Question #308474

A system vibrates according to the equation x ''(t)+9x(t)=6sin3t , where x is the displacement and t is the time. Determine a general solution for x(t) by using the method of undetermined coefficients .


1
Expert's answer
2022-03-10T03:55:34-0500

The auxiliary equation is m2+9=0    m=±3im^2+9 = 0 \implies m = \pm 3i. The complementary function is xc(t)=Acos3t+Bsin3tx_c (t) = A\cos 3t + B \sin 3t.


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)x_p (t)= t(A\cos3t + B\sin3t).


Now,

xp=t(3Asin3t+3Bcos3t)+Acos3t+Bsin3txp=t(9Acos3t9Bsin3t)+(3Asin3t+3Bcos3t)3Asin3t+3Bcos3t=t(9Acos3t9Bsin3t)+(6Asin3t+6Bcos3t)\begin{aligned} x_p' &= t(-3A \sin 3t + 3B \cos 3t) + A\cos 3t + B\sin 3t\\ x_p'' &= t(-9A\cos 3t -9B \sin 3t)+(-3A \sin 3t + 3B \cos 3t) -3A\sin 3t + 3B\cos3t\\ &= t(-9A\cos 3t -9B \sin 3t)+(-6A \sin 3t + 6B \cos 3t) \end{aligned}

Using this in the given differential equation, we get

xp+9xp=6sin3t9Atcos3t9Btsin3t6Asin3t+6Bcos3t+9Atcos3t+9Btsin3t=6sin3t6(Bcos3tAsin3t)=6sin3t\begin{aligned} x_p'' + 9x_p &= 6\sin3t\\ -9At\cos3t-9Bt\sin 3t-6A\sin3t\\+6B\cos3t +9At\cos3t+9Bt\sin3t &=6\sin 3t\\ 6(B\cos3t-A\sin 3t) & = 6\sin 3t \end{aligned}.


Equating the like coefficients, we get B=0,A=1B = 0, A =-1. Therefore, the particular solution is

xp=tcos3tx_p = -t\cos3t and the general solution is, x(t)=(At)cos3t+Bsin3tx(t) = (A-t)\cos3t+B\sin 3t.


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