Hier de oplossing compleet met bewijs. De op te lossen vergelijking luidt:
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\( \frac g{k^2}\ln(\frac {u \cos(\theta)-kx}{u \cos(\theta)})+x\tan(\theta)+\frac{gx}{ku\cos(\theta)}=0 \)
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Deel deze vergelijking links en rechts door
\( \frac{\mathrm{g}}{\mathrm{k}^2} \) en laat:
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\( a(\theta) = \frac{\mathrm{k}}{\mathrm{u} \cos(\theta)} \)
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\( b(\theta) = \frac{\mathrm{k}^2 \tan(\theta)}{\mathrm{g}} + \frac{\mathrm{k}}{\mathrm{u} \cos(\theta)} \)
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Dan hebben we:
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\( \ln (1 - a \, x) \, + \, b \, x = 0 \)
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\( \ln (1 - a \, x) \, + \, \frac{ ab \, x }{a} = 0 \)
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\( \ln (1 - a \, x) \, + \, \frac{ ab \, x \, - \, b}{a} = \, - \frac{b}{a} \)
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\( \ln \left ( (1 - a \, x) \, e^{\frac{ ab \, x \, - \, b}{a}} \right ) = \, \ln( e^{- \frac{b}{a} } ) \)
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\( (1 - a \, x) \, e^{\frac{ ab \, x \, - \, b}{a}} = \, e^{- \frac{b}{a} } \)
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\( (a \, x \, - \, 1) \, e^{\frac{ ab \, x \, - \, b}{a}} = \, - e^{- \frac{b}{a} } \)
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\( (ab \, x \, - \, b) \, e^{\frac{ ab \, x \, - \, b}{a}} = \, - b \, e^{- \frac{b}{a} } \)
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\( \frac{ab \, x \, - \, b}{a} \, e^{\frac{ ab \, x \, - \, b}{a}} = \, - \frac{ b \, e^{- \frac{b}{a} }}{a} \)
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\( \frac{ab \, x \, - \, b}{a} \, = \, \mathrm{W} \left ( - \frac{ b \, e^{- \frac{b}{a} }}{a} \right ) \)
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\( ab \, x \, - \, b \, = \, a \cdot \mathrm{W} \left ( - \frac{ b \, e^{- \frac{b}{a} }}{a} \right ) \)
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\( ab \, x \, = \, a \cdot \mathrm{W} \left ( - \frac{ b \, e^{- \frac{b}{a} }}{a} \right ) \, + \, b \)
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\( x \, = \, \frac{a \cdot \mathrm{W} \left ( - \frac{ b \, e^{- b/a }}{a} \right ) \, + \, b}{ab} \)
