Partial Fraction Expansion (Solving Linear ODE with Laplace)

Question

I have a linear ODE initial value problem that I performed the laplace transform on and simplified it a bit, however now I am stuck at the partial fraction expansion.

My workings image

Note that C and R are constants too as well as the A and B from the partial expansion part

Answer 1

Hint: Your Laplace transform is:

$$F(s)=\frac{C}{s(sR+1)}=\frac{\alpha}{s}+\frac{\beta}{sR+1}.$$

Quick way using Euler's trick: Now, use Euler's trick. Multiply the equation by $s$ and set $s=0$ (zero of $s$) to obtain $\alpha =C$ and then multiply the equation by $sR+1$ and set $s=-1/R$ (zero of $sR+1$) to obtain $\beta=-RC$. This trick is always applicable for single zeros.

Note, if you do not like the idea of setting $s$ equal to the poles then you can do it more formally by taking the limit to the pole.

The other method is not so quick and also more error prone:

$$\frac{C}{s(sR+1)}=\frac{\alpha}{s}+\frac{\beta}{sR+1}=\frac{\alpha(sR+1)+\beta s}{s(sR+1)}$$

$$\implies \frac{C+s\cdot 0}{s(sR+1)}=\frac{\alpha +s(\beta+\alpha R)}{s(sR+1)}$$

Now, compare the numerators to obtain two equations: $$C=\alpha \qquad \beta+\alpha R=0$$

Can you complete it from here?

Answer 2

$$\frac{C}{s(sR+1)}=\frac{A}{s} +\frac{B}{sR+1}=\frac{A (sR +1)+B s}{s(sR+1)}=\frac{A+(B+A R) s}{s(sR+1)}$$

$ \left\{ \begin{array}{l} A=C \\ B+A R=0 \\ \end{array} \right. $

$A=C;\;B=-CR$

$$\frac{C}{s(sR+1)}=\frac{C}{s} -\frac{CR}{sR+1}$$