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 formulae MATH and MATH

$\qquad $

Method 1 - Solving the DE in q

From the formula: MATH, we obtain:

MATH

On substituting, we have:

MATH

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 MATH

So $q=2+Ke^{-10t}$

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

So $q=2(1-e^{-10t}).$


6_RCex1__21.png

As MATH, $q\rightarrow 2$ C.

Now,

MATH


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

MATH

Since $t=0$, $q=0$, we have $K=-\ln 2$

So
MATH

Method 2:

We use the formulae MATH $\ $and MATH.

Now MATH

So:

MATH


MATH

Now
MATH

From here, we use $q(0)=0$ and obtain: $K_{1}=2.$

So $q=2(1-e^{-10t})$, as before. Also, as MATH, $q\rightarrow 2$ C.