Determine the Laplace transforms of the initial value problem (IVP)y′′−3y′+2y=4δ(t−3),y(0)=0,y′(0)=0𝑦″−3𝑦′+2𝑦=4𝛿(𝑡−3),𝑦(0)=0,𝑦′(0)=0and obtain an expression for Y(s)=L(y)(t)𝑌(𝑠)=𝐿(𝑦)(𝑡). Do not find the inverse Laplace transform of the resulting equation.

The Laplace transform of the given initial value problem can be found by taking the Laplace transform of each term separately.

The Laplace transform of y'' is s^2Y(s) - sy(0) - y'(0), which simplifies to s^2Y(s) since y(0) = y'(0) = 0.

The Laplace transform of -3y' is -3sY(s) + 3y(0), which simplifies to -3sY(s) since y(0) = 0.

The Laplace transform of 2y is 2Y(s).

The Laplace transform of 4δ(t-3) is 4e^(-3s) because the Laplace transform of δ(t-a) is e^(-as).

So, the Laplace transform of the given differential equation is s^2Y(s) - 3sY(s) + 2Y(s) = 4e^(-3s).

This simplifies to Y(s) = 4e^(-3s) / (s^2 - 3s + 2).

So, the Laplace transform of the solution to the initial value problem is Y(s) = 4e^(-3s) / ((s-1)(s-2)).