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5 Answers
- kbLv 74 years agoFavorite Answer
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!
- MyRankLv 64 years ago
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
Source(s): http://myrank.co.in/ - ComoLv 74 years ago
∫ dy / [ 1 - y ] = ∫ dt
- log [ 1 - y ] = t + C
log [ 1 - y ] = K - t
1 - y = e^(K - t)
y = 1 - e^(K - t)
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