We will solve this 3 ways, since it has a constant voltage source:

1 and 2: Solving the DE in q, as:

3. Using the formulas `V_C=V(1-e^(-t"/"RC))` and `i=V/R e^(-t"/"RC`.

Method 1 - Solving the DE in q

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

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

On substituting, we have:

`5(dq)/(dt)+1/0.02q=100`

`5(dq)/(dt)+50q=100`

`(dq)/(dt)+10q=20`

We can solve this DE 2 ways, since it is variables separable or we could do it as a linear DE. The algebra is easier if we do it as a linear DE.

Solving this differential equation as a linear DE, we have:

`"IF"=e^(10t`

So `qe^(10t)=int (e^(10t))20\ dt` `=2e^(10t)+K`

So `q=2+Ke^(-10t)`

Now, since `q(0)=0`, (that is, when `t=0`, `q=0`) this gives: `K=-2`.

So `q=2(1-e^(-10t))`.

0.10.20.30.40.50.60.511.52tqOpen image in a new page

Graph of `q=2(1-e^(-10t))`, solution of a differential equation.

As `t->oo`, `q->2\ "C"`.

Now,

`V_C=1/Cinti\ dt`

`=1/Cq`

`=1/0.02 2(1-e^(-10t))`

`=100(1-e^(10t)) `

For comparison, here is the solution of the DE using variables separable:

`(dq)/(dt)=10(2-q)`

`(dq)/(2-q)=10dt`

` -ln\ |2-q|=10t+K`

(We could continue and get the same expression as above.)

Since `t=0`, `q=0`, we have `K=-ln 2`.

So

`-ln\ |2-q|=10t-ln 2`

` -ln\ 2+ln\ |2-q|=-10t`

`(2-q)/2=e^(-10t)`

` 2-q=2e^(-10t)`

` q=2(1-e^(-10t))`

Method 2:

We use the formulae `V_C=V(1-e^(-t"/"RC))` and `i=V/R e^(-t"/"RC)`.

Now `1/(RC)=1/(5xx0.02)=10`

So:

`V_C=V(1-e^(-t"/"RC))` `=100(1-e^(-10t))`

`i=V/Re^(-t"/"RC)`

`=100/5e^(-t"/"0.1) `

`=20e^(-10t) `

Now

`q=inti\ dt`

`=int20e^(-10t)dt`

`=-2e^(-10t)+K_1`

From here, we use `q(0)=0` and obtain: `K_1=2.`

So `q=2(1-e^(-10t))`, as before. Also, as `t->oo`, `q->2\ "C"`.

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