MA2051 - Ordinary Differential Equations
Review For Exam II - Solutions 





REVIEW QUESTIONS




1. 
This problem analyzes .




  (a)


  The characteristic equation is 
  and so the roots are  and .  This gives two
  solutions  and .  To check 
  independence, compute the Wronskian and show that it is never zero.




  (b)

 
  To find a particular solution, use the method of undetermined
  coefficients and look for a solution of the form
  .  Plug this into the
  original equation and solve for A and B to obtain 
   and B=0.  Hence, a particular solution is
  .




  (c)

 
  The general solution is
  .  The initial
  data give you equations for  and .  Solving these
  equations give the solution
  .




  (d)


  The transient component is  (because this is
  the part that converges to zero as t goes to infinity).
  This is less than  for . 
  (Just set  and solve for t;
  it is a quadratic equation for .)




  (e)


  The limiting periodic solution (obtained in part (c)) is
  .  The period is  and the amplitude is .
  



  (f)

 
  Go back to the general solution in part (c) and choose the initial
  data so that .  The right choice is
   and .






2.
The characteristic equation has roots , so that
.  Square both sides ()
and simplify to obtain the
characteristic equation .  It follows that the
original differential equation was .


For part (b), .
Similarly, with a little differentiation, 
. 


3.
This problem analyzes a spring-mass system.




  (a)


  The system is undamped and there is no forcing term, so
  the mathematical model that you want is
  

  where m = 6 and k = 98.




  (b)

 
  The general solution is 
  
  where  
  is the natural frequency 
  and the period is  seconds.


  To find the amplitude of the motion, use the initial data to solve for
   and  to obtain .
  The amplitude is  meters.




  (c)


  The period is .  If you increase the mass
  by a factor of 4, you will double the period.
  

  The amplitude will not change with a change in the mass. (It depends
  only on the initial position in this problem.)




  (d)


  The new model is
  

  and the system will exhibit oscillations as long as the characteristic
  equation has complex roots.  This occurs for
  .  Plug in the numbers to find that you have oscillations
  for .






4. Solve a system of first-order equations.




  (a)

 
  The characteristic equation is
  

  The roots are  (where ). 
  The corresponding solution pairs are
  

  

  (You obtain  from , for example, by substituting
   back into the first differential equation in the system.)
  To verify independence, compute the Wronskian
  

  The Wronskian is never zero, so the solutions are linearly independent.
  The general solution pair is 
  
  where  and  are arbitrary constants.




  (b)

  
  The initial data give you  and  in the formula
  for the general solution.




  (c)


  Plot a few points to obtain a spiral heading away from the origin.




  (d)


  Differentiate the quantity  to obtain
  

  This last quantity is always positive and so, as soon as the
  solution curve is away from the stationary point ,
  its distance from the origin is  strictly increasing.






5. 
The first step is to write the second-order equation as a first-order
system:


Euler's method starts with  and  (the initial data).
Because you want the solution at t=1 and you are using
two steps, .
The next two steps are




Your approximation for  is .  From Problem #1,
the exact value is .
The absolute error is .  The relative
error is , or about 237%.  (Bigerror.)








  

    

      © 1996 by Will Brother.
      All rights Reserved. File last modified on April 25, 1996.