Consider a particle of mass $m$ and electric charge $q$, initially at rest (zero velocity, $v(0) = 0$), placed in a unidirectional potential $ψ(x) = −E_0qx \hspace{0.05in} \text{sin}(ωt)$. Assume that the particle experiences a spatially uniform electric force given by (for this one-dimensional system) as $F_E = −\frac{∂ψ}{∂x}$, and a nonlinear drag force in the from $F_D = δv^3$.
$(i)$ Use Newton’s second law to write the equation of motion for the particle’s velocity.
I have done the following
$F=ma=m\frac{d^2x}{dt^2}= F_D - F_E = -\frac{\partial \psi}{\partial x} -\delta v^3 $
$m\frac{dv}{dt}=+E_0q \hspace{0.05in} \text{sin}(ωt) -\delta v^3$
I do not know how to proceed further to set up an integration that will yield a differential equation. I have tried doing this so far:
$\int \frac{m}{E_0q \hspace{0.05in} \text{sin}(ωt) -\delta v^3} dv = \int 1 \cdot dt$
Since velocity is a function of time (and vice versa), I am stuck and do not know how to proceed.
$$ m\frac{dv}{dt}=+E_0q \hspace{0.05in} \sin(ωt) -\delta v^3 $$ is correct. Setting $\tilde t=ωt$, $v=v_0\tilde v$ results in $$ mωv_0\frac{d\tilde v}{d\tilde t}=+E_0q \hspace{0.05in} \sin(\tilde t) -\delta v_0^3\tilde v^3 $$ Now selecting $v_0$ so that $mωv_0=E_0q$. Dividing be this factor and combining the remaining constants in a new constant $\tilde δ$ gives the normalized equation $$ \frac{d\tilde v}{d\tilde t}=\sin(\tilde t) -\tilde \delta \tilde v^3. $$
Drop the tildes for shortness. The solution for $\delta=0$ is $v=1-\cos t$. The perturbation for $δ\approx 0$ is regular, thus the solutions will also vary by terms of size $δ$. Set $v=(1-\cos t)+δw$, then $$ \frac{dw}{dt}=-(1-\cos t)^3+O(δ)\implies w(t)=-t+3\sin t-\frac32(t+\sin t\cos t)+\sin t-\frac13\sin^3 t $$ This is valid as long as $|δt|\ll 1$.