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Question
Show that both functions y_1=e^(-2x) and y_2=xe^(-2x) are solutions to the differential equation y''+4y'+4y=0


This is what I came up with but was hoping you could check me and make sure I'm seeing this correctly? THANKS!! =)

r^2+4r+4=0
(r+2)(r+2)=0
r=-2(2x)

y_1=e^(-2x)
y_2=xe^(-2x)

So the general solution would be:

y=C_1 e^(-2x) + C_2 xe^(-2x)

Answer
Laura,

You are correct, simply from the fact that if you put in the general solution it works. However, you need to
explain how solving (r+2)(r+2) = 0 leads you to '-2(2x)'. For more maths/computing advice, visit:-

www.headingleysigns.com/maths

Bob.

Differential Equations

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Bob Marlow

Expertise

Questions on partial differential equations, particularly solution of time-dependent ones, and particularly parabolic. I have some knowledge of ODEs, but am out of practice with them.

Experience

My thesis was in this area.

Organizations
Headingley Signs

Publications
International Journal of Computers and Fluids, July 2011. Moving mesh methods for solving parabolic partial differential equations R. Marlow , M.E. Hubbard, P.K. Jimack.

Education/Credentials
PhD in computer science

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