Question

The Laplace transform of the function f(t) = t sin t is

• A. \( \frac{2s}{(s^2+1)^2} \)
• B. \( \frac{1}{s^2(s^2+1)} \)
• C. \( \frac{1}{s^2} + \frac{1}{(s^2+1)} \)
• D. \( \frac{1}{(s-1)^2+1} \)

Hint:

Laplace Transform Property. Multiplication by t: \(\mathcal{L\{t f(t)\ = - \frac{d{ds F(s)\), where \(F(s) = \mathcal{L\{f(t)\\). Remember \(\mathcal{L\{\sin(\omega t)\ = \omega/(s^2+\omega^2)\).

Answer 1

The Correct Option is A

We want to find \(\mathcal{L}\{t \sin t\}\).
We use the Laplace transform property for multiplication by t: $$ \mathcal{L}\{t^n f(t)\} = (-1)^n \frac{d^n}{ds^n} F(s) $$ where \(F(s) = \mathcal{L}\{f(t)\}\).
In this case, \(n=1\) and \(f(t) = \sin t\).
First, find \(F(s)\): $$ F(s) = \mathcal{L}\{\sin t\} = \frac{1}{s^2 + 1^2} = \frac{1}{s^2 + 1} $$ Now apply the property with \(n=1\): $$ \mathcal{L}\{t \sin t\} = (-1)^1 \frac{d}{ds} F(s) = -\frac{d}{ds} \left( \frac{1}{s^2 + 1} \right) $$ Use the quotient rule or chain rule for differentiation.
Let \(u = s^2+1\), \(\frac{d}{ds}(u^{-1}) = -u^{-2}\frac{du}{ds}\).
Here \(du/ds = 2s\).
$$ = - \left[ -(s^2+1)^{-2} (2s) \right] = - \left[ -\frac{2s}{(s^2+1)^2} \right] $$ $$ = \frac{2s}{(s^2+1)^2} $$ This matches option (1).