Given the values for R, L, C, and E, and noting that , we get the DE:
Let us solve the equivalent homogenous part first:
Assuming a solution of , we have the following characteristic equation:
Thus, the complementary solution is:
Now solve the non-homogeneous equation.
The technique I will use is known as the annihilator approach. I wasn't taught another way to solve these equations (besides Variation of Parameters, but I believe that it is unnecessary here).
First, we need to get the DE in Differential Operator Notation:
I will leave it for you (and others) to show that annihilates
Therefore, annihilates .
Applying the annihilator to both sides, we get:
Convert the equation so it has the form of the characteristic equation:
Thus the solution to the DE is:
where is the particular solution. The particular solution can't have arbitrary constants. To find and , which I will denote by and respectively, plug into the original DE.
You will find that
Therefore, the solution to the DE becomes:
Now apply the initial conditions .
Woah...that was long. I hope you can absorb all of this!!