We need to write the RHS of the DE in terms of unit step functions.

`(d^2i)/(dt^2)+2(di)/(dt)` `=u(t-10)-u(t-20)`

Now, taking Laplace transform of both sides gives us:

`(s^2I-s\ i(0)-i^(\ ')(0))+2(sI-i(0))` `=(e^(-10s))/s-(e^(-20s))/s`

`s^2I+2sI` `=1/s(e^(-10s)-e^(-20s))`

`(s^2+2s)I` `=1/s(e^(-10s)-e^(-20s))`

Solving for `I` gives:

`I=1/s((e^(-10s)-e^(-20s))/(s^2+2s))` `=1/(s^2(s+2))(e^(-10s)-e^(-20s))`

We need to find the Inverse Laplace of this expression. First, we concentrate on the `1/(s^2(s+2))` part and ignore the `(e^(-10s)-e^(-20s))` part for now.

Now, we find the partial fractions: `1/(s^2(s+2))` `=A/s+B/s^2+C/(s+2)`

Multiply both sides by `s^2(s+2)`:

`1=As(s+2)+B(s+2)+Cs^2`

`s = 0` gives `1 = 2B` gives `B=1/2`

`s=-2` gives `1 = 4C` gives `C=1/4`

`s=1` gives `1 = 3A + 3B + C` gives `A= -1/4`

So `1/(s^2(s+2))` `=-1/(4s)+1/(2s^2)+1/(4(s+2))`

Now, the inverse Laplace of this expression is:

`ccL^(-1){1/(s^2(s+2))}`

`=ccL^(-1){-1/(4s)+1/(2s^2)+1/(4(s+2))}`

`=-1/4+1/2t+1/4e^(-2t)`

So since

`I=1/(s^2(s+2))e^(-10s)-1/(s^2(s+2))e^(-20s)`,

then we have, using the Time-Displacement Theorem (see the Table of Laplace Transforms):

`i(t)=[-1/4*u(t-10)+1/2(t-10)*u(t-10)+1/4e^(-2(t-10))*u(t-10)]` `-[-1/4*u(t-20)+1/2(t-20)*u(t-20)+1/4e^(-2(t-20))*u(t-20)]`

`=1/4(2t-21+e^(-2(t-10)))*u(t-10)` `+1/4(41-2t-e^(-2(t-20)))*u(t-20)`

The graph of `i(t)` is as follows: