Question

The displacement of a certain forced oscillator can be modeled by the DE

$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{+}\mathbf{5}\frac{\mathbf{d}\mathbf{x}}{\mathbf{d}\mathbf{t}}\mathbf{+}\mathbf{6}\mathit{x}\mathbf{=}\mathit{c}\mathit{o}\mathit{s}\mathbf{}\mathbf{\left(}\mathit{t}\mathbf{\right)}$

(a) Find all solution of the DE.

(b) Describe the long-term behavior of the oscillator.

Short answer

(a) The solution is$x\left(t\right)=\frac{1}{10}\cos \left(t\right)+\frac{1}{10}\sin \left(t\right)+e^{-2t}{c}_1+e^{-3t}{c}_2$ .

(b) The solution of the differential equation does not exist as$t\to \infty$ .

Step-by-step solution

(a) Determine the auxiliary equation of d2xdt2+5dxdt+6x=cos (t) .

Consider, the differential equation$\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dt}}^2}+5\frac{\mathrm{dx}}{\mathrm{dt}}+6x=cos\left(t\right)$ .

Assume,$D^n=\frac{d^{n}x}{d{t}^n}$ , substitute the value$D^2$ for$\frac{d^{2}x}{d{t}^2}$ , D for$\frac{dx}{dt}$ and 1 for x in the equation$\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dt}}^2}+5\frac{\mathrm{dx}}{\mathrm{dt}}+6x=0$ as follows.

$$\begin{gathered} \frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dt}}^2}+5\frac{\mathrm{dx}}{\mathrm{dt}}+6x=0 \\ {D}^2+5D+6=0 \\ {D}^2+3D+2D+6=0 \\ D\left(D+3\right)+2\left(D+3\right)=0 \end{gathered}$$

Further, simplify the equation as follows.

$$\begin{gathered} D\left(D+3\right)+2\left(D+3\right)=0 \\ \left(D+3\right)+\left(D+3\right)=0 \end{gathered}$$

Determine the complementary solution of x(t) .

The solution of the differential equation is${\left(D\pm a\right)}^n=0isf\left(t\right)=e^{\mp at}\left\{\sum _{i=0}^{n-1}{c}_{i+1}{t}^i\right\}$ .

If a is the solution of$\left(D+m\right)=0$ , b is the solution of$\left(D+n\right)=0$ , c is the solution of$\left(D+n\right)=0$ then a+b+c is solution of$\left(D+m\right)\left(D+n\right)\left(D+o\right)=0$

Using the definition, the solution of the linear differential equation is defined as follows$\left(D+2\right)\left(D+3\right)=0$ .

$$\begin{gathered} x\left(t\right)=e^{-2t}\left\{\sum _{i=0}^{1-1}{c}_{i+1}{t}^i\right\}+e^{-2t}\left\{\sum _{i=0}^{1-1}{b}_{i+1}{t}^i\right\} \\ x\left(t\right)=e^{-2t}{c}_1+e^{-3t}{b}_1 \end{gathered}$$

As$b_i$ and$c_i$ are arbitrary constant, assume$b_1=c_2$ .

Substitute the values$c_2$ for$b_1$ in the equation$x\left(t\right)=e^{-2t}{c}_1+e^{-3t}{b}_1$ as follows.

$$\begin{gathered} x\left(t\right)=e^{-2t}{c}_1+e^{-3t}{b}_1 \\ x\left(t\right)=e^{-2t}{c}_1+e^{-3t}{c}_2 \end{gathered}$$

Therefore, the complimentary solution of the differential equation is$x_c\left(t\right)=e^{-2t}{c}_1+e^{-3t}{c}_2$

Determine the particular solution of x(t) .

As the particular solution of the differential equation of the form$x_p\left(t\right)=A\cos \left(t\right)+B\sin \left(t\right)$ .

Differentiate the equation$x\left(t\right)=A\cos \left(t\right)+B\sin \left(t\right)$ both side with respect to as follows.

$$\begin{gathered} \frac{d}{dt}\left\{x\left(t\right)\right\}=\frac{d}{dt}\left\{A\cos \left(t\right)+B\cos \left(t\right)\right\} \\ \frac{d}{dt}\left\{x\left(t\right)\right\}=-A\cos \left(t\right)+B\cos \left(t\right) \end{gathered}$$

Differentiate the equation$\frac{d}{dt}\left\{x\left(t\right)\right\}=-A\cos \left(t\right)+B\cos \left(t\right)$ both side with respect to t as follows.

$$\begin{gathered} \frac{d}{dt}\left(\frac{d}{dt}\left\{x\left(t\right)\right\}\right)=\frac{d}{dt}\left\{-A\cos \left(t\right)+B\cos \left(t\right)\right\} \\ \frac{d^2}{d{t}^2}\left\{x\left(t\right)\right\}=-A\cos \left(t\right)+B\cos \left(t\right) \end{gathered}$$

Substitute the valuerole="math" localid="1661143442250" $-A\cos \left(t\right)-B\sin \left(t\right)$ for$\frac{d^2}{d{t}^2}\left\{x\left(t\right)\right\}$ ,role="math" localid="1661143461574" $-A\sin \left(t\right)-B\cos \left(t\right)$ for$\frac{d}{dt}\left\{x\left(t\right)\right\}$ androle="math" localid="1661143473317" $A\cos \left(t\right)+B\sin \left(t\right)$ for$x\left(t\right)$ in the equation$\frac{d^{2}x}{d{t}^2}+5\frac{dx}{dt}+6x=\cos \left(t\right)$ as follows.

$$\begin{gathered} -A\cos \left(t\right)+B\sin \left(t\right)-5A\sin \left(t\right)+5B\cos \left(t\right)+6A\cos \left(t\right)+6B\sin \left(t\right)=\cos \left(t\right) \\ \left\{-A+5B+6A\right\}\cos \left(t\right)+\left\{-B+5A+6B\right\}\sin \left(t\right)=\cos \left(t\right) \\ \left\{5A+5B\right\}\cos \left(t\right)+\left\{-5A+5B\right\}\sin \left(t\right)=\cos \left(t\right) \end{gathered}$$

Compare the equation both side as follows.

$$\begin{gathered} 5A+5B=1 \\ -5A+5B=0 \end{gathered}$$

Add equations$5A+5B=1$and$-5A+5B=0$ as follows.

$$\begin{gathered} \left\{5A+5B\right\}+\left\{-5A+5B\right\}=1 \\ 10B=1 \\ B=\frac{1}{10} \end{gathered}$$

Substitute the value$\frac{1}{10}$ for B in the equation$5A+5B=1$ as follows.

$$\begin{gathered} 5A+5B=1 \\ 5A+5\left(\frac{1}{10}\right)=1 \\ A=\frac{1}{10} \end{gathered}$$

Substitute the values$\frac{1}{10}$ for B and$\frac{1}{10}$ for A in the equation$x_p\left(t\right)=A\cos \left(t\right)+B\sin \left(t\right)$ as follows.

$x_p\left(t\right)=\frac{1}{10}\cos \left(t\right)+\frac{1}{10}\sin \left(t\right)$

Determine the solution of x(t) .

As the solution of the differential equation is $x\left(t\right)=x_c\left(t\right)+x_p\left(t\right)$ , substitute the values$e^{-2t}{c}_1+e^{-3t}{c}_2$ for$x_c\left(t\right)$ and$\frac{1}{10}\cos \left(t\right)+\frac{1}{10}\sin \left(t\right)$ for$x_p\left(t\right)$ in the equation$x\left(t\right)=x_c\left(t\right)+x_p\left(t\right)$ as follows.

$$\begin{gathered} x\left(t\right)=x_c\left(t\right)+x_p\left(t\right) \\ x\left(t\right)=\frac{1}{10}\cos \left(t\right)+\frac{1}{10}\sin \left(t\right)+e^{-2t}{c}_1+e^{-3t}{c}_2 \end{gathered}$$

Hence, the solution of the linear differential equation$\frac{d^{2}x}{d{t}^2}+5\frac{dx}{dt}+6x=\cos \left(t\right)$ is$\frac{1}{10}\cos \left(t\right)+\frac{1}{10}\sin \left(t\right)+e^{-2t}{c}_1+e^{-3t}{c}_2$ .

(b) Determine the behavior of  in long term x(t)=110cos(t)+110sin(t)+e-2tc1+e-3tc2 .

Consider the solution$x\left(t\right)=\frac{1}{10}\cos \left(t\right)+\frac{1}{10}\sin \left(t\right)+e^{-2t}{c}_1+e^{-3t}{c}_2$ for the differential equation$\frac{d^{2}x}{d{t}^2}+5\frac{dx}{dt}+6x=\cos \left(t\right)$ .

Take limit$t\to \infty$ both side in the equation$x\left(t\right)=\frac{1}{10}\cos \left(t\right)+\frac{1}{10}\sin \left(t\right)+e^{-2t}{c}_1+e^{-3t}{c}_2$ as follows.

$$\begin{gathered} x\left(t\right)=\frac{1}{10}\cos \left(t\right)+\frac{1}{10}\sin \left(t\right)+e^{-2t}{c}_1+e^{-3t}{c}_2 \\ \underset{t\to \infty }{\lim }\left\{x\left(t\right)\right\}=\underset{t\to \infty }{\lim }\left\{\frac{1}{10}\cos \left(t\right)+\frac{1}{10}\sin \left(t\right)+e^{-2t}{c}_1+e^{-3t}{c}_2\right\} \\ =\underset{t\to \infty }{\lim }\left\{\frac{1}{10}\cos \left(t\right)\right\}+\underset{t\to \infty }{\lim }\left\{\frac{1}{10}\sin \left(t\right)\right\}+\underset{t\to \infty }{\lim }\left\{e^{-2t}{c}_1\right\}+\underset{t\to \infty }{\lim }\left\{e^{-3t}{c}_2\right\} \\ \underset{t\to \infty }{\lim }\left\{x\left(t\right)\right\}=\infty +\infty +c_1+c_2 \end{gathered}$$

Hence, the solution of the linear differential equation$\frac{d^{2}x}{d{t}^2}+5\frac{dx}{dt}+6x=\cos \left(t\right)$ in the long term does not exist.