The mass, $m\,$ grams, of a certain chemical during a reaction is modeled by a differential equation involving time $t$ , where $0≤t<π4\displaystyle 0 \le t < \frac{\pi}{4}$ .

Find the derivative with respect to $m\,$ of

\frac{1}{(1 + 2\ln m)^2}

Hence find the general solution to the differential equation

4\sec(2t) \frac{\text{d}m}{\text{d}t} = m(1 + 2\ln m)^3 \tan(2t)

for $m > e^{-1/2}$ .

Show that the particular solution of this differential equation for which the initial mass is 1 g (so $m = 1$ when $t = 0$ ) is given by

e^{A\sec t - \frac{1}{2}}

where $A\,$ is a constant to be found.

The mass, $m\,$ grams, of a certain chemical during a reaction is modeled by a differential equation involving time $t$ , where $0≤t<π4\displaystyle 0 \le t < \frac{\pi}{4}$ .

Find the derivative with respect to $m\,$ of

\frac{1}{(1 + 2\ln m)^2}

Hence find the general solution to the differential equation

4\sec(2t) \frac{\text{d}m}{\text{d}t} = m(1 + 2\ln m)^3 \tan(2t)

for $m > e^{-1/2}$ .

Show that the particular solution of this differential equation for which the initial mass is 1 g (so $m = 1$ when $t = 0$ ) is given by

e^{A\sec t - \frac{1}{2}}

where $A\,$ is a constant to be found.

1 Pure Mathematics

Question 465

Markscheme

1 Pure Mathematics

Question 465

Markscheme

1 Pure Mathematics

Question 465

1 Pure Mathematics Questions

1 Pure Mathematics

Question 465

1 Pure Mathematics Questions