University of Florida/Egm4313/s12.team4.Lorenzo/R1

Report 1

Problem 6

Problem 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 whether the principle of superposition can be applied.

Theory

Order

The order of an equation is determined by the highest derivative. In this report, the first derivative of the y variable is denoted as y′, and the second derivative is denoted as y′′ , and so on. This can be determined upon observation of the equation.

Linearity

An ordinary differential equation (ODE) is considered linear if it can be brought to the form: $y^{\prime }+p(x)y=q(x)$

Superposition

Superposition can be applied if when a homogeneous and particular solution of an original equation are added, that they are equivalent to the original equation. In this report, variables with a bar over them represent the addition of the homogeneous and particular solution's same variable. For example:

$y_p+y_h=\bar{y}$

Given

The following equations were given in the textbook on p. 3:

• 1.6a - $y^{″}=g=constant$

• 1.6b - $m{v}^{\prime }=mg-b{v}^2$

• 1.6c - $h^{\prime }=-k\sqrt{h}$

• 1.6d - $m{y}^{″}+ky=0$

• 1.6e - $y^{″}+{\omega }_0^{2}y=\cos \omega t,{\omega }_0=\omega$

• 1.6f - $L{I}^{″}+R{I}^{\prime }+\frac{1}{C}I=E^{\prime }$

• 1.6g - $EI{y}^{\omega }=f(x)$

• 1.6h - $L{\theta }^{″}+g\sin \theta =0$

Solution

1.6a

$y^{″}=g=constant$
order: 2nd
linear: yes
superposition: yes

The given equation can be algebraically modified as the following:

$y^{″}=g$

It can be split up into the following homogeneous and particular solutions:

$y_{h^{″}}=0$

$y_{p^{″}}=g$

Adding the two solutions:

$(y_{h^{″}}+y_{p^{″}})={\bar{y}}^{″}$

The solution resembles the original equation, therefore superposition is possible

1.6b

$m{v}^{\prime }=mg-b{v}^2$
order: 1st
linear: no
superposition: no

The given equation can be algebraically modified as the following:

$m{v}^{\prime }+b{v}^2=mg$

It can be split up into the following homogeneous and particular solutions:

$m{v}_{h^{\prime }}+b{v}_h^2=0$

$m{v}_{p^{\prime }}+b{v}_p^2=mg$

Adding the two solutions:

$m(v_{h^{\prime }}+v_{p^{\prime }})+b(v_h^2+v_p^2)=m{\bar{v}}^{\prime }+b(\stackrel{2}{\stackrel{‾}{v}})$

The solution cannot be algebraically modified to resemble the original equation, and therefore superposition is NOT possible as also proven in the class notes

1.6c

$h^{\prime }=-k\sqrt{h}$
order: 1st
linear: no
superposition: no

The given equation can be algebraically modified as the following:

$h^{\prime }+k\sqrt{h}=0$

It can be split up into the following homogeneous and particular solutions:

$h_{h^{\prime }}+k\sqrt{h_h}=0$

$h_{p^{\prime }}+k\sqrt{h_p}=0$

Adding the two solutions:

$(h_{h^{\prime }}+h_{p^{\prime }})+k(\sqrt{h_h}+\sqrt{h_p})={\bar{h}}^{\prime }+k\sqrt{\bar{h}}$

The solution cannot be algebraically modified to resemble the original equation, and therefore superposition is NOT possible

1.6d

$m{y}^{″}+ky=0$
order: 2nd
linear: yes
superposition: yes

The given equation can be algebraically modified as the following:

$y^{″}+\frac{k}{m}y=0$

It can be split up into the following homogeneous and particular solutions:

$y_{h^{″}}+\frac{k}{m}{y}_h=0$

$y_{p^{″}}+\frac{k}{m}{y}_p=0$

Adding the two solutions:

$(y_{h^{″}}+y_{p^{″}})+\frac{k}{m}(y_h+y_p)={\bar{y}}^{″}+\frac{k}{m}\bar{y}$

The solution resembles the original equation, therefore superposition is possible

1.6e

$y^{″}+{\omega }_0^{2}y=\cos \omega t,{\omega }_0=\omega$
order: 2nd
linear: yes
superposition: yes

The given equation can be algebraically modified as the following:

$y^{″}+{\omega }_o^{2}y=\cos (\omega t)$

It can be split up into the following homogeneous and particular solutions:

$y_{h^{″}}+{\omega }_o^2{y}_h=0$

$y_{p^{″}}+{\omega }_o^2{y}_p=\cos (\omega t)$

Adding the two solutions:

$(y_{h^{″}}+y_{p^{″}})+{\omega }_o^2(y_h+y_p)={\bar{y}}^{″}+{\omega }_o^2\bar{y}$

The solution resembles the original equation, therefore superposition is possible

1.6f

$L{I}^{″}+R{I}^{\prime }+\frac{1}{C}I=E^{\prime }$
order: 2nd
linear: yes
superposition: yes

The given equation can be algebraically modified as the following:

$I^{″}+\frac{R}{L}{I}^{\prime }+\frac{1}{LC}I=\frac{E^{\prime }}{L}$

It can be split up into the following homogeneous and particular solutions:

$I_{h^{″}}+\frac{R}{L}{I}_{h^{\prime }}+\frac{1}{LC}{I}_h=0$

$I_{p^{″}}+\frac{R}{L}{I}_{p^{\prime }}+\frac{1}{LC}{I}_p=\frac{E^{\prime }}{L}$

Adding the two solutions:

$(I_{h^{″}}+I_{p^{″}})+\frac{R}{L}(I_{h^{\prime }}+I_{p^{\prime }})+\frac{1}{LC}(I_h+I_p)={\bar{I}}^{″}+\frac{R}{L}{\bar{I}}^{\prime }+\frac{1}{LC}\bar{I}$

The solution resembles the original equation, therefore superposition is possible

1.6g

$EI{y}^{⁗}=f(x)$
order: 4th
linear: yes
superposition: yes

The given equation can be algebraically modified as the following:

$y^{⁗}-\frac{1}{EI}y=0$

It can be split up into the following homogeneous and particular solutions:

$y_{h^{⁗}}-\frac{1}{EI}{y}_h=0$

$y_{p^{⁗}}-\frac{1}{EI}{y}_p=0$

Adding the two solutions:

$(y_{h^{⁗}}+y_{p^{⁗}})-\frac{1}{EI}(y_h+y_p)={\bar{y}}^{⁗}-\frac{1}{EI}\bar{y}$

The solution resembles the original equation, therefore superposition is possible

1.6h

$L{\theta }^{″}+g\sin \theta =0$
order: 2nd
linear: no
superposition:no

The given equation can be algebraically modified as the following:

${\theta }^{″}+\frac{g}{L}\sin \theta =0$

It can be split up into the following homogeneous and particular solutions:

${\theta }_{h^{″}}+\frac{g}{L}\sin {\theta }_h=0$

${\theta }_{p^{″}}+\frac{g}{L}\sin {\theta }_p=0$

Adding the two solutions:

$({\theta }_{h^{″}}+{\theta }_{p^{″}})+\frac{g}{L}(\sin {\theta }_h+\sin {\theta }_p)={\bar{\theta }}^{″}+\frac{g}{L}\sin \bar{\theta }$

The solution cannot be algebraically modified to resemble the original equation, and therefore superposition is NOT possible