Question

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

Answer 1

@blues

Answer 2

\[y= 2e^{-x} + xe^{-x}\]
\[y'=\]
\[y''=\]

Answer 3

find the first and second derivatives ,

if \(y\) is a solution then

\[y'' + 2y' + y = 0\]

Answer 4

thank you :)

Answer 5

what do you get for the first and second derivative?

Answer 6

yes of course..,

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

Answer 7

ok for

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

Answer 8

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