Differential Equations question: Suppose y1(t) = 2e^(2t) + te^(2t) and y2 = -e^(5t) + te^(2t) are both solutions of the second order linear equation y''+p(t)y'+q(t)y=g(t). Which of the functions below is also a solution of the same equation?

The answer is y=2e^(5t) + te^(2t), but I'm not sure how. I think it's related to there being only one real repeated root due to the te^(t) parts of the solutions, but I'm not sure how to get the final answer. Anyone have any idea?

Question

Differential Equations question: Suppose y1(t) = 2e^(2t) + te^(2t) and y2 = -e^(5t) + te^(2t) are both solutions of the second order linear equation y''+p(t)y'+q(t)y=g(t). Which of the functions below is also a solution of the same equation?

The answer is y=2e^(5t) + te^(2t), but I'm not sure how. I think it's related to there being only one real repeated root due to the te^(t) parts of the solutions, but I'm not sure how to get the final answer. Anyone have any idea?

jamesj · Answer

Yes.  Remember that an inhomogeneous ODE like this one has two types or parts of a generation solution

y = y_particular + c1.y1 + c2.y2

where y1 and y2 are the (linearly independent) solutions of the homogeneous version of that equation.

Looking at the two solutions given to you, you can pick out the particular solution and a y1 and y2.  From that you can write down the general solution.  And from that general solution you can now check the other solutions they give you in order to see which fit the general form or not.

jamesj · Answer

first sentence.  Not 'generation solution', but 'general solution'

anonymous · Answer

what's tripping me up is y2(t). if there's one real repeated root, then shouldn't that mean that y2 should equal-e^(2t)  + te^(2t)? where does the -e^(5t) come into play?

jamesj · Answer

No. the t.exp(2t) appears in both solutions so it MUST be the particular solution, because no one of these solutions can only be a solution of the homogeneous equation.

anonymous · Answer

ah, got it. Thanks!