Simple Laplace Transform?

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Apply L to both sides (letting Y(s) = L{y}):
L{dy/dt} = L{1 - y}
==> s Y(s) - y(0) = L{1} - L{y}
==> s Y(s) - 4 = 1/s - Y(s).

Solve for Y(s):
(s + 1) Y(s) = 4 + 1/s = (4s + 1)/s
==> Y(s) = (4s + 1)/(s(s + 1))
==> Y(s) = 1/s + 3/(s + 1), by partial fractions.

Inverting yields y(t) = 1 + 3e^(-t).

I hope this helps!...Show more

yes...Show more

3. dy/dt = 1-y
L{dy/dt} = L{1-y}
Sy(s)-y(0) = L{1} – L{y}
(∵ f’(t)=sf(s)-f(0))
(∵ L{1} = 1/5)
y(s)(1+s) = 1/5+4
y(s) = 1+4s/s(1+s)
y(s) = A/s + B/(5+1)
y(s)=1/5+3/5+1
L-1{y(s)} = 1+3e-t...Show more

∫ dy / [ 1 - y ] = ∫ dt
- log [ 1 - y ] = t + C
log [ 1 - y ] = K - t
1 - y = e^(K - t)
y = 1 - e^(K - t)...Show more

(1 +4 s )/(s (1 + s))...Show more