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
Answered by Henry F. Maths tutor

3020 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Intergrate ln(x) with resepct to x


Find d^2y/dx^2 for y=4x^4−3x^3−6x^2+x


Find two values of k, such that the line y = kx + 2 is tangent to the curve y = x^2 + 4x + 3


solve 4^xe^(7x+5) = 21


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning