Question

Please help me find the Laplace transform of the given expression that has the properties y(0)=1 and y'(0)=-1:

y"+3y'

Answer 1

is the eq.y"+3y'=0

Answer 2

It is not an equation, just an expression

Answer 3

$$
\mathcal{L}\left\{y'(t)\right\}(s)=s Y(s) - y(0) \\
\mathcal{L}\left\{y''(t)\right\}(s)= s^2 Y(s) - s y(0) - y'(0) \\
$$
So, taking Laplace of y''+3y':
$$
\mathcal{L}\left\{y"\!+~3y'\right\}(s)= s^2 Y(s) - s y(0) - y'(0) +3\left(s Y(s) - y(0) \right )\\
=Y(s)(s^2+3s)-s+1-3s\\
=Y(s)(s^2+3s)-4s+1$$

Answer 4

Thanks for taking the time to do that but the answer in the book is:
\[(s^2+3s)F(s)-s-2\]

Answer 5

see typo -
$$
\mathcal{L}\left\{y"\!+~3y'\right\}(s)= s^2 Y(s) - s y(0) - y'(0) +3\left(s Y(s) - y(0) \right )\\
=Y(s)(s^2+3s)-s+1-\color{red}{3}\\
=Y(s)(s^2+3s)-s-2
$$

Answer 6

oh ok i see, thanks for working that out...

Answer 7

np