Q2E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

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 ${\mathit{y}}_{\mathbf{1}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{+}{\mathit{y}}_{\mathbf{2}}\mathbf{\left(}\mathit{t}\mathbf{\right)}$.

Step-by-Step Solution

Answer

a) When substituting cy in the given equation $my\text{'}\text{'}+by\text{'}+ky=0$, we will get cy  as a solution.

b) When substituting $y_1\left(t\right)+y_2\left(t\right)$ in the given equation $my\text{'}\text{'}+by\text{'}+ky=0$, we will get $y_1\left(t\right)+y_2\left(t\right)$ as a solution.

Step 1: Definition of Hooke’s Law

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 $\mathbf{\text{N/m}}$.

Step 2: Substituting the values

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.

Step 3: Verifying the linear 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.

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

Step 4: Substituting the values

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

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

  

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 $my\text{'}\text{'}+by\text{'}+ky=0$.