User:Egm6321.f10.team03/Hwk3

= Problem 1 - Motion of a Projectile with Air Resistance =

Given
Figure shows the Trajectory of a projectile (ex:Rocket):

Find

 * $$\left( 1 \right)$$ Drive Equation of Motion (EOM)
 * $$\left( 2 \right)$$ Particular case $$\displaystyle k = 0 $$ : Verify $$y\left( x \right)$$ is parabolla
 * $$\left( 3 \right)$$ Consider $$k\ne 0$$, $$\displaystyle {{v}_{x}}=0$$,


 * $$ (3.1) $$ Find $${{v}_{y}}\left( t \right)$$, $$\displaystyle y(t)$$ for $$ \displaystyle m$$ = constant
 * $$ (3.2) $$ Find $${{v}_{y}}\left( t \right)$$, $$\displaystyle y(t)$$ if $$m=m\left( t \right)$$

Solution
Part 1 Consider the trajectory of a projectile (ex. Rocket)


 * Various forces acting on the projectile at time 't' are:


 * 1) Weight of the projectile


 * {| style="width:100%" border="0"


 * $$\displaystyle W=mg$$
 * }
 * }
 * }


 * 2) Inertia force


 * {| style="width:100%" border="0"


 * $${{F}_{a}}=ma=m\cdot \frac{dv}{dt}$$ for particle with constant mass
 * $${{F}_{a}}=ma=m\cdot \frac{dv}{dt}$$ for particle with constant mass


 * }
 * }


 * 3) Air resistance which is proportional to the velocity of particle


 * {| style="width:100%" border="0"


 * $$\displaystyle {{F}_{D}}=k{{v}^{n}}$$
 * }
 * }
 * }

Now consider the force equilibrium in both horizontal and vertical direction


 * a) Force Equilibrium in horizontal direction:


 * {| style="width:100%" border="0"


 * $$\sum{{{F}_{H}}=0}$$
 * }
 * }
 * }


 * {| style="width:100%" border="0"


 * $$m\cdot \frac{d{{v}_{x}}}{dt}+k{{v}^{n}}\cos \alpha =0$$,
 * $$m\cdot \frac{d{{v}_{x}}}{dt}+k{{v}^{n}}\cos \alpha =0$$,


 * where $$\displaystyle {{v}_{x}}\to $$ horizontal component of velocity


 * }
 * }


 * {| style="width:100%" border="0"

$$m\cdot \frac{d{{v}_{x}}}{dt}=-k{{v}^{n}}\cos \alpha $$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * $$\displaystyle (Eq. 1.1)
 * }
 * }


 * b) Force Equilibrium in the vertical direction:


 * {| style="width:100%" border="0"


 * $$\sum{{{F}_{V}}=0}$$
 * }
 * }
 * }


 * {| style="width:100%" border="0"

$$m\cdot \frac{d{{v}_{y}}}{dt}+k{{v}^{n}}\sin \alpha +mg=0$$,


 * where $$\displaystyle {{v}_{y}}\to $$ vertical component of velocity


 * }
 * }


 * {| style="width:100%" border="0"

$$m\cdot \frac{d{{v}_{y}}}{dt}=-k{{v}^{n}}\sin \alpha -mg$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * $$\displaystyle (Eq. 1.2)
 * }
 * }

Part 2 


 * Particular case: When $$\displaystyle k=0$$

Eq.1.1 reduces to


 * {| style="width:100%" border="0"


 * $$m\frac{d{{v}_{x}}}{dt}=0\Rightarrow \frac{d{{v}_{x}}}{dt}=0$$
 * }
 * }
 * }

Integrating the above equation gives:


 * {| style="width:100%" border="0"

$$
 * $${{v}_{x}}\left( t \right)={{c}_{1}}$$
 * $$\displaystyle (Eq. 1.3)
 * $$\displaystyle (Eq. 1.3)
 * }
 * }

Apply 'initial condition' to determine integration constant,$$\displaystyle{{c}_{1}}$$


 * {| style="width:100%" border="0"


 * $${{v}_{x}}\left( t=0 \right)={{v}_}={{c}_{1}}$$
 * }
 * }
 * }

Now Eq.1.3 becomes:
 * {| style="width:100%" border="0"


 * $${{v}_{x}}\left( t \right)={{v}_}$$
 * }
 * }
 * }

Integrate the above equation to obtain ,$$\displaystyle x$$


 * {| style="width:100%" border="0"


 * $$x\left( t \right)={{v}_}t+{{c}_{2}}$$
 * }
 * }
 * }

Then 'Initial condition' is applied to determine ,$$\displaystyle{{c}_{2}}$$


 * {| style="width:100%" border="0"


 * $$x\left( t=0 \right)={{v}_}\times 0+{{c}_{2}}\Rightarrow {{c}_{2}}={{x}_{0}}$$
 * }
 * }
 * }

Therefore,


 * {| style="width:100%" border="0"


 * $$x\left( t \right)={{v}_}t+{{x}_{0}}$$
 * }
 * }
 * }

Now $$\displaystyle t$$, can be expressed in terms of $$\displaystyle x$$,


 * {| style="width:100%" border="0"

$$
 * $$t=\frac{x-{{x}_{0}}}$$
 * $$\displaystyle (Eq. 1.4)
 * $$\displaystyle (Eq. 1.4)
 * }
 * }

Similarly When $$\displaystyle k=0$$, Eq.1.2 reduces to


 * {| style="width:100%" border="0"


 * $$m\frac{d{{v}_{y}}}{dt}=-mg\Rightarrow \frac{d{{v}_{y}}}{dt}=-g$$
 * }
 * }
 * }

Integrate the above equation to evaluate, $$\displaystyle {{v}_{y}}$$


 * {| style="width:100%" border="0"

$$
 * $${{v}_{y}}\left( t \right)=-gt+{{c}_{3}}$$
 * $$\displaystyle (Eq. 1.5)
 * $$\displaystyle (Eq. 1.5)
 * }
 * }

Apply 'initial condition' to obtain $$\displaystyle {{c}_{3}}$$


 * {| style="width:100%" border="0"


 * $${{v}_{y}}\left( t=0 \right)={{v}_}={{c}_{3}}$$
 * }
 * }
 * }

Now Eq.1.5 becomes,


 * {| style="width:100%" border="0"


 * $${{v}_{t}}\left( t \right)=-gt+{{v}_}$$
 * }
 * }
 * }

Then integrate the above equation to determine, $$\displaystyle y $$


 * {| style="width:100%" border="0"


 * $$y\left( t \right)=\frac{-g{{t}^{2}}}{2}+{{v}_}t+{{c}_{4}}$$
 * }
 * }
 * }

$$\displaystyle {{c}_{4}}$$ is determined using 'initial condition' as:


 * {| style="width:100%" border="0"


 * $$y\left( t=0 \right)=\frac{-g\times 0}{2}+{{v}_}\times 0+{{c}_{4}}\Rightarrow {{c}_{4}}={{y}_{0}}$$
 * }
 * }
 * }

Therefore,


 * {| style="width:100%" border="0"


 * $$y\left( t \right)=\frac{-g{{t}^{2}}}{2}+{{v}_}t+{{y}_{0}}$$
 * }
 * }
 * }

Now Substitute Eq.1.4 for $$\displaystyle t$$ in the above equation;


 * {| style="width:100%" border="0"

$$y\left( x \right)=-\frac{g}{2}{{\left( \frac{x-{{x}_{0}}} \right)}^{2}}+{{v}_}\left( \frac{x-{{x}_{0}}} \right)+{{y}_{0}}$$ $$
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 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * $$\displaystyle (Eq. 1.6)
 * }
 * }

Eq.1.6 is in the form of a parabolic equation. Therefore $$\displaystyle y\left( x \right)$$ is parabola.

Part 3

Part 3.1: Constant mass, $$ k \neq0$$ Then
 * {| style="width:100%" border="0"

$$ m \frac{d {v}_{g}}{dt}=-k {{v}_{g}}^{n}-mg$$ || $$\displaystyle (Eq. 1.7.1) $$ Or
 * }
 * }
 * $$m \dot{v}=-k {v}^{n}-mg$$
 * Let
 * {| style="width:100%" border="0"

$$F \left(t,v, \dot{v} \right):=m \dot{v}+k {v}^{n}+mg=0= \frac{d \phi(t,v)}{dt}$$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.7.2) $$ Therefore
 * }
 * }
 * {| style="width:100%" border="0"

$$ \phi \left(t,v \right)=k$$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.7.3) $$ And equation 1.7.2 can be written in the form
 * }
 * }
 * {| style="width:100%" border="0"

$$F= \phi \left(x \right)+ \phi \left( y\right) y'=M+Ny'$$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.7.4) $$ Where
 * }
 * }
 * {| style="width:100%" border="0"

$$ \phi \left(x \right)=k {v}^{n}+mg$$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.7.5) $$ and
 * }
 * }
 * {| style="width:100%" border="0"

$$\phi \left(y \right)=m$$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.7.6) $$ Check the exactness of equation 1.7.4. First condition is satisfied as it is in the form
 * }
 * }
 * {| style="width:100%" border="0"

$$ F= \phi \left(x \right)+ \phi \left( y\right) y'=M+Ny'$$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.7.7) $$ Second condition is satisfied if
 * }
 * }
 * {| style="width:100%" border="0"

$$ {N}_{t}= {m}_{v}$$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.7.8) $$ Differentiating:
 * }
 * }
 * {| style="width:100%" border="0"

$$ {N}_{t}= \frac{d}{dt} \left(m \right)=0$$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.7.9) $$ and
 * }
 * }
 * {| style="width:100%" border="0"

$$ {M}_{v}= \frac{d}{dv} \left( k {v}^{n}+mg\right)=kn {v}^{n-1} \neq0 $$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.7.10) $$ Therefore equation 1.7.4 is not exact and must be made exact by finding an integration factor $$ h \left(t,v \right)$$
 * }
 * }
 * {| style="width:100%" border="0"

$$ h \left(t,v \right) \left[m \dot{v}+k {v}^{n}+mg\right]=0 $$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.7.11) $$ Choose an integrating factor as a function of time only
 * }
 * }
 * {| style="width:100%" border="0"

$$ h \left(t,v \right)=h \left(t \right)$$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.7.12) $$ Therefore $$ {h}_{v}=0$$ and
 * }
 * }
 * {| style="width:100%" border="0"

$$ {h}_{t}N+h \left( {N}_{t}- {M}_{v} \right)=0 $$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.7.13) $$ Rearranging and substituting
 * }
 * }
 * {| style="width:100%" border="0"

$$ \frac{1}{h}dh= -\frac{1}{m} \left[0-kn {v}^{n-1} \right]dt$$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.7.14) $$ Integrate equation 1.7.14 to get the integrating factor, $$h=exp \left( \frac{kn {v}^{n-1}}{m}t\right)$$. Multiply equation 1.7.2 by this gives us the exact form
 * }
 * }
 * {| style="width:100%" border="0"

$$F=exp \left( \frac{kn {v}^{n-1}}{m}t\right)v'+exp \left(  \frac{kn {v}^{n-1}}{m}t\right) \left[  \frac{k}{m} {v}^{n}+g\right]=0$$ Therefore
 * <p style="text-align:right;">$$\displaystyle (Eq. 1.7.15) $$
 * }
 * }
 * {| style="width:100%" border="0"

$${v}_{y} \left(t \right)= \frac{1}{exp \left( \frac{kn {v}^{n-1}}{m}t\right)}   \int^{t}-exp \left(  \frac{kn {v}^{n-1}}{m}\right)gds$$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.7.16) $$ To test this solution, assume $$n=1$$. Equation 1.7.16 then becomes
 * }
 * }


 * {| style="width:100%" border="0"

$$ {v}_{y} \left(t \right)= exp \left(- \frac{k}{m}t \right) \left[- \frac{mg}{k}exp \left( \frac{k}{m}t \right)+c \right]$$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.7.17) $$ Plug in initial conditions of $$ {t}_{0}=0$$, and $$ {v}_{y}={v}_{ {y}_{0}}$$ to obtain $$c= {v}_{ {y}_{0}}+ \frac{mg}{k}$$. Equation 1.7.17 becomes
 * }
 * }
 * {| style="width:100%" border="0"

$$ {v}_{y} \left(t \right)= \left( {v}_{ {y}_{0}}+ \frac{mg}{k} \right)exp \left(- \frac{k}{m}t \right)- \frac{mg}{k}$$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.7.18) $$ Integrate equation 1.7.18 to obtain
 * }
 * }
 * {| style="width:100%" border="0"

$$y \left(t \right)=- \frac{m}{k} \left( {v}_{ {y}_{0}}+ \frac{mg}{k} \right)exp \left(- \frac{k}{m}t \right)- \frac{mg}{k}t+ {c}_{2}$$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.7.19) $$ Plug in initial conditions of $$ {t}_{0}=0$$, and $$ y= {y}_{0}$$ to obtain $$$$. Equation 1.7.19 becomes
 * }
 * }


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$$ {c}_{2}= {y}_{0}- \frac{mg}{k}+ \frac{m}{k} \left( {v}_{ {y}_{0}}+ \frac{mg}{k}\right) \left[1-exp( \frac{k}{m}t \right] $$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.7.20) $$ Since equations 1.7.18 and 1.7.20 agree with common solutions found in undergraduate mechanics, the general case can be stated after introducing dummy variables $$s$$, and $$w$$
 * }
 * }


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$${v}_{y} \left(t \right)= \frac{1}{-exp \left( \frac{kn {v}^{n-1}}{m}t\right)}   \int^{t}-exp \left(  \frac{kn {v}^{n-1}}{m}\right) \left(g \right)ds $$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.7.21) $$ and
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$${y} \left(t \right)= \int_{ }^{t} {-exp \left( \frac{kn {v}^{n-1}}{m}w\right)}   \int^{w}-exp \left(  \frac{kn {v}^{n-1}}{m}s\right) \left(g \right)dsdw $$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.7.22) $$ Part 3.2 Find $$ {v}_{y} \left(t \right) $$ and $${y} \left(t \right)$$ for $$m=m\left(t \right)$$ To solve we first apply a force balance on the rocket
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$$ \frac{d \left(m {v}_{y} \right)}{dt}= -k{{v}_{y}}^{n}-mg$$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.8.1) $$ Use the conservation of linear momentum
 * }
 * }


 * {| style="width:100%" border="0"

$$d \left(m {v}_{y} \right)= md{v}_{y}+{v}_{y}dm$$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.8.2) $$ Plugging equation 1.8.2 into 1.8.1
 * }
 * }
 * {| style="width:100%" border="0"

$$m \frac{d{v}_{y}}{dt}+ {v}_{y} \frac{dm}{dt}=-k{{v}_{y}}^{n}-mg$$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.8.3) $$ Since the mass is reduced at a constant rate while the fuel is burned for $${t}_{0} \rightarrow {t}_{1}$$ the rate in equation 1.8.3 becomes a constant
 * }
 * }
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$$\frac{dm}{dt}= \frac{m- {m}_{0}}{ {t}_{1}}:=- \alpha $$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.8.4) $$ And equation 1.8.3 becomes
 * }
 * }
 * {| style="width:100%" border="0"

$$ m \frac{d{v}_{y}}{dt}- {v}_{y} \alpha=-k{{v}_{y}}^{n}-mg $$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.8.5) $$ Rearranging gives us
 * }
 * }
 * {| style="width:100%" border="0"

$$ F \left( \cdot \right)= \dot{ {v}_{y}}- \frac{ \alpha}{m}{v}_{y}+ \frac{k}{m} {{v}_{y}}^{n}+g $$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.8.6) $$ Check for exactness. Equation 1.8.6 is in the form to satisfy condition 1. Checking condition 2
 * }
 * }
 * {| style="width:100%" border="0"

$$ {N}_{t}= \frac{d}{dt} \dot{ {v}_{y}}=0 $$ ||
 * }
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$$ {M}_{{v}_{y}}= \frac{d}{d {v}_{y}} \left( \frac{k}{m} {{v}_{y}}^{n}- \frac{ \alpha}{m} {v}_{y}+g \right)= \frac{kn}{m} {{v}_{y}}^{n-1}- \frac{ \alpha}{m} \neq0$$ || Second condition is not satisfied. Make equation 1.8.6 exact using an integrating factor
 * }
 * }
 * {| style="width:100%" border="0"

$$hF \left( \cdot \right)=hM \left(t, {v}_{y} \right)+hN \left(t, {v}_{y} \right) \dot{ {v}_{y}}$$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.8.7) $$ Choose an integrating factor as a function of time only, such that $${h}_{v}=0 $$ and therefore
 * }
 * }


 * {| style="width:100%" border="0"

$$ {h}_{t}N+h \left( {N}_{x}- {M}_{y} \right)=0$$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.8.8) $$ Plugging in
 * }
 * }
 * {| style="width:100%" border="0"

$$ \frac{{h}_{t}}{h}= \frac{-1}{1} \left[0- \frac{kn{ {v}_{y}}^{n-1}}{m \left(t \right)}+ \frac{ \alpha}{m \left(t \right)} \right]$$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.8.9) $$ Integrating, using a dummy variable $$s$$
 * }
 * }
 * {| style="width:100%" border="0"

$$ \int_{}^{}\frac{1}{h}dh= \int_{}^{t} \left( \frac{kn{ {v}_{y} \left(s \right)}^{n-1}}{m \left(t \right)}+ \frac{ \alpha}{m \left(t \right)}\right)ds$$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.8.10) $$ Therefore
 * }
 * }
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$$ h= exp \left[ \int_{}^{t} \left( \frac{kn{ {v}_{y} \left(s \right)}^{n-1}}{m \left(s \right)}+ \frac{ \alpha}{m \left(s \right)}\right)ds\right]$$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.8.11) $$ Using a dummy variable $$w$$, a solution can be written as
 * }
 * }
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$$ {v}_{y} \left(t \right)= exp \left[ -\int_{}^{t} \left( \frac{kn{ {v}_{y} \left(s \right)}^{n-1}}{m \left(s \right)}+ \frac{ \alpha}{m \left(s \right)}\right)ds\right] \int_{}^{t} exp \left[ \int_{}^{s} \left( \frac{kn{ {v}_{y} \left(w \right)}^{n-1}}{m \left(w \right)}+ \frac{ \alpha}{m \left(w \right)}\right)dw\right]gds$$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.8.12) $$ and
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$$ y \left(t \right)= \int_{}^{t} {v}_{y} \left(s \right)ds$$ || <p style="text-align:right;">$$\displaystyle (Eq. 1.8.) $$ With varying mass and air resistance, the equations are typically solved by numerical methods.
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Contributing Members

 * Solved and posted part 1 & part 2 by Egm6321.f10.team3.Sudheesh 15:36, 4 October 2010 (UTC)
 * Proofread part 1 & 2 and posted part 3 by James Roark 19:48, 5 October 2010 (UTC)
 * Solved part 3 by Chris Cook --Egm6321.f10.team3.cook 22:10, 5 October 2010 (UTC)

= Problem 2- Equation of Motion of Pendulums Connected by a Spring =

Given
Shown in figure are the two Pendulums connected by a spring:

Find

 * $$\left( 1 \right)$$ Derive equation of motion:


 * {| style="width:100%" border="0"

$$
 * $${{m}_{1}}{{l}^{2}}{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}}=-k{{a}^{2}}\left( {{\theta }_{1}}-{{\theta }_{2}} \right)-{{m}_{1}}gl{{\theta }_{1}}+{{u}_{1}}l$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.1)
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.1)


 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $${{m}_{2}}{{l}^{2}}{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}}=-k{{a}^{2}}\left( {{\theta }_{2}}-{{\theta }_{1}}\right)-{{m}_{2}}gl{{\theta }_{2}}+{{u}_{2}}l$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.2)
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.2)
 * }
 * }


 * $$\left( 2 \right)$$ Write Eq.2.1 and Eq.2.2 in the form of [[media:2010_09_21_14_56_48.djvu|Mtg 13 (c),page2 ]], of:


 * {| style="width:100%" border="0"

$$
 * $$\underset{-}{\overset{\centerdot }{\mathop{x}}}\,=\underset{-}{\mathop{A}}\,\left( t \right)\underset{-}{\mathop{x}}\,\left( t \right)+\underset{-}{\mathop{B}}\,\left( t \right)\underset{-}{\mathop{u}}\,\left( t \right)$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.3)
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.3)
 * }
 * }


 * {| style="width:100%" border="0"

$$\underset{-}{\mathop{x}}\,={{\left[ \begin{matrix} {{\theta }_{1}} & \overset{\centerdot }{\mathop{{{\theta }_{1}}}}\, & {{\theta }_{2}} & \overset{\centerdot }{\mathop{{{\theta }_{2}}}}\, \\ \end{matrix} \right]}^{T}}$$ and $$\underset{-}{\mathop{u}}\,={{\left[ \begin{matrix} {{u}_{1}}l & {{u}_{2}}l \\ \end{matrix} \right]}^{T}}$$
 * Given
 * Given
 * }
 * }

Solution

 * $$\left( 1 \right)$$ Derive equation of motion:
 * (a) Consider Free Body Diagram of left pendulum:


 * For small angle:
 * {| style="width:100%" border="0"


 * $$\displaystyle l\sin {{\theta }_{1}}=l{{\theta }_{1}}$$ and $$\displaystyle l\cos {{\theta }_{1}}=l$$
 * }
 * }
 * }


 * {| style="width:100%" border="0"


 * $$acceleration, {{a}_{1}}=\frac{{{d}^{2}}\left( l{{\theta }_{1}} \right)}{d{{t}^{2}}}=l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}}$$
 * }
 * }
 * }


 * {| style="width:100%" border="0"


 * $$ Inertia force ={{m}_{1}}l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}}$$
 * }
 * }
 * }


 * {| style="width:100%" border="0"


 * $$\displaystyle Spring force =ka{{\theta }_{1}}-ka{{\theta }_{2}}=ka({{\theta }_{1}}-{{\theta }_{1}})$$
 * }
 * }
 * }

Now using D'Alembert's_principle, sum of the moments about pivot(A)is equal to zero


 * {| style="width:100%" border="0"


 * $$\sum{{{M}_{A}}=0\Rightarrow }\left( {{m}_{1}}l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}} \right)\cdot l+\left( ka({{\theta }_{1}}-{{\theta }_{2}}) \right)\cdot a+\left( {{m}_{1}}g \right)\cdot l{{\theta }_{1}}-\left( {{u}_{1}} \right)\cdot l=0$$
 * }
 * }
 * }


 * {| style="width:100%" border="0"

$$\Rightarrow {{m}_{1}}{{l}^{2}}{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}}=-k{{a}^{2}}({{\theta }_{1}}-{{\theta }_{2}})-{{m}_{1}}gl{{\theta }_{1}}+{{u}_{1}}l$$ $$
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 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.1)
 * }
 * }


 * (b) Consider Free Body Diagram of right pendulum:




 * For small angle:
 * {| style="width:100%" border="0"


 * $$\displaystyle l\sin {{\theta }_{2}}=l{{\theta }_{2}}$$ and $$\displaystyle l\cos {{\theta }_{2}}=l$$
 * }
 * }
 * }


 * {| style="width:100%" border="0"


 * $$acceleration, {{a}_{2}}=\frac{{{d}^{2}}\left( l{{\theta }_{2}} \right)}{d{{t}^{2}}}=l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}}$$
 * }
 * }
 * }


 * {| style="width:100%" border="0"


 * $$ Inertia force ={{m}_{2}}l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}}$$
 * }
 * }
 * }


 * {| style="width:100%" border="0"


 * $$\displaystyle Spring force =ka{{\theta }_{2}}-ka{{\theta }_{1}}=ka({{\theta }_{2}}-{{\theta }_{1}})$$
 * }
 * }
 * }

Using D'Alembert's_principle, sum of the moments about pivot(B)is equal to zero


 * {| style="width:100%" border="0"


 * $$\sum{{{M}_{B}}=0\Rightarrow }\left( {{m}_{2}}l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}} \right)\cdot l+\left( ka({{\theta }_{2}}-{{\theta }_{1}}) \right)\cdot a+\left( {{m}_{2}}g \right)\cdot l{{\theta }_{2}}-\left( {{u}_{2}} \right)\cdot l=0$$
 * }
 * }
 * }


 * {| style="width:100%" border="0"

$$\Rightarrow {{m}_{2}}{{l}^{2}}{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}}=-k{{a}^{2}}({{\theta }_{2}}-{{\theta }_{1}})-{{m}_{2}}gl{{\theta }_{2}}+{{u}_{2}}l$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.2)
 * }
 * }


 * $$\left( 2 \right)$$ Write Eq.2.1 and Eq.2.2 in the form of Eq.2.3(system of coupled equation):


 * Eq.2.1 can be rearranged as,


 * {| style="width:100%" border="0"

$$
 * $${{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}}=\frac{-\left( k{{a}^{2}}+{{m}_{1}}gl \right)}{m{}_{1}{{l}^{2}}}{{\theta }_{1}}+\frac{k{{a}^{2}}}{m{}_{1}{{l}^{2}}}{{\theta }_{2}}+\frac{{{u}_{1}}l}{m{}_{1}{{l}^{2}}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.4)
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.4)
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $${{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}}=\frac{k{{a}^{2}}}{m{}_{2}{{l}^{2}}}{{\theta }_{1}}+\frac{-\left( k{{a}^{2}}+{{m}_{2}}gl \right)}{m{}_{2}{{l}^{2}}}{{\theta }_{2}}+\frac{{{u}_{2}}l}{m{}_{2}{{l}^{2}}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.5)
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.5)
 * }
 * Now Eq.2.4 and Eq.2.5 can be put in the form of Eq.2.3 as:
 * Now Eq.2.4 and Eq.2.5 can be put in the form of Eq.2.3 as:


 * {| style="width:100%" border="0"


 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

$$\left[ \begin{matrix} {{\overset{\centerdot }{\mathop{\theta }}\,}_{1}} \\ {{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}} \\ {{\overset{\centerdot }{\mathop{\theta }}\,}_{2}} \\ {{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}} \\ \end{matrix} \right]=\underbrace{\left[ \begin{matrix} 0 & 1 & 0 & 0 \\   \frac{-\left( k{{a}^{2}}+{{m}_{1}}gl \right)}{m{}_{1}{{l}^{2}}} & 0 & \frac{k{{a}^{2}}}{m{}_{1}{{l}^{2}}} & 0  \\ 0 & 0 & 0 & 1 \\   \frac{k{{a}^{2}}}{m{}_{2}{{l}^{2}}} & 0 & \frac{-\left( k{{a}^{2}}+{{m}_{2}}gl \right)}{m{}_{2}{{l}^{2}}} & 0  \\ \end{matrix} \right]}_{\underset{-}{\mathop{A}}\,}\left[ \begin{matrix} {{\theta }_{1}} \\ {{\overset{\centerdot }{\mathop{\theta }}\,}_{1}} \\ {{\theta }_{2}} \\ {{\overset{\centerdot }{\mathop{\theta }}\,}_{2}} \\ \end{matrix} \right]+\underbrace{\left[ \begin{matrix} 0 & 0 \\   \frac{1}{m{}_{1}{{l}^{2}}} & 0  \\ 0 & 0 \\   0 & \frac{1}{m{}_{2}{{l}^{2}}}  \\ \end{matrix} \right]}_{\underset{-}{\mathop{B}}\,}\left[ \begin{matrix} {{u}_{1}}l \\ {{u}_{2}}l \\ \end{matrix} \right]$$

$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 2.6)
 * }
 * }


 * Where:


 * {| style="width:100%" border="0"

0 & 1 & 0 & 0 \\   \frac{-\left( k{{a}^{2}}+{{m}_{1}}gl \right)}{m{}_{1}{{l}^{2}}} & 0 & \frac{k{{a}^{2}}}{m{}_{1}{{l}^{2}}} & 0  \\ 0 & 0 & 0 & 1 \\   \frac{k{{a}^{2}}}{m{}_{2}{{l}^{2}}} & 0 & \frac{-\left( k{{a}^{2}}+{{m}_{2}}gl \right)}{m{}_{2}{{l}^{2}}} & 0  \\ \end{matrix} \right]$$
 * $$\underset{-}{\mathop{A}}\,=\left[ \begin{matrix}
 * $$\underset{-}{\mathop{A}}\,=\left[ \begin{matrix}


 * }
 * }


 * {| style="width:100%" border="0"

0 & 0 \\   \frac{1}{m{}_{1}{{l}^{2}}} & 0  \\ 0 & 0 \\   0 & \frac{1}{m{}_{2}{{l}^{2}}}  \\ \end{matrix} \right]$$
 * $$\underset{-}{\mathop{B}}\,=\left[ \begin{matrix}
 * $$\underset{-}{\mathop{B}}\,=\left[ \begin{matrix}
 * }
 * }

Contributing Members
Solved and posted by Egm6321.f10.team3.Sudheesh 15:39, 4 October 2010 (UTC)

Proofed by Kurt Schulze

Proofed by Egm6321.f10.team3.franklin 06:04, 6 October 2010 (UTC)

= Problem 3- Solution of State Equation (Control Engineering) =

Given

 * {| style="width:100%" border="0"

$$
 * $$ x'(t)=a(t)\cdot x(t)+b(t)\cdot u(t)$$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }

Find
Solve given equation by using Integrating Factor Method.

Solution
Let a(t) = a  &  b(t) = b


 * {| style="width:100%" border="0"

$$
 * $$ x'(t)=a\cdot x(t)+b\cdot u(t)$$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$
 * $$ x'(t)-a\cdot x(t)=b\cdot u(t)$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.1)
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.1)
 * }
 * }

We determine integrating factor from the coefficient of x(t) in Eq(3.1)


 * {| style="width:100%" border="0"

$$
 * $$ h(t) = e^{\int {-a \cdot dt}} = e^{-at} $$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }

Apply obtained integrating factor to Eq(3.1)


 * {| style="width:100%" border="0"

$$
 * $$ [x'(t)-a\cdot x(t)]\cdot e^{-at} = [b\cdot u(t)]\cdot e^{-at}$$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $$\underbrace{{x}'(t)\cdot {{e}^{-at}}-a\cdot x(t)\cdot {{e}^{-at}}}_{\because f(x)\cdot {g}'(x)+{f}'(x)\cdot g(x)={{\left( f(x)\cdot g(x) \right)}^{'}}}=b\cdot u(t)\cdot {{e}^{-at}}={{\left( f(x)\cdot g(x) \right)}^{'}}$$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$ $$
 * $$ [e^{-at}\cdot x(t)]' = b\cdot u(t)\cdot e^{-at}
 * $$ [e^{-at}\cdot x(t)]' = b\cdot u(t)\cdot e^{-at}
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }

Integrate between $$\displaystyle t_{0} $$ < $$\displaystyle \tau $$ < $$\displaystyle t $$


 * {| style="width:100%" border="0"

$$ $$
 * $$ \int\limits_^{t}{[e^{-a\tau}\cdot x(\tau)]'}d\tau = \int\limits_^{t} {b \cdot u(\tau)\cdot e^{-a\tau}}d\tau
 * $$ \int\limits_^{t}{[e^{-a\tau}\cdot x(\tau)]'}d\tau = \int\limits_^{t} {b \cdot u(\tau)\cdot e^{-a\tau}}d\tau
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$ $$
 * $$ e^{-at}\cdot x(t) - e^{-at_0}\cdot x(t_0) = \int\limits_^{t} {b \cdot u(\tau)\cdot e^{-a\tau}}d\tau
 * $$ e^{-at}\cdot x(t) - e^{-at_0}\cdot x(t_0) = \int\limits_^{t} {b \cdot u(\tau)\cdot e^{-a\tau}}d\tau
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$ $$
 * $$ e^{-at}\cdot x(t) = e^{-at_0}\cdot x(t_0) + \int\limits_^{t} {b \cdot u(\tau)\cdot e^{-a\tau}}d\tau
 * $$ e^{-at}\cdot x(t) = e^{-at_0}\cdot x(t_0) + \int\limits_^{t} {b \cdot u(\tau)\cdot e^{-a\tau}}d\tau
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$ $$
 * $$ x(t) = \frac {e^{-at_0}}{e^{-at}} \cdot x(t_0) + \frac {\int\limits_^{t} {b \cdot u(\tau)\cdot e^{-a\tau}}d\tau}{e^{-at}}
 * $$ x(t) = \frac {e^{-at_0}}{e^{-at}} \cdot x(t_0) + \frac {\int\limits_^{t} {b \cdot u(\tau)\cdot e^{-a\tau}}d\tau}{e^{-at}}
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$x(t) = e^{a(t-t_0)}\cdot x(t_0) + \int\limits_^{t} {e^{a(t-\tau)}}\cdot b \cdot u(\tau)d\tau$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 3.2)
 * }
 * }

Contributing Members

 * Solved and posted by-User:Egm6321.f10.team3.Hong SJ 21:18, 03 October 2010 (UTC)
 * Proof read by- Egm6321.f10.team3.Sudheesh 00:11, 5 October 2010 (UTC)
 * Proof read by-Egm6321.f10.team03.sigillo 13:57, 5 October 2010 (UTC)

=Problem 4: Expansion of a Taylor Series=

Given
Given an expression of $$\exp \left( x \right)$$

Find
Find the Taylor Series Expansion

Solution
In general, the Taylor Series can be defined as:


 * {|style="width:100%" border="0"

$$f\left( x \right)=\sum\limits_{n=0}^{\infty }{\frac{{{f}^{\left( n \right)}}\left( a \right)}{n!}}{{\left( x-a \right)}^{n}}$$
 * <p style="text-align:right">$$\,(Eq.4.1)$$
 * }

Where $$\displaystyle f^{\left( n \right)}$$ is the $$\displaystyle n$$th derivative of $$\displaystyle  f$$ evaluated at $$\displaystyle a$$. Expanding the series around point $$\displaystyle a=0$$ gives each $$\displaystyle  n$$th derivative of $$ \displaystyle  f$$ equal to zero, or $$\displaystyle f^{\left( n \right)}\left( 0 \right)=1$$. Thus, the Taylor series in this case can be defined as:


 * {|style="width:100%" border="0"

$$f\left( x \right)=\sum\limits_{n=0}^{\infty }{\frac{n!}}$$
 * <p style="text-align:right">$$\,(Eq.4.2)$$
 * }

Hence, for $$\displaystyle f\left( x \right)=\exp \left( x \right)$$, Taylor's series expansion is


 * {|style="width:100%" border="0"

$$\exp \left( x \right)=1+\frac{x}{1!}+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+...+\frac{x^{n}}{n!}$$
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * <p style="text-align:right">$$\,(Eq.4.3)$$
 * style= |
 * }

Contributing Members

 * Solved and posted by James Roark 23:02, 3 October 2010 (UTC)
 * Proof read by Egm6321.f10.team3.Sudheesh 00:36, 5 October 2010 (UTC)
 * Proof read byEgm6321.f10.team03.sigillo 14:05, 5 October 2010 (UTC)
 * Proof read byEgm6321.f10.team03.Hong SJ 11:24, 5 October 2010 (UTC)
 * Proof read by Egm6321.f10.team3.franklin 05:59, 6 October 2010 (UTC)

= Problem 5 - Generalization of a SC-L1-ODE-CC to SC-L1-ODE-VC =

Given
Given is the solution for $$\underline{x}(t)$$ of a first order linear ODE for a coupled system with constant coefficients (SC-L1-ODE-CC) out of the lecture notes in [[media:2010_09_23_13_57_17.djvu|Mtg 14 page 1-2, Eq.4]]:


 * {| style="width:100%" border="0"

$$\underline{x}(t)={{e}^{\underline{A}\cdot \left( t-{{t}_{0}} \right)}}\cdot \underline {x}({{t}_{0}})+\int\limits_^{t}{{{e}^{\underline{A}\cdot \left( t-\tau \right)}}\cdot \underline{B}\cdot \underline{u}\left( \tau  \right)d\tau }$$
 * <p style="text-align:right;">$$\,(Eq. 5.1) $$
 * style= |
 * }

Where $$\underline{x}$$ is a n x 1 vector, $$\underline{u}$$ a m x 1 vector, $$\underline{A}$$ is a n x n matrix and $$\underline{B}$$ is a n x m matrix.

Find
The general form for SC-L1-ODEs with varying coefficient (SC-L1-ODE-VC) has to be found based on the given SC-L1-ODE-CC

Solution
Following the approach of the integrating factor method and comparing the general form for L1-ODE-VC given in [[media:2010_09_23_13_57_17.djvu|Eq.3,page 1, Mtg 14 ]], it can be seen (and calculated through the intergration factor method - refer to Problem 3-) that the generalized form for (Eq. 5.1) can be achieved by integrating the argument of the exp(.)-function in the limits $$\,{{t}_{0}}$$ and $$\,t$$ for the first term, $$\,\tau $$ and $$\,t$$ for the second term:
 * {| style="width:100%" border="0"

$$\underline{x}(t)={{e}^{\int\limits_^{t}{\underline{A}(\tau )d\tau }}}\cdot \underline{x}({{t}_{0}})+\int\limits_^{t}{{{e}^{\int\limits_{\tau }^{t}{\underline{A}(s)ds}}}\cdot \underline{B}\cdot \underline{u}\left( \tau \right)d\tau }$$ Note: The integral of a matrix is calculated by integrating each element that composes the matrix.
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\,(Eq. 5.2) $$
 * style= |
 * }

Contributing Members

 * Solved by Michele Sigilló
 * Posted by Michele Sigilló
 * Proof read by Egm6321.f10.team3.Sudheesh 18:20, 5 October 2010 (UTC)

= Problem 6 - General SC-L1-ODE-CC with state transition matrix $$\underline{\Phi} $$ =

Given
The state equation for a coupled system like it's usual for a openloop control system is given as:
 * {| style="width:100%" border="0"

$$\underline{\dot{x}}(t)=\underline{A}\underline{x}(t)+\underline{B}\underline{u}(t)$$ The state transition matrix $$\underline{\Phi} $$ has the following properties: and where I is the identity matrix.
 * <p style="text-align:right;">$$\,(Eq. 6.1) $$
 * style= |
 * }

Find
With (Eq. 6.1) through (Eq. 6.3), Eq. (4) on pp. 14-1 of the lecture notes, has to be found:
 * {| style="width:100%" border="0"

$$\underline{x}(t)={{e}^{\underline{A}(t-{{t}_{0}})}}\cdot \underline{x}({{t}_{0}})+\int\limits_^{t}{{{e}^{\underline{A}(t-\tau )}}\underline{B}\underline{u}(\tau )\,d\tau }$$
 * <p style="text-align:right;">$$\,(Eq. 6.4) $$
 * style= |
 * }

Solution
In (Eq. 6.2) the homogeneous differential equation for $$\underline{\Phi} (t,{{t}_{0}})$$ is given. Thus replacing $$\underline{x}(t)$$ with $$\underline{\Phi} (t,{{t}_{0}})$$ in (Eq. 6.1) gives the inhomogeneous equation for $$\underline{\Phi} (t,{{t}_{0}})$$:
 * {| style="width:100%" border="0"

$$\dot{\underline{\phi} }(t,{{t}_{0}})=\underline{A}\underline{\phi} (t,{{t}_{0}})+\underline{B}\underline{u}(t)$$
 * <p style="text-align:right;">$$\,(Eq. 6.5) $$
 * style= |
 * }

Regrouping the equation above results in:
 * {| style="width:100%" border="0"

$$\underbrace{-\left( \underline{A}\underline{\phi} (t,{{t}_{0}})+\underline{B}\underline{u}(t) \right)}_{\underline{M}(\underline{\phi} ,\underline{u})}+\underbrace{1}_{\underline{N}(\underline{\phi} ,\underline{u})}\cdot \dot{\underline{\phi} }(t,{{t}_{0}})=0$$ (Eq. 6.6) meets the first condition of exactness, but the second condition $${{\underline{M}}_{\underline{\phi} }}={{\underline{N}}_{\underline{u}}}$$ needs to be checked:
 * <p style="text-align:right;">$$\,(Eq. 6.6) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${{\underline{M}}_{\underline{\phi} }}=-\underline{A}\ne 0={{\underline{N}}_{\underline{u}}}$$ The second condition is not met, thus a function $$h(t)$$ to make (Eq. 6.6) exact needs to be found. Therefore (Eq. 6.6) will be rewritten as:
 * <p style="text-align:right;">$$\,(Eq. 6.7) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$-\underline{A}\underline{\phi} (t,{{t}_{0}})+1\cdot \dot{\underline{\phi} }(t,{{t}_{0}})=\underline{B}\underline{u}(t)$$ The integrating factor $$h(t)$$ is then given as:
 * <p style="text-align:right;">$$\,(Eq. 6.8) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$h(t)={{e}^{\int\limits_^{t}{-\underline{A}\,ds}}}={{e}^{-\underline{A}(t-{{t}_{0}})}}$$ Multiplying $$h(t)$$ throughout (Eq. 6.8) gives:
 * <p style="text-align:right;">$$\,(Eq. 6.9) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$-\underline{A}\underline{\phi} (t,{{t}_{0}})\cdot {{e}^{-\underline{A}(t-{{t}_{0}})}}+{{e}^{-\underline{A}(t-{{t}_{0}})}}\cdot \dot{\underline{\phi} }(t,{{t}_{0}})=\underline{B}\underline{u}(t)\cdot {{e}^{-\underline{A}(t-{{t}_{0}})}}$$ which is equal to
 * <p style="text-align:right;">$$\,(Eq. 6.10) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\frac{d\left( {{e}^{-\underline{A}(t-{{t}_{0}})}}\cdot \underline{\phi} (t,{{t}_{0}}) \right)}{dx}=\underline{B}\underline{u}(t)\cdot {{e}^{-\underline{A}(t-{{t}_{0}})}}$$ Integrating (Eq. 6.11) on both sides between $${{t}_{0}}$$ and $$t$$ gives:
 * <p style="text-align:right;">$$\,(Eq. 6.11) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${{e}^{-\underline{A}(t-{{t}_{0}})}}\cdot \underline{\phi} (t,{{t}_{0}})-{{e}^{-\underline{A}({{t}_{0}}-{{t}_{0}})}}\cdot \underline{\phi} ({{t}_{0}},{{t}_{0}})={{e}^{\underline{A}{{t}_{0}}}}\cdot \int\limits_^{t}\cdot \underline{B}\underline{u}(t)\,d\tau $$ that is equal to
 * <p style="text-align:right;">$$\,(Eq. 6.12) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\underline{\phi} (t,{{t}_{0}})={{e}^{\underline{A}(t-{{t}_{0}})}}\cdot \underline{\phi} ({{t}_{0}},{{t}_{0}})+\int\limits_^{t}{{{e}^{\underline{A}(t-\tau )}}\cdot }\underline{B}\underline{u}(t)\,d\tau $$ Replacing $$\underline{\Phi} (t,{{t}_{0}})$$ with $$\underline{x}(t)$$ gives:
 * <p style="text-align:right;">$$\,(Eq. 6.13) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\underline{x}(t)={{e}^{\underline{A}(t-{{t}_{0}})}}\cdot \underline{x}({{t}_{0}})+\int\limits_^{t}{{{e}^{\underline{A}(t-\tau )}}\underline{B}\underline{u}(\tau )\,d\tau }$$ which is the same as the given solution in (Eq. 6.4)!
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * <p style="text-align:right;">$$\,(Eq. 6.13) $$
 * style= |
 * }

Contributing Members

 * Solved by Michele Sigilló
 * Posted by Michele Sigilló
 * Proof read by Egm6321.f10.team3.franklin 06:05, 6 October 2010 (UTC)
 * Proof read by James Roark 12:18, 6 October 2010 (UTC)

= Problem 7 - Roll Control of Rocket By Actuating Ailerons =

Given



 * {| style="width:100%" border="0"

$$
 * $$\overset{\centerdot }{\mathop{\phi }}\,=\omega $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.1)
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.1)
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $$\overset{\centerdot }{\mathop{\omega }}\,=\frac{-1}{\tau }\omega +\frac{Q}{\tau }\delta $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.2)
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.2)
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $$\overset{\centerdot }{\mathop{\delta }}\,=u$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.3)
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.3)
 * }
 * }

Find
Control behavior in form of


 * {| style="width:100%" border="0"

$$
 * $$\overset{\centerdot}{\mathop{x\left( t \right)=\underset{-}{\mathop{A}}\,}}\,\underset{-}{\mathop{x}}\,\left( t \right)+\underset{-}{\mathop{B}}\,\underset{-}{\mathop{u}}\,\left( t \right)$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.4)
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.4)


 * }
 * }

Solution

 * Integrating Eq.7.3 gives,


 * {| style="width:100%" border="0"

$$
 * $$\displaystyle \delta =ut$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.5)
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.5)
 * }
 * Substitute Eq.7.5 for $$\displaystyle \delta $$, in Eq.7.2:
 * Substitute Eq.7.5 for $$\displaystyle \delta $$, in Eq.7.2:


 * {| style="width:100%" border="0"

$$
 * $$\overset{\centerdot }{\mathop{\omega }}\,=\frac{-1}{\tau }\omega +\frac{Q}{\tau }ut$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.6)
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.6)
 * }
 * }


 * Now integrate Eq.7.6:


 * {| style="width:100%" border="0"

$$
 * $$\omega =\frac{-1}{\tau }\omega t+\frac{Q}{2\tau }u{{t}^{2}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.7)
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.7)
 * }
 * }


 * Similarly integrating Eq.7.1 gives,


 * {| style="width:100%" border="0"

$$
 * $$\displaystyle \phi =\omega t$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.8)
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.8)
 * }
 * }


 * Substitute Eq.7.8 in Eq.7.7 implies,


 * {| style="width:100%" border="0"

$$
 * $$\omega =\frac{-1}{\tau }\phi +\frac{Q{{t}^{2}}}{2\tau }u$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.9)
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.9)
 * }
 * }


 * From Eq.7.1 and Eq.7.9,


 * {| style="width:100%" border="0"

$$
 * $$\overset{\centerdot }{\mathop{\phi }}\,=\frac{-1}{\tau }\phi +\frac{Q{{t}^{2}}}{2\tau }u$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.10)
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.10)
 * }
 * }


 * Now, Eq.7.3, Eq.7.6 and Eq.7.10 can be written in the required form as,


 * {| style="width:100%" border="0"


 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |

$$\left[ \begin{matrix} \overset{\centerdot }{\mathop{\phi }}\, \\ \overset{\centerdot }{\mathop{\omega }}\, \\ \overset{\centerdot }{\mathop{\delta }}\, \\ \end{matrix} \right]=\left[ \begin{matrix} \frac{-1}{\tau } & 0 & 0 \\ 0 & \frac{-1}{\tau } & 0 \\ 0 & 0 & 0 \\ \end{matrix} \right]\left[ \begin{matrix} \phi  \\ \omega  \\ \delta  \\ \end{matrix} \right]+\left[ \begin{matrix} \frac{Q{{t}^{2}}}{2\tau } \\ \frac{Qt}{\tau } \\ 1 \\ \end{matrix} \right]\left[ u \right]$$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq. 7.11)
 * }
 * }


 * where,


 * {| style="width:100%" border="0"

\frac{-1}{\tau } & 0 & 0 \\ 0 & \frac{-1}{\tau } & 0 \\ 0 & 0 & 0 \\ \end{matrix} \right]$$
 * $$\underset{-}{\mathop{A}}\,=\left[ \begin{matrix}
 * $$\underset{-}{\mathop{A}}\,=\left[ \begin{matrix}


 * }
 * }


 * {| style="width:100%" border="0"

\frac{Q{{t}^{2}}}{2\tau } \\ \frac{Qt}{\tau } \\ 1 \\ \end{matrix} \right]$$
 * $$\underset{-}{\mathop{B}}\,=\left[ \begin{matrix}
 * $$\underset{-}{\mathop{B}}\,=\left[ \begin{matrix}

|-
 * }


 * {| style="width:100%" border="0"

\phi  \\ \omega  \\ \delta  \\ \end{matrix} \right]$$ |-
 * $$\underset{-}{\mathop{x}}\,=\left[ \begin{matrix}
 * $$\underset{-}{\mathop{x}}\,=\left[ \begin{matrix}
 * }


 * {| style="width:100%" border="0"

|-
 * $$\underset{-}{\mathop{u}}\,=\left[ u \right]$$
 * $$\underset{-}{\mathop{u}}\,=\left[ u \right]$$
 * }

Contributing Members

 * Solved by: Egm6321.f10.team03 19:29, 4 October 2010 (UTC).Kurt Schulze
 * Posted by: Egm6321.f10.team3.Sudheesh 18:59, 4 October 2010 (UTC)
 * Proof read by Egm6321.f10.team03.sigillo 14:30, 5 October 2010 (UTC)
 * Proof read by Egm6321.f10.team3.franklin 05:58, 6 October 2010 (UTC)

= Problem 8- Discuss the search for h =

Given
From the lecture notes of Mtg [|Mtg 17]


 * {| style="width:100%" border="0"

$$\displaystyle {h}_{x} + {h}_{y} {P} = 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right;">$$\displaystyle (1) p. 17-1
 * }

Find
Without assuming a priory that $$ h = const$$, discuss the search for the solution of $$ (1) p. 17-1$$

Solution
To find $$ h(x,y)$$ we begin by rearranging $$ (1) p. 17-1 $$ to find


 * {| style="width:100%" border="0"

$$\displaystyle {h}_{x} = - {h}_{y}P $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right;">$$\displaystyle Eq. (2)
 * }

Next we will use the property that


 * {| style="width:100%" border="0"

$$\displaystyle {h}_{xy} = {h}_{yx} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right;">$$\displaystyle Eq. (3)
 * }

and see if it holds. First we take evaluate $$ {h}_{xy}$$


 * {| style="width:100%" border="0"

$$ {h}_{x}= - {h}_{y} \underbrace_{:= {y}^{'}}
 * style="width:95%" |
 * style="width:95%" |

\Rightarrow

{h}_{xy} = - {h}_{yy}{y}^{'} - {h}_{y} {y}^{''} $$ $$
 * <p style="text-align:right;">$$\displaystyle Eq. (4)
 * }

By inserting $$ Eq. (4) $$ into $$ Eq. (3)$$ we see that this property does not hold


 * {| style="width:100%" border="0"

$$ {h}_{xy} = - {h}_{yy}{y}^{'} - {h}_{y} {y}^{''} \neq {h}_{yx} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right;">$$\displaystyle Eq. (5)
 * }

So for $$ (1) p. 17-1 $$ to be true, $$ {h}_{x} = {h}_{y} = 0$$.

Therefore $$ h = const $$

Contributing Members
Solved and Posted By Chris Cook --Egm6321.f10.team3.cook 22:08, 5 October 2010 (UTC)

= Problem 9  -  Check Exactness of Differential Equation =

Given

 * {| style="width:100%" border="0"

$$ $$
 * $$ \underbrace{\phi_p}_{f}\cdot y'' + \underbrace{\phi_y \cdot y' + \phi_x}_{g} = 0
 * $$ \underbrace{\phi_p}_{f}\cdot y'' + \underbrace{\phi_y \cdot y' + \phi_x}_{g} = 0
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$ $$
 * $$ \underbrace{(15p^4\cdot cos (x^2))}_{\phi_p}\cdot y'' + \underbrace{(6xy^2)}_{\phi_y} \cdot y' + \underbrace{(-6xp^5\cdot sin(x^2) + 2y^3)}_{\phi_x} = 0
 * $$ \underbrace{(15p^4\cdot cos (x^2))}_{\phi_p}\cdot y'' + \underbrace{(6xy^2)}_{\phi_y} \cdot y' + \underbrace{(-6xp^5\cdot sin(x^2) + 2y^3)}_{\phi_x} = 0
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$ $$
 * $$ \therefore f = \phi_p = 15p^4\cdot cos (x^2)
 * $$ \therefore f = \phi_p = 15p^4\cdot cos (x^2)
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $$ \therefore g = \phi_y \cdot y' + \phi_x = (6xy^2)\cdot y' + (-6xyp^5\cdot sin(x^2) + 2y^3)$$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }

Find

 * {| style="width:100%" border="0"

$$ $$
 * $$ (1) f_{xx} + 2p\cdot f_{xy} + p^2\cdot f_{yy} = g_{xp} + p\cdot g_{yp}- g_y
 * $$ (1) f_{xx} + 2p\cdot f_{xy} + p^2\cdot f_{yy} = g_{xp} + p\cdot g_{yp}- g_y
 * <p style="text-align:right;">$$\displaystyle Eq(9.1)
 * }
 * }


 * {| style="width:100%" border="0"

$$ $$
 * $$ (2) f_{xp} + p\cdot f_{yp} + 2\cdot f_{y} = g_{pp}
 * $$ (2) f_{xp} + p\cdot f_{yp} + 2\cdot f_{y} = g_{pp}
 * <p style="text-align:right;">$$\displaystyle Eq(9.2)
 * }
 * }

Solution

 * {| style="width:100%" border="0"

$$
 * $$\displaystyle for Eq(9.1)$$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $$ f_{xx} = \frac{\partial }{\partial x} (\frac{\partial }{\partial x} (15p^4\cdot cos(x^2)))$$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$ $$
 * $$\displaystyle = \frac{\partial }{\partial x} (-2x\cdot sin(x^2)\cdot 15p^4)
 * $$\displaystyle = \frac{\partial }{\partial x} (-2x\cdot sin(x^2)\cdot 15p^4)
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$ $$
 * $$\displaystyle = \frac{\partial }{\partial x} (-30p^4\cdot x\cdot sin(x^2))
 * $$\displaystyle = \frac{\partial }{\partial x} (-30p^4\cdot x\cdot sin(x^2))
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$ $$
 * $$\displaystyle  = -30p^4\cdot x \cdot 2x \cdot cos(x^2) + (-30p^4\cdot sin(x^2))
 * $$\displaystyle  = -30p^4\cdot x \cdot 2x \cdot cos(x^2) + (-30p^4\cdot sin(x^2))
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$ $$
 * $$\displaystyle  = -60p^4\cdot x^2 \cdot cos(x^2) - 30p^4\cdot sin(x^2)
 * $$\displaystyle  = -60p^4\cdot x^2 \cdot cos(x^2) - 30p^4\cdot sin(x^2)
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $$\displaystyle f_{xy} = \frac{\partial }{\partial y} (\frac{\partial }{\partial x} (15p^4\cdot cos(x^2)))$$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$ $$
 * $$\displaystyle = \frac{\partial }{\partial y} (-2x\cdot sin(x^2)\cdot 15p^4)
 * $$\displaystyle = \frac{\partial }{\partial y} (-2x\cdot sin(x^2)\cdot 15p^4)
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$ $$
 * $$\displaystyle = \frac{\partial }{\partial y} (-30p^4\cdot x\cdot sin(x^2))
 * $$\displaystyle = \frac{\partial }{\partial y} (-30p^4\cdot x\cdot sin(x^2))
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$ $$
 * $$\displaystyle = 0
 * $$\displaystyle = 0
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $$\displaystyle f_{yy} = \frac{\partial }{\partial y} (\frac{\partial }{\partial y} (15p^4\cdot cos(x^2)))$$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $$\displaystyle  = \frac{\partial }{\partial y} (0)$$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $$\displaystyle = 0 $$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $$\displaystyle g_{xp} = \frac{\partial }{\partial p} (\frac{\partial }{\partial x} (6xy^2p - 6xp^5\cdot sin(x^2) + 2y^3))$$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $$\displaystyle = \frac{\partial }{\partial p} (6y^2p-(6p^5x\cdot 2x\cdot cos(x^2) + 6p^5\cdot sin(x^2)))$$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $$\displaystyle = 6y^2 - 60x^2\cdot cos(x^2)\cdot p^4 - 30sin(x^2)\cdot p^4$$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $$\displaystyle g_{yp} = \frac{\partial }{\partial p} (\frac{\partial }{\partial y} (6xy^2p - 6xp^5\cdot sin(x^2) + 2y^3))$$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $$\displaystyle = \frac{\partial }{\partial p} (12xyp - 0 +6y^2)$$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $$\displaystyle = 12xy $$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $$\displaystyle g_{y} = \frac{\partial }{\partial y} (6xy^2p - 6xp^5\cdot sin(x^2) + 2y^3))$$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $$\displaystyle = 12xyp + 6y^2$$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }

Apply obtained factors to Eq(9.1)


 * {| style="width:100%" border="0"

$$ $$
 * $$\displaystyle L.H.S = -60p^4x^2\cdot cos(x^2) - 30p^4\cdot sin(x^2)
 * $$\displaystyle L.H.S = -60p^4x^2\cdot cos(x^2) - 30p^4\cdot sin(x^2)
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$ $$
 * $$\displaystyle R.H.S = (6y^2 - 60x^2\cdot cos(x^2)\cdot p^4 - 30sin(x^2)\cdot p^4)+ p(12xy) - (12xyp + 6y^2)
 * $$\displaystyle R.H.S = (6y^2 - 60x^2\cdot cos(x^2)\cdot p^4 - 30sin(x^2)\cdot p^4)+ p(12xy) - (12xyp + 6y^2)
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $$\displaystyle = -60p^4x^2\cdot cos(x^2) - 30p^4\cdot sin(x^2)$$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$\displaystyle \therefore L.H.S = R.H.S$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$$$
 * style= |
 * }


 * {| style="width:100%" border="0"

$$
 * $$\displaystyle for Eq(9.2)$$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $$\displaystyle f_{xp} = \frac{\partial }{\partial p} (\frac{\partial }{\partial x} (15p^4\cdot cos(x^2)))$$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$ $$
 * $$\displaystyle = \frac{\partial }{\partial p} (-2x\cdot sin(x^2)\cdot 15p^4)
 * $$\displaystyle = \frac{\partial }{\partial p} (-2x\cdot sin(x^2)\cdot 15p^4)
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$ $$
 * $$\displaystyle = \frac{\partial }{\partial p} (-30p^4\cdot x\cdot sin(x^2))
 * $$\displaystyle = \frac{\partial }{\partial p} (-30p^4\cdot x\cdot sin(x^2))
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$ $$
 * $$\displaystyle = -120x\cdot sin(x^2)\cdot p^3
 * $$\displaystyle = -120x\cdot sin(x^2)\cdot p^3
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $$\displaystyle f_{yp} = \frac{\partial }{\partial p} (\frac{\partial }{\partial y} (15p^4\cdot cos(x^2)))$$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $$\displaystyle = \frac{\partial }{\partial p} (0)$$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $$\displaystyle  = 0$$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $$\displaystyle f_{y} = \frac{\partial }{\partial y} (15p^4\cdot cos(x^2)))$$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $$\displaystyle = 0 $$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $$\displaystyle g_{pp} = \frac{\partial }{\partial p} (\frac{\partial }{\partial p} (6xy^2p - 6xp^5\cdot sin(x^2) + 2y^3))$$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $$\displaystyle g_{pp} = \frac{\partial }{\partial p} (6xy^2 - 30x/cdot sin(x^2)\cdot p^4)$$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$
 * $$\displaystyle -120x\cdot sin(x^2)\cdot p^3$$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }

Apply obtained factors to Eq(9.2)


 * {| style="width:100%" border="0"

$$ $$
 * $$\displaystyle L.H.S = -120x\cdot sin(x^2)\cdot p^3
 * $$\displaystyle L.H.S = -120x\cdot sin(x^2)\cdot p^3
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$ $$
 * $$\displaystyle R.H.S = -120x\cdot sin(x^2)\cdot p^3
 * $$\displaystyle R.H.S = -120x\cdot sin(x^2)\cdot p^3
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }


 * {| style="width:100%" border="0"

$$\displaystyle \therefore L.H.S = R.H.S$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$$$
 * style= |
 * }

Contributing Members

 * Solved and posted by-User:Egm6321.f10.team3.Hong SJ 17:09, 04 October 2010 (UTC)
 * Proof read by Kurt Schulze
 * Proof read by Egm6321.f10.team3.Sudheesh 18:27, 6 October 2010 (UTC)

= Problem 10 - Generate exact N2-ODE's - Reverse Engineering=

Given
Given is the following function from meeting 17
 * {| style="width:100%" border="0"

$$\phi \left( x,y,p \right)=3{{p}^{5}}\cos \left( {{x}^{2}} \right)+2x{{y}^{3}}=k$$ and Eq. (2) from meeting 17
 * <p style="text-align:right;">$$\,(Eq. 10.1) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\left( 6x{{y}^{2}} \right)\cdot {y}'+2{{y}^{3}}=\left( {{h}_{y}}+0 \right){y}'+{{h}_{x}}$$ on page 17-1 and 17-3 respectively in the lecture notes.
 * <p style="text-align:right;">$$\,(Eq. 10.2) $$
 * style= |
 * }

Find
Using (Eq. 10.2) the function $$\,h\left( x,y \right)$$ has to be found in order to get again the given function $$\,\phi $$ (reverse engineering) assuming $${{h}_{y}}\cdot {y}'=2{{y}^{3}}$$.

Solution
Applying the 'separation of variables'-method to the given assumption, $$\,h\left( x,y \right)$$ can be calculated:
 * {| style="width:100%" border="0"

$${{h}_{y}}\cdot dy=2{{y}^{3}}\cdot dx$$
 * <p style="text-align:right;">$$\,(Eq. 10.3) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\int{{{h}_{y}}\cdot dy=\int{2{{y}^{3}}\cdot dx}}$$
 * <p style="text-align:right;">$$\,(Eq. 10.4) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$h\left( x,y \right)=2{{y}^{3}}x+{{k}_{1}}(y)$$ Take derivative of $$\,h\left( x,y \right)$$ with respect to $$\,x$$ gives:
 * <p style="text-align:right;">$$\,(Eq. 10.5) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\,{{h}_{x}}=2{{y}^{3}}$$ At the same time it can be seen from (Eq. 10.2) that $$\,{{h}_{x}}$$ is equal to $$\left( 6x{{y}^{2}} \right)\cdot {y}'$$, so that the following is valid:
 * <p style="text-align:right;">$$\,(Eq. 10.6) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${{h}_{x}}=2{{y}^{3}}=6x{{y}^{2}}\cdot {y}'$$ What implies that:
 * <p style="text-align:right;">$$\,(Eq. 10.7) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${y}'=\frac{y}{3x}$$ and
 * <p style="text-align:right;">$$\,(Eq. 10.8) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\,{{k}_{1}}(y)=0$$
 * <p style="text-align:right;">$$\,(Eq. 10.9) $$
 * style= |
 * }

We know from Eq. (1) of page 17-3 in meeting 17


 * {| style="width:100%" border="0"


 * $$\phi =h\left( x,y \right)+\int{f(x,y,p)dp}$$
 * <p style="text-align:right;">$$\,(Eq. 10.10) $$
 * style= |
 * }
 * style= |
 * }

For the given problem,


 * {| style="width:100%" border="0"


 * $$\int{f(x,y,p)dp}=3{{p}^{5}}\cos {{x}^{2}}$$ from page 17-3 meeting 17
 * style= |
 * }
 * }

Now, inserting the found experssion in (Eq. 10.5) with $$\,{{k}_{1}}(y)=0$$ into Eq. 10.10 gives:
 * {| style="width:100%" border="0"

$$\phi \left( x,y,p \right)=3{{p}^{5}}\cos \left( {{x}^{2}} \right)+2x{{y}^{3}}=k$$ Which is the same as the given function $$\,\phi \left( x,y,p \right)$$
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * <p style="text-align:right;">$$\,(Eq. 10.11) $$
 * style= |
 * }

Contributing Members

 * Solved by Michele Sigilló
 * Posted by Michele Sigilló
 * Proof read by Egm6321.f10.team3.Sudheesh 18:14, 5 October 2010 (UTC)

=Contributing Team Members=
 * Kurt Schulze, solved 7, proofed 2 and 9.
 * Michele Sigilló, solved 5,6,10, proofed 3,4,7.
 * Sung Joo, Hong, solved 3, 9, proofed 4.
 * Sudheesh Thiyyakkandi, Solved and posted 1(part1& part2), 2, Posted 7, proofed 3,4,5,9,10.
 * Adam Franklin, proofed 2, 4, 6, and 7.