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I'm stuck in finding the general solution for this DE because I'm not sure what technique I should use:

$$\frac{dv}{dx}=\frac{g}{v}-\frac{k}{m}$$

Where g, k and m are constants.

I'm not sure if I can really use substitution method here. I've also tried long division but I always end up having a higher degree of the numerator. Partial fractions also doesn't seem to work as the denominator is not factorable when considered as a separable equation.

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  • $\begingroup$ Do you want to find a solution or a differential equation? It is not clear from your question. $\endgroup$ Feb 3, 2017 at 16:21
  • $\begingroup$ Sorry for the confusion. Yes I want to find the general solution for this, I just want to really know what technique I should use. $\endgroup$ Feb 3, 2017 at 16:23
  • $\begingroup$ This isn't the correct equation of motion for a free-falling object. $\endgroup$
    – Mark Viola
    Feb 3, 2017 at 17:31

3 Answers 3

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The correct equation of motion for a free-falling object as written in the OP is incorrect. Rather, we have

$$m\frac{dv}{dt}=mg-kv^2 \tag 1$$

Dividing by $v=\frac{dx}{dt}$ reveals that

$$\frac{dv}{dx}=\frac{g}{v}-\frac{k}m v$$

Then, we have

$$\int_{x_0}^x dx' = \frac{m}{k}\int_{v_0}^v \frac{\nu}{\frac{mg}k-\nu^2}\,d\nu \tag 2$$

where we note that the integration limits cannot embed $v=\sqrt{mg/k}=v_{\infty}$, which is the terminal velocity. Aside, in $(2)$, $v_0$ is the speed of the object when $x=x_0$.

The solution to $(2)$ is given by

$$x=x_0+\frac{m}{2k}\log\left|\frac{v_{\infty}^2-v_0^2}{v_{\infty}^2-v^2}\right| \tag 3$$

Unfortunately, $(3)$ does not provide much insight into the dynamics of the free-falling object. It provides only a relationship between the position of the object and its speed at that position.

It is better to rearrange $(1)$ as

$$\int_{0}^t d\tau =\frac{m}{k}\int_{v(0)}^{v(t)} \frac{1}{v_{\infty}^2-\nu^2}\,d\nu \tag 4$$

We can evaluate the integral on the right-hand side of $(4)$ by enforcing the substitution $\nu = v_{\infty}\tanh(z)$. Proceeding we find that

$$t=\sqrt{\frac{m}{kg}} \left(\text{artanh}(v(t)/v_{\infty})-\text{artanh}(v(0)/v_{\infty})\right)\tag 5$$

Solving $(5)$ for $v(t)$ reveals

$$v(t)=v_{\infty}\left(\frac{\frac{1+v(0)/v_{\infty}}{1-v(0)/v_{\infty}}e^{2t/\sqrt{mg/k}}-1}{\frac{1+v(0)/v_{\infty}}{1-v(0)/v_{\infty}}e^{2t/\sqrt{mg/k}}+1}\right)$$

which provides a more useful form than $(3)$. For example, we see from $(5)$ that if $v(0)>v_{\infty}$, then $v(t)$ decreases to $v_{\infty}$ while if $v(0)<v_{\infty}$, then $v(t)$ increases to $v_{\infty}$.

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark $\endgroup$
    – Mark Viola
    Feb 15, 2017 at 20:10
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This equation is separable. You write:

$$dv = \left(\frac{g}{v} -\frac{k}{m}\right)\cdot dx $$ $$ \Rightarrow \frac{dv}{\frac{g}{v} -\frac{k}{m}} = dx$$ $$ \Rightarrow \int_{v_0}^{v} \frac{dv}{\frac{g}{v} -\frac{k}{m}} = \int_0^xdx$$ $$\Rightarrow \int_{v_0}^{v} \frac{v}{g-\frac{k}{m}v} dv = x $$

Now you have to look up this integral on some integral table and solve for $x$.

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  • $\begingroup$ So I can let some variable u to the denominator then integrate by substitution? $\endgroup$ Feb 3, 2017 at 16:33
  • $\begingroup$ Yes. You can define $u = g-\frac{k}{m}v$. Then you have $du= \frac{k}{m}dv$ and $v=\frac{g - u}{\frac{k}{m}}$. $\endgroup$ Feb 3, 2017 at 16:35
  • $\begingroup$ Is this the only way I can find the general solution or is still there another technique? $\endgroup$ Feb 3, 2017 at 16:37
  • $\begingroup$ The ODE is not the correct equation of motion for a free-falling mass. $\endgroup$
    – Mark Viola
    Feb 4, 2017 at 0:56
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The answer on your question is given by @JakobElias. But I want to give some extra information.

When we have an oject falling vertically and we have drag we can use the following system of equations:

$$ \begin{cases} \text{g}=\frac{\text{d}^2\space\text{h}\left(t\right)}{\text{d}\space t^2}+\frac{\rho\cdot\text{A}\cdot\text{C}_\text{d}}{2\cdot\text{m}}\cdot\left(\frac{\text{d}\space\text{h}\left(t\right)}{\text{d}\space t}\right)^2\\ \\ \text{V}_\text{T}=\sqrt{\frac{2\cdot\text{m}\cdot\text{g}}{\rho\cdot\text{A}\cdot\text{C}_\text{d}}}\\ \\ \text{C}_\text{d}=\frac{\text{F}_\text{d}}{\frac{1}{2}\cdot\rho_\text{m}\cdot\text{u}^2\cdot\text{A}_\text{m}} \end{cases}\tag1 $$

Where $\text{h}$ is the fall distance, $\rho$ is the fluid density, $\text{m}$ is the mass, $\text{C}_\text{d}$ is the drag coefficient, $\text{A}$ is the projected surface area, $\text{g}$ is the gravitational acceleration, $t$ is time, $\text{V}_\text{T}$ is the terminal velocity, $\text{F}_\text{d}$ is the drag force, $\rho_\text{m}$ is the mass density, $\text{u}$ is the characteristic speed and $\text{A}_\text{m}$ is the area.

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