Question

Second Order Equations-Constant Coefficient Homogeneous Basic: Problem 3 (3 points) y ′′+3y ′−18y=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

The characteristic polynomial for the differential equation is r^2 + 3r - 18 = 0. The roots of the characteristic polynomial are r = -6 and r = 3. The basic solutions for the differential equation are y₁(t) = e^(3t) and y₂(t) = e^(-6t).

Step-by-step explanation:

(a) Characteristic polynomial:

The characteristic polynomial for the given differential equation y ′′+3y ′−18y=0 is obtained by substituting r for the variable y in the equation. This gives us the characteristic polynomial as r^2 + 3r - 18 = 0.

(b) Roots of the characteristic polynomial:

To find the roots, we can factorize the characteristic polynomial as (r + 6)(r - 3) = 0. Therefore, the roots of the characteristic polynomial are r = -6 and r = 3.

(c) Basic Solutions:

The basic solutions for the differential equation y ′′+3y ′−18y=0 are y₁(t) = e^(3t) and y₂(t) = e^(-6t), where e denotes the exponential function.