Question

Solve using Laplace transforms:
y''+3y'+2y=6e^(-t)

Answer 1

I've got this:
L{y"}+3L{y'}+2L{y}=6L{e^(-t)}
s^2-sy(0)-y'(0)+3[sL{y}-y(0)]+2L{y}=6L...

Answer 2

I also have y(0)=1 and y'(0)=2, so I can plug those in

s^2-s-2+3sL{y}-3+2L{y}=...

Answer 3

or s^2-s+3sL{y}+2L{y}-5=...

Answer 4

L{y''} = s^2Y(s)- y'(0) -s y(0)

3L{y'} = 3s Y(s)-3y(0)

2L{y} = 2 Y(s)

6L{e^-t} = 6/(s+1)

s^2Y(s)- 2 -s +3s Y(s)-3 +2 Y(s) = 6/(s+1)

Y(s)(s^2 +3s +2) = 6/(s+1) +5+s

\[Y(s)=\frac{\frac{6}{s+1}+5+s}{s^2+3s+2}\]

Answer 5

..you rock. thank you.