User:Egm4313.s12.team3.pliskow/R1work

=Problem 6: Classifying ODE's and the Superposition Principle=

Problem Statement
For each of the following ODE's, determine the order and linearity. Then, determine if the superposition principle can be applied.


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$$\displaystyle {y}''=g $$     (6.1)
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$$\displaystyle m{v}'=mg-bv^{2} $$     (6.2)
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$$\displaystyle {h}'=-k\sqrt{h} $$     (6.3)
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$$\displaystyle m{y}''+ky=0 $$     (6.4)
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$$\displaystyle {y}''+\omega_0^{2}y=cos(\omega t)\;\;,\;\;\omega_0 \approx \omega $$     (6.5)
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$$\displaystyle L{I}''+R{I}'+\frac{1}{C}I={E}' $$     (6.6)
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$$\displaystyle EI{y}=f(x) $$     (6.7)
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$$\displaystyle L{\theta}''+gsin\theta =0 $$     (6.8)
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Problem Solution
For information regarding the order and linearity of ordinary differential equations, see class notes, sec1. For information regarding the application of the superposition principle, see class notes, sec2.

Problem 6.1

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$$\displaystyle {y}''=g $$     (6.1)
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This is a second order, linear ODE.
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 * }

For a non-homogeneous equation, the superposition principle applies if the sum of the homogeneous solution and the particular solution is also a solution to the original ODE. This is tested below.

Homogeneous problem:
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$$\displaystyle {y_h}''=0 $$     (1)
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Particular problem:
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$$\displaystyle {y_p}''=g $$     (2)
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We define the sum of the homogeneous solution and particular solution:
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$$\displaystyle \bar{y}=y_h+y_p $$     (3)
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Summing (1) and (2), we obtain the following:


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$$\displaystyle {y_h}+{y_p}=g $$     (4)
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$$\displaystyle {({y_h}'+{y_p}')}'=g $$     (5)
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$$\displaystyle {(y_h+y_p)}''=g $$     (6)
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$$\displaystyle {\bar{y}}''=g $$     (7)
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From equation (7), we can see that $$ \bar{y} $$ is also a solution to the original ODE, and thus the superposition principle can be applied.
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Problem 6.2

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$$\displaystyle m{v}'=mg-bv^{2} $$     (6.2)
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This is a second order, non-linear ODE.
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 * }

For a non-homogeneous equation, the superposition principle applies if the sum of the homogeneous solution and the particular solution is also a solution to the original ODE. This is tested below.

Homogeneous problem:
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$$\displaystyle m{v_h}'+bv_h^{2}=0 $$     (1)
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Particular problem:
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$$\displaystyle m{v_p}'+bv_p^{2}=mg $$     (2)
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We define the sum of the homogeneous solution and particular solution:
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$$\displaystyle \bar{v}=v_h+v_p $$     (3)
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Summing (1) and (2), we obtain the following:


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$$\displaystyle m({v_h}'+{v_p}')+b(v_h^{2}+v_p^{2})=mg $$     (4)
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$$\displaystyle m{(v_h+v_p)}'+b(v_h^{2}+v_p^{2})=mg $$     (5)
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$$\displaystyle m{\bar{v}}'+b(v_h^{2}+v_p^{2})=mg $$     (6)
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We cannot substitute any further since $$ b(v_h^{2}+v_p^{2}) \neq b(v_h+v_p)^{2} $$.


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$$ \bar{v} $$ is NOT also a solution to the original ODE, and thus the superposition principle can't be applied.
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Problem 6.3

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$$\displaystyle {h}'=-k\sqrt{h} $$     (6.3)
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This is a first order, non-linear ODE.
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For a homogeneous equation, the superposition principle applies if a linear combination of any two solutions is also a solution to the original ODE. This is tested below.

Suppose $$h_1$$ is one solution:
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$$\displaystyle {h_1}'+k\sqrt{h_1}=0 $$     (1)
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And $$h_2$$ is another solution:
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$$\displaystyle {h_2}'+k\sqrt{h_2}=0 $$     (2)
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We define any linear combination of these solutions:
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$$\displaystyle \bar{h}=c_1h_1+c_2h_2 $$     (3)
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Summing (1) and (2), we obtain the following:


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$$\displaystyle {h_1}'+{h_2}'+k\sqrt{h_1}+k\sqrt{h_2}=0 $$     (4)
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$$\displaystyle {(h_1+h_2)}'+k(\sqrt{h_1}+\sqrt{h_2})=0 $$     (5)
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$$\displaystyle {\bar{h}}'+k(\sqrt{h_1}+\sqrt{h_2})=0 $$     (6)
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We cannot substitute any further since $$ k(\sqrt{h_1}+\sqrt{h_2})) \neq k\sqrt{h_1+h_2} $$.


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$$ \bar{h} $$ is NOT also a solution to the original ODE, and thus the superposition principle can't be applied.
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Problem 6.4

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$$\displaystyle m{y}''+ky=0 $$     (6.4)
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This is a second order, linear ODE.
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 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

For a homogeneous equation, the superposition principle applies if a linear combination of any two solutions is also a solution to the original ODE. This is tested below.

Suppose $$y_1$$ is one solution:
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$$\displaystyle m{y_1}''+ky_1=0 $$     (1)
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And $$y_2$$ is another solution:
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$$\displaystyle m{y_2}''+ky_2=0 $$     (2)
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We define any linear combination of these solutions:
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$$\displaystyle \bar{y}=c_1y_1+c_2y_2 $$     (3)
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Summing (1) and (2), we obtain the following:


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$$\displaystyle m({y_1}+{y_2})+k(y_1+y_2)=0 $$     (4)
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$$\displaystyle m{(y_1+y_2)}''+k(y_1+y_2)=0 $$     (5)
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$$\displaystyle m{\bar{y}}''+k\bar{y}=0 $$     (6)
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From equation (6), we can see that $$ \bar{y} $$ is also a solution to the original ODE, and thus the superposition principle can be applied.
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Problem 6.5

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$$\displaystyle {y}''+\omega_0^{2}y=cos(\omega t)\;\;,\;\;\omega_0 \approx \omega $$     (6.5)
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This is a second order, linear ODE.
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 * <p style="text-align:right">
 * }

For a non-homogeneous equation, the superposition principle applies if the sum of the homogeneous solution and the particular solution is also a solution to the original ODE. This is tested below.

Homogeneous problem:
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$$\displaystyle {y_h}''+\omega_0^{2}y_h=0 $$     (1)
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Particular problem:
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$$\displaystyle {y_p}''+\omega_0^{2}y_p=cos(\omega t) $$ (2)
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We define the sum of the homogeneous solution and particular solution:
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$$\displaystyle \bar{y}=y_h+y_p $$     (3)
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Summing (1) and (2), we obtain the following:


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$$\displaystyle {y_h}+{y_p}+\omega_0^{2}y_h+\omega_0^{2}y_p=cos(\omega t) $$ (4)
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$$\displaystyle {(y_h+y_p)}''+\omega_0^{2}(y_h+y_p)=cos(\omega t) $$ (5)
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$$\displaystyle {\bar{y}}''+\omega_0^{2}(\bar{y})=cos(\omega t) $$ (6)
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From equation (6), we can see that $$ \bar{y} $$ is also a solution to the original ODE, and thus the superposition principle can be applied.
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Problem 6.6

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$$\displaystyle L{I}''+R{I}'+\frac{1}{C}I={E}' $$     (6.6)
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This is a second order, linear ODE.
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 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

For a non-homogeneous equation, the superposition principle applies if the sum of the homogeneous solution and the particular solution is also a solution to the original ODE. This is tested below.

Homogeneous problem:
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$$\displaystyle L{I_h}''+R{I_h}'+\frac{1}{C}I_h=0 $$     (1)
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Particular problem:
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$$\displaystyle L{I_p}''+R{I_p}'+\frac{1}{C}I_p={E}' $$     (2)
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We define the sum of the homogeneous solution and particular solution:
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$$\displaystyle \bar{I}=I_h+I_p $$     (3)
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Summing (1) and (2), we obtain the following:


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$$\displaystyle L({I_h}+{I_p})+R({I_h}'+{I_p}')+\frac{1}{C}(I_h+I_p)={E}' $$     (4)
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$$\displaystyle L{(I_h+I_p)}''+R{(I_h+I_p)}'+\frac{1}{C}(I_h+I_p)={E}' $$     (5)
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$$\displaystyle L{\bar{I}}''+R{\bar{I}}'+\frac{1}{C}\bar{I}={E}' $$     (6)
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From equation (6), we can see that $$ \bar{I} $$ is also a solution to the original ODE, and thus the superposition principle can be applied.
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Problem 6.7

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$$\displaystyle EI{y}=f(x) $$     (6.7)
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This is a fourth order, linear ODE.
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 * <p style="text-align:right">
 * }

For a non-homogeneous equation, the superposition principle applies if the sum of the homogeneous solution and the particular solution is also a solution to the original ODE. This is tested below.

Homogeneous problem:
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$$\displaystyle EI{y_h}=0 $$     (1)
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Particular problem:
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$$\displaystyle EI{y_p}=f(x) $$     (2)
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We define the sum of the homogeneous solution and particular solution:
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$$\displaystyle \bar{y}=y_h+y_p $$     (3)
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Summing (1) and (2), we obtain the following:


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$$\displaystyle EI({y_h}'+{y_p}')=f(x) $$     (4)
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$$\displaystyle EI{(y_h+y_p)}=f(x) $$     (5)
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$$\displaystyle EI{\bar{y}}=f(x) $$     (6)
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From equation (6), we can see that $$ \bar{y} $$ is also a solution to the original ODE, and thus the superposition principle can be applied.
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Problem 6.8

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$$\displaystyle L{\theta}''+gsin\theta =0 $$     (6.8)
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This is a second order, non-linear ODE.
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 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

For a homogeneous equation, the superposition principle applies if a linear combination of any two solutions is also a solution to the original ODE. This is tested below.

Suppose $$\theta_1$$ is one solution:
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$$\displaystyle L{\theta_1}''+gsin\theta_1 =0 $$     (1)
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And $$\theta_2$$ is another solution:
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$$\displaystyle L{\theta_2}''+gsin\theta_2 =0 $$     (2)
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We define any linear combination of these solutions:
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$$\displaystyle \bar{\theta}=c_1\theta_1+c_2\theta_2 $$     (3)
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Summing (1) and (2), we obtain the following:


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$$\displaystyle L({\theta_1}+{\theta_2})+g(sin\theta_1+sin\theta_2)=0 $$     (4)
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 * }


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$$\displaystyle L{(\theta_1+\theta_2)}''+g(sin\theta_1+sin\theta_2)=0 $$     (5)
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 * }


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$$\displaystyle L{\bar{\theta}}''+g(sin\theta_1+sin\theta_2)=0 $$     (6)
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 * }

We cannot substitute any further since $$ sin\theta_1+sin\theta_2 \neq sin(\theta_1+\theta_2) $$.


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$$ \bar{\theta} $$ is NOT also a solution to the original ODE, and thus the superposition principle can't be applied.
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=Contributing Members for This Project=