Solving a Differential Equation with L, R, E(t), and i(0) as Constants

[hola]

I am stumped... here is the problem:
Solve the DE using the following:
L and R are constants

$$L\frac{di}{dt} + Ri = E(t)$$

$$i(0) = i_0$$

$$E(t) = E_0*sin(wt)$$

Here is my work so far:

I got the integrating factor to become $$e^{Rt/L}$$. But now:

$$\frac{d(e^{\frac{Rt}{L}}*i)}{dt} = e^{\frac{Rt}{L}}\frac{E_0}{L}*sin(wt)$$

But I am stuck from there. Help would be appreciated.

 

[Corneo]

Since you have

$$\mu(t) = e^{\frac {R}{L}t}$$

and you also have

$$(\mu(t)i)' = \mu(t) \frac {E_0}{L} \sin (\omega t)$$

You can integrate both sides and divide by $\mu(t)$

Hence
$$i(t) = \frac {\int \mu(s) \frac {E_0}{L} \sin (\omega s) ds}{\mu(t)}$$

I switched the t to a s in the numerator to avoid confusion. After you integrate the numerator, you can replace the s with a t.

 

[hola]

That's the problem... I can't integrate it.

 

[quantumdude]

Integrate it by parts. Let $u=\sin(\omega t)$ and let $dv=exp\left(\frac{Rt}{L})$.

You'll have to integrate by parts twice and then algebraically solve for the integral. This integral actually pops up all the time in second order dynamic systems.

 

[Tide]

You can either integrate by parts or, if you're comfortable with complex analysis, you can note that

$$\int e^{at} \sin \omega t dt = I am \int e^{(a + i \omega) t} dt$$

and extract the imaginary part after performing the integration.

 

[hola]

I got some really messy answer... is that ok?

 

[quantumdude]

That depends on the answer!

Why don't you post what you did so we can see it?