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. $\mathbf{y}\mathbf{"}\mathbf{-}\mathbf{ty}\mathbf{\text{'}}\mathbf{-}\mathbf{y}\mathbf{=}\mathbf{0}$

Mass–spring oscillator equation is given as:

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.

(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 $\mathbf{k}\mathbf{=}\mathbf{-}{\mathbf{t}}^2\mathbf{<}\mathbf{0}$.

(c).

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

To verify: Whetherthe 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 $4+2\mathrm{cost}>1$.

(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 $\mathrm{k}=-1<0$.