User:Egm4313.s12.team18.windham

=Alissa Windham=

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Problem 1.6
For each ODE in Fig. 2 in K2011 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 applies.

Solution

Tower of Pisa


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$$ \displaystyle { y{}''=g } $$ (1)
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Order: Second Linear Principle of Superposition applies:


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$$ \displaystyle { y_{h}{}''=0} $$ (1.1)
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$$ \displaystyle { y_{p}{}''=g} $$ (1.2)
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$$ \displaystyle {(y_{h}+y_{p})=\bar{y}} $$ (1.3)
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$$ \displaystyle { (y_{h}+y_{p}){}''=g} $$ (1.4)
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$$ \displaystyle {\mathbf{\bar{ y}{}''=g}} $$ Is the same as $$ \displaystyle { \mathbf{y{}''=g}} $$ (1.5)
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Parachutist


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$$ \displaystyle {mv{}'+bv^{2}=mg} $$ (2)
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Order: First Non-Linear Principle of Superposition does not apply:


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$$ \displaystyle { mv_{h}{}'+bv_{h}^{2}=0} $$ (2.1)
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$$ \displaystyle { mv_{p}{}'+bv_{p}^{2}=mg} $$ (2.2)
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$$ \displaystyle {(v_{h}+v_{p})=\bar{v}} $$ (2.3)
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$$ \displaystyle { m(v_{h}+v_{p}){}'+b(v_{h}^{2}+v_{p}^{2})=mg} $$ (2.4)
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$$ \displaystyle {\mathbf{ m\bar{v}{}'+b(v_{h}^{2}+v_{p}^{2})=mg} } $$ Is not the same as $$ \displaystyle { \mathbf{mv{}'+bv^{2}=mg}} $$ (2.5)
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Out-flowing Water


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$$ \displaystyle {h{}'=-k\sqrt{h}} $$ (3)
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Order: First Non-Linear Principle of Superposition does not apply:


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$$ \displaystyle { h_{h}{}'+k\sqrt{h_{h}}=0} $$ (3.1)
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$$ \displaystyle { h_{p}{}'+k\sqrt{h_{p}}=0} $$ (3.2)
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$$ \displaystyle {(h_{h}+h_{p})=\bar{h}} $$ (3.3)
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$$ \displaystyle { (h_{h}+h_{p}){}'+k(\sqrt{h_{h}}+\sqrt{h_{p}})=0} $$ (3.4)
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$$ \displaystyle {\mathbf{\bar{h}{}'+k(\sqrt{h_{h}}+\sqrt{h_{p}})=0 }} $$ Is not the same as $$ \displaystyle {\mathbf{ h{}'+k\sqrt{h}=0}} $$ (3.5)
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Vibrating Mass on a String


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$$ \displaystyle {my{}''+ky=0} $$ (4)
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Order:2nd Linear Principle of Superposition applies:


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$$ \displaystyle {my_{h}{}''+ky_{h}=0} $$ (4.1)
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$$ \displaystyle {my_{p}{}''+ky_{p}=0} $$ (4.2)
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$$ \displaystyle {(y_{p}+y_{h})=\bar{y}} $$ (4.3)
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$$ \displaystyle {m(y_{p}+y_{h}){}''+k(y_{p}+y_{h})=0} $$ (4.4)
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$$  \displaystyle {\mathbf{m\bar{y}''+k\bar{y}=0 }} $$ is the same as $$ \displaystyle { \mathbf{my{}''+ky=0}} $$     (4.5)
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Beats of a Vibrating Sytem


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$$ \displaystyle {y{}''+\omega _{0}^{2}y=\cos( \omega t)} $$ (5)
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Order:2nd Linear Principle of Superposition applies:


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$$ \displaystyle {y_{h}{}''+\omega _{0}^{2}y_{h}=0} $$ (5.1)
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$$ \displaystyle {y_{p}{}''+\omega _{0}^{2}y_{p}=\cos( \omega t)} $$ (5.2)
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$$ \displaystyle {(y_{p}+y_{h})=\bar{y}} $$ (5.3)
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$$ \displaystyle {(y_{p}+y_{h}){}''+\omega _{0}^{2}(y_{p}+y_{h})=\cos( \omega t)} $$ (5.4)
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$$ \displaystyle {\mathbf{\bar{y}{}''+\omega _{0}^{2}\bar{y}=\cos( \omega t)}} $$ Is the same as $$ \displaystyle {\mathbf{ y{}''+\omega _{0}^{2}y=\cos( \omega t)}} $$ (5.5)
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Current I in RLC Circuit


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$$ \displaystyle {LI{}''+RI{}'+\frac{1}{c}I=E{}'} $$ (6)
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Order:2nd Linear Principle of Superposition applies:


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$$ \displaystyle {LI_{h}{}''+RI_{h}{}'+\frac{1}{c}I_{h}=0} $$ (6.1)
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$$ \displaystyle {LI_{p}{}''+RI_{p}{}'+\frac{1}{c}I_{p}=E{}'} $$ (6.2)
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$$ \displaystyle {(I_{p}+I_{h})=\bar{I}} $$ (6.3)
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$$ \displaystyle {L(I_{p}+I_{h}){}''+R(I_{p}+I_{h}){}'+\frac{1}{c}(I_{p}+I_{h})=E{}'} $$ (6.4)
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$$ \displaystyle {\mathbf{L\bar{I}{}''+R\bar{I}{}'+\frac{1}{c}\bar{I}=E{}' }} $$ is the same as $$ \displaystyle {\mathbf{ LI{}''+RI{}'+\frac{1}{c}I=E{}'}} $$ (6.5)
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Deformation of a Beam


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$$ \displaystyle {EIy^{iv}=f(x)} $$ (7)
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Order:Fourth Linear Principle of Superposition applies:


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$$ \displaystyle {EIy_{h}^{iv}=0} $$ (7.1)
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$$ \displaystyle {EIy_{p}^{iv}=f(x)} $$ (7.2)
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$$ \displaystyle {(y_{p}+y_{h})=\bar{y}} $$ (7.3)
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$$ \displaystyle {EI(y_{h}+y_{p})^{iv}=f(x)} $$ (7.4)
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$$ \displaystyle {\mathbf{EI\bar{y}^{iv}=f(x)} } $$ Is the same as $$ \displaystyle {\mathbf{ EIy^{iv}=f(x)}} $$ (7.5)
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Pendulum


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$$ \displaystyle {L\Theta {}''+gsin(\Theta )=0} $$ (8)
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Order:Second Non-Linear Principle of Superposition does not apply:


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$$ \displaystyle {L\Theta_{h} {}''+gsin(\Theta_{h} )=0} $$ (8.1)
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$$ \displaystyle {L\Theta_{p} {}''+gsin(\Theta_{p} )=0} $$ (8.2)
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$$ \displaystyle {(\Theta_{p}+\Theta_{h})=\bar{\Theta}} $$ (8.3)
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$$ \displaystyle {L(\Theta_{p}+\Theta_{h}){}''+gsin(\Theta_{p}+\Theta_{h})=0} $$ (8.4)
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$$ \displaystyle {\mathbf{L\bar{\Theta} {}+gsin(\Theta_{p}+\Theta_{h} )=0}} $$ Is not the same as $$ \mathbf{L\Theta {}+gsin(\Theta )=0} $$
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Problem 2.3
Solve problems 3 and 4 on page 59 in the textbook, Kreyszig 2011. "Find a general solution for the ODE and check by substitution."

3.$$ \displaystyle { y{}''+6y{}'+8.96y=0 } $$

4.$$ \displaystyle {y{}''+4y{}'+\left ( \pi ^{2}+4 \right )y=0} $$

Solution
Background information: For homogeneous linear ODE's with constant coefficients, a general equation can be described as, $$ \displaystyle {y{}''+ay{}'+by=0.} $$

The function $$ \displaystyle {y=e^{\lambda x}} $$ may be tried as a solution for the general equation above.

After substituting this equation and its derivatives: $$ \displaystyle {y{}'=\lambda e^{\lambda x}} $$ and $$ \displaystyle {y{}''=\lambda ^{2}e^{\lambda x}} $$ into the general equation,the result is

$$ \displaystyle {(\lambda ^{2}+a\lambda +b)e^{\lambda x}=0} $$

and the characteristic equation is

$$ \displaystyle {(\lambda ^{2}+a\lambda +b)=0} $$ which means that if $$ \displaystyle {\lambda} $$ is a solution of this characteristic equation, then the exponential function is a solution of the general equation. If we solve this characteristic equation for $$ \displaystyle {\lambda} $$, the result is$$ \displaystyle {\lambda _{1,2}=\frac{1}{2}(-a\pm \sqrt{a^{^{2}}-4b})} $$

Therefore there are 3 cases when it comes to finding the solution to Homogeneous Linear ODE's with CC:

Case 1:    Two real roots if  $$ \displaystyle {a^{^{2}}-4b>0} $$

Case 2:    A real double root if $$\displaystyle {a^{^{2}}-4b=0} $$

Case 3: Complex conjugate roots if $$\displaystyle {a^{^{2}}-4b=0} $$

Problem 3 page 59 :
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$$ \displaystyle { y{}''+6y{}'+8.96y=0 } $$ (1) The discriminant for this equation (1) is:
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$$ \displaystyle { \sqrt{6^{^{2}}-4(8.96)}=.16} $$ which is >0 and therefore a case 1 (2)
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$$ \displaystyle {\lambda _{1}=-2.8, \lambda _{2}=-3.2} $$ and $$ \displaystyle {y _{1}=e^{-2.8x}, y _{2}=e^{-3.2x}} $$ (3)
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The general solution to (1) is


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$$ \displaystyle {y=c_{1}e^{-2.8x}+c_{2}e^{^{-3.2x}}} $$ (4) Check this solution by plugging in
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$$ \displaystyle {y _{1}=e^{-2.8x}} $$ and $$ \displaystyle {y _{2}=e^{-3.2x}} $$ (5) into (1). The results for plugging in the above equations(5) and their derivatives:
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$$ \displaystyle {y_{1}=e^{-2.8x},y_{1}{}'=-2.8e^{-2.8x},y_{1}{}''=7.84e^{-2.8x}} $$ and $$ \displaystyle {y_{2}=e^{-3.2x},y_{2}{}'=-3.2e^{-3.2x},y_{2}{}''=10.24e^{-3.2x}} $$ (6) into equation (1) respectively are:
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$$ \displaystyle {e^{-2.8x}(7.84-(6)(2.8)+8.96)=0} $$ and $$ \displaystyle {e^{-3.2x}(10.24-(6)(3.2)+8.96)=0} $$
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(7)
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Both of these reduce to
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$$ \displaystyle {0=0} $$ (8)
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Problem 4 page 59:
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$$ \displaystyle {y{}''+4y{}'+\left ( \pi ^{2}+4 \right )y=0} $$ (9)
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The discriminant for this equation is:


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$$ \displaystyle {4^{2}-4(\pi ^{2}+4)=-4\pi^{2}} $$ (10) which is a case 3 and solutions to the characteristic equation are:
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$$ \displaystyle {\lambda _{1}=-\frac{1}{2}a+i\omega, \lambda _{2}=-\frac{1}{2}a-i\omega} $$ (11) and our particular solutions for characteristic equation for this problem are:
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$$ \displaystyle {\lambda _{1}=-\frac{1}{2}(4)+i\pi, \lambda _{2}=-\frac{1}{2}4-i\pi} $$ (12)
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The general solution to (9) is


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$$ \displaystyle {y=e^{-\frac{4}{2}x}(A\cos \pi x+Bsin\pi x)} $$ (13) Check this solution by plugging in
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$$ \displaystyle {y _{1}=e^{-\frac{4}{2}x}\cos \pi x} $$ and $$ \displaystyle {y _{2}=e^{-\frac{4}{2}x}\sin \pi x} $$ (14) into (1). The results for plugging in the equations (14) and their derivatives:
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$$ \displaystyle {y_{1}=e^{-2x}\cos \pi x, y_{1}{}'=-\pi e^{-2x}-2e^{-2x}\cos \pi x, y_{1}{}''=e^{-2x}((4-\pi ^{2})\cos \pi x+4\pi \sin \pi x)} $$ (15)
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and
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$$ \displaystyle {y_{2}=e^{-2x}\sin \pi x, y_{2}{}'=e^{-2x}(\pi \cos \pi x-2\sin \pi x), y_{2}{}''=e^{-2x}(-4\pi \cos \pi x+(4-\pi ^{2})\sin \pi x)} $$ (16)  into equation (1) respectively are:
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$$ \displaystyle {e^{-2x}((4-\pi ^{2})\cos \pi x+4\sin \pi x+4(-\pi \sin \pi x-2\cos \pi x)+(\pi ^{2}+4\cos \pi x)=0} $$ (17)
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$$ \displaystyle {e^{-2x}(-4\pi \cos \pi x+(4-\pi ^{2})\sin \pi x)+4e^{-2x}(\pi \cos \pi x-2\sin \pi x)+(\pi ^{2}+4)e^{-2x}\sin \pi x=0} $$ (18) Both of these reduce to
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$$ \displaystyle {0=0} $$ (19)
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Solution
Background information:

Basic Rule- if r(x) in $$ \displaystyle {y{}''+ay{}'+by=r(x)} $$ is one of the functions in the first column in table 2.1, choose $$ \displaystyle {y_{p}} $$ in the same line and determine its undetermined coefficients by substituting $$ \displaystyle {y_{p}} $$ and its derivatives into the above equation.