60.2.365 problem 942

Internal problem ID [10952]
Book : Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section : Chapter 1, Additional non-linear first order
Problem number : 942
Date solved : Monday, January 27, 2025 at 10:30:33 PM
CAS classification : [[_1st_order, _with_linear_symmetries]]

\begin{align*} y^{\prime }&=-\frac {y^{2}+2 y x +x^{2}+{\mathrm e}^{\frac {2 \left (x -y\right )^{3} \left (x +y\right )^{3}}{-y^{2}+x^{2}-1}}}{-y^{2}-2 y x -x^{2}+{\mathrm e}^{\frac {2 \left (x -y\right )^{3} \left (x +y\right )^{3}}{-y^{2}+x^{2}-1}}} \end{align*}

✓ Solution by Maple

Time used: 0.254 (sec). Leaf size: 43

dsolve(diff(y(x),x) = -(y(x)^2+2*x*y(x)+x^2+exp(2*(x-y(x))^3*(x+y(x))^3/(-y(x)^2+x^2-1)))/(-y(x)^2-2*x*y(x)-x^2+exp(2*(x-y(x))^3*(x+y(x))^3/(-y(x)^2+x^2-1))),y(x), singsol=all)
 

\[ y = {\mathrm e}^{\operatorname {RootOf}\left (-\textit {\_Z} +\int _{}^{{\mathrm e}^{2 \textit {\_Z}}-2 x \,{\mathrm e}^{\textit {\_Z}}}\frac {1}{{\mathrm e}^{\frac {2 \textit {\_a}^{3}}{\textit {\_a} +1}}+\textit {\_a}}d \textit {\_a} +c_{1} \right )}-x \]

✓ Solution by Mathematica

Time used: 2.307 (sec). Leaf size: 349

DSolve[D[y[x],x] == (-E^((2*(x - y[x])^3*(x + y[x])^3)/(-1 + x^2 - y[x]^2)) - x^2 - 2*x*y[x] - y[x]^2)/(E^((2*(x - y[x])^3*(x + y[x])^3)/(-1 + x^2 - y[x]^2)) - x^2 - 2*x*y[x] - y[x]^2),y[x],x,IncludeSingularSolutions -> True]
 

\[ \text {Solve}\left [\int _1^{y(x)}\left (-\frac {2 K[2]}{-x^2+\exp \left (\frac {2 (x-K[2])^3 (x+K[2])^3}{x^2-K[2]^2-1}\right )+K[2]^2}-\int _1^x\left (\frac {2 K[1] \left (-2 K[2]-\exp \left (\frac {2 (K[1]-K[2])^3 (K[1]+K[2])^3}{K[1]^2-K[2]^2-1}\right ) \left (\frac {6 (K[1]+K[2])^2 (K[1]-K[2])^3}{K[1]^2-K[2]^2-1}+\frac {4 K[2] (K[1]+K[2])^3 (K[1]-K[2])^3}{\left (K[1]^2-K[2]^2-1\right )^2}-\frac {6 (K[1]+K[2])^3 (K[1]-K[2])^2}{K[1]^2-K[2]^2-1}\right )\right )}{\left (K[1]^2-\exp \left (\frac {2 (K[1]-K[2])^3 (K[1]+K[2])^3}{K[1]^2-K[2]^2-1}\right )-K[2]^2\right )^2}-\frac {1}{(K[1]+K[2])^2}\right )dK[1]+\frac {1}{x+K[2]}\right )dK[2]+\int _1^x\left (\frac {1}{K[1]+y(x)}-\frac {2 K[1]}{K[1]^2-\exp \left (\frac {2 (K[1]-y(x))^3 (K[1]+y(x))^3}{K[1]^2-y(x)^2-1}\right )-y(x)^2}\right )dK[1]=c_1,y(x)\right ] \]