- #1
PNGeng
- 17
- 0
The question is:
Find a second order linear equation which has y1=-3e^(2t) and y2=e^(2t)+2te^(2t) as two of its particular solutions.
Attempt at a solution:
Since it's a repeated root problem, we know r=2, therefore the characteristic equation must look like (r-2)^2=0
r^2-4r+4=0 ----------------------- > y''-4y'+4y=0
Is this problem really that easy? I have a feeling I'm misinterpreting the question cause this is way too trivial.
Find a second order linear equation which has y1=-3e^(2t) and y2=e^(2t)+2te^(2t) as two of its particular solutions.
Attempt at a solution:
Since it's a repeated root problem, we know r=2, therefore the characteristic equation must look like (r-2)^2=0
r^2-4r+4=0 ----------------------- > y''-4y'+4y=0
Is this problem really that easy? I have a feeling I'm misinterpreting the question cause this is way too trivial.
