1]^{k+2}+k+2}{K[1]}dK[1]\right )dK[2]\right )} \\ y(x)\to \frac {x^{-k} \left (\frac {\exp \left (\int _1^x-\frac {-a \tan ^m(K[1]) K[1]^{k+2}+k+2}{K[1]}dK[1]\right )}{\int _1^x\exp \left (\int _1^{K[2]}-\frac {-a \tan ^m(K[1]) K[1]^{k+2}+k+2}{K[1]}dK[1]\right )dK[2]}+\frac {k+1}{x}\right )}{k+1} \\ \end{align*}