Q15E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Use the mass-spring oscillator analogy to decide whether all solutions to each of the following differential equations are bounded as $\mathbf{t}\to \mathbf{+}\mathrm{\infty }$

1. $\mathbf{y}\mathbf{"}\mathbf{+}{\mathbf{t}}^2\mathbf{y}\mathbf{=}\mathbf{0}$
   
2. $\mathbf{y}\mathbf{"}\mathbf{-}{\mathbf{t}}^2\mathbf{y}\mathbf{=}\mathbf{0}$
   
3. $\mathbf{y}\mathbf{"}\mathbf{+}{\mathbf{y}}^5\mathbf{=}\mathbf{0}$
   
4. $\mathbf{y}\mathbf{"}\mathbf{+}{\mathbf{y}}^6\mathbf{=}\mathbf{0}$
   
5. $\mathbf{y}\mathbf{"}\mathbf{+}(\mathbf{4}\mathbf{+}\mathbf{2}\mathbf{cost})\mathbf{y}\mathbf{=}\mathbf{0}$(Mathieu’s equation)
6. $\mathbf{y}\mathbf{"}\mathbf{+}\mathbf{ty}\mathbf{\text{'}}\mathbf{+}\mathbf{y}\mathbf{=}\mathbf{0}$
7. $\mathbf{y}\mathbf{"}\mathbf{-}\mathbf{ty}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{0}$

Step-by-Step Solution

Answer

a. Therefore, the given equation is bounded.

  

b. Hence, the given equation is unbounded.

 

c. So, the given equation is bounded.

 

d. Thereafter, the given equation is unbounded.

 

e. Consequently, the given equation is bounded.

 

f. Thus, the given equation is bounded.

 

g. Then, the given equation is unbounded.

Step 1: Mass–spring oscillator equation

Mass–spring oscillator equation is given as:

$\begin{array}{c}{\mathbf{F}}_{\mathrm{ext}}\mathbf{=}\left[\mathbf{inertia}\right]\mathbf{y}\mathbf{"}\mathbf{+}\left[\mathbf{damping}\right]\mathbf{y}\mathbf{\text{'}}\mathbf{+}\left[\mathbf{stiffness}\right]\mathbf{y}\\ \mathbf{=}\mathbf{my}\mathbf{"}\mathbf{+}\mathbf{by}\mathbf{\text{'}}\mathbf{+}\mathbf{ky}\dots \left(1\right)\end{array}$ 

 

The rule for the bounded equation: Just based on stiffness we can decide whether it is bounded or not. If the stiffness is k > 0 then it is bounded and if k < 0 then it is unbounded.

Step 2: Verify the equations

(a).

 

The given equation is $\mathrm{y}"+{\mathrm{t}}^2\mathrm{y}=0.$

 

To verify: Whether the given equation is bounded or not.

 

Compare the given equation with equation (1).

 

Then, $\mathrm{m}=1,\mathrm{b}=0$ and $\mathrm{k}={\mathrm{t}}^2$.

 

So, the given equation is bounded because $\mathrm{k}={\mathrm{t}}^2>0$.

(b).

 

The equation is $\mathrm{y}"-{\mathrm{t}}^2\mathrm{y}=0$.

 

To verify: Whether the given equation is bounded or not.

 

Compare the given equation with equation (1).

 

Then, $\mathrm{m}=1,\mathrm{b}=0$ and $\mathrm{k}=-{\mathrm{t}}^2$.

 

Thus, the given equation is unbounded because $\mathbf{k}\mathbf{=}\mathbf{-}{\mathbf{t}}^2\mathbf{<}\mathbf{0}$.

Step 3: Verify the equation

(c).

 

The equation is $\mathrm{y}"+{\mathrm{y}}^5=0$.

 

To verify: Whether the given equation is bounded or not.

 

Compare the given equation with equation (1).

 

Then,$\mathrm{m}=1,\mathrm{b}=0$ and $\mathrm{k}={\mathrm{y}}^4$.

 

So, the given equation is bounded because $\mathrm{k}={\mathrm{y}}^4>0$.

(d).

 

Given that, $\mathrm{y}"+{\mathrm{y}}^6=0$.

 

To verify: whether the given equation is bounded or not.

 

Compare the given equation with equation (1).

 

Then,$\mathrm{m}=1,\mathrm{b}=0$ and $\mathrm{k}={\mathrm{y}}^5$.

Hence, the given equation is unbounded because $\mathrm{k}={\mathrm{y}}^5>0$ of for $\mathrm{y}>0$ and ${\mathrm{y}}^5<0$ for $\mathrm{y}<0$.

(e).

Given that,$\mathrm{y}"+(4+2\mathrm{cost})\mathrm{y}=0$ (Mathieu’s equation).

 

To verify: whether the given equation is bounded or not.

 

Compare the given equation with equation (1).

 

Then,$\mathrm{m}=1,\mathrm{b}=0$ and $\mathrm{k}=4+2\mathrm{cost}$.

 

So, the given equation is bounded because $\mathrm{k}=4+2\mathrm{cost}>0$.

 We know that $-1\le \mathrm{cost}\le 1$, then $4+2\mathrm{cost}>1$.

Step4: Verify the equation

(f).

 

Given that, $\mathrm{y}"+\mathrm{ty}\text{'}+\mathrm{y}=0$ .

 

To verify: whether the given equation is bounded or not.

 

Compare the given equation with equation (1).

 

Then, $\mathrm{m}=1,\mathrm{b}=\mathrm{t}$ and $\mathrm{k}=1$.

 

Consequently, the given equation is bounded because $\mathrm{k}=1>0$.

(g).

 

Given that, $\mathrm{y}"-\mathrm{ty}\text{'}-\mathrm{y}=0$.

 

To verify: whether the given equation is bounded or not.

 

Compare the given equation with equation (1).

 

Then,$\mathrm{m}=1,\mathrm{b}=-\mathrm{t}$ and $\mathrm{k}=-1$.

 

Therefore, the given equation is unbounded because $\mathrm{k}=-1<0$.