Question

Constant Coefficient Homogeneous Basic: Problem 2 (3 points) y ′′ +y ′ −2y=0

a.Give the characteristic polynomial for the differential equation. (use r as your variable.)
b. List the roots of the characteristic polynomial for the differential equation.
c. List the basic solutions for the differential equation, i.e. y₁(t),y₂ (t).

Answer 1

To find the characteristic polynomial, assume a solution of the form y(t) = e^(rt). The roots of the characteristic equation are r = 1 and r = -2. The basic solutions are y₁(t) = e^(t) and y₂(t) = e^(-2t).

Step-by-step explanation:

To find the characteristic polynomial for the given differential equation y'' + y' - 2y = 0, we assume a solution of the form y(t) = e^(rt). Substituting this into the differential equation, we get the characteristic equation r^2 + r - 2 = 0.

To find the roots of the characteristic equation, we factor it as (r - 1)(r + 2) = 0. So, the roots are r = 1 and r = -2.

Using the roots, we can find the basic solutions for the differential equation. The basic solutions are y₁(t) = e^(t) and y₂(t) = e^(-2t).