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anonymous
 4 years ago
Determine whether y(x) = (2e^x) + (xe^x) is a solution of y'' + 2y' + y = 0
anonymous
 4 years ago
Determine whether y(x) = (2e^x) + (xe^x) is a solution of y'' + 2y' + y = 0

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UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.1\[y= 2e^{x} + xe^{x}\] \[y'=\] \[y''=\]

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.1find the first and second derivatives , if \(y\) is a solution then \[y'' + 2y' + y = 0\]

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.1what do you get for the first and second derivative?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes of course.., y' (x) = 2e^(x) + e^(x)  xe^(x) = e^(x)  xe^(x) right??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok for y'' (x) = e^(x)  e^(x) + xe^(x) = ex^x

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.1thats 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
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