How to Effectively Solve for the Particular Solution in Differential Equations?

[joker2014]

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

Is it only by guessing and that's it? I still can't answer these questions 100% correctly. I would like to have your advises to find the best way to solve this complicated undertemind coeff.

 

[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.

 

[Mark44]

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

Is it only by guessing and that's it? I still can't answer these questions 100% correctly. I would like to have your advises to find the best way to solve this complicated undertemind coeff.

For relatively simple problems like the ones above, the method of undetermined coefficients works pretty well.
1. y'' + 4y = 9te^t + 4
Homogenous problem: y'' + 4y = 0
Solution set basis: {cos(2t), sin(2t)}
Nonhomogeneous problem: y'' + 4y = 9te^t + 4
Particular solution set basis: {e^t, te^t, 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 te^t, our particular solution has to include both e^t and te^t. If the right side had included t²e^t, we would have needed to have e^t, te^t, and t²e^t 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).