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Please help. Can't start it

2007-08-07 16:04:40 · 2 answers · asked by Marvin 1 in Science & Mathematics Mathematics

The current in an LC circuit is given by I′′(t) + 4I(t) = g(t), where g(t) is zero for
the first 5 seconds and then equal to 120 sin 60
2 t, then g(t) is zero again. Find the
current at time t.

2007-08-07 19:15:37 · update #1

The current in an LC circuit is given by I′′(t) + 4I(t) = g(t), where g(t) is zero for
the first 5 seconds and then equal to 120 sin (60t/2pi) for ten seconds ,then g(t) is zero again. Find the
current at time t.

2007-08-07 19:17:14 · update #2

2 answers

take the Laplace of both sides.
for g(t)
use step function
sin(60t/(2Pi))[u(t-5)-u(t-10)]
then take the inverse laplace

2007-08-07 18:33:45 · answer #1 · answered by John 5 · 0 0

I assume you just want some help in getting started. I am not an expert at differential equations but might I suggest using some sort of sin function since the function on the right involves the sin?

The second derivative of the sin is -sin so the left hand side will also involve just the sin.

Something like I = a sin(bt) so I'' = -ab^2 sin(bt)
And the equation becomes:
(-ab^2 + 4a) sin(bt) = sin(60t/2pi) for t>5 and t <15

and b = 60/2pi (-ab^2 + 4a) = 1 so a=1/(1 - b^2)

So if t <=5 or t >= 15 then I = 0
Otherwise I = a sin(bt) with b = 60/2pi , a=1/(1 - b^2)

2007-08-07 17:20:22 · answer #2 · answered by Captain Mephisto 7 · 0 1

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