I have to find the ODE of an object movement with mass $m$ and acceleration $a\neq0$ given $$x(0)=x'(0)=0$$ and there's a force against the movement $$F=\alpha\cdot x'$$ with $\alpha\in\mathbb{R},\alpha\neq0$.
The only answer I can think about is $$m\cdot x''=\alpha x'$$ but with these initial conditions, I'd have $$x(t)=0$$ which it doesn't make any sense.
The force opposes the motion so the correct equation is: $$ mx''=-\alpha x' $$ or: $$ \frac{x''}{x'}=-\frac{\alpha }{m} $$ which can be written as: $$ (\ln x')'=-\frac{\alpha }{m} $$ With a 1st integral: $$x'(t)=x'_0e^{-\frac{\alpha}{m}(t-t_0)} $$ and a second integral $$x (t)=x_0+x'_0\frac{m}{\alpha}\big[1-e^{-\frac{\alpha}{m}(t-t_0)}\big] $$ If there is no motion at $t=t_0$, that is $x'_0=0$, you'll get $x(t)=x_0$