Here's the question you clicked on:
gerryliyana
Determine whether y(x) = (2e^-x) + (xe^-x) is a solution of y'' + 2y' + y = 0
\[y= 2e^{-x} + xe^{-x}\] \[y'=\] \[y''=\]
find the first and second derivatives , if \(y\) is a solution then \[y'' + 2y' + y = 0\]
what do you get for the first and second derivative?
yes of course.., y' (x) = -2e^(-x) + e^(-x) - xe^(-x) = -e^(-x) - xe^(-x) right??
ok for y'' (x) = e^(-x) - e^(-x) + xe^(-x) = ex^-x
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