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Unformatted text preview: ary equation:
is 0, ﬁnd pp= )dD dS+= thedp =25p +4100 = or of the= 84 ±+i. p = 100.
Corresponding dtp ? dt e somet), e sin(3t). diﬀerent 2
cos(3 solutions for
dt values of p .
These are linearly independent on IR by Proposition 3.37 in Course Notes.
(b) Associated homogeneous equation: p00 + 8p0 + 25pp=(t) = c e 4t cos(3t) + c e 4t sin(3t).
General solution of associated homogeneous DE: h 64 11
Auxiliary equation: 2 + 8 + (t) = 0 ) ) =p (t8+=2 K . 00 = 4 ± 3i.
RHS of inhomogeneous DE: F 25 = 100
Corresponding solutions: e 4t25p =t100 )sin(3t).
Substituting into DE yields: cos(3 ), e
p = 4.
These are linearly independent on IR by p(t) = c1 e 4t 3.37 in) Course4t sin(3t) + 4.
General solution of inhomogeneous DE: Proposition cos(3t + c2 e Notes.
General solution of associated homogeneous = ( ph (t) =2c1 e 44tcos(3tt) +3c2 e 4t 2 )e 4t).
p0 (t) DE: 4c1 +3c )e t cos(3) ( c1 +4c sin(3 sin(3t).
RHS ofICs: p(0) = p0 ) c1 F (4)= p100 ) =...
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This document was uploaded on 02/09/2014.
- Winter '14