Question 970844
1) solve for X: 8 dx/dt + 4 given x=0 @ t=0 
I assume you mean 8 * (dx/dt) + 4 = 0
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The general rule of Laplace transforms is

L[x'] = sL[x(t)] - x(0)

So taking the laplace transform of both sides 
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L[8 * x' + 4] = L[0]
L[8x'] + L[4] = L[0]
note that L[c] = c/s and L[0] = 0
8L[x'] + 4/s = 0
8*(sL[x(t)] - x(0)) + 4/s = 0
since we are given x(0) = 0
8sL[x(t)] + 4/s = 0
divide both sides of = by 4
2sL[x(t)] + 1/s = 0
2sL[x(t)] = -1/s
divide both sides of = by s
2L[x(t)] = -1/(s^2)
divide both sides of = by 2
L[x(t)] = -1/(2*s^2)
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2)  L[f(t)]=F(s)=∫_0^∞ e^(-st) f(t)dt to find the transform of the voltage shown in an exponential decay for t≥0. 
voltage shown?