Given that dx/dt = (1+2x)*4e^(-2t) and x = 1/2 when t = 0, show that ln[2/(1+2x)] = 8[1 - e^(-2t)]
1/(1+2x) dx = 4e^(-2t) dt Integrate both sides: ln[2/(1+2x)] = -8e^(-2t) + c input x = 1/2, t = 0: ln(2/2) = -8*(1) + c ln 1 = 0, so c = 8ln[2/1+2x] = 8[1-e^(-2t)]
HF
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