Finding Particular solution

1. Mar 31, 2015

joker2014

I have an exam in 2 days and Im still getting confused on how to find yp , the particular solution.
for example
y''+ 4y = 9te^t+ 4
or tsin(2t) + 2

Is it only by guessing and thats it? I still cant answer these questions 100% correctly. I would like to have your advises to find the best way to solve this complicated undertemind coeff.

2. Mar 31, 2015

the_wolfman

You can always use variation of parameters or Laplace transforms to find the particular solution.

The method of undetermined coefficients is nice because it requires a minimal amount of work when the form of the solution is obvious. It also helps develp a intuition to the solution of various differential equations. Which is useful later on. That being said, it takes practice to develop a good intuition.

3. Apr 1, 2015

Staff: Mentor

For relatively simple problems like the ones above, the method of undetermined coefficients works pretty well.
1. y'' + 4y = 9tet + 4
Homogenous problem: y'' + 4y = 0
Solution set basis: {cos(2t), sin(2t)}
Nonhomogeneous problem: y'' + 4y = 9tet + 4
Particular solution set basis: {et, tet, 1}
Particular solution: $y_p = Ae^t + Bte^t + C$
General solution: $y = c_1cos(2t) + c_2sin(2t) + Ae^t + Bte^t + C$
It's worth noting here that the "forcing function" (the right side of the nonhomogeneous problem) has no solutions in common with those of the homogeneous problem. It's also worth noting that since the right side includes tet, our particular solution has to include both et and tet. If the right side had included t2et, we would have needed to have et, tet, and t2et in the particular solution.

2. y'' + 4y = tsin(2t) + 4
Homogenous problem: y'' + 4y = 0 (same as before)
Solution set basis: {cos(2t), sin(2t)} (same as before)
Nonhomogeneous problem: y'' + 4y = tsin(2t) + 4
Particular solution set basis: {{s}cos(2t)[/s], sin(2t), tcos(2t), tsin(2t), 1}
I have the four sin/cos terms because of the tsin(2t) term in the forcing function. I have two of them lined out because cos(2t) and sin(2t) are solutions to the homogeneous problem, so couldn't possibly be solutions of the nonhomogeneous problem.
Particular solution: $y_p = Atcos(2t) + Btsin(2t) + C$
General solution: $y = c_1cos(2t) + c_2sin(2t) + Atcos(2t) + Btsin(2t) + C$

If the right side of the nonhomogeneous problem had included t2sin(2t), the basis for the nonhomog. problem would have needed to included tcos(2t), tsin(2t), t2cos(2t), and t2sin(2t).