User:Egm4313.s12.team6.berthoumieux

Statement
For each ODE in Fig.2 in K 2011 p.3 (except the last one involving a system of 2 ODEs), determine the order, linearity (or lack of), and show wether the principle of superposition can be applied.

Given
$$(1) {y}''=g=constant$$

$$(2) m{v}'=mg-bv^{2}$$

$$(3) {h}'=-k\sqrt{h}$$

$$(4) m{y}''+ky=0$$

$$(5) y''+\omega_{0}^2=cos(\omega t)$$,  $$\omega_{0}=\omega$$

$$(6) L{I}''+R{I}'+\frac{1}{c}I={E}'$$

$$(7) EIy^{iv}=\mathit{f(x)}$$

$$(8) \mathit{L{\theta}''+gsin{\theta}=0}$$

Solution
$$(1) {y}''=g=constant$$

Order:2nd order. The highest derivative is a 2nd derivative on the y.

Linearity:Linear.

Superposition:Yes.

$$(2) m{v}'=mg-bv^{2}$$

Order:1st order. The highest derivative is a 1st on the v.

Linearity:Non-linear.

Superposition:No.

$$(3) {h}'=-k\sqrt{h}$$

Order:1st order

Linearity:Non-linear.

Superposition:No.

$$(4) m{y}''+ky=0$$

Order:2nd order.

Linearity:Linear.

Superposition:Yes.

$$(5) y''+\omega_{0}^2=cos(\omega t)$$,  $$\omega_{0}=\omega$$

Order:2nd order

Linearity:Linear.

Superposition:Yes.

$$(6) L{I}''+R{I}'+\frac{1}{c}I={E}'$$

Order:2nd order

Linearity:Linear.

Superposition:Yes.

$$(7) EIy^{iv}=\mathit{f(x)}$$

Order:4th oder

Linearity:Linear.

Superposition:Yes.

$$(8) \mathit{L{\theta}''+gsin{\theta}=0}$$

Order:2nd order

Linearity:Non-linear.

Superposition:Yes.

Statement
Realize spring-mass-dashpot systems in series as shown in the figure.

With characteristic: $${\lambda}^2-10{\lambda}+25=0$$

And with double real root at  $${\lambda}=-3$$ Find the values of k, c, m.

Solution
First, we write down the equations we can figure out from kinematics and from kinetics.

Kinematics: $$y=y_{k}+y_{c} $$

Kinetics: $$m{y}''+f_{l}=f(t)$$

$$f_{l}=f_{k}=f_{c}$$

Next, we can find any relations that exist:

$$f_{k}=ky_{k} $$

$$f_{c}=c{y_{c}}' $$

Since we already established that $$y=y_{k}+y_{c} $$  , we can say that $${y}={y_{k}} + {y_{c}}'' $$

We also established that $$f_{k}=f_{c}$$  so we can say that $$ky_{y}=c{y_{c}}'n \to {y_{c}}'=\frac{k}{c}y_{k}$$

Next, we plug in the value for $${y_{c}}'$$  :

$${y}={y_{k}}+ {({y_{c}}')}'={y_{k}}''+\frac{k}{c}{y_{k}}'$$

Finally, plug $${y}''$$  back into the kinetics equation:

$$m({y_{k}}''+\frac{k}{c}{y_{k}}')+ky_{k}=f(t)$$

$$m{y_{k}}''+m\frac{k}{c}{y_{k}}'+ky_{k}=f(t)$$

We are given that $${\lambda}=-3$$ so the characteristic equation must be:

$$(\lambda-(-3))^{2}=0$$

$$\lambda^{2}+6\lambda+9=0$$

This relates to the second order ODE:

$${y}''+6{y}'+9y=0$$

Comparing this to the following equation:

$$m{y_{k}}''+m\frac{k}{c}{y_{k}}'+ky_{k}=f(t)$$

We find that $$m=1$$ $$k=9$$ $$c=\frac{9}{6}=\frac{3}{2}$$