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 .



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

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  • $\begingroup$ 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



  • $\begingroup$ 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

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 $v_0$?

  • $\begingroup$ 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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