If ${\mathit{F}}_{\mathit{e}\mathit{x}\mathit{t}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{0}$, equation (3) becomes $\mathit{m}\mathit{y}\text{'}\text{'}\mathbf{+}\mathit{b}\mathit{y}\text{'}\mathbf{+}\mathit{k}\mathit{y}\mathbf{=}\mathbf{0}$. For this equation, verify the following:

(a) If $\mathit{y}\mathbf{\left(}\mathit{t}\mathbf{\right)}$is a solution so is $\mathit{c}\mathit{y}\mathbf{\left(}\mathit{t}\mathbf{\right)}$, for any constant c.

(b) If${\mathit{y}}_{\mathbf{1}}\mathbf{\left(}\mathit{t}\mathbf{\right)}$and${\mathit{y}}_{\mathbf{2}}\mathbf{\left(}\mathit{t}\mathbf{\right)}$are solutions, so is their sum ${\mathit{y}}_{\mathbf{1}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{+}{\mathit{y}}_{\mathbf{2}}\mathbf{\left(}\mathit{t}\mathbf{\right)}$.

Hooke’s law states that the strain of the material is proportional to the applied stress within the elastic limit of that material.

$\mathit{F}\mathbf{=}\mathbf{-}\mathit{k}\mathit{x}$

In the equation, F is the force, x is the extension length, k is the constant of proportionality known as spring constant in $\mathbf{\text{N/m}}$.

Given equation is $my\text{'}\text{'}+by\text{'}+ky=0$, now substitute cy in the given equation.

$$\begin{gathered} cym\left(cy\right)\text{'}\text{'}+b\left(cy\right)\text{'}+k\left(cy\right)=0 \\ cmy\text{'}\text{'}+cby\text{'}+cky=0 \\ c\left(my\text{'}\text{'}+by\text{'}+ky\right)=0 \\ my\text{'}\text{'}+by\text{'}+ky=0 \end{gathered}$$

Therefore $cy\left(t\right)$is a solution of given equation.

Given $y_1\left(t\right)$and $y_2\left(t\right)$are the solutions of the given equation $my\text{'}\text{'}+by\text{'}+ky=0$. Now you have to verify that the linear combination of $y_1\left(t\right)$and $y_2\left(t\right)$in the given equation.

$$\begin{gathered} my\text{'}\text{'}+by\text{'}+ky=0 \\ m\left(y_1\left(t\right)+y_2\left(t\right)\right)\text{'}\text{'}+b\left(y_1\left(t\right)+y_2\left(t\right)\right)\text{'}+k\left(y_1\left(t\right)+y_2\left(t\right)\right)=0 \\ \left[m{y}_1\left(t\right)\text{'}\text{'}+b{y}_1\left(t\right)\text{'}+k{y}_1\left(t\right)\right]+\left[m{y}_2\left(t\right)\text{'}\text{'}+b{y}_2\left(t\right)\text{'}+k{y}_2\left(t\right)\right]=0 \dots \left(a\right) \end{gathered}$$

Given $y_1\left(t\right)$and $y_2\left(t\right)$are the solutions, so

role="math" localid="1664097230506" $$\begin{gathered} m{y}_1\left(t\right)\text{'}\text{'}+b{y}_1\left(t\right)\text{'}+k{y}_1\left(t\right)=0 \dots \left(b\right) \\ m{y}_2\left(t\right)\text{'}\text{'}+b{y}_2\left(t\right)\text{'}+k{y}_2\left(t\right)=0 \dots \left(c\right) \end{gathered}$$

Substitute (b) and (c) in ,

$$\begin{gathered} 0+0=0 \\ 0=0 \end{gathered}$$

So, we get LHS=RHS.

Therefore$y_1\left(t\right)+y_2\left(t\right)$ is a solution of given equation.

Hence $cy\left(t\right)$and $y_1\left(t\right)+y_2\left(t\right)$are the solutions of the given equations $my\text{'}\text{'}+by\text{'}+ky=0$.