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Mathematics 21 Online
OpenStudy (anonymous):

Characteristic equation problem http://i.imgur.com/7jZcz.jpg

OpenStudy (anonymous):

Can some confirm that the characteristic equation from this is 4r^2 + 8r + 13 =0 Just need to double check that I'm right!

OpenStudy (anonymous):

sorry 4r^2 + 7r + 13 =0

OpenStudy (anonymous):

in this case we have a euler equation. these are ODE's in form ax^2y''+bxy'+cy=0 where a, b and c are any constants. now for these equations, our characteristic equation is ar(r-1)+br+c=0 so we'll have 4r(r-1)+8r+13=0 4r^2-4r+8r+13=0 4r^2+4r+13=0 is our characteristic equation.

OpenStudy (anonymous):

@anonymoustwo44 Ah I see, I think my general solution is for the equation r^2 + 8r + 13 =0 the 4 constant changes everything.

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