# Stumped- DE problem

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.

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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.

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

Tom Mattson
Staff Emeritus
Gold Member
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
Homework Helper
You can either integrate by parts or, if you're comfortable with complex analysis, you can note that

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

and extract the imaginary part after performing the integration.

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

Tom Mattson
Staff Emeritus
That depends on the answer! 