The auxiliary equation is m2+9=0⟹m=±3i. The complementary function is xc(t)=Acos3t+Bsin3t.
Since the term on the right-hand side is linearly dependent on the complementary function, the particular solution will be of the form xp(t)=t(Acos3t+Bsin3t).
Now,
xp′xp′′=t(−3Asin3t+3Bcos3t)+Acos3t+Bsin3t=t(−9Acos3t−9Bsin3t)+(−3Asin3t+3Bcos3t)−3Asin3t+3Bcos3t=t(−9Acos3t−9Bsin3t)+(−6Asin3t+6Bcos3t)
Using this in the given differential equation, we get
xp′′+9xp−9Atcos3t−9Btsin3t−6Asin3t+6Bcos3t+9Atcos3t+9Btsin3t6(Bcos3t−Asin3t)=6sin3t=6sin3t=6sin3t.
Equating the like coefficients, we get B=0,A=−1. Therefore, the particular solution is
xp=−tcos3t and the general solution is, x(t)=(A−t)cos3t+Bsin3t.
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