A physical system is found to have the following differential equation with all initial condition being zero

Question

A physical system is found to have the following differential equation with all initial condition being zero $$ {{\rm d}^{2}{\rm X}\left(t\right) \over {\rm d}t^{2}} + 4\,{{\rm d}{\rm X}\left(t\right) \over {\rm d}t} + 3{\rm X}\left(t\right) = {\rm U}\left(t\right) \quad \mbox{where}\quad {\rm U}\left(t\right) = 2\cos\left(3t\right) $$ Find the value of ${\rm X}\left(t\right)$.

Answer 1

Method 1: Undetermined Coefficients

The homogeneous is:

$$m^2 + 4m + 3 \rightarrow m_{1,2} = -1, -3$$

This gives:

$$X_h(t) = c_1 e^{-t} + c_2 e^{-3t}$$

For the particular, choose $x_p(t) = a \cos 3t + b \sin 3t$ and solve for the constants.

We get $a = -\dfrac{1}{15}, b = \dfrac{2}{15}$.

We have:

$$x(t) = x_h(t) + x_p(t) = c_1 e^{-t} + c_2 e^{-3t} -\dfrac{1}{15} \cos 3t + \dfrac{2}{15} \sin 3t$$

You can now use the initial conditions $x'(0) = 0$ and $x(0) = 0$, yielding:

$$c_1 = -\dfrac{1}{10}, c_2 = \dfrac{1}{6}$$

Method 2: Laplace Transforms

• $\mathscr{L}~ (x''(t)) = s^2 x(s) - s x(0) - x'(0) = s^2 x(s)$ (due to initial conditions)
• $\mathscr{L}~ (4 x'(t)) = 4(s x(s) - x'(0)) = 4 s x(s)$ (due to initial conditions)
• $\mathscr{L}~ (3 x(t)) = 3 x(s)$
• $\mathscr{L}~ (2 \cos 3t) = \dfrac{2s}{s^2 + 9}$

Now simplify, solve for $x(s)$, do partial fractions and find the inverse Laplace Transform.