Solve the initial-value problem (d^2y)/(dt^2) + 8(dy)/(dt) + 16y = 0, y(1) = 0, y'(1) = 1. Answe

Name: Solve the initial-value problem (d^2y)/(dt^2) + 8(dy)/(dt) + 16y = 0, y(1) = 0, y'(1) = 1. Answer: y(t) =
Uploaded: 2023-04-25T18:10:04-08:00
Duration: 1 min 12 s
Channel: Manisha Sarker
Description: Solve the initial-value problem (d^2y)/(dt^2) + 8(dy)/(dt) + 16y = 0, y(1) = 0, y'(1) = 1. Answer: y(t) =

Step 1: Write the characteristic equation for y'' + 8y' + 16y = 0: r^2 + 8r + 16 = 0. Factor: (r

Question

Solve the initial-value problem \frac{d^2y}{dt^2} + 8\frac{dy}{dt} + 16y = 0, y(1) = 0, y'(1) = 1. \text{Answer: } y(t) =

          Solve the initial-value problem \frac{d^2y}{dt^2} + 8\frac{dy}{dt} + 16y = 0, y(1) = 0, y'(1) = 1. \text{Answer: } y(t) =

Solve the initial-value problem (d^2y)/(dt^2) + 8(dy)/(dt) + 16y = 0, y(1) = 0, y'(1) = 1. Answer: y(t) =

Added by John J.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Manisha Sarker

04/25/2023

Step 1

Factor: (r + 4)^2 = 0, so r = -4 is a double root. Show more…

Show all steps

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Solve the initial-value problem: d²y/dt² + 8(dy/dt) + 16y = 0, y(1) = 0, y'(1) = 1. Solve the initial-value problem: d²y/dt² + 2(dy/dt) + 16y = 0, y(1) = 0, y'(1) = 1. Answer: y(t) = 0