Finding the general solution to a differential equation

Click For Summary
The discussion centers on solving the differential equation d²y/dt² + 4dy/dt + 20y = e^(-2t)(sin4t + cos4t). The homogeneous solution is identified as y = k1e^(-2t)cos4t + k2e^(-2t)sin4t. The user attempted to find a particular solution using the form ae^(-2+4i) but encountered difficulties. It is suggested that a suitable particular solution might be yp = te^(-2t)(Ccos(4t) + Dsin(4t), although the user initially believes this approach also fails. The conversation emphasizes the importance of checking work for errors in the solution process.
clarineterr
Messages
12
Reaction score
0

Homework Statement


\frac{d^{2}y}{dt} +4\frac{dy}{dt}+20y=e^{-2t}(sin4t+cos4t)

Homework Equations





The Attempt at a Solution



The solution to the homogeneous equation: \frac{d^{2}y}{dt} +4\frac{dy}{dt}+20y=0 is

y= k1e^{-2t}cos4t +k2e^{-2t}sin4t

Then I guessed ae^{-2+4i} as a possible solution and it didn't work, and that's where I'm stuck.
 
Physics news on Phys.org
Since the complementary solution yc= e-2t(Acos(4t) + Bsin(4t)) is repeated in the non-homogeneous term, for your particular solution try

yp = te-2t(Ccos(4t) + Dsin(4t))
 
Nope...didn't work. :(
 
clarineterr said:
Nope...didn't work. :(

Yes, it does work. Check your work or show it here if you can't find your error.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 20 ·
Replies
20
Views
2K
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
7
Views
2K