Question

Solve the following differential equation (a) or pair of equations (b) using the Laplace transform and the given initial conditions and inputs.

(D²+ 4D + 4 + y(t)= (D+1)x(t) y(0⁻)=2, y′(0⁻)=0, x(t)=e⁻ᵗu(t)
(D+2)y1​(t)−(D+1)y2​(t)=0

Answer 1

The differential equation is solved by applying the Laplace transform, incorporating the initial conditions, finding the transformed function Y(s), and then using an inverse Laplace transform to obtain y(t).

Step-by-step explanation:

To solve the given differential equation using the Laplace transform, we start by taking the Laplace transform of both sides of the equation. The equation is given as:

(D² + 4D + 4)y(t) = (D+1)x(t)

with the initial conditions y(0⁻)=2 and y′(0⁻)=0, and the input x(t)=e⁻¹t⋅u(t).

Applying the Laplace transform, we denote L{y(t)} as Y(s), L{y′(t)} as sY(s) - y(0⁻), and L{y″(t)} as s²Y(s) - sy(0⁻) - y′(0⁻). Substituting the initial conditions and solving for Y(s), we can then use the inverse Laplace transform to find y(t).

Note that the function x(t)=e⁻¹t⋅u(t) has a Laplace transform of X(s) = \frac{1}{s+1}. The corresponding Laplace transformed equation is then solved for Y(s) and the inverse Laplace transform is used to find y(t).

For the pair of equations involving y1(t) and y2(t), we would convert them into the s-domain and solve the resulting algebraic equations to find Y1(s) and Y2(s). Then, the inverse Laplace transform would provide y1(t) and y2(t).