anonymous
  • anonymous
Determine whether y(x) = (2e^-x) + (xe^-x) is a solution of y'' + 2y' + y = 0
Differential Equations
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@blues
UnkleRhaukus
  • UnkleRhaukus
\[y= 2e^{-x} + xe^{-x}\] \[y'=\] \[y''=\]
UnkleRhaukus
  • UnkleRhaukus
find the first and second derivatives , if \(y\) is a solution then \[y'' + 2y' + y = 0\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
thank you :)
UnkleRhaukus
  • UnkleRhaukus
what do you get for the first and second derivative?
anonymous
  • anonymous
yes of course.., y' (x) = -2e^(-x) + e^(-x) - xe^(-x) = -e^(-x) - xe^(-x) right??
anonymous
  • anonymous
ok for y'' (x) = e^(-x) - e^(-x) + xe^(-x) = ex^-x
UnkleRhaukus
  • UnkleRhaukus
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

Looking for something else?

Not the answer you are looking for? Search for more explanations.