60.1.22 problem 22
Internal
problem
ID
[10036]
Book
:
Differential
Gleichungen,
E.
Kamke,
3rd
ed.
Chelsea
Pub.
NY,
1948
Section
:
Chapter
1,
linear
first
order
Problem
number
:
22
Date
solved
:
Monday, January 27, 2025 at 06:19:02 PM
CAS
classification
:
[_Riccati]
\begin{align*} y^{\prime }-y^{2}-y \sin \left (2 x \right )-\cos \left (2 x \right )&=0 \end{align*}
✓ Solution by Maple
Time used: 0.004 (sec). Leaf size: 96
dsolve(diff(y(x),x) - y(x)^2 -y(x)*sin(2*x) - cos(2*x)=0,y(x), singsol=all)
\[
y = \frac {\sin \left (x \right ) \left (\operatorname {HeunC}\left (1, \frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right ) c_{1} +2 \cos \left (x \right ) \left (\operatorname {HeunCPrime}\left (1, \frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right ) \cos \left (x \right ) c_{1} +\operatorname {HeunCPrime}\left (1, -\frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right )\right )\right )}{c_{1} \cos \left (x \right ) \operatorname {HeunC}\left (1, \frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right )+\operatorname {HeunC}\left (1, -\frac {1}{2}, -\frac {1}{2}, -1, \frac {7}{8}, \frac {\cos \left (2 x \right )}{2}+\frac {1}{2}\right )}
\]
✓ Solution by Mathematica
Time used: 2.100 (sec). Leaf size: 238
DSolve[D[y[x],x] - y[x]^2 -y[x]*Sin[2*x] - Cos[2*x]==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*}
y(x)\to \frac {\tan (x) \left (\int _1^{\cos (x)}\frac {\exp \left (-2 \int _1^{K[3]}\frac {K[1]-2 K[1]^3}{2-2 K[1]^2}dK[1]\right )}{K[3]^2}dK[3]+\sec (x) \exp \left (-2 \int _1^{\cos (x)}\frac {K[1]-2 K[1]^3}{2-2 K[1]^2}dK[1]\right )+c_1\right )}{\int _1^{\cos (x)}\frac {\exp \left (-2 \int _1^{K[3]}\frac {K[1]-2 K[1]^3}{2-2 K[1]^2}dK[1]\right )}{K[3]^2}dK[3]+c_1} \\
y(x)\to \tan (x) \\
y(x)\to \frac {\tan (x) \sec (x) \exp \left (-2 \int _1^{\cos (x)}\frac {K[1]-2 K[1]^3}{2-2 K[1]^2}dK[1]\right )}{\int _1^{\cos (x)}\frac {\exp \left (-2 \int _1^{K[3]}\frac {K[1]-2 K[1]^3}{2-2 K[1]^2}dK[1]\right )}{K[3]^2}dK[3]}+\tan (x) \\
\end{align*}