We will solve this 2 ways:

1. Solving in *q*.

2. Using Scientific Notebook.

[**NOTE: **We cannot use the formulae `V_C=V(1-e^(-t"/"RC))` and `i=V/re^(-t"/"RC)`, since the voltage source is **not constant**.]

From the formula: `Ri+1/Cint i\ dt=V`, we obtain:

`R(dq)/(dt)+1/Cq=V`

Since `R=10`, `C=4xx10^-3`, and `V=85 cos 150t`, we have:

`10(dq)/(dt)+1/(4xx10^-3)q=85 cos 150t`

`10(dq)/(dt)+250q=85 cos 150t`

`(dq)/(dt)+25q=8.5 cos 150t`

Now, we can solve this differential equation in *q* using the **linear DE** process as follows:

`"IF"=e^(25t)`

`e^(25t)q=inte^(25t)8.5 cos 150t dt` `=8.5inte^(25t)cos\ 150t dt`

Then we use the integration formula (found in our standard integral table):

`inte^(mt)cos\ nt\ dt=` `(e^(mt))/(m^2+n^2)(m cos nt+n sin nt)`

We obtain:

`e^(25t)q=8.5inte^(25t)\ cos\ 150t\ dt`

`=8.5(e^(25t))/23125(25 cos 150t+` `{:150e^(25t) sin 150t)`

`=0.0092e^(25t) cos 150t+` `0.055e^(25t) sin 150t+K`

Dividing throughout by `e^(25t` gives:

`q=0.0092 cos 150t+` `0.055 sin 150t+` `Ke^(-25t)`

We now need to find `K`:

`q(0)=-0.05` means `K=-0.05-0.0092` `=-0.0592`

So this gives us:

`q=0.0092 cos 150t+` `0.055 sin 150t-` `0.0592e^(-25t)`

We set up the differential equation and the initial conditions in a **matrix** (not a table) as follows:

`(dq)/(dt)+25q=8.5 cos 150t`

`q(0)=-0.05`

Choosing **Solve ODE - Exact** from the **Compute** menu gives:

Exact solution is:

`q(t)=0.0092 cos 150t+` `0.055 sin 150t-` `0.059e^(-25t)`

The graph for `q(t)` is as follows:

Graph of current `q` at time `t`. It's in steady state by around `t=0.12`.

We are also asked to find the current. We simply differentiate the expression for *q*:

`i` `=d/(dt)(0.0092 cos 150t+` `{:0.055 sin 150t-0.0592e^(-25t))` `=-1.38 sin 150t+` `8.25 cos 150t+` `1.48e^(-25t)`

The graph for *i*(*t*):

Graph of current `i` at time `t`. It's also in steady state by around `t=0.12`.

Easy to understand math videos:

MathTutorDVD.com