A function $$y(t)$$ satisfies the following differential equation : $${{dy\left( t \right)} \over {dt}} + y\left( t \right) = \delta \left( t \right)$$

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

Consider the function $$F\left( s \right) = {5 \over {s\left( {{s^2} + 3s + 2} \right)}}$$ Where $$F(s)$$ is the Laplace transform of the function $$f(t).$$ The initial value of $$f(t)$$ is equal to

Consider the function $$F\left( s \right) = {5 \over {s\left( {{s^2} + 3s + 2} \right)}}$$ Where $$F(s)$$ is the Laplace transform of the function $$f(t).$$ The initial value of $$f(t)$$ is equal to

Consider the function $$F\left( s \right) = {5 \over {s\left( {{s^2} + 3s + 2} \right)}}$$ Where $$F(s)$$ is the Laplace transform of the function $$f(t).$$ The initial value of $$f(t)$$ is equal to