# User:Egm6321.f10.team03/Hwk3

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# Problem 1 - Motion of a Projectile with Air Resistance

## Given

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

 \begin{align} &k,n \in \mathbb{R}\\ &m = \text{mass of the particle}\\ &g = \text{acceleration of gravity} \end{align}

## 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
 $\displaystyle W=mg$
2) Inertia force
 ${{F}_{a}}=ma=m\cdot \frac{dv}{dt}$ for particle with constant mass
3) Air resistance which is proportional to the velocity of particle
 $\displaystyle {{F}_{D}}=k{{v}^{n}}$

Now consider the force equilibrium in both horizontal and vertical direction

a) Force Equilibrium in horizontal direction:
 $\sum{{{F}_{H}}=0}$
$m\cdot \frac{d{{v}_{x}}}{dt}+k{{v}^{n}}\cos \alpha =0$,
where $\displaystyle {{v}_{x}}\to$ horizontal component of velocity
 $m\cdot \frac{d{{v}_{x}}}{dt}=-k{{v}^{n}}\cos \alpha$ $\displaystyle (Eq. 1.1)$

b) Force Equilibrium in the vertical direction:
 $\sum{{{F}_{V}}=0}$
$m\cdot \frac{d{{v}_{y}}}{dt}+k{{v}^{n}}\sin \alpha +mg=0$,
where $\displaystyle {{v}_{y}}\to$ vertical component of velocity
 $m\cdot \frac{d{{v}_{y}}}{dt}=-k{{v}^{n}}\sin \alpha -mg$ $\displaystyle (Eq. 1.2)$

#### Part 2

Particular case: When $\displaystyle k=0$

Eq.1.1 reduces to

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

Integrating the above equation gives:

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

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

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

Now Eq.1.3 becomes:

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

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

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

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

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

Therefore,

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

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

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

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

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

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

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

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

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

Now Eq.1.5 becomes,

 ${{v}_{t}}\left( t \right)=-gt+{{v}_{{{y}_{0}}}}$

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

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

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

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

Therefore,

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

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

 $y\left( x \right)=-\frac{g}{2}{{\left( \frac{x-{{x}_{0}}}{{{v}_{{{x}_{0}}}}} \right)}^{2}}+{{v}_{{{y}_{0}}}}\left( \frac{x-{{x}_{0}}}{{{v}_{{{x}_{0}}}}} \right)+{{y}_{0}}$ $\displaystyle (Eq. 1.6)$

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

 $m \frac{d {v}_{g}}{dt}=-k {{v}_{g}}^{n}-mg$ $\displaystyle (Eq. 1.7.1)$

Or

$m \dot{v}=-k {v}^{n}-mg$
Let
 $F \left(t,v, \dot{v} \right):=m \dot{v}+k {v}^{n}+mg=0= \frac{d \phi(t,v)}{dt}$ $\displaystyle (Eq. 1.7.2)$

Therefore

 $\phi \left(t,v \right)=k$ $\displaystyle (Eq. 1.7.3)$

And equation 1.7.2 can be written in the form

 $F= \phi \left(x \right)+ \phi \left( y\right) y'=M+Ny'$ $\displaystyle (Eq. 1.7.4)$

Where

 $\phi \left(x \right)=k {v}^{n}+mg$ $\displaystyle (Eq. 1.7.5)$

and

 $\phi \left(y \right)=m$ $\displaystyle (Eq. 1.7.6)$

Check the exactness of equation 1.7.4. First condition is satisfied as it is in the form

 $F= \phi \left(x \right)+ \phi \left( y\right) y'=M+Ny'$ $\displaystyle (Eq. 1.7.7)$

Second condition is satisfied if

 ${N}_{t}= {m}_{v}$ $\displaystyle (Eq. 1.7.8)$

Differentiating:

 ${N}_{t}= \frac{d}{dt} \left(m \right)=0$ $\displaystyle (Eq. 1.7.9)$

and

 ${M}_{v}= \frac{d}{dv} \left( k {v}^{n}+mg\right)=kn {v}^{n-1} \neq0$ $\displaystyle (Eq. 1.7.10)$

Therefore equation 1.7.4 is not exact and must be made exact by finding an integration factor $h \left(t,v \right)$

 $h \left(t,v \right) \left[m \dot{v}+k {v}^{n}+mg\right]=0$ $\displaystyle (Eq. 1.7.11)$

Choose an integrating factor as a function of time only

 $h \left(t,v \right)=h \left(t \right)$ $\displaystyle (Eq. 1.7.12)$

Therefore ${h}_{v}=0$ and

 ${h}_{t}N+h \left( {N}_{t}- {M}_{v} \right)=0$ $\displaystyle (Eq. 1.7.13)$

Rearranging and substituting

 $\frac{1}{h}dh= -\frac{1}{m} \left[0-kn {v}^{n-1} \right]dt$ $\displaystyle (Eq. 1.7.14)$

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

 $F=exp \left( \frac{kn {v}^{n-1}}{m}t\right)v'+exp \left( \frac{kn {v}^{n-1}}{m}t\right) \left[ \frac{k}{m} {v}^{n}+g\right]=0$ $\displaystyle (Eq. 1.7.15)$

Therefore

 ${v}_{y} \left(t \right)= \frac{1}{exp \left( \frac{kn {v}^{n-1}}{m}t\right)} \int^{t}-exp \left( \frac{kn {v}^{n-1}}{m}\right)gds$ $\displaystyle (Eq. 1.7.16)$

To test this solution, assume $n=1$. Equation 1.7.16 then becomes

 ${v}_{y} \left(t \right)= exp \left(- \frac{k}{m}t \right) \left[- \frac{mg}{k}exp \left( \frac{k}{m}t \right)+c \right]$ $\displaystyle (Eq. 1.7.17)$

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

 ${v}_{y} \left(t \right)= \left( {v}_{ {y}_{0}}+ \frac{mg}{k} \right)exp \left(- \frac{k}{m}t \right)- \frac{mg}{k}$ $\displaystyle (Eq. 1.7.18)$

Integrate equation 1.7.18 to obtain

 $y \left(t \right)=- \frac{m}{k} \left( {v}_{ {y}_{0}}+ \frac{mg}{k} \right)exp \left(- \frac{k}{m}t \right)- \frac{mg}{k}t+ {c}_{2}$ $\displaystyle (Eq. 1.7.19)$

Plug in initial conditions of ${t}_{0}=0$, and $y= {y}_{0}$ to obtain . Equation 1.7.19 becomes

 ${c}_{2}= {y}_{0}- \frac{mg}{k}+ \frac{m}{k} \left( {v}_{ {y}_{0}}+ \frac{mg}{k}\right) \left[1-exp( \frac{k}{m}t \right]$ $\displaystyle (Eq. 1.7.20)$

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 $w$

 ${v}_{y} \left(t \right)= \frac{1}{-exp \left( \frac{kn {v}^{n-1}}{m}t\right)} \int^{t}-exp \left( \frac{kn {v}^{n-1}}{m}\right) \left(g \right)ds$ $\displaystyle (Eq. 1.7.21)$

and

 ${y} \left(t \right)= \int_{ }^{t} {-exp \left( \frac{kn {v}^{n-1}}{m}w\right)} \int^{w}-exp \left( \frac{kn {v}^{n-1}}{m}s\right) \left(g \right)dsdw$ $\displaystyle (Eq. 1.7.22)$

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

 $\frac{d \left(m {v}_{y} \right)}{dt}= -k{{v}_{y}}^{n}-mg$ $\displaystyle (Eq. 1.8.1)$

Use the conservation of linear momentum

 $d \left(m {v}_{y} \right)= md{v}_{y}+{v}_{y}dm$ $\displaystyle (Eq. 1.8.2)$

Plugging equation 1.8.2 into 1.8.1

 $m \frac{d{v}_{y}}{dt}+ {v}_{y} \frac{dm}{dt}=-k{{v}_{y}}^{n}-mg$ $\displaystyle (Eq. 1.8.3)$

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

 $\frac{dm}{dt}= \frac{m- {m}_{0}}{ {t}_{1}}:=- \alpha$ $\displaystyle (Eq. 1.8.4)$

And equation 1.8.3 becomes

 $m \frac{d{v}_{y}}{dt}- {v}_{y} \alpha=-k{{v}_{y}}^{n}-mg$ $\displaystyle (Eq. 1.8.5)$

Rearranging gives us

 $F \left( \cdot \right)= \dot{ {v}_{y}}- \frac{ \alpha}{m}{v}_{y}+ \frac{k}{m} {{v}_{y}}^{n}+g$ $\displaystyle (Eq. 1.8.6)$

Check for exactness. Equation 1.8.6 is in the form to satisfy condition 1. Checking condition 2

 ${N}_{t}= \frac{d}{dt} \dot{ {v}_{y}}=0$
 ${M}_{{v}_{y}}= \frac{d}{d {v}_{y}} \left( \frac{k}{m} {{v}_{y}}^{n}- \frac{ \alpha}{m} {v}_{y}+g \right)= \frac{kn}{m} {{v}_{y}}^{n-1}- \frac{ \alpha}{m} \neq0$

Second condition is not satisfied. Make equation 1.8.6 exact using an integrating factor

 $hF \left( \cdot \right)=hM \left(t, {v}_{y} \right)+hN \left(t, {v}_{y} \right) \dot{ {v}_{y}}$ $\displaystyle (Eq. 1.8.7)$

Choose an integrating factor as a function of time only, such that ${h}_{v}=0$ and therefore

 ${h}_{t}N+h \left( {N}_{x}- {M}_{y} \right)=0$ $\displaystyle (Eq. 1.8.8)$

Plugging in

 $\frac{{h}_{t}}{h}= \frac{-1}{1} \left[0- \frac{kn{ {v}_{y}}^{n-1}}{m \left(t \right)}+ \frac{ \alpha}{m \left(t \right)} \right]$ $\displaystyle (Eq. 1.8.9)$

Integrating, using a dummy variable $s$

 $\int_{}^{}\frac{1}{h}dh= \int_{}^{t} \left( \frac{kn{ {v}_{y} \left(s \right)}^{n-1}}{m \left(t \right)}+ \frac{ \alpha}{m \left(t \right)}\right)ds$ $\displaystyle (Eq. 1.8.10)$

Therefore

 $h= exp \left[ \int_{}^{t} \left( \frac{kn{ {v}_{y} \left(s \right)}^{n-1}}{m \left(s \right)}+ \frac{ \alpha}{m \left(s \right)}\right)ds\right]$ $\displaystyle (Eq. 1.8.11)$

Using a dummy variable $w$, a solution can be written as

 ${v}_{y} \left(t \right)= exp \left[ -\int_{}^{t} \left( \frac{kn{ {v}_{y} \left(s \right)}^{n-1}}{m \left(s \right)}+ \frac{ \alpha}{m \left(s \right)}\right)ds\right] \int_{}^{t} exp \left[ \int_{}^{s} \left( \frac{kn{ {v}_{y} \left(w \right)}^{n-1}}{m \left(w \right)}+ \frac{ \alpha}{m \left(w \right)}\right)dw\right]gds$ $\displaystyle (Eq. 1.8.12)$

and

 $y \left(t \right)= \int_{}^{t} {v}_{y} \left(s \right)ds$ $\displaystyle (Eq. 1.8.)$

With varying mass and air resistance, the equations are typically solved by numerical methods.

## 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:

 \begin{align} &k= \text{spring constant}\\ &{{m}_{1}},{{m}_{2}}= \text{mass of pendulums}\\ &{{\theta }_{1}}, {{\theta }_{2}} =\text{Rotation angles}\\ &{{u}_{1}}, {{u}_{2}} = \text{control forces} \end{align}

## Find

$\left( 1 \right)$ Derive equation of motion:
 ${{m}_{1}}{{l}^{2}}{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}}=-k{{a}^{2}}\left( {{\theta }_{1}}-{{\theta }_{2}} \right)-{{m}_{1}}gl{{\theta }_{1}}+{{u}_{1}}l$ $\displaystyle (Eq. 2.1)$
 ${{m}_{2}}{{l}^{2}}{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}}=-k{{a}^{2}}\left( {{\theta }_{2}}-{{\theta }_{1}}\right)-{{m}_{2}}gl{{\theta }_{2}}+{{u}_{2}}l$ $\displaystyle (Eq. 2.2)$
$\left( 2 \right)$ Write Eq.2.1 and Eq.2.2 in the form of Mtg 13 (c),page2 , of:
 $\underset{-}{\overset{\centerdot }{\mathop{x}}}\,=\underset{-}{\mathop{A}}\,\left( t \right)\underset{-}{\mathop{x}}\,\left( t \right)+\underset{-}{\mathop{B}}\,\left( t \right)\underset{-}{\mathop{u}}\,\left( t \right)$ $\displaystyle (Eq. 2.3)$
 Given $\underset{-}{\mathop{x}}\,={{\left[ \begin{matrix} {{\theta }_{1}} & \overset{\centerdot }{\mathop{{{\theta }_{1}}}}\, & {{\theta }_{2}} & \overset{\centerdot }{\mathop{{{\theta }_{2}}}}\, \\ \end{matrix} \right]}^{T}}$ and $\underset{-}{\mathop{u}}\,={{\left[ \begin{matrix} {{u}_{1}}l & {{u}_{2}}l \\ \end{matrix} \right]}^{T}}$

## Solution

$\left( 1 \right)$ Derive equation of motion:
(a) Consider Free Body Diagram of left pendulum:
For small angle:
 $\displaystyle l\sin {{\theta }_{1}}=l{{\theta }_{1}}$ and $\displaystyle l\cos {{\theta }_{1}}=l$
 $acceleration, {{a}_{1}}=\frac{{{d}^{2}}\left( l{{\theta }_{1}} \right)}{d{{t}^{2}}}=l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}}$
 $Inertia force ={{m}_{1}}l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}}$
 $\displaystyle Spring force =ka{{\theta }_{1}}-ka{{\theta }_{2}}=ka({{\theta }_{1}}-{{\theta }_{1}})$

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

 $\sum{{{M}_{A}}=0\Rightarrow }\left( {{m}_{1}}l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}} \right)\cdot l+\left( ka({{\theta }_{1}}-{{\theta }_{2}}) \right)\cdot a+\left( {{m}_{1}}g \right)\cdot l{{\theta }_{1}}-\left( {{u}_{1}} \right)\cdot l=0$
 $\Rightarrow {{m}_{1}}{{l}^{2}}{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}}=-k{{a}^{2}}({{\theta }_{1}}-{{\theta }_{2}})-{{m}_{1}}gl{{\theta }_{1}}+{{u}_{1}}l$ $\displaystyle (Eq. 2.1)$
(b) Consider Free Body Diagram of right pendulum:
For small angle:
 $\displaystyle l\sin {{\theta }_{2}}=l{{\theta }_{2}}$ and $\displaystyle l\cos {{\theta }_{2}}=l$
 $acceleration, {{a}_{2}}=\frac{{{d}^{2}}\left( l{{\theta }_{2}} \right)}{d{{t}^{2}}}=l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}}$
 $Inertia force ={{m}_{2}}l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}}$
 $\displaystyle Spring force =ka{{\theta }_{2}}-ka{{\theta }_{1}}=ka({{\theta }_{2}}-{{\theta }_{1}})$

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

 $\sum{{{M}_{B}}=0\Rightarrow }\left( {{m}_{2}}l{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}} \right)\cdot l+\left( ka({{\theta }_{2}}-{{\theta }_{1}}) \right)\cdot a+\left( {{m}_{2}}g \right)\cdot l{{\theta }_{2}}-\left( {{u}_{2}} \right)\cdot l=0$
 $\Rightarrow {{m}_{2}}{{l}^{2}}{{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}}=-k{{a}^{2}}({{\theta }_{2}}-{{\theta }_{1}})-{{m}_{2}}gl{{\theta }_{2}}+{{u}_{2}}l$ $\displaystyle (Eq. 2.2)$
$\left( 2 \right)$ 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,
 ${{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}}=\frac{-\left( k{{a}^{2}}+{{m}_{1}}gl \right)}{m{}_{1}{{l}^{2}}}{{\theta }_{1}}+\frac{k{{a}^{2}}}{m{}_{1}{{l}^{2}}}{{\theta }_{2}}+\frac{{{u}_{1}}l}{m{}_{1}{{l}^{2}}}$ $\displaystyle (Eq. 2.4)$
 ${{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}}=\frac{k{{a}^{2}}}{m{}_{2}{{l}^{2}}}{{\theta }_{1}}+\frac{-\left( k{{a}^{2}}+{{m}_{2}}gl \right)}{m{}_{2}{{l}^{2}}}{{\theta }_{2}}+\frac{{{u}_{2}}l}{m{}_{2}{{l}^{2}}}$ $\displaystyle (Eq. 2.5)$
Now Eq.2.4 and Eq.2.5 can be put in the form of Eq.2.3 as:
 $\left[ \begin{matrix} {{\overset{\centerdot }{\mathop{\theta }}\,}_{1}} \\ {{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{1}} \\ {{\overset{\centerdot }{\mathop{\theta }}\,}_{2}} \\ {{\overset{\centerdot \centerdot }{\mathop{\theta }}\,}_{2}} \\ \end{matrix} \right]=\underbrace{\left[ \begin{matrix} 0 & 1 & 0 & 0 \\ \frac{-\left( k{{a}^{2}}+{{m}_{1}}gl \right)}{m{}_{1}{{l}^{2}}} & 0 & \frac{k{{a}^{2}}}{m{}_{1}{{l}^{2}}} & 0 \\ 0 & 0 & 0 & 1 \\ \frac{k{{a}^{2}}}{m{}_{2}{{l}^{2}}} & 0 & \frac{-\left( k{{a}^{2}}+{{m}_{2}}gl \right)}{m{}_{2}{{l}^{2}}} & 0 \\ \end{matrix} \right]}_{\underset{-}{\mathop{A}}\,}\left[ \begin{matrix} {{\theta }_{1}} \\ {{\overset{\centerdot }{\mathop{\theta }}\,}_{1}} \\ {{\theta }_{2}} \\ {{\overset{\centerdot }{\mathop{\theta }}\,}_{2}} \\ \end{matrix} \right]+\underbrace{\left[ \begin{matrix} 0 & 0 \\ \frac{1}{m{}_{1}{{l}^{2}}} & 0 \\ 0 & 0 \\ 0 & \frac{1}{m{}_{2}{{l}^{2}}} \\ \end{matrix} \right]}_{\underset{-}{\mathop{B}}\,}\left[ \begin{matrix} {{u}_{1}}l \\ {{u}_{2}}l \\ \end{matrix} \right]$ $\displaystyle (Eq. 2.6)$
Where:
 $\underset{-}{\mathop{A}}\,=\left[ \begin{matrix} 0 & 1 & 0 & 0 \\ \frac{-\left( k{{a}^{2}}+{{m}_{1}}gl \right)}{m{}_{1}{{l}^{2}}} & 0 & \frac{k{{a}^{2}}}{m{}_{1}{{l}^{2}}} & 0 \\ 0 & 0 & 0 & 1 \\ \frac{k{{a}^{2}}}{m{}_{2}{{l}^{2}}} & 0 & \frac{-\left( k{{a}^{2}}+{{m}_{2}}gl \right)}{m{}_{2}{{l}^{2}}} & 0 \\ \end{matrix} \right]$
 $\underset{-}{\mathop{B}}\,=\left[ \begin{matrix} 0 & 0 \\ \frac{1}{m{}_{1}{{l}^{2}}} & 0 \\ 0 & 0 \\ 0 & \frac{1}{m{}_{2}{{l}^{2}}} \\ \end{matrix} \right]$

## 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

 $x'(t)=a(t)\cdot x(t)+b(t)\cdot u(t)$ $\displaystyle$

## Find

Solve given equation by using Integrating Factor Method.

## Solution

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

 $x'(t)=a\cdot x(t)+b\cdot u(t)$ $\displaystyle$
 $x'(t)-a\cdot x(t)=b\cdot u(t)$ $\displaystyle (Eq. 3.1)$

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

 $h(t) = e^{\int {-a \cdot dt}} = e^{-at}$ $\displaystyle$

Apply obtained integrating factor to Eq(3.1)

 $[x'(t)-a\cdot x(t)]\cdot e^{-at} = [b\cdot u(t)]\cdot e^{-at}$ $\displaystyle$

 $\underbrace{{x}'(t)\cdot {{e}^{-at}}-a\cdot x(t)\cdot {{e}^{-at}}}_{\because f(x)\cdot {g}'(x)+{f}'(x)\cdot g(x)={{\left( f(x)\cdot g(x) \right)}^{'}}}=b\cdot u(t)\cdot {{e}^{-at}}={{\left( f(x)\cdot g(x) \right)}^{'}}$ $\displaystyle$

 $[e^{-at}\cdot x(t)]' = b\cdot u(t)\cdot e^{-at}$ $\displaystyle$

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

 $\int\limits_{{{t}_{0}}}^{t}{[e^{-a\tau}\cdot x(\tau)]'}d\tau = \int\limits_{{{t}_{0}}}^{t} {b \cdot u(\tau)\cdot e^{-a\tau}}d\tau$ $\displaystyle$

 $e^{-at}\cdot x(t) - e^{-at_0}\cdot x(t_0) = \int\limits_{{{t}_{0}}}^{t} {b \cdot u(\tau)\cdot e^{-a\tau}}d\tau$ $\displaystyle$
 $e^{-at}\cdot x(t) = e^{-at_0}\cdot x(t_0) + \int\limits_{{{t}_{0}}}^{t} {b \cdot u(\tau)\cdot e^{-a\tau}}d\tau$ $\displaystyle$
 $x(t) = \frac {e^{-at_0}}{e^{-at}} \cdot x(t_0) + \frac {\int\limits_{{{t}_{0}}}^{t} {b \cdot u(\tau)\cdot e^{-a\tau}}d\tau}{e^{-at}}$ $\displaystyle$
 $x(t) = e^{a(t-t_0)}\cdot x(t_0) + \int\limits_{{{t}_{0}}}^{t} {e^{a(t-\tau)}}\cdot b \cdot u(\tau)d\tau$ $\displaystyle (Eq. 3.2)$

## 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:

 $f\left( x \right)=\sum\limits_{n=0}^{\infty }{\frac{{{f}^{\left( n \right)}}\left( a \right)}{n!}}{{\left( x-a \right)}^{n}}$ $\,(Eq.4.1)$

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:

 $f\left( x \right)=\sum\limits_{n=0}^{\infty }{\frac{{{x}^{n}}}{n!}}$ $\,(Eq.4.2)$

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

 $\exp \left( x \right)=1+\frac{x}{1!}+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\frac{x^{4}}{4!}+...+\frac{x^{n}}{n!}$ $\,(Eq.4.3)$

## 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 Mtg 14 page 1-2, Eq.4:

 $\underline{x}(t)={{e}^{\underline{A}\cdot \left( t-{{t}_{0}} \right)}}\cdot \underline {x}({{t}_{0}})+\int\limits_{{{t}_{0}}}^{t}{{{e}^{\underline{A}\cdot \left( t-\tau \right)}}\cdot \underline{B}\cdot \underline{u}\left( \tau \right)d\tau }$ $\,(Eq. 5.1)$

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 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:

 $\underline{x}(t)={{e}^{\int\limits_{{{t}_{o}}}^{t}{\underline{A}(\tau )d\tau }}}\cdot \underline{x}({{t}_{0}})+\int\limits_{{{t}_{0}}}^{t}{{{e}^{\int\limits_{\tau }^{t}{\underline{A}(s)ds}}}\cdot \underline{B}\cdot \underline{u}\left( \tau \right)d\tau }$ $\,(Eq. 5.2)$

Note: The integral of a matrix is calculated by integrating each element that composes the matrix.

## 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:

 $\underline{\dot{x}}(t)=\underline{A}\underline{x}(t)+\underline{B}\underline{u}(t)$ $\,(Eq. 6.1)$

The state transition matrix $\underline{\Phi}$ has the following properties:

 $\frac{d\underline{\phi} (t,{{t}_{0}})}{dt}=\underline{A}\underline{\phi} (t,{{t}_{0}})$ $\,(Eq. 6.2)$

and

 $\displaystyle\underline{\phi} ({{t}_{0}},{{t}_{0}})=I$ $\,(Eq. 6.3)$

where I is the identity matrix.

## Find

With (Eq. 6.1) through (Eq. 6.3), Eq. (4) on pp. 14-1 of the lecture notes, has to be found:

 $\underline{x}(t)={{e}^{\underline{A}(t-{{t}_{0}})}}\cdot \underline{x}({{t}_{0}})+\int\limits_{{{t}_{0}}}^{t}{{{e}^{\underline{A}(t-\tau )}}\underline{B}\underline{u}(\tau )\,d\tau }$ $\,(Eq. 6.4)$

## 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 $\underline{\Phi} (t,{{t}_{0}})$:

 $\dot{\underline{\phi} }(t,{{t}_{0}})=\underline{A}\underline{\phi} (t,{{t}_{0}})+\underline{B}\underline{u}(t)$ $\,(Eq. 6.5)$

Regrouping the equation above results in:

 $\underbrace{-\left( \underline{A}\underline{\phi} (t,{{t}_{0}})+\underline{B}\underline{u}(t) \right)}_{\underline{M}(\underline{\phi} ,\underline{u})}+\underbrace{1}_{\underline{N}(\underline{\phi} ,\underline{u})}\cdot \dot{\underline{\phi} }(t,{{t}_{0}})=0$ $\,(Eq. 6.6)$

(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:

 ${{\underline{M}}_{\underline{\phi} }}=-\underline{A}\ne 0={{\underline{N}}_{\underline{u}}}$ $\,(Eq. 6.7)$

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:

 $-\underline{A}\underline{\phi} (t,{{t}_{0}})+1\cdot \dot{\underline{\phi} }(t,{{t}_{0}})=\underline{B}\underline{u}(t)$ $\,(Eq. 6.8)$

The integrating factor $h(t)$ is then given as:

 $h(t)={{e}^{\int\limits_{{{t}_{0}}}^{t}{-\underline{A}\,ds}}}={{e}^{-\underline{A}(t-{{t}_{0}})}}$ $\,(Eq. 6.9)$

Multiplying $h(t)$ throughout (Eq. 6.8) gives:

 $-\underline{A}\underline{\phi} (t,{{t}_{0}})\cdot {{e}^{-\underline{A}(t-{{t}_{0}})}}+{{e}^{-\underline{A}(t-{{t}_{0}})}}\cdot \dot{\underline{\phi} }(t,{{t}_{0}})=\underline{B}\underline{u}(t)\cdot {{e}^{-\underline{A}(t-{{t}_{0}})}}$ $\,(Eq. 6.10)$

which is equal to

 $\frac{d\left( {{e}^{-\underline{A}(t-{{t}_{0}})}}\cdot \underline{\phi} (t,{{t}_{0}}) \right)}{dx}=\underline{B}\underline{u}(t)\cdot {{e}^{-\underline{A}(t-{{t}_{0}})}}$ $\,(Eq. 6.11)$

Integrating (Eq. 6.11) on both sides between ${{t}_{0}}$ and $t$ gives:

 ${{e}^{-\underline{A}(t-{{t}_{0}})}}\cdot \underline{\phi} (t,{{t}_{0}})-{{e}^{-\underline{A}({{t}_{0}}-{{t}_{0}})}}\cdot \underline{\phi} ({{t}_{0}},{{t}_{0}})={{e}^{\underline{A}{{t}_{0}}}}\cdot \int\limits_{{{t}_{0}}}^{t}{{{e}^{-\underline{A}\tau }}}\cdot \underline{B}\underline{u}(t)\,d\tau$ $\,(Eq. 6.12)$

that is equal to

 $\underline{\phi} (t,{{t}_{0}})={{e}^{\underline{A}(t-{{t}_{0}})}}\cdot \underline{\phi} ({{t}_{0}},{{t}_{0}})+\int\limits_{{{t}_{0}}}^{t}{{{e}^{\underline{A}(t-\tau )}}\cdot }\underline{B}\underline{u}(t)\,d\tau$ $\,(Eq. 6.13)$

Replacing $\underline{\Phi} (t,{{t}_{0}})$ with $\underline{x}(t)$ gives:

 $\underline{x}(t)={{e}^{\underline{A}(t-{{t}_{0}})}}\cdot \underline{x}({{t}_{0}})+\int\limits_{{{t}_{0}}}^{t}{{{e}^{\underline{A}(t-\tau )}}\underline{B}\underline{u}(\tau )\,d\tau }$ $\,(Eq. 6.13)$

which is the same as the given solution in (Eq. 6.4)!

## 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

 \begin{align} &{\delta} = \text{Aileron angle/deflection}\\ &{\phi} = \text{Roll angle}\\ &{\omega} = \text{Roll angular velocity}\\ &Q = \text{Aileron effectiveness}\\ &{\tau} = \text{Roll time constant}\\ &u = \text{Command signal to Ailerons}\\ \end{align}
 $\overset{\centerdot }{\mathop{\phi }}\,=\omega$ $\displaystyle (Eq. 7.1)$
 $\overset{\centerdot }{\mathop{\omega }}\,=\frac{-1}{\tau }\omega +\frac{Q}{\tau }\delta$ $\displaystyle (Eq. 7.2)$
 $\overset{\centerdot }{\mathop{\delta }}\,=u$ $\displaystyle (Eq. 7.3)$

## Find

Control behavior in form of

 $\overset{\centerdot}{\mathop{x\left( t \right)=\underset{-}{\mathop{A}}\,}}\,\underset{-}{\mathop{x}}\,\left( t \right)+\underset{-}{\mathop{B}}\,\underset{-}{\mathop{u}}\,\left( t \right)$ $\displaystyle (Eq. 7.4)$

## Solution

Integrating Eq.7.3 gives,
 $\displaystyle \delta =ut$ $\displaystyle (Eq. 7.5)$
Substitute Eq.7.5 for $\displaystyle \delta$, in Eq.7.2:
 $\overset{\centerdot }{\mathop{\omega }}\,=\frac{-1}{\tau }\omega +\frac{Q}{\tau }ut$ $\displaystyle (Eq. 7.6)$
Now integrate Eq.7.6:
 $\omega =\frac{-1}{\tau }\omega t+\frac{Q}{2\tau }u{{t}^{2}}$ $\displaystyle (Eq. 7.7)$
Similarly integrating Eq.7.1 gives,
 $\displaystyle \phi =\omega t$ $\displaystyle (Eq. 7.8)$
Substitute Eq.7.8 in Eq.7.7 implies,
 $\omega =\frac{-1}{\tau }\phi +\frac{Q{{t}^{2}}}{2\tau }u$ $\displaystyle (Eq. 7.9)$
From Eq.7.1 and Eq.7.9,
 $\overset{\centerdot }{\mathop{\phi }}\,=\frac{-1}{\tau }\phi +\frac{Q{{t}^{2}}}{2\tau }u$ $\displaystyle (Eq. 7.10)$
Now, Eq.7.3, Eq.7.6 and Eq.7.10 can be written in the required form as,
 $\left[ \begin{matrix} \overset{\centerdot }{\mathop{\phi }}\, \\ \overset{\centerdot }{\mathop{\omega }}\, \\ \overset{\centerdot }{\mathop{\delta }}\, \\ \end{matrix} \right]=\left[ \begin{matrix} \frac{-1}{\tau } & 0 & 0 \\ 0 & \frac{-1}{\tau } & 0 \\ 0 & 0 & 0 \\ \end{matrix} \right]\left[ \begin{matrix} \phi \\ \omega \\ \delta \\ \end{matrix} \right]+\left[ \begin{matrix} \frac{Q{{t}^{2}}}{2\tau } \\ \frac{Qt}{\tau } \\ 1 \\ \end{matrix} \right]\left[ u \right]$ $\displaystyle (Eq. 7.11)$
where,
 $\underset{-}{\mathop{A}}\,=\left[ \begin{matrix} \frac{-1}{\tau } & 0 & 0 \\ 0 & \frac{-1}{\tau } & 0 \\ 0 & 0 & 0 \\ \end{matrix} \right]$
 $\underset{-}{\mathop{B}}\,=\left[ \begin{matrix} \frac{Q{{t}^{2}}}{2\tau } \\ \frac{Qt}{\tau } \\ 1 \\ \end{matrix} \right]$
 $\underset{-}{\mathop{x}}\,=\left[ \begin{matrix} \phi \\ \omega \\ \delta \\ \end{matrix} \right]$
 $\underset{-}{\mathop{u}}\,=\left[ u \right]$

## 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 [17]

 $\displaystyle {h}_{x} + {h}_{y} {P} = 0$ $\displaystyle (1) p. 17-1$

## 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

 $\displaystyle {h}_{x} = - {h}_{y}P$ $\displaystyle Eq. (2)$

Next we will use the property that

 $\displaystyle {h}_{xy} = {h}_{yx}$ $\displaystyle Eq. (3)$

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

 ${h}_{x}= - {h}_{y} \underbrace{{P}}_{:= {y}^{'}} \Rightarrow {h}_{xy} = - {h}_{yy}{y}^{'} - {h}_{y} {y}^{''}$ $\displaystyle Eq. (4)$

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

 ${h}_{xy} = - {h}_{yy}{y}^{'} - {h}_{y} {y}^{''} \neq {h}_{yx}$ $\displaystyle Eq. (5)$

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

$\underbrace{\phi_p}_{f}\cdot y'' + \underbrace{\phi_y \cdot y' + \phi_x}_{g} = 0$
 $\displaystyle$
$\underbrace{(15p^4\cdot cos (x^2))}_{\phi_p}\cdot y'' + \underbrace{(6xy^2)}_{\phi_y} \cdot y' + \underbrace{(-6xp^5\cdot sin(x^2) + 2y^3)}_{\phi_x} = 0$
 $\displaystyle$
$\therefore f = \phi_p = 15p^4\cdot cos (x^2)$
 $\displaystyle$
$\therefore g = \phi_y \cdot y' + \phi_x = (6xy^2)\cdot y' + (-6xyp^5\cdot sin(x^2) + 2y^3)$
 $\displaystyle$

## Find

$(1) f_{xx} + 2p\cdot f_{xy} + p^2\cdot f_{yy} = g_{xp} + p\cdot g_{yp}- g_y$
 $\displaystyle Eq(9.1)$

$(2) f_{xp} + p\cdot f_{yp} + 2\cdot f_{y} = g_{pp}$
 $\displaystyle Eq(9.2)$

## Solution

 $\displaystyle for Eq(9.1)$ $\displaystyle$
 $f_{xx} = \frac{\partial }{\partial x} (\frac{\partial }{\partial x} (15p^4\cdot cos(x^2)))$ $\displaystyle$
 $\displaystyle = \frac{\partial }{\partial x} (-2x\cdot sin(x^2)\cdot 15p^4)$ $\displaystyle$
 $\displaystyle = \frac{\partial }{\partial x} (-30p^4\cdot x\cdot sin(x^2))$ $\displaystyle$
 $\displaystyle = -30p^4\cdot x \cdot 2x \cdot cos(x^2) + (-30p^4\cdot sin(x^2))$ $\displaystyle$
 $\displaystyle = -60p^4\cdot x^2 \cdot cos(x^2) - 30p^4\cdot sin(x^2)$ $\displaystyle$

 $\displaystyle f_{xy} = \frac{\partial }{\partial y} (\frac{\partial }{\partial x} (15p^4\cdot cos(x^2)))$ $\displaystyle$
 $\displaystyle = \frac{\partial }{\partial y} (-2x\cdot sin(x^2)\cdot 15p^4)$ $\displaystyle$
 $\displaystyle = \frac{\partial }{\partial y} (-30p^4\cdot x\cdot sin(x^2))$ $\displaystyle$
 $\displaystyle = 0$ $\displaystyle$
 $\displaystyle f_{yy} = \frac{\partial }{\partial y} (\frac{\partial }{\partial y} (15p^4\cdot cos(x^2)))$ $\displaystyle$
 $\displaystyle = \frac{\partial }{\partial y} (0)$ $\displaystyle$
 $\displaystyle = 0$ $\displaystyle$
 $\displaystyle g_{xp} = \frac{\partial }{\partial p} (\frac{\partial }{\partial x} (6xy^2p - 6xp^5\cdot sin(x^2) + 2y^3))$ $\displaystyle$
 $\displaystyle = \frac{\partial }{\partial p} (6y^2p-(6p^5x\cdot 2x\cdot cos(x^2) + 6p^5\cdot sin(x^2)))$ $\displaystyle$
 $\displaystyle = 6y^2 - 60x^2\cdot cos(x^2)\cdot p^4 - 30sin(x^2)\cdot p^4$ $\displaystyle$
 $\displaystyle g_{yp} = \frac{\partial }{\partial p} (\frac{\partial }{\partial y} (6xy^2p - 6xp^5\cdot sin(x^2) + 2y^3))$ $\displaystyle$
 $\displaystyle = \frac{\partial }{\partial p} (12xyp - 0 +6y^2)$ $\displaystyle$
 $\displaystyle = 12xy$ $\displaystyle$
 $\displaystyle g_{y} = \frac{\partial }{\partial y} (6xy^2p - 6xp^5\cdot sin(x^2) + 2y^3))$ $\displaystyle$
 $\displaystyle = 12xyp + 6y^2$ $\displaystyle$

Apply obtained factors to Eq(9.1)

 $\displaystyle L.H.S = -60p^4x^2\cdot cos(x^2) - 30p^4\cdot sin(x^2)$ $\displaystyle$
 $\displaystyle R.H.S = (6y^2 - 60x^2\cdot cos(x^2)\cdot p^4 - 30sin(x^2)\cdot p^4)+ p(12xy) - (12xyp + 6y^2)$ $\displaystyle$
 $\displaystyle = -60p^4x^2\cdot cos(x^2) - 30p^4\cdot sin(x^2)$ $\displaystyle$

 $\displaystyle \therefore L.H.S = R.H.S$

 $\displaystyle for Eq(9.2)$ $\displaystyle$
 $\displaystyle f_{xp} = \frac{\partial }{\partial p} (\frac{\partial }{\partial x} (15p^4\cdot cos(x^2)))$ $\displaystyle$
 $\displaystyle = \frac{\partial }{\partial p} (-2x\cdot sin(x^2)\cdot 15p^4)$ $\displaystyle$
 $\displaystyle = \frac{\partial }{\partial p} (-30p^4\cdot x\cdot sin(x^2))$ $\displaystyle$
 $\displaystyle = -120x\cdot sin(x^2)\cdot p^3$ $\displaystyle$
 $\displaystyle f_{yp} = \frac{\partial }{\partial p} (\frac{\partial }{\partial y} (15p^4\cdot cos(x^2)))$ $\displaystyle$
 $\displaystyle = \frac{\partial }{\partial p} (0)$ $\displaystyle$
 $\displaystyle = 0$ $\displaystyle$
 $\displaystyle f_{y} = \frac{\partial }{\partial y} (15p^4\cdot cos(x^2)))$ $\displaystyle$
 $\displaystyle = 0$ $\displaystyle$
 $\displaystyle g_{pp} = \frac{\partial }{\partial p} (\frac{\partial }{\partial p} (6xy^2p - 6xp^5\cdot sin(x^2) + 2y^3))$ $\displaystyle$
 $\displaystyle g_{pp} = \frac{\partial }{\partial p} (6xy^2 - 30x/cdot sin(x^2)\cdot p^4)$ $\displaystyle$
 $\displaystyle -120x\cdot sin(x^2)\cdot p^3$ $\displaystyle$

Apply obtained factors to Eq(9.2)

 $\displaystyle L.H.S = -120x\cdot sin(x^2)\cdot p^3$ $\displaystyle$
 $\displaystyle R.H.S = -120x\cdot sin(x^2)\cdot p^3$ $\displaystyle$
 $\displaystyle \therefore L.H.S = R.H.S$

## 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

 $\phi \left( x,y,p \right)=3{{p}^{5}}\cos \left( {{x}^{2}} \right)+2x{{y}^{3}}=k$ $\,(Eq. 10.1)$

and Eq. (2) from meeting 17

 $\left( 6x{{y}^{2}} \right)\cdot {y}'+2{{y}^{3}}=\left( {{h}_{y}}+0 \right){y}'+{{h}_{x}}$ $\,(Eq. 10.2)$

on page 17-1 and 17-3 respectively in the lecture notes.

## 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:

 ${{h}_{y}}\cdot dy=2{{y}^{3}}\cdot dx$ $\,(Eq. 10.3)$
 $\int{{{h}_{y}}\cdot dy=\int{2{{y}^{3}}\cdot dx}}$ $\,(Eq. 10.4)$
 $h\left( x,y \right)=2{{y}^{3}}x+{{k}_{1}}(y)$ $\,(Eq. 10.5)$

Take derivative of $\,h\left( x,y \right)$ with respect to $\,x$ gives:

 $\,{{h}_{x}}=2{{y}^{3}}$ $\,(Eq. 10.6)$

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:

 ${{h}_{x}}=2{{y}^{3}}=6x{{y}^{2}}\cdot {y}'$ $\,(Eq. 10.7)$

What implies that:

 ${y}'=\frac{y}{3x}$ $\,(Eq. 10.8)$

and

 $\,{{k}_{1}}(y)=0$ $\,(Eq. 10.9)$

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

 $\phi =h\left( x,y \right)+\int{f(x,y,p)dp}$ $\,(Eq. 10.10)$

For the given problem,

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

Now, inserting the found experssion in (Eq. 10.5) with $\,{{k}_{1}}(y)=0$ into Eq. 10.10 gives:

 $\phi \left( x,y,p \right)=3{{p}^{5}}\cos \left( {{x}^{2}} \right)+2x{{y}^{3}}=k$ $\,(Eq. 10.11)$

Which is the same as the given function $\,\phi \left( x,y,p \right)$

## Contributing Members

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