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i need another check \[y'' + 9y = \sec (3t)\] i got \[y = C_1 \cos (3t) + C_2 \sin (3t) + \cos (3t) \ln |\cos (t)| + \frac{t\sin (3t)}{3}\] is that right?
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ok first for complementary solution m^2+9=0 m=+-3i so yc=(c1cos(3t)+c2sin(3t)
so complementary is correct:)
now for particular did u use variation of parameters?
wait..i changed my mind...should be 1/9 cos (3t) ln (cos t0
and yes i used variation
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ok let me check. i am just checking not solving :)
uhmm yeah sure
yes you are 100% correct
except for the 1/9 with the log term
it should be 1/9cos(3t)lncos(t)
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thanks!!!
yw:)
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