A couple of puzzles from Russia

A generation (or two) ago, the most interesting textbooks in India came from the USSR - they were typically written by the best mathematicians and scientists of that country, and mixed a sense of fun into the excitement the authors obviously felt about their subjects. Some of these books are again becoming available, but through foreign publishers and not at the old prices of a few rupees per book! One that has been reborn is a slim volume titled "Functions and Graphs" by I.M. Gelfand and two others. Here are two puzzles from this book:

1. An honest merchant knew one arm of the weighing scales he was using was slightly longer than the other. In an effort to be fair, he decided to weigh half the merchandise to each buyer on one pan, and half on the other. Did he gain or lose as a result?
2. Seven matchboxes are arranged in a circle. The first contains 19 matches, the second 9, and then 26, 8, 18, 11, and 14 respectively. Matches can be moved from any box to an adjacent box. How should they be shifted so that all the boxes have an equal number and we have shifted as few matches as possible?

The First Quiz

Students of the Ordinary Differential Equations course had their first quiz on Monday, August 29, 2011 - exactly one week into their first semester. There was some consternation till it was pointed out that an early quiz is necessarily an easy quiz.

Here are the questions and their solutions:

1. Give the order and degree of $ 2x^2y^{\prime\prime} -3y^\prime +y=0$.
   
   Solution: The highest order derivative present is of second order ($y^{\prime\prime}$), so the ODE has order 2. The highest order derivative is present with degree 1, so the ODE has degree 1.
   
   
2. 
   
   
   1. Verify that $y(x)= ce^{-x}+2$ is a general solution to $y^\prime +y =2$.
      
      Solution: We calculate $y^\prime = -ce^{-x}$. Now we substitute $y$ and $y^\prime$ into the ODE:
      \[ LHS = y^\prime +y = -ce^{-x} + ce^{-x}+2 = 2 = RHS\]
      So the ODE is satisfied and the given $y$ is a general solution.
   2. Find the particular solution to this ODE given that $y=3.2$ when $x=0$.
      
      Solution: Substitute this pair of values into the general solution:
      \[ce^{-0} + 2 = 3.2\]
      We solve this for $c$ and get $c=1.2$. So the particular solution for the given initial condition is $y= 1.2 e^{-x}+2$.
      
   3. Graph this solution.
      
      Solution: