# Perhaps delta function or inverse Laplace transform?

1. Aug 30, 2005

### EvLer

Hello everyone,
i have this question and not even sure how to approach it:

$$\frac {di}{dt}+4i+3\int_{0^-}^t{i(z)dz = 12(t-1)u(t-1)$$

and $$i(0^-) = 0$$

find $$i(t)$$

last topics we covered were laplace transforms (and inverse) and dirac delta function.
At least some hint to get me started would be a great help.

EDIT:
oh, and again u(t) = 1 for t >= 0 and u(t) = 0 elsewhere.

Last edited: Aug 30, 2005
2. Aug 30, 2005

### BobG

I would get rid of the integral by taking the derivative of the entire equation - which gives you a second order differential equation and leads you into the LaPlace transform.

3. Aug 30, 2005

### Tide

Laplace transform the equation directly. To find the transform of the integral, just do an integration by parts. That will avoid complications on the right side. :)

4. Aug 31, 2005

### EvLer

thanks for replies, as i looked further through the book, we actually have an entry in the table for this integral, but what do I do with $$i$$ for Laplace transform? it does not have u(t)...

5. Sep 2, 2005

### Tide

Ev,

I presume that your goal is to solve for i(t). After performing the Laplace transforms, you will have an algebraic equation for I(s). The right side will have two terms. One will relate to the initial value of i and the other will be a product of an algebraic quantity with the Laplace transform of the integral containing u. You should have no difficulty inverting the first part and you should be able to do something with the second using the convolution theorem for Laplace transforms.

Let us know what you end up with!