Differential Equation (Falling body with fluid drag square of velocity)

Asked Aug 29 '20 at 03:57

Active Aug 29 '20 at 04:34

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I recently solved a differential equation for a body in free-fall. $$\frac{dv}{dt} + \left(\frac km\right)v = g$$

This is first-degree, non-separable with an integrating factor.

However, most physicists know that the fluid drag $\left(\frac km\right)v$ goes as the square of the velocity. The equation takes the form: $$\frac{dv}{dt} + \left(\frac km\right)v^2 = g$$

How do you solve this equation, and what type is it? It doesn't appear to be a "Bernoulli".

edited Aug 29 '20 at 04:34

Aiden Chow

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asked Aug 29 '20 at 03:57

Frederick

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It is separable: $\frac{dv}{g-\left(\frac km\right)v^2}= dt$. Integrate both sides. – Conifold Aug 29 '20 at 04:50
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Yes @Conifold is right. It is indeed separable. You can use $\int{\frac{1}{a^2-x^2}dx}=\frac{ln(|\frac{x+a}{x-a}|)}{2a}$ – DatBoi Aug 29 '20 at 04:55
Thank you. I should have caught this. LOL – Frederick Aug 30 '20 at 05:46

Differential Equation (Falling body with fluid drag square of velocity)

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