Answer to Question #308474 in Differential Equations for Rhulani

Question

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 .

Answer 1

The auxiliary equation is $m^2+9 = 0 \implies m = \pm 3i$. The complementary function is $x_c (t) = A\cos 3t + B \sin 3t$.

Since the term on the right-hand side is linearly dependent on the complementary function, the particular solution will be of the form $x_p (t)= t(A\cos3t + B\sin3t)$.

Now,

$\begin{aligned} x_p' &= t(-3A \sin 3t + 3B \cos 3t) + A\cos 3t + B\sin 3t\\ x_p'' &= t(-9A\cos 3t -9B \sin 3t)+(-3A \sin 3t + 3B \cos 3t) -3A\sin 3t + 3B\cos3t\\ &= t(-9A\cos 3t -9B \sin 3t)+(-6A \sin 3t + 6B \cos 3t) \end{aligned}$

Using this in the given differential equation, we get

$\begin{aligned} x_p'' + 9x_p &= 6\sin3t\\ -9At\cos3t-9Bt\sin 3t-6A\sin3t\\+6B\cos3t +9At\cos3t+9Bt\sin3t &=6\sin 3t\\ 6(B\cos3t-A\sin 3t) & = 6\sin 3t \end{aligned}$.

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

$x_p = -t\cos3t$ and the general solution is, $x(t) = (A-t)\cos3t+B\sin 3t$.