How Do You Find the Particular Solution for Second Order Linear Equations?

[hbomb]

I need some help on finding the general solution.

I can find the complimentary solution, I'm having trouble finding the particular solution. Can anyone give me any tips.

y"+9y=t^2e^(3t)+6
y"-2y'-3y=-3te^-t
y"+2y'=3+4sin(2t)

 

[Pyrrhus]

What method are you supposed to employ? Undetermined Coefficients or Variation of parameters.

 

[hbomb]

Undetermined coefficients

 

[Pyrrhus]

1) (At^2 +Bt + C)e^{3t} + D

2) Pay attention to the complimentary solution, because there's already e^{-t}, therefore you must use:

Ate^{-t} + Bt^{2} e^{-t}

3) Pay attention to the complimentary solution, because there's already a constant, therefore you must use:

At + B \sin 2t + C \cos 2t

 

[hbomb]

I don't mean to be asking too much, but could you list the special cases for putting together the form for the particular solutions.

 

[Pyrrhus]

I don't mean to be asking too much, but could you list the special cases for putting together the form for the particular solutions.

Well, they come from inspecting independent term.

 

[HallsofIvy]

The types of functions you can expect to find as solutions to "linear differential equations with constant coefficients are:
exponentials: e^ax
powers of x: xⁿ
sine and cosine: sin(ax) and cos(ax)
and products of those.
Generally speaking, if the right hand side of the equation has
exponentials, try the same exponentials.
power of x, try that and lower powers of x
sine and/or cosine, try sine AND cosine
products of those, products of what you would try for each

BUT if one of those is also a solution to the homogeneous equation, you will have to multiply by x.