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UPSC 2015 Maths Optional Paper 1 Q7a-i — Step-by-Step Solution

6 marks · Section B

Inverse Laplace transform · ODEs · asked 2× in 13 yrs · Read the full method →

Question

Obtain Laplace inverse transform of $\left\{\ln\!\left(1+\dfrac{1}{s^2}\right)+\dfrac{s}{s^2+25}e^{-\pi s}\right\}.$

Technique

Two standard tools: second-shift theorem for $e^{-\pi s}$; "$\mathcal L\{tf(t)\}=-F'(s)$" identity for the log term, identifying $-F'(s)=2(1/s-s/(s^2+1))$.

Solution

Split into two pieces.

Step 1 — $\mathcal L^{-1}\!\left\{\dfrac{s}{s^2+25}e^{-\pi s}\right\}$

Standard pair: $\mathcal L^{-1}\!\left\{\dfrac{s}{s^2+25}\right\}=\cos 5t$. With the $e^{-\pi s}$ multiplier and second shifting theorem:

where $u$ is the Heaviside step.

So

Note $\cos(5t-5\pi)=\cos(5t)\cos(5\pi)+\sin(5t)\sin(5\pi)=-\cos 5t$ (since $\cos 5\pi=-1$, $\sin 5\pi=0$).

So this piece $=-\cos(5t)\,u(t-\pi)$.

Step 2 — $\mathcal L^{-1}\!\left\{\ln\!\left(1+\dfrac{1}{s^2}\right)\right\}=\mathcal L^{-1}\!\left\{\ln\!\dfrac{s^2+1}{s^2}\right\}$

Use the identity $\mathcal L\{f(t)/t\}=\int_s^\infty F(\sigma)\,d\sigma$, or equivalently if $F(s)=\int_s^\infty G(\sigma)\,d\sigma$ then $\mathcal L^{-1}\{F(s)\}=\dfrac{g(t)}{t}$ where $g(t)=\mathcal L^{-1}\{G(s)\}$.

Differentiate $F(s)=\ln(s^2+1)-\ln(s^2)$:

So $-F'(s)=\dfrac{2}{s}-\dfrac{2s}{s^2+1}=\mathcal L\{2\}-\mathcal L\{2\cos t\}=\mathcal L\{2-2\cos t\}$.

Now, $\mathcal L\{tf(t)\}=-F'(s)$. Setting $f(t)$ such that $\mathcal L\{tf(t)\}=\mathcal L\{2-2\cos t\}$:

So

Step 3 — Combine

Answer