$\mathrm{DE1}:=\left\{\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)\=y\left(x\right)z\left(x\right)\,\frac{\ⅆ}{\ⅆx}z\left(x\right)\=y\left(x\right)x\right\}$

$\mathrm{DEplot3d}\left(\mathrm{DE1}\,\left[y\left(x\right)\,z\left(x\right)\right]\,x\=-2..2\,\left[\left[y\left(0\right)\=1\,\mathrm{D}\left(y\right)\left(0\right)\=2\,z\left(0\right)\=1\right]\right]\,y\=-4..4\,\mathrm{obsrange}\=\mathrm{true}\,\mathrm{stepsize}\=0.05\,\mathrm{iterations}\=3\,\mathrm{orientation}\=\left[-124\,72\right]\right)$

$\mathrm{DE1}≔\left\{\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)=y\left(x\right)z\left(x\right)\,\frac{\ⅆ}{\ⅆx}z\left(x\right)=y\left(x\right)x\right\}$

$\mathrm{DE6}:=\left\{\mathrm{D}\left(x\right)\left(t\right)\=y\left(t\right)\,\mathrm{D}\left(y\right)\left(t\right)\=-x\left(t\right)-y\left(t\right)\right\}$

$\mathrm{DEplot3d}\left(\mathrm{DE6}\,\left[x\left(t\right)\,y\left(t\right)\right]\,t\=0..10\,\left[\left[x\left(0\right)\=0\,y\left(0\right)\=1\right]\,\left[x\left(0\right)\=0\,y\left(0\right)\=0.5\right]\right]\,\mathrm{scene}\=\left[t\,x\left(t\right)\,y\left(t\right)\right]\,\mathrm{stepsize}\=0.1\,\mathrm{linecolour}\=t-\sqrt{t}\,\mathrm{method}\=\mathrm{classical}\left[\mathrm{rk2}\right]\right)$

$\mathrm{DE6}≔\left\{\mathrm{D}\left(x\right)\left(t\right)=y\left(t\right)\,\mathrm{D}\left(y\right)\left(t\right)=-x\left(t\right)-y\left(t\right)\right\}$

$\mathrm{DE7}:=\left\{\mathrm{D}\left(x\right)\left(t\right)\=y\left(t\right)\,\mathrm{D}\left(y\right)\left(t\right)\=-x\left(t\right)-y\left(t\right)\right\}$

$\mathrm{DEplot3d}\left(\mathrm{DE7}\,\left[x\left(t\right)\,y\left(t\right)\right]\,t\=0..10\,\left[\left[x\left(0\right)\=0\,y\left(0\right)\=1\right]\,\left[x\left(0\right)\=0\,y\left(0\right)\=0.5\right]\right]\,\mathrm{scene}\=\left[t\,x\left(t\right)\,y\left(t\right)\right]\,\mathrm{stepsize}\=0.1\,\mathrm{linecolor}\=t-\sqrt{t}\,\mathrm{method}\=\mathrm{dverk78}\right)$

$\mathrm{DE7}≔\left\{\mathrm{D}\left(x\right)\left(t\right)=y\left(t\right)\,\mathrm{D}\left(y\right)\left(t\right)=-x\left(t\right)-y\left(t\right)\right\}$

$\mathrm{DEplot3d}\left(\left\{\frac{\ⅆ}{\ⅆt}x\left(t\right)\=y\left(t\right)\,\frac{\ⅆ}{\ⅆt}y\left(t\right)\=-\sin \left(x\left(t\right)\right)-\frac{y\left(t\right)}{10}\right\}\,\left[x\left(t\right)\,y\left(t\right)\right]\,t\=-10..10\,\left[\left[x\left(0\right)\=1\,y\left(0\right)\=1\right]\right]\,\mathrm{stepsize}\=0.15\,x\=-1..1.5\,\mathrm{obsrange}\=\mathrm{false}\right)$

$\mathrm{DEplot3d}\left(\left\{\frac{\ⅆ}{\ⅆt}x\left(t\right)\=y\left(t\right)\,\frac{\ⅆ}{\ⅆt}y\left(t\right)\=-\sin \left(x\left(t\right)\right)-\frac{y\left(t\right)}{10}\right\}\,\left[x\left(t\right)\,y\left(t\right)\right]\,t\=-10..10\,\left[\left[x\left(0\right)\=1\,y\left(0\right)\=1\right]\right]\,\mathrm{stepsize}\=0.15\,x\=-1..1.5\,\mathrm{obsrange}\=\mathrm{true}\right)$

$$\begin{gathered} \mathrm{DEplot3d}\left(\left\{\frac{\ⅆ}{\ⅆt}x\left(t\right)\=y\left(t\right)\,\frac{\ⅆ}{\ⅆt}y\left(t\right)\=-\sin \left(x\left(t\right)\right)\right\}\,\left[x\left(t\right)\,y\left(t\right)\right]\,t\=0..10\, \left[\left[x\left(0\right)\=0\,y\left(0\right)\=0.5\right]\,\left[x\left(0\right)\=0\,y\left(0\right)\=1\right]\, \\ \left[x\left(0\right)\=0\,y\left(0\right)\=1.8\right]\, \left[x\left(0\right)\=-2\mathrm{\pi }\,y\left(0\right)\=1\right]\,\left[x\left(0\right)\=2\mathrm{\pi }\,y\left(0\right)\=0.5\right]\,\left[x\left(0\right)\=-2\mathrm{\pi }\,y\left(0\right)\=2.1\right]\, \\ \left[x\left(0\right)\=2 \mathrm{\π}\,y\left(0\right)\=-2.1\right]\right]\, \mathrm{stepsize}\=0.2\,\mathrm{title}\=''Pendulum\; Vibrations''\,\mathrm{orientation}\=\left[0\,90\right]\,\mathrm{method}\=\mathrm{classical}\left[\mathrm{abmoulton}\right]\,\mathrm{corrections}\=3\,\mathrm{axes}\=\mathrm{normal}\,\mathrm{linecolor}\=\sin \left(t\right)-t\right) \end{gathered}$$