## 4 thoughts on “3.11”

1. Redacted says:

Why is the particular solution equal to zero in the second part of the problem?

1. Redacted says:

Oh. Is it because for t>1 f(t) is technically equal to zero, therefore, there can be no particular solution?

1. tutorpaul says:

Yes, when the right hand side of the equation of motion is 0 the particular solution is also 0. This can be explained in the context of $f(t)$ being constant as well. When the right hand side is a constant, say $f(t) = h$, the particular solution is

When $h=0$, $y_p = 0$!