whats the general answer of y""+2y"+y=0

Question

anonymous · Answer

homogenous solution
 $$y^{iv}+2y^{II}+y$$
         write characteristic equations
               r^4+2r^2+1=0
          solve for r

anonymous · Answer

how to solve for r in test without calculator

anonymous · Answer

solve it like you are solving quadratic; try factorng

anonymous · Answer

i dont think it has any real root.

anonymous · Answer

(r^2    +1 )(r^2    +1)

mr.math · Answer

What kind of notation are you using imran, iv and II? :P

anonymous · Answer

superbowl notation

anonymous · Answer

y=0 is one solution..
get others by factoring y^3+y+1=0

anonymous · Answer

it should be sin/cos

jamesj · Answer

sam, no.

And yes, there are no real roots.  But that doesn't phase us, because for every complex root, its conjugate is also a root and we know (don't we?) how to put the corresponding solutions back together into real functions.

anonymous · Answer

yes ..i got u..

anonymous · Answer

I meant in terms of sin and cos

jamesj · Answer

the only other thing you need to know is how to deal with repeated roots.

For example, what are the solutions of
y'' - 2y + y = 0 ?

The corresponding characteristic equation is $ r^2 - 2r + 1 =0 $, i.e., r = 1 twice.  The general solution is
$$ y(x) = c_1e^x + c_2xe^x $$

anonymous · Answer

ah.... i know that but the real problem in test is finding the roots
y=c1Cosx + c2Sinx +x(c3Cosx + c4Sinx)

anonymous · Answer

that's it , I believe