Question

how to find y(s) y''+3y'+2y = 5 y(0)=-1 , y'(0)=2
with laplace

Answer 1

start by taking the laplace transform of both sides

Answer 2

L{y'' + 3y' + 2y} = L{5}

[L{y}s^2 - sy(0) - y'(0)] + 3[L{y}s - y(0)] + 2[L{y}] = 5/s

Answer 3

plug values for y(0) and y'(0)

Answer 4

im stuck in replace initial conditios :S

Answer 5

Just plug them in after you take the laplace of each side. Then solve the LHS for L{y}

Answer 6

i have (s^2+3s+2)y(s) + S - 2 - 3 = 5/S correct ?

Answer 7

I think that it should be +3 on the left not -3

Answer 8

I have L{y}(s^2+3s+2) = 5/s - 1 - s

Answer 9

The solution that I came up with is \[y = -5e^{-t}+{3 \over 2}e^{-2t} + {5 \over 2}\]

Answer 10

thanks