Question

In Problems 27-36solve the given initial-value problem.

$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{+}{\mathit{\omega }}^{\mathbf{2}}\mathit{x}\mathbf{=}{\mathit{F}}_{\mathbf{0}}\mathit{s}\mathit{i}\mathit{n}\mathit{\omega }\mathit{t}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathit{x}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{x}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$

Short answer

$x\left(t\right)=\frac{F_0}{2{\omega }^2}\sin \omega t-\frac{F_0}{2\omega }t\cos \omega t$

Step-by-step solution

Complementary function

Consider the differential equation$\frac{d^{2}x}{d{t}^2}+{\omega }^{2}x=F_0\sin \omega t,x\left(0\right)=0,x\text{'}\left(0\right)=0 ······\left(1\right)$

Homogeneous equation associated with differential equation (1) is,

$\frac{d^{2}x}{d{t}^2}+{\omega }^{2}x=0$

Let$x\left(t\right)=e^{mt}$ . Then,$\frac{dx}{dt}=m{e}^{mt},\frac{d^{2}x}{d{t}^2}=m^2{e}^{mt}$

$$\begin{gathered} \frac{d^{2}x}{d{t}^2}+{\omega }^{2}x=0 \\ {m}^2{e}^{mt}+{\omega }^2{e}^{mt}=0 \\ \left(m^2+{\omega }^2\right)e^{mt}=0 \\ {m}^2+{\omega }^2=0 \\ m=\pm i\omega \end{gathered}$$

So,$x_c\left(t\right)=c_1\cos \omega t+c_2\sin \omega t$

Particular solution

Let particular solution has the form,

$$\begin{gathered} d{x}_p=t(A\cos \omega t+B\sin \omega t)\backslash \mathrm{hfill} \\ {x}_p\text{'}=t(-A\omega \sin \omega t+B\omega \cos \omega t)+(A\cos \omega t+B\sin \omega t) \\ {x}_p\text{'}\text{'}=t\left(-A{\omega }^2\cos \omega t-B{\omega }^2\sin \omega t\right)+(-A\omega \sin \omega t+B\omega \cos \omega t)+(-A\omega \sin \omega t+B\omega \cos \omega t) \end{gathered}$$

As$x_p$is a solution of the equation (i) we have that,

$$\begin{gathered} x_p^{\text{'}\text{'}}+{\omega }^2{x}_p=F_0\sin \omega t \\ t\left(-A{\omega }^2\cos \omega t-B{\omega }^2\sin \omega t\right)+(-A\omega \sin \omega t+B\omega \cos \omega t)+(-A\omega \sin \omega t+B\omega \cos \omega t) \\ +{\omega }^{2}t(A\cos \omega t+B\sin \omega t)=F_0\sin \omega t \\ -2A\omega \sin \omega t+2B\omega \cos \omega t=F_0\sin \omega t \end{gathered}$$

Find value of coefficient

By equating the coefficients of the corresponding elements on both sides get,

$-2A\omega =F_0,2B\omega =0$

This gives us$A=-\frac{F_0}{2\omega },B=0$

Therefore the particular solution is,

$$\begin{gathered} x_p=t\left(-\frac{F_0}{2\omega }·\cos \omega t+0·\sin \omega t\right) \\ =-\frac{F_0}{2\omega }t\cos \omega t \end{gathered}$$

The general solution is:

$$\begin{gathered} x\left(t\right)=x_c\left(t\right)+x_p\left(t\right) \\ x\left(t\right)=c_1\cos \omega t+c_2\sin \omega t-\frac{F_0}{2\omega }t\cos \omega t \end{gathered}$$

Apply initial condition

Differentiating we have:

$$\begin{gathered} x\left(t\right)=c_1\cos \omega t+c_2\sin \omega t-\frac{F_0}{2\omega }t\cos \omega t \\ {x}^{\text{'}}\left(t\right)=-c_1\omega \sin \omega t+c_2\omega \cos \omega t-\frac{F_0}{2\omega }(-t\omega \sin \omega t+\cos \omega t) \end{gathered}$$

Use$x\left(0\right)=0$gives,

$$\begin{gathered} x\left(0\right)=c_1\cos \omega \left(0\right)+c_2\sin \omega \left(0\right)-\frac{F_0}{2\omega }\left(0\right)\cos \omega \left(0\right) \\ 0=c_1\cos \left(0\right)+c_2\sin \left(0\right)-\frac{F_0}{2\omega }\left(0\right)\cos \left(0\right) \\ 0=c_1 \end{gathered}$$

For$x\text{'}\left(0\right)=0$

$$\begin{gathered} x\text{'}\left(0\right)=-c_1\omega \sin \omega \left(0\right)+c_2\omega \cos \omega \left(0\right)-\frac{F_0}{2\omega }(-(0\left)\omega \sin \omega \right(0)+\cos \omega (0\left)\right) \\ 0=-c_1\omega \sin \left(0\right)+c_2\omega \cos \left(0\right)-\frac{F_0}{2\omega }(-(0\left)\sin \right(0)+\cos (0\left)\right) \\ \frac{F_0}{2\omega }=c_2 \end{gathered}$$

So, the solution is the required solution is $\mathit{x}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{F}}_{\mathbf{0}}}{\mathbf{2}{\mathbf{\omega }}^{\mathbf{2}}}\mathit{s}\mathit{i}\mathit{n}\mathit{\omega }\mathit{t}\mathbf{-}\frac{{\mathbf{F}}_{\mathbf{0}}}{\mathbf{2}\mathbf{\omega }}\mathit{t}\mathit{c}\mathit{o}\mathit{s}\mathit{\omega }\mathit{t}$