Now `V_R(t)=u(t)-u(t-2)`

To solve this, we need to work in voltages, not current.

We start with .`Ri+1/Cinti\ dt=V`

The voltage across a capacitor is given by `v=1/Cinti\ dt`.

It follows that `C(dv)/(dt)=i`.

So for this example we have:

`RC(dv)/(dt)+v=u(t)-u(t-2)`

Substituting known values:

`RC=10^6xx10^-6=1`

Then

`(dv)/(dt)+v=u(t)-u(t-2)`

Taking Laplace Transform of both sides: `(sV-v_0)+V=1/s-(e^(-2s))/s`

Since `v_0=0`, we have:

`(s+1)V=1/s-(e^(-2s))/s`

`{: (V,=1/(s(s+1))-(e^(-2s))/(s(s+1))),(,=(1/s-1/(s+1))-(e^(-2s)/s-(e^(-2s))/(s+1))),(,=1/s-1/(s+1)-(e^(-2s))/s+(e^(-2s))/(s+1)):}`

So, taking inverse Laplace

`v=u(t)-e^(-t)*u(t)-u(t-2)+e^(-t+2)*u(t-2)`

NOTE: For the part: `ccL^(-1){(e^(-2s))/(s+1)}`, we use:

`ccL^(-1){e^(-as)G(s)}=g(t-a)*u(t-a)`

So we have:

`G(s)=1/(s+1)`

`g(s)=e^(-t)`

`a=2`

`{: (ccL^(-1){(e^(-2s))/(s+1)},=g(t-a)*u(t-a)),(,=e^(-(t-2))*u(t-2)),(,=e^(-t+2)*u(t-2)):}`

Solution Using Scientific Notebook

1. To find the Inverse Laplace:

`ccL^(-1){1/(s(s+1))-(e^(-2s))/(s(s+1))}` `=1-e^(-t)-"Heaviside"(t-2)(1-e^(-t+2))`

2. To solve the original DE:

`{:((dv)/(dt)+v=u(t)-u(t-2)),(v(0)=0):}`

Exact solution for v(t):

`v(t)="Heaviside"(t)` `-e^(-t)"Heaviside"(t)` `-"Heaviside"(t-2)` `+"Heaviside"(t-2)e^(-t+2)`

To see what this means, we could write it as follows:

`{:(v(t)=(1-e^(-t))u(t)+(-1+e^(-t+2))*u(t-2)),(=(1-e^(-t))*u(t)+(-1+e^(-t)e^2)*u(t-2)),(=(1-e^(-t))*u(t)+(-1+e^(-t))*u(t-2)+e^(-t)(e^2-1)*u(t-2)),(=(1-e^(-t))*(u(t)-u(t-2))+e^(-t)(e^2-1)*u(t-2)):}`

To get an even better idea what our expression for v(t) means, we graph it as follows: