11.8   ODE No. 1920

$$\boxed{{\displaystyle \{\frac{\mathrm{d}}{\mathrm{d}t}x\left(t\right)=-y\left(t\right)+x\left(t\right)({\left(x\left(t\right)\right)}^2+{\left(y\left(t\right)\right)}^2-1),\frac{\mathrm{d}}{\mathrm{d}t}y\left(t\right)=x\left(t\right)+y\left(t\right)({\left(x\left(t\right)\right)}^2+{\left(y\left(t\right)\right)}^2-1)\}}}$$

1. Problem in Latex
2. Mathematica input
3. Maple input

Mathematica: cpu = 0.081510 (sec), leaf count = 52 $$\text{DSolve}[\{x^{\prime }(t)=x(t)(x(t{)}^2+y(t{)}^2-1)-y(t),y^{\prime }(t)=y(t)(x(t{)}^2+y(t{)}^2-1)+x(t)\},\{x(t),y(t)\},t]$$

Maple: cpu = 2.294 (sec), leaf count = 250 $$\{[\{x\left(t\right)=0\},\{y\left(t\right)=0\}],[\{x\left(t\right)=\mathit{O}\mathit{D}\mathit{E}\mathit{S}\mathit{o}\mathit{l}\mathit{S}\mathit{t}\mathit{r}\mathit{u}\mathit{c}(\mathit{\_}\mathit{a},[\{\left(\frac{\mathrm{d}}{\mathrm{d}\mathit{\_}\mathit{a}}\mathit{\_}\mathit{b}\left(\mathit{\_}\mathit{a}\right)\right)\mathit{\_}\mathit{b}\left(\mathit{\_}\mathit{a}\right)-\frac{1}{2{\mathit{\_}\mathit{a}}^3}(6{\left(\mathit{\_}\mathit{b}\left(\mathit{\_}\mathit{a}\right)\right)}^2{\mathit{\_}\mathit{a}}^2+4{\mathit{\_}\mathit{a}}^3\mathit{\_}\mathit{b}\left(\mathit{\_}\mathit{a}\right)-4{\mathit{\_}\mathit{a}}^4+6\mathit{\_}\mathit{a}\mathit{\_}\mathit{b}\left(\mathit{\_}\mathit{a}\right)+4{\mathit{\_}\mathit{a}}^2+1-\sqrt{-64{\mathit{\_}\mathit{a}}^6{\left(\mathit{\_}\mathit{b}\left(\mathit{\_}\mathit{a}\right)\right)}^2-64\mathit{\_}\mathit{b}\left(\mathit{\_}\mathit{a}\right){\mathit{\_}\mathit{a}}^7-16{\mathit{\_}\mathit{a}}^8+64{\mathit{\_}\mathit{a}}^3{\left(\mathit{\_}\mathit{b}\left(\mathit{\_}\mathit{a}\right)\right)}^3+128{\mathit{\_}\mathit{a}}^4{\left(\mathit{\_}\mathit{b}\left(\mathit{\_}\mathit{a}\right)\right)}^2+48{\mathit{\_}\mathit{a}}^5\mathit{\_}\mathit{b}\left(\mathit{\_}\mathit{a}\right)+48{\left(\mathit{\_}\mathit{b}\left(\mathit{\_}\mathit{a}\right)\right)}^2{\mathit{\_}\mathit{a}}^2+64{\mathit{\_}\mathit{a}}^3\mathit{\_}\mathit{b}\left(\mathit{\_}\mathit{a}\right)+16{\mathit{\_}\mathit{a}}^4+12\mathit{\_}\mathit{a}\mathit{\_}\mathit{b}\left(\mathit{\_}\mathit{a}\right)+8{\mathit{\_}\mathit{a}}^2+1})=0\},\{\mathit{\_}\mathit{a}=x\left(t\right),\mathit{\_}\mathit{b}\left(\mathit{\_}\mathit{a}\right)=\frac{\mathrm{d}}{\mathrm{d}t}x\left(t\right)\},\{t=\int {\left(\mathit{\_}\mathit{b}\left(\mathit{\_}\mathit{a}\right)\right)}^{-1}\mathrm{d}\mathit{\_}\mathit{a}+\mathit{\_}\mathit{C}\mathit{1},x\left(t\right)=\mathit{\_}\mathit{a}\}])\},\{y\left(t\right)=\frac{\left(\frac{{\mathrm{d}}^2}{\mathrm{d}{t}^2}x\left(t\right)\right){\left(x\left(t\right)\right)}^2+2{\left(x\left(t\right)\right)}^3-2{\left(x\left(t\right)\right)}^2\frac{\mathrm{d}}{\mathrm{d}t}x\left(t\right)-3x\left(t\right){\left(\frac{\mathrm{d}}{\mathrm{d}t}x\left(t\right)\right)}^2-x\left(t\right)-\frac{\mathrm{d}}{\mathrm{d}t}x\left(t\right)}{2{\left(x\left(t\right)\right)}^2+4x\left(t\right)\frac{\mathrm{d}}{\mathrm{d}t}x\left(t\right)+1}\}]\}$$