This requires a bit of setup, so bear with me. I'm working through a physics problem. The physics itself is irrelevant, but the setup gives you a system of differential equations. The details in how I solve it are also not very important. My question pertains to the last part. You can skip ahead if you want; I only list my work to be exhaustive, in case it does matter. I'm reasonably confident that my work is right.
$$m\dot{v_x} = qE_0+\frac{q}{c}v_yB_0 \tag{1}$$ $$m\dot{v_y} = -\frac{q}{c}v_xB_0 \tag{2}$$
To solve this, I differentiate 2 and solve for $\dot{v_x}$:
$$v_x = -\frac{mc}{qB_0}\dot{v_y} \tag{3}$$ $$\rightarrow \dot{v_x} = -\frac{mc}{qB_0}\ddot{v_y} \tag{4}$$
Plugging this into 2, I get:
$$m(-\frac{mc}{qB_0}\ddot{v_y}) = qE_0+\frac{q}{c}v_yB_0$$ $$\rightarrow \ddot{v_y} + \frac{q^2B_0^2}{m^2c^2}v_y = -\frac{q^2E_0B_0}{m^2c} \tag{5}$$
The homogenous solution vs a choice of particular solution yields:
$$v_y = A\cos{(\frac{qB_0}{mc}t)} + B\sin{(\frac{qB_0}{mc}t)} + C \tag{6}$$
Differentiating (to check the validity of the homogenous solution) and plugging that into 4 to solve for the particular solution:
$$\ddot{v_y} = -\frac{q^2B_0^2}{m^2c^2}(A\cos{(\frac{qB_0}{mc}t)} + B\sin{(\frac{qB_0}{mc}t)}) \\ \rightarrow -\frac{q^2B_0^2}{m^2c^2}(A\cos{(\frac{qB_0}{mc}t)} + B\sin{(\frac{qB_0}{mc}t)}) + \frac{q^2B_0^2}{m^2c^2}(A\cos{(\frac{qB_0}{mc}t)} + B\sin{(\frac{qB_0}{mc}t)} + C) = -\frac{q^2E_0B_0}{m^2c} \\ \frac{q^2B_0^2}{m^2c^2}C = -\frac{q^2E_0B_0}{m^2c}$$ $$\rightarrow C = -\frac{E_0c}{B_0} \tag{7}$$
From here, the initial conditions are:
$$v_x(0) = v_y(0) = 0 \tag{8}$$
Combining $(6)$, $(7)$, and $(8)$:
$$v_y(0) = A\cos{(\frac{qB_0}{mc}(0))} + B\sin{(\frac{qB_0}{mc}(0))} - \frac{E_0c}{B_0} \\ = A - \frac{E_0c}{B_0} = 0$$ $$\rightarrow A = \frac{E_0c}{B_0} \tag{9}$$
From here is the crux of the issue. I can now solve for $v_x$ in two different ways, and this presents my issue. The first way is using $(3)$:
$$v_y = \frac{E_0c}{B_0}\cos{(\frac{qB_0}{mc}t)} + B\sin{(\frac{qB_0}{mc}t)} - \frac{E_0c}{B_0} \\ \rightarrow \dot{v_y} = -\frac{qE_0}{m}\sin{(\frac{qB_0}{mc}t)} + B\frac{qB_0}{mc}\cos{(\frac{qB_0}{mc}t)} \\ \rightarrow v_x = -\frac{mc}{qB_0}(-\frac{qE_0}{m}\sin{(\frac{qB_0}{mc}t)} + B\frac{qB_0}{mc}\cos{(\frac{qB_0}{mc}t)})$$ $$\rightarrow v_x = \frac{E_0c}{B_0}\sin{(\frac{qB_0}{mc}t)} - B\cos{(\frac{qB_0}{mc}t)}) \tag{10}$$
From $(10)$, it's plain to see that $B$ would be $0$. However, an alternative approach using $(1)$ yields a contradiction:
$$m\dot{v_x} = qE_0 + \frac{q}{c}v_yB_0$$ $$\rightarrow mv_x = qE_0t + \frac{qB_0}{c}\int(v_y(t))dt \tag{11} \\ = qE_0t + \frac{qB_0}{c}(\frac{E_0mc^2}{qB_0^2}\sin{(\frac{qB_0}{mc}t)} - B\frac{mc}{qB_0}\cos{(\frac{qB_0}{mc}t)} - \frac{E_0c}{B_0}t) + D$$ $$\rightarrow v_x = \frac{E_0c}{B_0}\sin{(\frac{qB_0}{mc}t)} - B\cos{(\frac{qB_0}{mc}t)} + D \tag{12}$$
These two equations, $(10)$ and $(12)$, are the exact same aside from the integration constant $D$, which begs the question - should $(11)$ have contained the definite integral $\int_0^t(v_y(t'))dt'$ instead, but the above is still a solution to $(1)$ despite not being a solution to $(2)$. I get that the indefinite vs definite thing is the issue, here, but what is the explanation for how equation $(2)$ seemingly limits the possibilities for the solution to $v_x$ to specifically $D = 0$. I'm struggling to come up with an intuition for why that makes sense.