UPSC 2018 Maths Optional Paper 1 Q5d-ii — Step-by-Step Solution

5 marks · Section B

Inverse Laplace transform · ODEs · asked 2× in 14 yrs · Read the full method →

Question

Find the inverse Laplace transform of $\dfrac{5s^2+3s-16}{(s-1)(s-2)(s+3)}$ .

Partial-fraction decomposition (cover-up / residue rule) over three distinct simple poles, then $\mathcal L^{-1}\{(s-a)^{-1}\}=e^{at}$ .

Write

\frac{5s^2+3s-16}{(s-1)(s-2)(s+3)}=\frac{A}{s-1}+\frac{B}{s-2}+\frac{C}{s+3}.

Use the cover-up rule (residue at each simple pole).

At $s=1$ : $A=\dfrac{5(1)+3(1)-16}{(1-2)(1+3)}=\dfrac{5+3-16}{(-1)(4)}=\dfrac{-8}{-4}=2.$

At $s=2$ : $B=\dfrac{5(4)+3(2)-16}{(2-1)(2+3)}=\dfrac{20+6-16}{(1)(5)}=\dfrac{10}{5}=2.$

At $s=-3$ : $C=\dfrac{5(9)+3(-3)-16}{(-3-1)(-3-2)}=\dfrac{45-9-16}{(-4)(-5)}=\dfrac{20}{20}=1.$

Hence

\frac{5s^2+3s-16}{(s-1)(s-2)(s+3)}=\frac{2}{s-1}+\frac{2}{s-2}+\frac{1}{s+3}.

Using $\mathcal L^{-1}\{1/(s-a)\}=e^{at}$ :

\boxed{\;\mathcal L^{-1}\!\left\{\frac{5s^2+3s-16}{(s-1)(s-2)(s+3)}\right\}=2e^{t}+2e^{2t}+e^{-3t}.\;}