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How do I find the solution of "$v$" from the following differential equation:

$$\frac{dv_x}{dt}= \alpha v$$ Where, $v_x$=velocity in x direction.

$v$=resultant velocity of the body .

$\alpha=constant$.

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closed as unclear what you're asking by Yves Daoust, Clarinetist, user284331, uniquesolution, Rodrigo de Azevedo Jan 2 '18 at 20:31

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ Are you sure you know nothing about the relation of $v_1, v_2$? Do you know something about the position $s$ (where $v=\dot s$)? Perhaps you know acceleration is only horizontal for some reason? Or $v_2$ is changed by gravity? Otherwise one would have to assume that $v_2$ is a constant, or that $v_1(t)=v_1(0)-\alpha \int_0^t v_2(\tau)\mathrm d \tau$. $\endgroup$ – Dan Robertson Jan 2 '18 at 15:05
  • $\begingroup$ I doubt that you can say more than $v_2=-(1/\alpha)({\rm d}v_1/{\rm d}t)$ because the acceleration in the $x$-direction is independent from $v_2$. $\endgroup$ – Gerhard S. Jan 2 '18 at 15:05
  • $\begingroup$ Unless you have more relation between them... We have acceleration in the $x$ direction depends on $y$ which implies that the place in the $y$ direction depends on $x$ so I guess that this question supposed to be answered using energy and not force $\endgroup$ – Holo Jan 2 '18 at 15:22
  • $\begingroup$ Silently changing the statement is a bad idea. It makes previous comments and answers nonsensical. $\endgroup$ – Yves Daoust Jan 2 '18 at 15:31
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Trivially,

$$v=\frac1\alpha\frac{dv_0}{dt}.$$

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  • $\begingroup$ I suppose we need to find $v_2(t)$..! $\endgroup$ – samjoe Jan 2 '18 at 15:14
  • $\begingroup$ @samjoe: oooops, right. This is trivial then ! $\endgroup$ – Yves Daoust Jan 2 '18 at 15:16
  • $\begingroup$ Yes haha, I don't know what Op is upto. cheers :) $\endgroup$ – samjoe Jan 2 '18 at 15:16
  • $\begingroup$ @samjoe I'm guessing he is trying to find relation between energy and force... Or OP just forgot some information $\endgroup$ – Holo Jan 2 '18 at 15:28
  • $\begingroup$ @Holo: let me doubt that. I see no kinetic nor potential energy terms, and a force proportional to a speed evokes viscous friction. $\endgroup$ – Yves Daoust Jan 2 '18 at 15:34
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Yes, If you are solving for $v$ then you need not integrate because it is isolated and is not being differentiated. Dividing by $\alpha$ will yield the solution

$$v = \frac{1}{\alpha} \frac{dv_0}{dt},$$

Did you mean you wanted to solve for $v_0$?

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  • $\begingroup$ α is a constant here,not acceleration... and the body moves in my plane only while force acts along z direction... $\endgroup$ – Nehal Samee Jan 2 '18 at 18:38

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