Question

A function y(t) satisfies the following differential equation:
$\frac{dy\left(t\right)}{dt}+y\left(t\right)=\delta \left(t\right)$
where $\delta \left(t\right)$ is the delta function. Assuming zero initial condition, and denoting the unit step function by u(t), y(t) can be of the form

• A. $e^t$
• B. $e^{-t}$
• C. $e^{t}u\left(t\right)$
• D. $e^{-t}u\left(t\right)$

Answer 1

The correct option is D $e^{-t}u\left(t\right)$

Taking (L.T.) on both sides

$Y\left(s\right)(s+1)=1$

$\therefore Y\left(s\right)=\frac{1}{s+1}$

Taking inverse laplace transform

$y\left(t\right)=e^{-t}u\left(t\right)$