Please ignore my writing. Here is the question:
"A charged particle having charge $q$ and mass $m$ is projected into a region of electric field $E$ with its intial velocity $v$ perpendicular to $E$. If throughout the motion, the particle experiences a dissipative force of constant magnitude $qE$ and directed opposite to its velocity then find its speed when it has turned through an angle of $90$ degrees, given that magnitude of $v=6$ m/s."
So I wrote the general equations of motion but ended up with a vector differential equation I have no idea how one approaches, the solution to this problem does not utilize this and this question could have been solved via a different method but I got curious on how one approaches such a differential equation to obtain velocity as purely a function of time. WLOG, Assuming that E acts in negative y and intial velocity is in positive x, I wrote these equations
$$\vec{a}=\frac{-qE}{m}\hat{j}-\frac{qE}{m}\hat{v}$$ $$\vec{v}= \int \vec{a} dt$$ $$\vec{v}-6\hat{i}= -\frac{qEt}{m}\vec{j}-\int_{0}^{t}\frac{qE\hat{v}}{m}dt$$
As the question is about direction, parametrize the velocity in polar coordinates, $\vec v = v\hat v$, $v$ a scalar and $\hat v=(\cos\theta,\sin\theta)$ a unit vector. Then for the length $v$ of $\vec v$ one has the differential equation $$ v\dot v=\vec v^T\dot{\vec v}=-v_y-v \\\implies \dot v=-\sin\theta-1, $$ after changing the time scale in such a way that $qE/m=1$. The angle equation is $$ v^2\dot\theta = \det(\vec v,\dot{\vec v}) = -\det(\vec v,\hat j)-\det(\vec v,\hat v)=-v_x=-v\cos\theta \\\implies \dot\theta = -\frac{\cos\theta}{v} $$ Taken together this gives a separable equation $$ \frac{\dot v}{v}=-\frac{1+\sin\theta}{\cos\theta}\dot\theta=-\frac{\cos\theta}{1-\sin\theta}\dot\theta\\ \implies v=C(1-\sin\theta) $$
So initially $\theta=0$ which is then decreasing to negative values. It becomes perpendicular for $\theta=-\frac\pi2$. At that point $v=6m/s=2C$. Thus at $\theta=0$ one has $v=C=3m/s$.