Na wat proberen vermoed ik dat dit werkt: [imath] \mathrm{U}(x) = \frac{1}{2} \mathrm{R} \, x^2 \, + \, \frac{\mathrm{R} \alpha}{\beta^2} \cdot \sin(\beta \, x) \, - \, \frac{\mathrm{R} \alpha}{\beta} \cdot x \, \cos(\beta x) [/imath]
Even zien:
[imath] \frac{ \mathrm{d} U }{ \mathrm{d} x } = \mathrm{R} \, x \, + \, \frac{\mathrm{R} \alpha}{\beta} \cdot \cos(\beta \, x) \, - \, \frac{\mathrm{R} \alpha}{\beta} \cdot ( \cos(\beta x) \, - \, x \cdot \sin(\beta x) \cdot \beta) [/imath]
[imath] \frac{ \mathrm{d} U }{ \mathrm{d} x } = \mathrm{R} \, x \, + \, \frac{\mathrm{R} \alpha}{\beta} \cdot \cos(\beta \, x) \, - \, \frac{\mathrm{R} \alpha}{\beta} \cdot \cos(\beta x) \, + \, \frac{\mathrm{R} \alpha}{\beta} \cdot x \cdot \sin(\beta x) \cdot \beta [/imath]
[imath] \frac{ \mathrm{d} U }{ \mathrm{d} x } = \mathrm{R} \, x \, + \, \mathrm{R} \, \alpha \cdot x \cdot \sin(\beta x) [/imath]
[imath] \frac{ \mathrm{d} U }{ \mathrm{d} x } = \mathrm{R} \cdot ( x \, + \, \alpha \cdot x \cdot \sin(\beta x) ) [/imath]
[imath] \frac{ \mathrm{d} U }{ \mathrm{d} x } = \mathrm{R} \cdot (1 \, + \, \alpha \cdot \sin(\beta x) ) \cdot x [/imath]
[imath] - \frac{ \mathrm{d} U }{ \mathrm{d} x } = - \mathrm{R} \cdot (1 \, + \, \alpha \cdot \sin(\beta x) ) \cdot x [/imath]
[imath] \mathrm{F}(x) = - \frac{ \mathrm{d} U }{ \mathrm{d} x } [/imath]
Heel fijn - dus voor [imath] \beta \neq 0 [/imath] levert de voorgestelde functie [imath] U = \mathrm{U}(x) [/imath] een geldige potentiaal.