# Thread: non homogeneous second order, should be simple

1. ## non homogeneous second order, should be simple

Find the particular solution to:
y" - 6y' + 9y = 17e^(3t)

(The homogeneous solution is y_1 = e^(3t), after finding the roots)

For the particular solution, I try:
y = Ae^(3t) ....... then y' = 3Ae^(3t)...... y" = 9Ae^(3t)

Substitute that in.....

9Ae^(3t) - 6(3Ae^(3t)) + 9(Ae^(3t)) = 17e^(3t)
(simplified)
9Ae^(3t) - 18Ae^(3t) + 9Ae^(3t) = 17e^(3t)
simplify this further, and everything on the left hand side cancels....which means
0 = 17e^(3t) which is not possible since e is asymptotic at y = 0.

What do I do!?

(edit)

Same thing happens when I tried y = At(e^(3t)). Everything on the left hand side still cancels out after I substitute

2. Originally Posted by plopony
Find the particular solution to:
y" - 6y' + 9y = 17e^(3t)

(The homogeneous solution is y_1 = e^(3t), after finding the roots)

For the particular solution, I try:
y = Ae^(3t) ....... then y' = 3Ae^(3t)...... y" = 9Ae^(3t)

Substitute that in.....

9Ae^(3t) - 6(3Ae^(3t)) + 9(Ae^(3t)) = 17e^(3t)
(simplified)
9Ae^(3t) - 18Ae^(3t) + 9Ae^(3t) = 17e^(3t)
simplify this further, and everything on the left hand side cancels....which means
0 = 17e^(3t) which is not possible since e is asymptotic at y = 0.

What do I do!?

(edit)

Same thing happens when I tried y = At(e^(3t)). Everything on the left hand side still cancels out after I substitute
Since you have two repeated roots of $3$ and $3$, the complimentary solution is

$y_1 = e^{3t}, y_2 = t e^{3t}$ so you'll need to try $y_p = At^2 e^{3t}$ .

3. thank you, it would have been logical for me to keep following the pattern eh? thanks again