Consider the initial value problem y′+5y=⎧⎩⎨⎪⎪0110 if 0≤t<3 if 3≤t<5 if 5≤t<[infinity],y(0)=4. y′+5y={0 if 0≤t<311 if 3≤t<50 if 5≤t<[infinity],y(0)=4. Take the Laplace transform of both sides of the given differential equation to create the corresponding algebraic equation. Denote the Laplace transform of y(t)y(t) by Y(s)Y(s). Do not move any terms from one side of the equation to the other (until you get to part (b) below).

It looks like the ODE is

with the initial condition of .

Rewrite the right side in terms of the unit step function,

In this case, we have

The Laplace transform of the step function is easy to compute:

So, taking the Laplace transform of both sides of the ODE, we get

Solve for :

We can split the first term into partial fractions:

If , then .

If , then .

Take the inverse transform of both sides, recalling that

where is the Laplace transform of the function . We have

We then end up with

step-by-step explanation: