61.11.11 problem 37
Internal
problem
ID
[12211]
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
:
37
Date
solved
:
Tuesday, January 28, 2025 at 07:30:18 PM
CAS
classification
:
[_Riccati]
\begin{align*} \left (a \tan \left (\lambda x \right )+b \right ) y^{\prime }&=y^{2}+k \tan \left (\mu x \right ) y-d^{2}+k d \tan \left (\mu x \right ) \end{align*}
✓ Solution by Maple
Time used: 0.009 (sec). Leaf size: 351
dsolve((a*tan(lambda*x)+b)*diff(y(x),x)=y(x)^2+k*tan(mu*x)*y(x)-d^2+k*d*tan(mu*x),y(x), singsol=all)
\[
y = \frac {-\left (\sec \left (\lambda x \right )^{2}\right )^{\frac {a d}{\lambda \left (a^{2}+b^{2}\right )}} \left (a \tan \left (\lambda x \right )+b \right )^{-\frac {2 a d}{\lambda \left (a^{2}+b^{2}\right )}} {\mathrm e}^{\frac {\lambda k \left (a^{2}+b^{2}\right ) \left (\int \frac {\tan \left (\mu x \right )}{a \tan \left (\lambda x \right )+b}d x \right )-2 \arctan \left (\tan \left (\lambda x \right )\right ) b d}{\lambda \left (a^{2}+b^{2}\right )}}-d \left (\int \left (a \tan \left (\lambda x \right )+b \right )^{\frac {\left (-a^{2}-b^{2}\right ) \lambda -2 a d}{\lambda \left (a^{2}+b^{2}\right )}} \left (\sec \left (\lambda x \right )^{2}\right )^{\frac {a d}{\lambda \left (a^{2}+b^{2}\right )}} {\mathrm e}^{\frac {\lambda k \left (a^{2}+b^{2}\right ) \left (\int \frac {\tan \left (\mu x \right )}{a \tan \left (\lambda x \right )+b}d x \right )-2 \arctan \left (\tan \left (\lambda x \right )\right ) b d}{\lambda \left (a^{2}+b^{2}\right )}}d x -c_{1} \right )}{\int \left (a \tan \left (\lambda x \right )+b \right )^{\frac {\left (-a^{2}-b^{2}\right ) \lambda -2 a d}{\lambda \left (a^{2}+b^{2}\right )}} \left (\sec \left (\lambda x \right )^{2}\right )^{\frac {a d}{\lambda \left (a^{2}+b^{2}\right )}} {\mathrm e}^{\frac {\lambda k \left (a^{2}+b^{2}\right ) \left (\int \frac {\tan \left (\mu x \right )}{a \tan \left (\lambda x \right )+b}d x \right )-2 \arctan \left (\tan \left (\lambda x \right )\right ) b d}{\lambda \left (a^{2}+b^{2}\right )}}d x -c_{1}}
\]
✓ Solution by Mathematica
Time used: 26.191 (sec). Leaf size: 800
DSolve[(a*Tan[\[Lambda]*x]+b)*D[y[x],x]==y[x]^2+k*Tan[\[Mu]*x]*y[x]-d^2+k*d*Tan[\[Mu]*x],y[x],x,IncludeSingularSolutions -> True]
\[
\text {Solve}\left [\int _1^x\frac {e^{-\int _1^{K[2]}\frac {\sec (\mu K[1]) (2 d \cos (\lambda K[1]-\mu K[1])+2 d \cos (\lambda K[1]+\mu K[1])+k \sin (\lambda K[1]-\mu K[1])-k \sin (\lambda K[1]+\mu K[1]))}{2 (b \cos (\lambda K[1])+a \sin (\lambda K[1]))}dK[1]} (d \cos (\lambda K[2]-\mu K[2])-y(x) \cos (\lambda K[2]-\mu K[2])+d \cos (\lambda K[2]+\mu K[2])+k \sin (\lambda K[2]-\mu K[2])-k \sin (\lambda K[2]+\mu K[2])-\cos (\lambda K[2]+\mu K[2]) y(x))}{k \mu (b \cos (\lambda K[2]-\mu K[2])+b \cos (\lambda K[2]+\mu K[2])+a \sin (\lambda K[2]-\mu K[2])+a \sin (\lambda K[2]+\mu K[2])) (d+y(x))}dK[2]+\int _1^{y(x)}\left (\frac {e^{-\int _1^x\frac {\sec (\mu K[1]) (2 d \cos (\lambda K[1]-\mu K[1])+2 d \cos (\lambda K[1]+\mu K[1])+k \sin (\lambda K[1]-\mu K[1])-k \sin (\lambda K[1]+\mu K[1]))}{2 (b \cos (\lambda K[1])+a \sin (\lambda K[1]))}dK[1]}}{k \mu (d+K[3])^2}-\int _1^x\left (\frac {e^{-\int _1^{K[2]}\frac {\sec (\mu K[1]) (2 d \cos (\lambda K[1]-\mu K[1])+2 d \cos (\lambda K[1]+\mu K[1])+k \sin (\lambda K[1]-\mu K[1])-k \sin (\lambda K[1]+\mu K[1]))}{2 (b \cos (\lambda K[1])+a \sin (\lambda K[1]))}dK[1]} (-\cos (\lambda K[2]-\mu K[2])-\cos (\lambda K[2]+\mu K[2]))}{k \mu (d+K[3]) (b \cos (\lambda K[2]-\mu K[2])+b \cos (\lambda K[2]+\mu K[2])+a \sin (\lambda K[2]-\mu K[2])+a \sin (\lambda K[2]+\mu K[2]))}-\frac {e^{-\int _1^{K[2]}\frac {\sec (\mu K[1]) (2 d \cos (\lambda K[1]-\mu K[1])+2 d \cos (\lambda K[1]+\mu K[1])+k \sin (\lambda K[1]-\mu K[1])-k \sin (\lambda K[1]+\mu K[1]))}{2 (b \cos (\lambda K[1])+a \sin (\lambda K[1]))}dK[1]} (d \cos (\lambda K[2]-\mu K[2])-K[3] \cos (\lambda K[2]-\mu K[2])+d \cos (\lambda K[2]+\mu K[2])-\cos (\lambda K[2]+\mu K[2]) K[3]+k \sin (\lambda K[2]-\mu K[2])-k \sin (\lambda K[2]+\mu K[2]))}{k \mu (d+K[3])^2 (b \cos (\lambda K[2]-\mu K[2])+b \cos (\lambda K[2]+\mu K[2])+a \sin (\lambda K[2]-\mu K[2])+a \sin (\lambda K[2]+\mu K[2]))}\right )dK[2]\right )dK[3]=c_1,y(x)\right ]
\]