Use the Laplace transform to solve the given initial-value problem. y' + 3y = f(t), y(0) = 0
where f(t) = t, 0 ≤ t < 1 0, t ≥ 1

The solution to the given Initial - Value - Problem is

Step-by-step explanation:

y' + 3y = f(t).................(1)

f(t) = t      when 0 ≤ t < 1

f(t) = 0     when t ≥ 1

Step 1: Take the Laplace transform of the LHS of equation (1)

That is L(y' + 3y) = sY(s) + 3Y(s) = Y(s)[s + 3]..............(*)

Step 2: Get an expression for f(t)

For f(t) = t      when 0 ≤ t < 1

f₁(t) = t (1 - u(t - 1)) ( there is a time shift of the unit step)

For f(t) = 0     when t ≥ 1

f₂(t) = 0(u(t-1))

f(t) = f₁(t) + f₂(t)

f(t) = t - t u(t-1)................(2)

Step 3: Taking the Laplace transform of equation (2)

...............(**)

Step 4: Equating * and **

.......................(3)

Since y(t) is the solution we are looking for we need to find the Inverse Laplace Transform of equation (3) by first breaking every  fraction into partial fraction:

We can rewrite equation (3) by representing the fractions by their partial fractions.

................(4)

step 5: Take the inverse Laplace transform of equation (4)

Simplifying the above equation:

i am unable to solve this problem

step-by-step explanation: