Question

In Problems 3–8, determine whether the given function is a solution to the given differential equation.

$\mathbf{x}\mathbf{=}\mathbf{2}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{t}\mathbf{-}\mathbf{3}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{t}\mathbf{ }\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{x}\mathbf{=}\mathbf{0}$

Short answer

The given function is a solution to the given differential equation.

Step-by-step solution

Differentiating the given equation w.r.t. (with respect to) x

Firstly, we will differentiate $\mathrm{x}=2\cos \mathrm{t}-3\sin \mathrm{t}$with respect to t,

$\mathrm{x}\text{'}=-2\sin \mathrm{t}-3\cos \mathrm{t}$

Again, differentiating with respect to t,

$$\begin{gathered} \mathrm{x}\text{'}\text{'}=-2\cos \mathrm{t}-3\left(-\sin \mathrm{t}\right) \\ \mathrm{x}\text{'}\text{'}=-2\cos \mathrm{t}+3\sin \mathrm{t} \end{gathered}$$

Simplification

Putting the values from step 1 in the L.H.S. (Left-hand side) of the given differential equation,

$$\begin{gathered} \mathrm{x}\text{'}\text{'}+\mathrm{x}=-2\cos \mathrm{t}+3\sin \mathrm{t}+2\cos \mathrm{t}-3\sin \mathrm{t} \\ \mathrm{x}\text{'}\text{'}+\mathrm{x}=0 \end{gathered}$$

which is the same as the R.HS. (Right-hand side) of the given differential equation.

Hence, $\mathbf{x}\mathbf{=}\mathbf{2}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{t}\mathbf{-}\mathbf{3}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{t}\mathbf{ }$is a solution to the differential equation $\mathbf{x}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathbf{x}\mathbf{=}\mathbf{0}$.