Take the Laplace transform of the IVP

Question

-4 dy/dt - y = 0 y(0) = -8

I need to use Y for laplace transform of y

( ) = Y{0}

So Y = ( ) / (s - )

I found that y(t) = -8e^(-t/4)

Answer 1

First note the conventional solution of -4dy/dt − y = 0; y(0) = -8. Write dy/dt = -y/4 or dy = -ydt/4 or dy/-y = dt/4.

Rewrite as dy/y = -dt/4 and integrate to ln |y| = -t/4 + constant. This then gives y = Ce^(-t/4) where C is equal to e^constant. For y(0) = -8, obtain -8 = Ce^(-0/4) or C = -8 and y(t) = -8e^-(t/4).

For the Laplace solution, write dy/dt as y' and take the Laplace Transforms of both sides of -4y' − y = 0:

-4L{y'} − L{y} = L{0}.

For y(0) = -8 and Y(s) = L{y}, write -4[sY(s) − -8] − Y(s) = 0 or -4sY(s) − 32 − Y(s) = 0.

Then (-4s − 1)Y(s) = 32 or Y(s) = -32/(4s + 1) or -8/(s + 1/4). Reform as Y(s) = -8/(s − -1/4) and set down the Inverse Laplace Transform of Y(s) as y(t) = L^-1{Y(s)} or L^-1{-8/(s − -1/4)} or -8L^-1{1/(s − -1/4)}.

A Table of Laplace Transforms ties L^-1{1/(s − -1/4)} to e^-(t/4) which yields y(t) = -8e^-(t/4) as in the conventional solution above.