Take the Laplace transform -4 dy/dt - y = 0 y(0) = -8

Question

Take the Laplace transform of the IVP

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

I need to use Y for laplace transform of y

( ) = Y{0}

So Y = (-8) / (s - 1/4 )

and y(t) = -8e^(-t/4)

I tried sY+8+1/4Y = Y{0} but it says I am incorrect, I also tried the bold part and that is incorrect.

I also tried this

dy/dt + 1/4y = 0;

sY(s) + 8 + 1/4Y(s) = 0; Y(s) = - 8/(s + 1/4);

y(t) = L^-1(- 8/(s + 1/4) = - 8e^{- t/4}

Yefim S. keeps giving me the wrong answer and does not read my comments, he just keeps posting the same incorrect answer.

Answer 1

L{f(n)}= s^nY(s)− s^(n−1)y(0)−s^(n−2)y′(0)−⋯

-4dy/dt - y = 0 <— remember it’s really 0t here

Laplace applied individually to each term. Each term will become n+1 terms (n being the order of the derivative), stopping at the term where you hit an exponent of 0.

-4L{dy/dt} - L{y}= L{0t}

For the first term n= 1 (1st derivative) so we only need the first 2 terms of the transform

-4[s^1Y(s) + s^(1-1)y(0)]

= -4sY(s) - y(0)

For the second term n = 0 (not a derivative)

-1[s^0Y(s)]

= -1Y(s)

For the right side, (and in general for any #t, your laplace is #/s^2) we have 0t, which turns into 0/s^2

= 0 it remains 0

so we have -4sY(s) - y(0) - Y(s) = 0 plug in y(0) = -8

-4sY(s) - (-8) - Y(s) = 0 move the 8 to the other side

-4sY(s) - Y(s) = -8 I’m going to multiply everything by -1 to make it look a little better

4sY(s) + Y(s) = 8 factor out GCF which is Y(s)

Y(s)(4s+1) = 8

Y(s) = 8/(4s+1)

We need the denominator in the form s-a: factor by 4

8/4(s + 1/4) and we need it to say s-a so we will write it as a double negative

8/4(s - -1/4)

we can see that a = -1/4

The 8/4 can divide and make a 2

Y(s) = 2/(s- -1/4)

when the Laplace is in the form 1/s-a, the solutions is

f(t) = e^at

f(t) = 2e^(-1/4 t)