Solving Ri(t)+Ld$\frac{di(t)}{dt}=V_{m}sin(wt+\theta)

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Can anyone help me with this?
Ri(t) + L\frac{di(t)} {dt}=V_{m}sin(wt+\theta)
 
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Hint:
What is the integrating factor here?
 
\exp\int\frac{R}{L}dt?
 
Which can be written more simply as..:?(assuming R, L constants, that is)
 
\exp\int{x}dt
 
Thread 'Direction Fields and Isoclines'

Thread 'Direction Fields and Isoclines'

I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
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