Originally Posted by

**MarkFL** No, I was just answering your question about the constant factor. I haven't solved the given ODE. Let's take a look:

$\displaystyle L\frac{dI}{dt}+RI=E_0\sin(\omega t)$

Divide through by $\displaystyle L$ to put into standard linear form:

$\displaystyle \frac{dI}{dt}+\frac{R}{L}I= \frac{E_0}{L}\sin(\omega t)$

Compute the integrating factor:

$\displaystyle \mu(t)=\exp\left(\frac{R}{L}\int\,dt\right)= e^{\Large\frac{R}{L}t}$

Multiplying the ODE by $\displaystyle \mu(t)$ there results:

$\displaystyle e^{\Large\frac{R}{L}t}\frac{dI}{dt}+\frac{R}{L}e^{ \Large\frac{R}{L}t}I= \frac{E_0}{L}e^{\Large\frac{R}{L}t}\sin(\omega t)$

Rewrite the LHS as the derivative of a product:

$\displaystyle \frac{d}{dt}\left(e^{\Large\frac{R}{L}t}I\right)= \frac{E_0}{L}e^{\Large\frac{R}{L}t}\sin(\omega t)$

Integrate with respect to $\displaystyle t$:

$\displaystyle e^{\Large\frac{R}{L}t}I= \frac{E_0}{L}\cdot\frac{e^{\Large\frac{R}{L}t}} {\left(\dfrac{R}{L}\right)^2+\omega^2} \left(\frac{R}{L}\sin(\omega t)- \omega\cos(\omega t)\right)+c_1$

Hence:

$\displaystyle I(t)= \frac{E_0}{L}\cdot \frac{\dfrac{R}{L}\sin(\omega t)- \omega\cos(\omega t)}{\left(\dfrac{R}{L}\right)^2+ \omega^2} +c_1e^{-\Large\frac{R}{L}t}$