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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)

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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