Question

Given that $\mathbf{x}\left(t\right)\mathbf{=}{\mathbf{c}}_{\mathbf{1}}\mathbf{cos}\mathbf{}\mathbf{\omega t}\mathbf{+}{\mathbf{c}}_{\mathbf{2}}\mathbf{sin}\mathbf{}\mathbf{\omega t}$is the general solution of $\mathbf{x}\mathbf{\text{'}\text{'}}\mathbf{+}{\mathbf{\omega }}^{\mathbf{2}}\mathbf{x}\mathbf{=}\mathbf{0}$ on the interval $\left(-\infty ,\infty \right)$ , show that a solution satisfying the initial conditions $\mathbf{x}\left(0\right)\mathbf{=}{\mathbf{x}}_{\mathbf{0}}\mathbf{,}\mathbf{x}\mathbf{\text{'}}\left(0\right)\mathbf{=}{\mathbf{x}}_{\mathbf{1}}$is given by

$\mathbf{x}\left(t\right)\mathbf{=}{\mathbf{x}}_{\mathbf{0}}\mathbf{cos}\mathbf{}\mathbf{\omega t}\mathbf{+}\frac{{\mathbf{x}}_{\mathbf{1}}}{\mathbf{\omega }}\mathbf{sin}\mathbf{}\mathbf{\omega t}$

Short answer

Proved

Step-by-step solution

Derive the given equation under initial condition using intervals

We are given solution $\mathrm{x}\left(\mathrm{t}\right)$

Use initial condition $\mathrm{x}\left(0\right)={\mathrm{x}}_0$to evaluate one constant.

${\mathrm{c}}_1\cos \mathrm{\omega }.0+{\mathrm{c}}_2\sin \mathrm{\omega }.0={\mathrm{x}}_0\Rightarrow {\mathrm{c}}_1={\mathrm{x}}_0$

Calculate $\mathrm{x}\text{'}\left(\mathrm{t}\right)$.

role="math" localid="1667902796864" $\mathrm{x}\text{'}\left(\mathrm{t}\right)=-{\mathrm{\omega c}}_1\sin \mathrm{\omega t}+{\mathrm{\omega c}}_2\cos \mathrm{\omega t}$

Use initial condition $\mathrm{x}\text{'}\left(0\right)={\mathrm{x}}_1$to evaluate other constant.

$-\omega c_1\sin \omega .0+\omega c_2\cos \omega .0=x_1\Rightarrow c_2=\frac{x_1}{\omega }$

Plug in calculated constants back into solution $\mathrm{x}\left(\mathrm{t}\right)$to obtain particular solution

$\mathrm{x}\left(\mathrm{t}\right)={\mathrm{x}}_0\cos \mathrm{\omega t}+\frac{{\mathrm{x}}_1}{\mathrm{\omega }}\sin \mathrm{\omega t}$

Hence, it is shown that a solution satisfying the initial conditions