The mass, m m\,m grams, of a certain chemical during a reaction is modeled by a differential equation involving time ttt, where 0≤t<π4\displaystyle 0 \le t < \frac{\pi}{4}0≤t<4π.
Find the derivative with respect to m m\,m of
1(1+2lnm)2 \frac{1}{(1 + 2\ln m)^2} (1+2lnm)21Hence find the general solution to the differential equation
4sec(2t)dmdt=m(1+2lnm)3tan(2t) 4\sec(2t) \frac{\text{d}m}{\text{d}t} = m(1 + 2\ln m)^3 \tan(2t) 4sec(2t)dtdm=m(1+2lnm)3tan(2t)for m>e−1/2m > e^{-1/2}m>e−1/2.
Show that the particular solution of this differential equation for which the initial mass is 1 g (so m=1m = 1m=1 when t=0t = 0t=0) is given by
m=eAsect−12 m = e^{A\sec t - \frac{1}{2}} m=eAsect−21where A A\,A is a constant to be found.
