How do I manipulate algebraic fractions with an addition in the denominator? My lecturer has given me some notes to study and I can't follow one of the steps...
I need to find the inverse laplace transform of $$\frac3{s(0.1s+1)}\;.$$ The notes do the following:
$$L^{-1}\left[\frac3{s(0.1s+1)}\right] = L^{-1}\left[\frac1s-\frac{0.1}{0.1s+1}\right]$$
That's the step I don't follow.
It then goes on to equal $3-3e^{-10t}$. I have an inkling that the $3$ has been missed in the step above and then reapplied to the final answer? But even aside from the missing $3$, I still don't fully understand what he has done.
Any help would be appreciated!
 A: You’re right about the $3$; the righthand side of the second displayed line should read $$L^{-1}\left[\frac3{s(0.1s+1)}\right] = 3L^{-1}\left[\frac1s-\frac{0.1}{0.1s+1}\right]$$ or $$L^{-1}\left[\frac3{s(0.1s+1)}\right] = L^{-1}\left[\frac3s-\frac{0.3}{0.1s+1}\right]\;.$$
This simply uses the algebraic identity $$\frac1{s(0.1s+1)} = \frac1s-\frac{0.1}{0.1s+1}\;.\tag{1}$$ You can check this by performing the subtraction on the righthand side and verifying that you get the fraction on the lefthand side:
$$\begin{align*}
\frac1s-\frac{0.1}{0.1s+1}&=\frac1s\cdot\frac{0.1s+1}{0.1s+1}-\frac{0.1}{0.1s+1}\cdot\frac{s}s\\
&=\frac{0.1s+1}{s(0.1s+1)}-\frac{0.1s}{s(0.1s+1)}\\
&=\frac{0.1s+1-0.1s}{s(0.1s+1)}\\
&=\frac1{s(0.1s+1)}\;.
\end{align*}$$
The process of starting with the lefthand side of $(1)$ and finding the righthand side known as decomposing the lefthand side into partial fractions. Since you’ve tagged this (algebra-precalculus), I don’t know whether you’ll actually be required to perform such decompositions, or merely to be able to verify them. (I’m also a bit surprised to see inverse Laplace transforms in a precalculus setting.)
A: It's called partial fraction decomposition.
$$\rm \frac{1}{(as+b)(cs+d)}=\frac{\square}{as+b}+\frac{\triangle}{cs+d}$$
Setting up constraints on $\rm \square,\triangle$ to make the above true for $\rm s$ arbitrary,
$$\rm (c\square+a\triangle)s+(d\square+b\triangle)=0s+1 \implies\quad \begin{cases} \rm c\square+a\triangle=0 \\ \rm d\square+b\triangle=1\end{cases}$$
which is a linear system with solution
$$\rm \begin{pmatrix}\square \\ \triangle\end{pmatrix}=\begin{pmatrix}c & \rm a \\ \rm d & \rm b\end{pmatrix}^{-1}\begin{pmatrix}0 \\ 1\end{pmatrix}=\frac{1}{ad-bc}\begin{pmatrix}a \\ \rm -c\end{pmatrix}.$$
There is a much simpler case, though:
$$\rm \frac{1}{s(s+h)}=\frac{1}{h}\frac{h}{s(s+h)}=\frac{1}{h}\frac{(s+h)-s}{s(s+h)}=\frac{1}{h}\left(\frac{1}{s}-\frac{1}{s+h}\right).$$
Note that the most general form can be brought into this form:
$$\rm \frac{m}{(as+b)(cs+d)}=\frac{m}{ac(s+b/a)(s+d/c)}=\frac{m}{ac}\frac{1}{r(r+d/c-b/a)},$$
where $\rm r=s+b/a$. Thus we have
$$\rm \frac{3}{s(0.1s+1)}=\frac{30}{s(s+10)}=\frac{30}{10}\left(\frac{1}{s}-\frac{1}{s+10}\right)=\frac{3}{s}-\frac{3}{s+10}.$$
(There is indeed a factor of $3$ missing in the notes, apparently.)
