Nonhomogeneous with constant coefficients equation

[Nok1]

y'' - 3y' + 2y = e^t + t²

r = 1, 2 -> y_c = c₁e^t+c₂e^2t

y_p1 = Ate^t since Ae^t is a linear combination of our solution to y_c.

y_p2 = At²+Bt+C
y'_p2 = 2At+B
y''_p2 = 2A

via substitution we have

2A-3(2At+B)+2(At²+Bt+C) = t²

by isolating terms:

• 2At² = t²
  2A = 1 -> A = 1/2
• -6At + 2Bt = 0
  2B=3 -> B = 3/2
• 2A-3B+2C = 0
  1 - 9/2 + 2C = 0 -> C = -3.5/2

y_general= y_c+y_p1+y_p2


Question 1: How do I determine the A for the Ate^t term? same way?
Question 2: Can anyone confirm that I am doing the problem correctly, or am on the right track at all?

Thanks

 

[Office_Shredder]

You're almost there, but you should solve for y_p1 and y_p2 as one function, not two.

You have a linear equation in y, so if you find y_c which is the general solution of the homogenous equation, you only need to find one y_p which is a solution to your non-homogenous equation. Then your general solution will be of the form y_p + y_c. Since you have a t² and an e^t we would guess that y_p has t² and e^t terms in it. Looking at the equation, you realize that once you take the derivative of t² you have some t terms that you need to cancel, so you should probably include a t in your y_p guess. Similarly for a constant term. Also, e^t is in the homogenous solution, so you know that part will disappear. So you guess you need a te^t instead. Thus your guess for y_p will be

y_p = Ate^t + Bt² + Ct + D

This looks what you would have for y_p1 + y_p2 so you were doing this in a sideways fashion, but got a bit lost in the process. So exactly as you did for y_p2 you plug your y_p into your differential equation, and you collect terms. You'll get a coefficient times te^t in terms of A,B,C and D that needs to be zero, a coefficient in front of e^t in terms of A,B,C,D that needs to be 1, a coefficient in front of t² that needs to be 1, a coefficient in front of t that needs to be 0 and a constant coefficient that needs to be 0. Solve the system of equations (if more than one solution exists, just pick anyone since it doesn't make a difference)

 

[Nok1]

You're almost there, but you should solve for y_p1 and y_p2 as one function, not two.

It shouldn't matter in the end, as it should come out to be the same? I mean, y_p=y_p1+y_p2 right? The reason I'm splitting it appart is because it's easier to solve for (for me at least) than when everything is combined. But yeah, y=y_c+y_p

So I guess the correct answer continue something like this:
Solving for y_p1:
y_p1=Ate^t
y'_p1=Ate^t+Ae^t
y''_p1=Ate^t+2Ae^t

• Ate^t+2Ae^t-3(Ate^t+Ae^t)+2(Ate^t)=e^t (which is trivial to solve)
  Ate^t-3Ate^t+2Ate^t = 0
  0=0 heh.
• 2Ae^t-3Ae^t=e^t
  -1A = 1 -> A = -1

So then using that,
y_p=-te^t+1/2t²+3/2t-3.5/2

Which means that the General solution is:
y_general=c₁e^t+c₂e^2t-te^t+1/2t²+3/2t-3.5/2

 

[HallsofIvy]

It really doesn't matter whether you treat e^t+ t² as a single function or as two separate functions.

You do the Ate^t exactly like the At²+ Bt+ C:

If y= Ate^t, then y'= Ae^t+ Ate^t and y"= 2Ae^t+ Ate^t. Putting those into the equation, y"- 3y'+ 2y= (2Ae^t+ Ate^t)- 3(Ae^t+ Ate^t)+ 2(Ate^t)= -3Ae^t. The terms involving a multiplication by t have canceled just as we would expect. -Ae^t= e^t so A= -1. y= -te^t is a particular solution.

 

[Nok1]

It really doesn't matter whether you treat e^t+ t² as a single function or as two separate functions.

You do the Ate^t exactly like the At²+ Bt+ C:

If y= Ate^t, then y'= Ae^t+ Ate^t and y"= 2Ae^t+ Ate^t. Putting those into the equation, y"- 3y'+ 2y= (2Ae^t+ Ate^t)- 3(Ae^t+ Ate^t)+ 2(Ate^t)= -3Ae^t. The terms involving a multiplication by t have canceled just as we would expect. -Ae^t= e^t so A= -1. y= -te^t is a particular solution.

Thank you very much for your help :).