Solve y' + 2y = 3e-4t +5, starting with y(0) = 0. What happens as t...

 

Transcribed Image Text:

Solve y' + 2y = 3e-4t +5, starting with y(0) = 0. What happens as t → ∞? Solve and sketch the solution. (a) y' + y = 2e-t, y(0) = 3 (b) y + y= 2 cost, y(0) = 3 (c) y'+y=2H(t - 4), y(0) = 3 (d) y' + y = 28(t-4), y(0) = 3 Solve y' + y = 38(t-4) +2H(t-5). What happens as t→ ∞o? Compare wit y + y = 2. Solve y' + 3y = 3H(t - 1) (1 -e¹-t). Determine the long-term behavior. Solve y' + 2y = 3e-4t +5, starting with y(0) = 0. What happens as t → ∞? Solve and sketch the solution. (a) y' + y = 2e-t, y(0) = 3 (b) y + y= 2 cost, y(0) = 3 (c) y'+y=2H(t - 4), y(0) = 3 (d) y' + y = 28(t-4), y(0) = 3 Solve y' + y = 38(t-4) +2H(t-5). What happens as t→ ∞o? Compare wit y + y = 2. Solve y' + 3y = 3H(t - 1) (1 -e¹-t). Determine the long-term behavior.

Answer rating: 100% (QA)

Lets solve the given differential equations one by one a y 2y 3et 5 y0 0 This is a linear firstorder ... View the full answer