Use Laplace transforms to solve the initial value problem d^2y/dt^2 + dy/dt − 6y = 12e^−t , y(0) = 1, y'(0) = −2.

Question

Use Laplace transforms to solve the initial value problem

d^2y/dt^2 + dy/dt − 6y = 12e^−t , y(0) = 1, y'(0) = −2.

Answer 1

ÿ + ý - 6y = 12 e^-t y = f(t), f(0) = 1, f '(0) = -2

^{Take Laplace Transform of each.}

s²F(s) - sf(0) - f '(0) + sF(s) - f(0) - 6F(s) = 12/(s + 1)

Substitute f(0) = 1, f '(0) = -2

s²F(s) - s + 2 + sF(s) - 1 - 6F(s) = 12/(s + 1)

Gather all the F(s) to the left side; carry all other terms to the right side.

(s² + s - 6) F(s) = 12/(s + 1) + s - 1

Isolate F(s).

F(s) = 12/(s+1)(s+3)(s-2) + (s - 1)/(s+3)(s-2)

Change to partial fractions.

12/(s+1)(s+3)(s-2) = -2/(s+1) + 1.2/(s+3) + 1.5/(s - 2)

(s-1)/(s+3)(s - 2) = 0.25/(s-2) + 0.8/(s+3)

Substitute and combine the terms for F(s).

F(s) = -2/(s+1) + 2/ (s+3) + 1.75/(s-2)

Take the inverse Laplace Transform.

f(t) = -2e^-t + 2e ^-3t + 1.75 e ^-2t

Plain

Bash

C++

C#

CSS

Diff

HTML/XML

Java

Javascript

Markdo