Question

I am do not know how to proceed as to how to show that y = xe^{-2x} is a solution of the differential equation y'' + 4y' + 4y = 0.

This is what I said: (attempt):
http://i1084.photobucket.com/albums/j409/QRAWarrior/MATA36-IMG-006.png

Answer 1

the characteristic equation would be:
r^2 +4r +4 = 0
--> r = -2

solution:
\[y= c e^{-2x}\]

Answer 2

I got r = -2.

Answer 3

But I have c1, and I have c2.

Answer 4

sorry didn't look at pic first

Answer 5

yes what you said works, if c1=0 and c2 = 1 then the given function is a particular solution to that DE

Answer 6

But does that really "show"?

Answer 7

It says to "show" this

Answer 8

oh i see, well then you need to take 1st and 2nd derivatives and show it equals 0

Answer 9

y' = e^(-2x) -2xe^(-2x)

y'' = -4e^(-2x) +4xe^(-2x)

--> y'' +4y' +4y = 0
--> -4e^(-2x) +4xe^(-2x) + 4e^(-2x)-8xe^(-2x) +4xe^(-2x) = 0
--> 0 = 0

Answer 10

yep!

they tell you what "y" is... and they give you an equation that contains y' and y''. You simply take the derivative of y twice to get y' and y''. then plug that in! if it works you have shown that it is infact a solution

Answer 11

The characteristic root r=-2 was a double roots, hence the two
linearly independent solutions are \( e^{-2 x}\) and \( x e^{-2 x}\)

Answer 12

To show it's true, substitute it back in and see that the equation holds.