Bring the chapter together: compute and invert transforms with the table and the shift rules, solve an initial value problem in three steps, write a switched forcing with steps, respond to an impulse, and read a product of transforms as a convolution. No worked example sits above the answers.

A translator’s exam: a passage in, its translation out, and the grammar checked on the way. Knowing which rule a phrase needs is most of the work.

Decide first what is asked. Transforming: linearity, the table, $e^{at}f\mapsto F(s-a)$ and $u(t-c)f(t-c)\mapsto e^{-cs}F(s)$. Solving: transform, solve for $Y$, split into partial fractions, invert. A factor $e^{-cs}$: invert the rest and delay it by $c$. A product $F(s)G(s)$: the convolution $f*g$.

• $\mathcal{L}\{e^{at}\}=\tfrac{1}{s-a}$, $\mathcal{L}\{t^{n}\}=\tfrac{n!}{s^{n+1}}$, $\mathcal{L}\{\sin\omega t\}=\tfrac{\omega}{s^{2}+\omega^{2}}$, $\mathcal{L}\{\cos\omega t\}=\tfrac{s}{s^{2}+\omega^{2}}$.
• $\mathcal{L}\{u(t-c)\}=\tfrac{e^{-cs}}{s}$ and .
• A jump in the forcing is a jump in $y''$; an impulse is a jump in $y'$.
• $(f*g)(t)=\int_{0}^{t}f(\tau)g(t-\tau)\,d\tau$ and .