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.
Practise AQA A Level Maths 1.11 H: Integration with exam-style questions for A Level Maths. 436 questions covering 1.11.1 Fundamental Theorem of Calculus, 1.11.2 Integrating standard functions, 1.11.3 Definite integrals and areas, 1.11.4 Integration as the limit of a sum (A-level only), 1.11.5 Integration by substitution and by parts (A-level only), 1.11.6 Integration using partial fractions (A-level only), 1.11.7 Differential equations with separable variables (A-level only), and 1.11.8 Interpreting solutions of differential equations (A-level only), matched to the AQA A Level Maths (7357) specification and written in Paper 1, Paper 2 and Paper 3 style. Every question includes a full worked solution and mark scheme, so you can see where marks are awarded rather than just whether you got the answer right.
