Question

Use the general solution of $\mathbf{x}\mathbf{\text{'}\text{'}}\mathbf{+}{\mathbf{\omega }}^{\mathbf{2}}\mathbf{x}\mathbf{=}\mathbf{0}$given in Problem 7 to show that a solution satisfying the initial conditions $\mathbf{x}\left(t_0\right)\mathbf{=}{\mathbf{x}}_{\mathbf{0}}\mathbf{,}\mathbf{x}\mathbf{\text{'}}\left(t_0\right)\mathbf{=}{\mathbf{x}}_{\mathbf{1}}$is the solution given in Problem 7 shifted by an amount ${\mathbf{t}}_{\mathbf{0}\mathbf{}}\mathbf{:}\mathbf{}\mathbf{x}\left(t\right)\mathbf{=}{\mathbf{x}}_{\mathbf{0}}\mathbf{cos\omega }\left(t-t_0\right)\mathbf{+}\frac{{\mathbf{x}}_{\mathbf{1}}}{\mathbf{\omega }}\mathbf{sin\omega }\left(t-t_0\right)$.

Short answer

Proved

Step-by-step solution

Find the first equation satisfying the first constant

The initial conditions are:

$$\begin{gathered} \mathrm{x}\left({\mathrm{t}}_0\right)={\mathrm{x}}_0 \\ \mathrm{x}\text{'}\left({\mathrm{t}}_0\right)={\mathrm{x}}_1 \end{gathered}$$

Substituting the first condition in the general solution,

$$\begin{gathered} x_0=c_1\cos \omega t_0+c_2\sin \omega t_0 \\ {c}_1\cos \omega t_0=x_0-c_2\sin \omega t_0 \\ {c}_1=\frac{x_0-c_2\sin \omega t_0}{\cos \omega t_0} \end{gathered}$$

Find the second equation satisfying the first constant

Determining the first derivative of the general solution,

$\mathrm{x}\text{'}\left({\mathrm{t}}_0\right)=-\mathrm{\omega }{\mathrm{c}}_1\sin {\mathrm{\omega t}}_0+\mathrm{\omega }{\mathrm{c}}_2\cos {\mathrm{\omega t}}_0$

Substituting the second condition in the general solution,

$$\begin{gathered} x_1=-{\mathrm{\omega c}}_1\sin {\mathrm{\omega t}}_0+{\mathrm{\omega c}}_2\cos {\mathrm{\omega t}}_0 \\ {\mathrm{\omega c}}_1\sin {\mathrm{\omega t}}_0=-x_1+{\mathrm{\omega c}}_2\cos {\mathrm{\omega t}}_0 \\ {c}_1=\frac{-x_1+{\mathrm{\omega c}}_2\cos {\mathrm{\omega t}}_0}{\omega \sin \omega t_0} \end{gathered}$$

Find the value for second constant of the general solution

Comparing the two equations satisfying the first constant,

$$\begin{gathered} \frac{x_0-c_2\sin {\mathrm{\omega t}}_0}{\cos \omega t_0}=\frac{-x_1+{\mathrm{\omega c}}_2\cos {\mathrm{\omega t}}_0}{\omega \sin \omega t_0} \\ {\mathrm{\omega c}}_2{\cos }^2{\mathrm{\omega t}}_0-x_1\cos {\mathrm{\omega t}}_0=x_0\omega \sin {\mathrm{\omega t}}_0-{\mathrm{\omega c}}_2{\sin }^2{\mathrm{\omega t}}_0 \\ {\mathrm{\omega c}}_2\left({\cos }^2{\mathrm{\omega t}}_0+{\sin }^2{\mathrm{\omega t}}_0\right)=x_0\omega \sin {\mathrm{\omega t}}_0+x_1\cos \omega t_0 \\ {\mathrm{\omega c}}_2\left(1\right)=x_0\omega \sin {\mathrm{\omega t}}_0+x_1\cos \omega t_0 \end{gathered}$$

Solving for ${\mathrm{c}}_2$,

${\mathrm{c}}_2=\frac{{\mathrm{x}}_0\mathrm{\omega }\sin {\mathrm{\omega t}}_0+{\mathrm{x}}_1\cos {\mathrm{\omega t}}_0}{\mathrm{\omega }}$

Find the value for first constant of the general solution

Substituting the value of ${\mathrm{c}}_2$in one of the equations satisfying the first constant,

$$" width="9" height="19" role="math">c1=x0x0ωsinωt0+x1cosωt0ωsinωt0cosωt0=x0ω1-sin2ωt0-x1sinωt0cosωt0ωcosωt0

Find the value general solution satisfying the initial condition

Substituting the values of the constants in the general solution,

$$\begin{gathered} x\left(t\right)=\left(\frac{{\mathrm{x}}_0\mathrm{\omega }\left(1-{\sin }^2{\mathrm{\omega t}}_0\right)-{\mathrm{x}}_1\sin {\mathrm{\omega t}}_0\cos {\mathrm{\omega t}}_0}{\mathrm{\omega cos}{\mathrm{\omega t}}_0}\right)\cos \omega t+\left(\frac{{\mathrm{x}}_0\mathrm{\omega sin}{\mathrm{\omega t}}_0+{\mathrm{x}}_1\cos {\mathrm{\omega t}}_0}{\omega }\right)\sin \omega t \\ =\frac{x_0\mathrm{\omega cos}{\mathrm{\omega t}}_0-x_1\sin \omega t_0}{\omega }\cos \omega t+\frac{{\mathrm{x}}_0\mathrm{\omega sin}{\mathrm{\omega t}}_0+x_1\cos {\mathrm{\omega t}}_0}{\omega }\sin \omega t \\ =\frac{x_0\mathrm{\omega cos}{\mathrm{\omega t}}_0\cos \mathrm{\omega t}-x_1\sin \omega t_0\cos \mathrm{\omega t}+{\mathrm{x}}_0\mathrm{\omega sin}{\mathrm{\omega t}}_0\sin \omega t+x_1\cos {\mathrm{\omega t}}_0\sin \omega t}{\omega } \\ =x_0\cos {\mathrm{\omega t}}_0\cos \mathrm{\omega t}+x_0\sin \omega t_0\sin \mathrm{\omega t}-\frac{x_1}{\omega }\sin {\mathrm{\omega t}}_0\cos \omega t+\frac{x_1}{\omega }\cos {\mathrm{\omega t}}_0\sin \omega t \end{gathered}$$

Simplifying it further,

$$\begin{gathered} x\left(t\right)=x_0\left(\cos {\mathrm{\omega t}}_0\cos \mathrm{\omega t}+\sin \omega t_0\sin \mathrm{\omega t}\right)-\frac{x_1}{\omega }\left(\sin {\mathrm{\omega t}}_0\cos \omega t+\cos {\mathrm{\omega t}}_0\sin \omega t\right) \\ =x_0\cos \omega \left(t-t_0\right)+\frac{x_1}{\omega }\sin \omega \left(t-t_0\right) \end{gathered}$$

localid="1668489622024" $\mathbf{}\mathbf{Hence}\mathbf{,}\mathbf{}\mathbf{a}\mathbf{}\mathbf{solution}\mathbf{}\mathbf{satisfying}\mathbf{}\mathbf{the}\mathbf{}\mathbf{initial}\mathbf{-}\mathbf{value}\mathbf{}\mathbf{problem}\mathbf{}\mathbf{is}\mathbf{}\mathbf{}\mathbf{}\mathbf{x}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathbf{x}}_{\mathbf{0}\mathbf{}}\mathbf{cos}\mathbf{}\mathbf{\omega }\mathbf{(}\mathbf{t}\mathbf{-}{\mathbf{t}}_{\mathbf{0}}\mathbf{)}\mathbf{+}\frac{{\mathbf{x}}_{\mathbf{1}}}{\mathbf{\omega }}\mathbf{sin}\mathbf{}\mathbf{\omega }\mathbf{(}\mathbf{t}\mathbf{-}{\mathbf{t}}_{\mathbf{0}}\mathbf{)}\mathbf{.}$