MA2051 - Ordinary Differential Equations
Sample Exam #3 - Solutions


Second Exam - Orginally Given 1996 A Term



    1. Since tex2html_wrap_inline279 N/m, the governing equation is tex2html_wrap_inline281 , y(0) = 0.1, y'(0) = 0.
    2. The mass oscillates because 0.2 y'' + 80 y = 0 has a characteristic equation with pure imaginary roots: tex2html_wrap_inline289 . The general solution includes tex2html_wrap_inline291 and tex2html_wrap_inline293 . The period of the oscillations is tex2html_wrap_inline295 or tex2html_wrap_inline297 s.

      If m = 0.8, then tex2html_wrap_inline301 . The natural frequency is reduced to 10 rad/s; the period increases to tex2html_wrap_inline303 s.

    3. Since a steady state is constant, tex2html_wrap_inline305 , and tex2html_wrap_inline307 . This steady state is reasonable because a) it is negative -- the spring sags when the weight is added -- and because b) tex2html_wrap_inline305 represents a balance between the spring force due to the extension tex2html_wrap_inline311 and the force of gravity -mg.
    4. The mass oscillates so long as the characteristic equation of my'' + p y' + k y = 0 has complex roots; i.e., so long as tex2html_wrap_inline317 . Hence, oscillations are possible for tex2html_wrap_inline319 N-m/s.

      The 0.8 kg mass would oscillate over the same range of p values as m = 0.2 kg because tex2html_wrap_inline325 still holds with m = 0.8 kg; or you could argue that m = 0.8 kg oscillates over a larger range of p values, tex2html_wrap_inline333 .


    1. Characteristic equations yields r = -1, -4. Substituting tex2html_wrap_inline337 in the first equation yields tex2html_wrap_inline339 . Substituting tex2html_wrap_inline341 in the first equation yields tex2html_wrap_inline343 . A general solution is

      displaymath277

    2. Use tex2html_wrap_inline345 , tex2html_wrap_inline347 , tex2html_wrap_inline349 :

      eqnarray268


    1. Characteristic equations: tex2html_wrap_inline351 . Hence, tex2html_wrap_inline353 .
    2. Need a particular solution. Use undetermined coefficients: tex2html_wrap_inline355 . Find tex2html_wrap_inline357 forces A = 4 and tex2html_wrap_inline361 . Hence, tex2html_wrap_inline363 .
    3. Use the initial conditions in the previous general solution to find tex2html_wrap_inline365 , tex2html_wrap_inline367 . Hence, tex2html_wrap_inline369 , tex2html_wrap_inline371 , and the solution of the initial-value problem is tex2html_wrap_inline373 .