Note: There's a similar problem to this (https://math.stackexchange.com/questions/2796694/how-to-derive-an-equation-for-terminal-velocity-assuming-air-resistance-is-some#:~:text=Taking%20proper%20sign%20of%20air,velocity%20as%20an%20asymptotic%20value.). However, I have a different approach and the above will not solve my problem.
Consider the differential equation $$ma = -mg-cv^2$$ As stated in the above similar problem as well. What I did is differentiation (instead of solving directly). Considering initial conditions, I obtained $$m\frac{da}{dt} = -2cva$$ $$\ln{a} = -2cvt/m+C_1$$ $$a = ge^{\frac{-2cvt}{m}}$$ Assuming $v=v_0$ at $t=0$, integration yields $$v = -\frac{mg}{2cv}e^{\frac{-2cv}{m}t} + (v_0+\frac{mg}{2cv})$$ This suggests that the terminal velocity, which is obtained when $t=\inf$, is dependent on $v_0$, inconsistent with theory.
Where did I do wrong?
This differential equation does not have a terminal velocity, the acceleration is always greater than $g$ in downward direction, thus the velocity in that direction always increasing.
Obviously this is not physical. For the physical equation the friction force has to be in the opposite direction to the velocity, so $$ ma = -mg - c|v|v=-mg-cv^2\hat v $$ where in the one-dimensional case the direction unit vector is $\hat v=sign(v)$.