Mastering Mathematics Smartly
by Wee Wen Shih

Let us look at some common ways to form differential equations (DEs) from given problem descriptions.

• Problems that involve "growth and decay": DE will take the form of "dy / dt = ky, where k > 0 for growth and k < 0 for decay".
  Check out this useful link.
  
  
• Problems that apply "Newton's law of cooling": DE will take the form of "dT / dt = -k(T - T0), where T0 is the temperature of the surroundings".
  Check out this useful link.
  
  
• Problems that involve "mixing" and "money": DE will take the form of "Total Rate = Rate In - Rate Out".
  Check out these: link 1 and link 2 (refer to the 2nd example).
  
  
• Problems that involve "population": DE will take the form of "Total Rate = Birth Rate - Death Rate".
  Check out this useful link.

It is necessary for us to pay attention to keywords such as "rate of increase", "rate of decrease", "directly proportional to", "inversely proportional to", etc. when we form DEs.

This question was asked in SgForums.

A certain radioactive material is known to decay at a rate propotional to the amount present. A block of this substance having mass of 100g originally is observed. After 40 hours, its mass reduces to 80g. Let m denote the mass of the radioactive material present at anytime t.

(a) Write down the differential equation which describes the rate of change m with respect to t.

(b) Solve the differential equation for m in terms of t.

(c) Find the mass of the substance after 80 hours.

 

This was my response:

  It is possible to proceed if we try to analyse the problem statement carefully.

  1. "decay at a rate proportional to the amount present" means:

       dm/dt = km, where k is the constant of proportionality.

  2. "having mass of 100g originally" means:

      m = 100 when t = 0.

  3. "After 40 hours, its mass reduces to 80g" means:

      m = 80 when t = 40.

  In part (a), we have dm/dt = km as our differential equation.

  In part (b), we solve the differential equation by the method of variable separable (i.e put all the m's on one side and all the t's on the other).

  integral 1/m dm = integral k dt

  ln m = kt + c

  m = e^(kt + c) = A e^(kt), where A = e^c.

  Now, we will use the information from points 2 and 3 to solve for A and then k, which I'll leave you to continue. Answers: A = 100, k = 1/40 ln (4/5).

  In (c), just find m when t = 80.