Determine whether y(x) = (2e^-x) + (xe^-x) is a solution of y'' + 2y' + y = 0

Question

anonymous · Answer

@blues

unklerhaukus · Answer

$$y= 2e^{-x} + xe^{-x}$$
$$y'=$$
$$y''=$$

unklerhaukus · Answer

find the first and second derivatives ,

if $y$ is a solution then 

$$y'' + 2y' + y = 0$$

anonymous · Answer

thank you :)

unklerhaukus · Answer

what do you get for the first and second derivative?

anonymous · Answer

yes of course..,

y' (x) = -2e^(-x) + e^(-x) - xe^(-x) = -e^(-x) - xe^(-x) 
right??

anonymous · Answer

ok for 

y'' (x) = e^(-x) - e^(-x) + xe^(-x) = ex^-x

unklerhaukus · Answer

thats right but i think you typed the very last bit the wrong way around.

now substitute these into 

$$y′′+2y′+y=0$$

$$ (xe^{−x})+2(−e^{−x}−xe^{−x})+(2e−x+xe^{−x})=0$$



if y is a solution then 
this will simplify to  a statement that is always true