User:Egm6321.f10.team2.oztekin/HW3

= Problem 1: Derive EOM for the particle =

From the lecture slide Mtg 13-1.

Given
The picture was given in the MT13-1 for the moving particle :



For the dynamic objects Newton's Second law cites that: The total force applied on a body is equal to the time derivative of linear momentum of the body.See Newton's Law of Motion According to this law :
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$$  \displaystyle \vec{F}=\frac{\mathrm{d} (m\vec{v})}{\mathrm{d} t} $$ (Eq 1.1)
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If we manipulate this equation for the x and y axis;
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$$  \displaystyle \sum F_{x}=ma_{x} $$     (Eq 1.2)
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$$  \displaystyle \sum F_{y}=ma_{y} $$     (Eq 1.3)
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m: mass of the particle

k:air resistence constant

g:acceleration of gravity

Find
Case 1:Derive the EOM's for the moving particle with air resistance.

Case 2:k=0 no air resistance.

Case 3:Exist air resistance.

Case 1
For the x axis:
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$$  \displaystyle a_{x}=\frac{\mathrm{d} v_{x}}{\mathrm{d} t} $$ (Eq 1.4)
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$$  \displaystyle F_{air_x}=kv^{n}cos(\alpha ) $$     (Eq 1.5) Then now we can say that according to Newton's Law: Total force acting on the particle=Particle momentum change. Since there is only force that acting on the particle on x axis is just the x component of air resistance.By tis way we can concluded that:
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$$  \displaystyle m\frac{\mathrm{d} v_{x}}{\mathrm{d} t}=-kv^{n}cos(\alpha ) $$     (Eq 1.6)
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Similarly if we write EOM for the y axis
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$$  \displaystyle F_{air_y}=kv^{n}sin(\alpha ) $$     (Eq 1.7)
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$$  \displaystyle a_{y}=\frac{\mathrm{d} v_{y}}{\mathrm{d} t} $$ (Eq 1.8)
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And total forces acting on the particle are y companent of air resistance and gravity force.


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$$  \displaystyle m\frac{\mathrm{d} v_{y}}{\mathrm{d} t}=-kv^{n}sin(\alpha )-mg $$     (Eq 1.9)
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Case 2
For k=0;
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$$  \displaystyle m\frac{\mathrm{d} v_{x}}{\mathrm{d} t}=0 $$     (Eq 1.10)
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$$  \displaystyle m\frac{\mathrm{d} v_{y}}{\mathrm{d} t}=-mg $$     (Eq 1.11)
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$$  \displaystyle \int dv_{x}=\int dt $$ (Eq 1.12)
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$$  \displaystyle v=V_{xo} $$     (Eq 1.13)
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$$  \displaystyle x(t)=\int V_{x0}dt+x(0) $$     (Eq 1.14)
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$$  \displaystyle x(t)=V_{x0}t+x(0) $$     (Eq 1.15)
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$$  \displaystyle \int dv_{y}=-\int gdt $$     (Eq 1.16)
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$$  \displaystyle v_{y}=-gt+V_{y0} $$     (Eq 1.17)
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$$  \displaystyle x(t)=\int (-gt+V_{y0})+x(0) $$     (Eq 1.18)
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$$  \displaystyle x(t)=-\frac{gt^{2}}{2}+V_{y0}t+y(0) $$     (Eq 1.19)
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Case 3.1: constant mass
To alleviate notion, we define,


 * $$\displaystyle z\equiv {{v}_{y}}$$

Then the original equation can be written as,


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$$ \displaystyle m{z}'=-k{{z}^{n}}-mg$$ (Eq 1.21)
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This nonlinear first order ODE can be solved by the separation of variables. By some manipulation, we can write Eq 1.21 as,


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$$ \displaystyle -\frac{m}{mg+k{{z}^{n}}}dz=dt$$ (Eq 1.22)
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Then we integrate Eq 1.22, we have,


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$$ \displaystyle \int{\frac{m}{mg+k{{z}^{n}}}dz=C-t}$$ (Eq 1.23)
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where $$C$$ is some constant.

If you want to use Euler interation factor method, we can still get the same answer as Eq 1.23.

The original equation can be written as,


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$$ \displaystyle k{{z}^{n}}+mg+m{z}'=0$$ (Eq 1.24)
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It's easy to see that the Eq 1.24 is not exact. So we expect we can multiple an integrating factor $$\displaystyle h$$ to make it exact.

Define:


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& M\equiv k{{z}^{n}}+mg \\ & N\equiv m \\ \end{align} \right.$$

The prerequisite that Eq 1.24 being exact is,


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$$ \displaystyle \frac{\partial (hM)}{\partial z}=\frac{\partial (hN)}{\partial t}$$ (Eq 1.25)
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Thus,


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$$ \displaystyle {{h}_{t}}N-{{h}_{z}}M+h({{N}_{t}}-{{M}_{z}})=0$$ (Eq 1.26)
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If we assume $$\displaystyle h$$ is a function only of $$\displaystyle z$$, meaning $$\displaystyle {{h}_{t}}=0$$, then substituting into Eq 1.26 $$\displaystyle M$$ and $$\displaystyle N$$, we have


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$$ \displaystyle \frac{h}=\frac{1}{M}({{N}_{t}}-{{M}_{z}})=\frac{-kn{{z}^{n-1}}}{k{{z}^{n}}+mg}$$ (Eq 1.27)
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By integrating Eq 1.27, we have,


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$$ \displaystyle h=\frac{1}{k{{z}^{n}}+mg}$$ (Eq 1.28)
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We can see the assumption "$$\displaystyle h$$ is a function only of $$\displaystyle z$$" is satisfied.

Then we mutiple $$\displaystyle h$$ to both sides of Eq 1.24, we have,


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$$ \displaystyle 1+\frac{m}{k{{z}^{n}}+mg}{z}'=0$$ (Eq 1.29)
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We can easily see that Eq 1.29 is indeed exact. Because it's exact we can rewrite Eq 1.29 as,


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$$ \displaystyle \frac{d}{dt}(t+\int{\frac{m}{mg+k{{z}^{n}}}dz})=0$$ (Eq 1.30)
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That means,


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$$ \displaystyle t+\int{\frac{m}{mg+k{{z}^{n}}}dz}=C$$ (Eq 1.31)
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or


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$$ \displaystyle \int{\frac{m}{mg+k{{z}^{n}}}dz=C-t}$$ (Eq 1.32)
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where $$C$$ is some constant.

Case 3.2: varying mass
If mass of the rocket becomes variable, then the original equation can be written as:
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$$ \displaystyle k{{z}^{n}}+m(t)g+m(t){z}'=0$$ (Eq 1.33)
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I believe this equation is unlikely to be made exact because the assumption "$$\displaystyle h$$ is a function only of $$\displaystyle z$$" won't be satisfied neither will "$$\displaystyle h$$ is a function only of $$\displaystyle t$$". Without any of these two crucial assumptions, it is very difficult to find an appropriate Euler integrating factor. And also the variable mass makes the ODE's variables unseparable. I think we should turn to other method(numerical method?) for the solution.

= Problem 2 =

From the lecture slide Mtg 13-3

Given
The pendulum problem was shown in MT13-3 as

Angular momentum of a particle is gives as :
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$$  \displaystyle \vec{L}=\vec{r}\times \vec{p} $$     (Eq 2.1) Where r=the position of the particle from the origin
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p=The linear momentum of a particle.

Find
Derive the EOM.

Solution
For deriving EOM's we think in angular basis. Since we are interested in angular momentum we can enlarge Newton's Law for linear momentum into angular momentum.According to Newton's Law Total moment acting on the particle=Total angular momentum chance of particle.So


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$$  \displaystyle \vec{r}\times \frac{\mathrm{d}(m\vec{v)}}{\mathrm{d} t}=\sum \vec{r}\times \vec{F} $$     (Eq 2.2)
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$$  \displaystyle \sum \vec{M}=\sum \vec{r}\times \vec{F} $$     (Eq 2.3)
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Total moment acting on the particle mass of m1=Spring moment+gravity force moment+control force moment.

Spring moment


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$$  \displaystyle \vec{M}_{spring}=\vec{a}\times k\vec{a}(sin\theta _{1}-sin\theta _{2}) $$     (Eq 2.4)
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Gravity force moment


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$$  \displaystyle \vec{M}_{gravity}=\vec{l}sin\theta _{1}\times m\vec{g} $$     (Eq 2.5) For small angular displacements
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$$  \displaystyle sin\theta _{1}\approx \theta _{1} $$     (Eq 2.6)
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Control Force moment
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$$  \displaystyle \vec{M}_{control}=\vec{l}\times \vec{u_{1}} $$     (Eq 2.7)
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If we subtitute all term in angular momentum equation we concluded


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$$  \displaystyle m_{1}l^{2}\ddot{\theta _{1}}=-ka^{2}(\theta _{1}-\theta _{2})-m_{1}gl\theta_{1}+u_{1}l $$     (Eq 2.8)
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Similarly for the second pendulum


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$$  \displaystyle m_{2}l^{2}\ddot{\theta _{2}}=-ka^{2}(\theta _{2}-\theta _{1})-m_{2}gl\theta_{2}+u_{2}l $$     (Eq 2.9)
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We can write this equations in matrix form for the SC_L1_ODEs.The form can be shown as basically:


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$$  \displaystyle \dot{X}_{4\times 1}=A(t)_{4\times 4}.X_{4\times 1}+B(t)_{4\times 2}.U_{2\times 1} $$     (Eq 2.10)
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$$  \displaystyle X=\begin{bmatrix} \theta _{1}\\ \dot{\theta _{1}}\\ \theta _{2}\\ \dot{\theta _{2}}\\
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\end{bmatrix} $$     (Eq 2.11)
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$$  \displaystyle \left[ \begin{matrix} \overset{.}{\mathop}\, \\ \overset{..}{\mathop}\, \\ \overset{.}{\mathop}\, \\ \overset{..}{\mathop}\, \\ \end{matrix} \right]=\left[ \begin{matrix} 0 \\   -\frac{k{{a}^{2}}}-\frac{g}{l}  \\ 0 \\   \frac{k{{a}^{2}}}  \\ \end{matrix}\begin{matrix} 1 \\   0  \\   0  \\   0  \\ \end{matrix}\begin{matrix} 0 \\   \frac{k{{a}^{2}}}  \\ 0 \\   -\frac{k{{a}^{2}}}-\frac{g}{l}  \\ \end{matrix}\begin{matrix} 0 \\   0  \\   1  \\   0  \\ \end{matrix} \right]\left[ \begin{matrix} {{\theta }_{1}} \\ \overset{.}{\mathop}\, \\ {{\theta }_{2}} \\ \overset{.}{\mathop}\, \\ \end{matrix} \right]+\left[ \begin{matrix} 0 \\   \frac{1}{{{m}_{1}}l}  \\ 0 \\   0  \\ \end{matrix}\begin{matrix} 0 \\   0  \\   0  \\   \frac{1}{{{m}_{2}}l}  \\ \end{matrix} \right]\left[ \begin{matrix} {{u}_{1}} \\ {{u}_{2}} \\ \end{matrix} \right] $$     (Eq 2.11)
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= Problem 3 =

From the lecture slide Mtg 14-1

Given
Linear first order ODE:


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$$ \displaystyle \dot{x}(t)=a(t)x(t)+b(t)u(t)$$ (Eq 3.1)
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Initial Condition:


 * $$\displaystyle {{t}_{0}}$$ and $$\displaystyle x({{t}_{0}})$$.

Find
$$\displaystyle x(t)$$ when $$\displaystyle a(t)$$ and $$\displaystyle b(t)$$ are constants and varying coefficients respetively.

Case of Varying Coefficients
First assume $$\displaystyle a(t)$$ and $$\displaystyle b(t)$$ are varying coefficients since constants are only a particular case.

The Eq 3.1 can be written as:


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$$ \displaystyle \dot{x}(t)-a(t)x(t)=b(t)u(t)$$ (Eq 3.2)
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Thus we have the Euler integrating factor:


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$$ \displaystyle h(t)=\exp \int_ – ^{t}{-a(s)ds}$$ (Eq 3.3)
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Or:


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$$ \displaystyle h(t)=\exp \int_^{t}{-a(s)ds}+{{k}_{1}}$$ (Eq 3.4)
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where $$\displaystyle {{t}_{M}}$$ is the lower limit of integral with respect to integration constant $$\displaystyle {{k}_{1}}$$.

Then we have $$\displaystyle x(t)$$:


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$$ \displaystyle \begin{align} x(t) & =\exp \int_ – ^{t}{a(s)ds*\int_ – ^{t}{\left[ \exp \int_ – ^{\tau }{-a(s)ds} \right]}}*bu(\tau )d\tau \\ & =\exp \int_^{t}{a(s)ds}*\left\{ \int_^{t}{\left[ \exp \int_^{\tau }{-a(s)ds} \right]}*b(\tau )u(\tau )d\tau +{{k}_{2}} \right\} \end{align}$$ (Eq 3.5)
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where $$\displaystyle {{t}_{N}}$$ is the lower limit of integral with respect to integration constant $$\displaystyle {{k}_{2}}$$. Note that $$\displaystyle {{k}_{1}}$$ disappeared because Patterson and Rasul have proved that it is unnecessary. In fact $$\displaystyle {{t}_{M}}$$ will also be canceled in the following procedure.

To derive $$\displaystyle {{t}_{N}}$$ and $$\displaystyle {{k}_{2}}$$, we should substituting the initial condition into Eq 3.5:


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$$ \displaystyle x({{t}_{o}})=\exp \int_^{a(s)ds}*\left\{ \int_^{\left[ \exp \int_^{\tau }{-a(s)ds} \right]}*b(\tau )u(\tau )d\tau +{{k}_{2}} \right\}$$ (Eq 3.6)
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then we get $$\displaystyle {{k}_{2}}$$:


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$$ \displaystyle {{k}_{2}}=\frac{x({{t}_{0}})}{\exp \int_^{a(s)ds}}-\int_^{\left[ \exp \int_^{\tau }{-a(s)ds} \right]}*b(\tau )u(\tau )d\tau $$ (Eq 3.7)
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Substituting $$\displaystyle {{k}_{2}}$$ back into Eq 3.5, we derive:


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$$ \displaystyle \begin{align} x(t) & =\exp \int_^{t}{a(s)ds}*\left\{ \int_^{t}{\left[ \exp \int_^{\tau }{-a(s)ds} \right]}*b(\tau )u(\tau )d\tau +\frac{x({{t}_{0}})}{\exp \int_^{a(s)ds}}-\int_^{\left[ \exp \int_^{\tau }{-a(s)ds} \right]}*b(\tau )u(\tau )d\tau \right\} \\ & =x({{t}_{0}})\exp \left[ \int_^{t}{a(s)ds}-\exp \int_^{a(s)ds} \right]+\exp \int_^{t}{a(s)ds}*\left\{ \int_^{t}{\left[ \exp \int_^{\tau }{-a(s)ds} \right]}*b(\tau )u(\tau )d\tau -\int_^{\left[ \exp \int_^{\tau }{-a(s)ds} \right]}*b(\tau )u(\tau )d\tau \right\} \\ & =x({{t}_{0}})\exp \int_^{t}{a(s)ds}+\exp \int_^{t}{a(s)ds}*\int_^{t}{\left[ \exp \int_{\tau }^{a(s)ds} \right]}*b(\tau )u(\tau )d\tau \\ & =x({{t}_{0}})\exp \int_^{t}{a(s)ds}+\exp \int_^{t}{a(s)ds}*\int_^{t}{\left[ \exp (\int_{\tau }^{t}{a(s)ds+\int_{t}^{a(s)ds)}} \right]}*b(\tau )u(\tau )d\tau \\ & =x({{t}_{0}})\exp \int_^{t}{a(s)ds}+\exp \int_^{t}{a(s)ds}*\exp \int_{t}^{a(s)ds*}\int_^{t}{\left[ \exp \int_{\tau }^{t}{a(s)ds} \right]}*b(\tau )u(\tau )d\tau \\ & =x({{t}_{0}})\exp \int_^{t}{a(s)ds}+\int_^{t}{\left[ \exp \int_{\tau }^{t}{a(s)ds} \right]}*b(\tau )u(\tau )d\tau \end{align}$$ (Eq 3.8)
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So we have got the $$\displaystyle x(t)$$ when $$\displaystyle a(t)$$ and $$\displaystyle b(t)$$ are varying coefficients:


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$$ \displaystyle x(t)=x({{t}_{0}})\exp \int_^{t}{a(s)ds}+\int_^{t}{\left[ \exp \int_{\tau }^{t}{a(s)ds} \right]}*b(\tau )u(\tau )d\tau $$ (Eq 3.9)
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Verifying of result
Note that the derivation rule of $$\displaystyle \frac{d}{dt}\int_^{t}{f(\tau )d\tau }$$ between that of $$\displaystyle \frac{d}{dt}\int_^{t}{f(\tau ,t)d\tau }$$ are totaly different:


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$$ \displaystyle \frac{d}{dt}\int_^{t}{f(\tau )d\tau =}f(t)$$, (Eq 3.10)
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However,


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$$ \displaystyle \frac{d}{dt}\int_^{t}{f(\tau ,t)d\tau =}f(t,t)+\int_^{t}{\frac{\delta f(\tau ,t)}{\delta t}d\tau }$$ (Eq 3.11)
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More generally,
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$$ \displaystyle \frac{d}{dt}\int_{b(t)}^{a(t)}{f(\tau ,t)d\tau =}\frac{da(t)}{dt}f\left( a(t),t \right)-\frac{db(t)}{dt}f\left( b(t),t \right)+\int_{b(t)}^{a(t)}{\frac{\delta f(\tau ,t)}{\delta t}d\tau }$$ (Eq 3.12)
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Thus by Eq 3.11,


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$$ \displaystyle
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\begin{align} \dot{x}(t) & ={{\left[ x({{t}_{0}})\exp \int_^{t}{a(s)ds} \right]}^{\prime }}+{{\left\{ \int_^{t}{\left[ \exp \int_{\tau }^{t}{a(s)ds} \right]}*b(\tau )u(\tau )d\tau \right\}}^{\prime }} \\ & =x({{t}_{0}})\exp \int_^{t}{a(s)ds}*a(t)+\exp \int_{t}^{t}{a(s)ds}*b(t)u(t)+\int_^{t}{\left[ \exp \int_{\tau }^{t}{a(s)ds} \right]}*a(t)b(\tau )u(\tau )d\tau \\ & =a(t)\left\{ x({{t}_{0}})\exp \int_^{t}{a(s)ds}+\int_^{t}{\left[ \exp \int_{\tau }^{t}{a(s)ds} \right]}*b(\tau )u(\tau )d\tau \right\}+b(t)u(t) \\ & =a(t)x(t)+b(t)u(t) \end{align}$$ (Eq 3.13)
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Thus the result is veryfied. See also Team2 discussion on Mtg 14.

Case of Constant Coefficients
As constant coefficients is only a particular case of varying coefficients, we can just replace $$\displaystyle a(s)$$ by $$\displaystyle a$$ in the Eq 3.9,


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$$ \displaystyle \begin{align} x(t) & =x({{t}_{0}})\exp \int_^{t}{a(s)ds}+\int_^{t}{\left[ \exp \int_{\tau }^{t}{a(s)ds} \right]}*b(\tau )u(\tau )d\tau \\ & =x({{t}_{0}})[\exp a(t-{{t}_{0}})]+\int_^{t}{\left[ \exp a(t-\tau ) \right]}*b(\tau )u(\tau )d\tau \end{align}$$ (Eq 3.14)
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Then this is the answer to the case of constant coefficent.

Another way to solve the Case of Constant Coefficients(by Team2. Yaakov)
The problem is time independent; therefore, we can rewrite the initial equation as:


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$$ \displaystyle \dot{x}(t)-ax(t)=bu(t)$$ (Eq 3.15)
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 * }

To calculate the integrating factor we find


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

$$ \displaystyle \exp (\int{-adt)=\exp (-at})$$ (Eq 3.16)
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 * style="width:95%" |
 * <p style="text-align:right">
 * }

Now we can multiply the entire equation 3.16 with the integrating factor


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

$$ \displaystyle \exp (-at)\dot{x}(t)-\exp (-at)ax(t)=\exp (-at)bu(t)$$ (Eq 3.17)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

It is noteworthy to realize that the LHS of the equation is equal to $$\displaystyle [\exp (-at)x(t)]'$$

Integrating both sides of the new equation from $$\displaystyle {{t}_{0}}$$ to $$\displaystyle t$$ and changing the variables from $$\displaystyle t$$ to $$\displaystyle \tau $$ for integration will yield us the following equation:


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

$$ \displaystyle \int\limits_^{t}{[\exp (-a\tau )x(\tau )]'=}\int\limits_^{t}{\exp (-a\tau )bu(\tau )dt}$$ (Eq 3.18)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \displaystyle \exp (-at)x(t)-\exp (-a{{t}_{0}})x({{t}_{0}})=\int\limits_^{t}{\exp (-a\tau )bu(\tau )dt}$$ (Eq 3.19)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Solving the equation for $$x(t)$$ will yield us to the proof:


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

$$ \displaystyle x(t)=[\exp a(t-{{t}_{0}})]x({{t}_{0}})+\int\limits_^{t}{[\exp a(t-\tau )]bu(\tau )d\tau }$$ (Eq 3.20)
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 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

Exponantial Of Matrix
Taylor series are consist of infinite components of function and its derivatives.The series are centered on any number that is interested in.


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

$$  \displaystyle f(a)+\frac{{f}'(a)}{1!}\left ( x-a \right )+\frac{{f}''(a)}{2!}\left ( x-a \right )^{2}+\frac{f^{3}(a)}{3!}\left ( x-a \right )^{3}........... $$     (Eq 3.21)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

For f(x)=expx can be characterized as the same way.


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

$$  \displaystyle e^{x}=\sum_{n=0}^{\infty }\frac{x^{n}}{n!}=1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+........... $$     (Eq 3.22)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

And if we substitute matrix A into x we concluded that:


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

$$  \displaystyle e^{A}_{n\times n}=\sum_{n=0}^{\infty }\frac{x^{n}}{n!}=I_{n\times n}+A_{n\times n}+\frac{A_{n\times n}^{2}}{2!}+\frac{A_{n\times n}^{3}}{3!}+........... $$     (Eq 3.23)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Same observation for Bryson, A.E., and Ho, Y.C Book's Eq.
The solution of SC_N1_ODE is given in the book as follows ;


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

$$  \displaystyle X(t)=\phi (t,t_{0})X(t_{0})+\int_{t_{0}}^{t}\phi (t,t_{0})B(\tau )U(\tau )d\tau $$     (Eq 3.24)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

So we can see that


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

$$  \displaystyle \phi (t,t_{0})_{n\times n}=expA(t-t_{0})_{n\times n} $$ (Eq 3.25)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle \dot{X(t)}=A\phi (t,t_{0})x(t_{0})+\frac{\mathrm{d} t}{\mathrm{d} t}\phi (t,t)B(t)U(t)+\int_{t_{0}}^{t}\frac{\mathrm{d} }{\mathrm{d} t}\phi (t,\tau )B(\tau )U(\tau )d\tau $$     (Eq. 3.26)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle \dot{X(t)}=A\phi (t,t_{0})x(t_{0})+B(t)U(t)+A\int_{t_{0}}^{t}\varphi (t,\tau )B(\tau )U(\tau )d\tau $$     (Eq 3.27)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle \dot{X}(t)-AX(t)=B(t )U(t) $$     (Eq. 3.28)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Since zou has drived comprehensively in his discussion page we concluded briefly just to show that book's equation yields exactly the same equtiation we drived in our lecture.

= Problem 4 =

From the lecture slide Mtg 14-2

Given
Solution of system of coupled linear first order ODE of constant coefficient:


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

$$ \displaystyle \underline{x}(t)=[\exp (t-{{t}_{0}})\underline{A}]\underline{x}({{t}_{0}})+\int_^{t}{\left[ \exp (t-\tau )\underline{A} \right]\underline{B}\underline{u}(\tau )}d\tau $$ (Eq 4.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Find
Solution of System of coupled linear first order ODE of varying coefficient

Solution
Just turn constant cofficient matrices $$\underline{A}$$ and $$\underline{B}$$ in Eq 4.1 into varying coefficient matrices case, and via similarity to Eq 3.9, we have,


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

$$ \displaystyle \underline{x}(t)=[\exp \int_^{t}{\underline{A}(s)ds}]\underline{x}({{t}_{0}})+\int_^{t}{\left[ \exp \int_{\tau }^{t}{\underline{A}(s)ds} \right]\underline{B}(\tau )\underline{u}(\tau )}d\tau $$ (Eq 4.2)
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 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

= Problem 5 =

From the lecture slide Mtg 15-1

Given
System of Linear first order ODE:


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

$$ \displaystyle \underline(t)=\underline{A}\underline{x}(t)+\underline{B}\underline{u}(t)$$ (Eq 5.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

where $$\displaystyle \underline{A}$$ and $$\displaystyle \underline{B}$$ are constant matrices.

Initial condition:


 * $$\displaystyle {{t}_{0}}$$ and $$\displaystyle \underline{x}({{t}_{0}})$$.

Find
$$\displaystyle \underline{x}(t)$$

Solution
From Eq 5.1 we derive Euler integrating factor,


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

$$ \displaystyle \underline{h}(t)=\exp \int_ – ^{t}{-\underline{A}ds}=\exp (\underline{M}-t\underline{A})$$ (Eq 5.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

where $$\displaystyle \underline{M}$$ is a matrix consists of constants.

Thus we get $$\displaystyle \underline{x}(t)$$,


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

$$ \displaystyle \underline{x}(t)=\exp (t\underline{A}-\underline{M})*\left\{ \int_^{t}{\left[ \exp (M-\tau \underline{A}) \right]\underline{B}\underline{u}(\tau )d\tau +\underline{N}} \right\}$$ (Eq 5.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

where $$\displaystyle {{t}_{N}}$$ is lower limit of integral with respect to $$\displaystyle \underline{N}$$ which is also a matrix consists of constants.

To derive $$\displaystyle \underline{N}$$ we should use the initial condition,


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

$$ \displaystyle \underline{x}({{t}_{0}})=\exp (t\underline{A}-\underline{M})*\left\{ \int_^{\left[ \exp (M-\tau \underline{A}) \right]\underline{B}\underline{u}(\tau )d\tau +\underline{N}} \right\}$$ (Eq 5.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Since,


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

$$ \displaystyle \exp (t\underline{A}-\underline{M})*\exp (M-t\underline{A})=1$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Then $$\displaystyle \exp (M-t\underline{A})$$ are the inverse matrix of $$\displaystyle \exp (t\underline{A}-\underline{M})$$.

Then, by manipulation of Eq 5.4, we have,


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

$$ \displaystyle \underline{N}=\exp (M-{{t}_{0}}\underline{A})\underline{x}({{t}_{0}})-\int_^{\left[ \exp (M-\tau \underline{A}) \right]\underline{B}\underline{u}(\tau )}d\tau $$ (Eq 5.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Substituting back to Eq 5.4, we have


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

$$ \displaystyle \begin{align} \underline{x}(t) & =\exp (t\underline{A}-\underline{M})*\left\{ \int_^{t}{\left[ \exp (M-\tau \underline{A}) \right]\underline{B}\underline{u}(\tau )d\tau +\exp (M-{{t}_{0}}\underline{A})\underline{x}({{t}_{0}})-\int_^{\left[ \exp (M-\tau \underline{A}) \right]\underline{B}\underline{u}(\tau )}d\tau } \right\} \\ & =\exp (t\underline{A}-\underline{M}+M-{{t}_{0}}\underline{A})\underline{x}({{t}_{0}})+\exp (t\underline{A}-\underline{M})*\int_^{t}{\left[ \exp (M-\tau \underline{A}) \right]\underline{B}\underline{u}(\tau )}d\tau \\ & =[\exp (t-{{t}_{0}})\underline{A}]\underline{x}({{t}_{0}})+\int_^{t}{\left[ \exp (t-\tau )\underline{A} \right]\underline{B}\underline{u}(\tau )}d\tau \end{align}$$ (Eq 5.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Thus we have derive $$\displaystyle \underline{x}(t)$$:


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

$$ \displaystyle \underline{x}(t)=[\exp (t-{{t}_{0}})\underline{A}]\underline{x}({{t}_{0}})+\int_^{t}{\left[ \exp (t-\tau )\underline{A} \right]\underline{B}\underline{u}(\tau )}d\tau $$ (Eq 5.7)
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 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

=Problem 6:Discuss h=

Given
Example given in MT15-3 as


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

$$  \displaystyle x({y}')^{2}+y{y}'+xy{y}''=0 $$     (Eq 6.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle g(x,y,p)=x({y}')^{2}+y{y}' $$     (Eq 6.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle f(x,y,p)=xy $$     (Eq 6.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle h_{x}+h_{y}p=0 $$     (Eq 6.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Find
Without assuming h=const. discuss other solutions.

Solution
For the condition given Eq.6.4 both terms must be zero. So


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

$$  \displaystyle h_{x}=0 $$     (Eq 6.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle h_{y}p=0 $$     (Eq 6.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

And we know that h=h(x,y).Possible outcomes for the derivations;


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

$$  \displaystyle h_{x}=h(x,y) $$     (Eq 6.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle h_{y}=h(x,y) $$     (Eq 6.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

By this way we can not find common h for stisfy Eq.6.4

=Problem 7:SC-L1-ODE-CC=

Given

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

$$ \displaystyle \dot{\phi}=\omega $$ (Eq 7.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle \dot{\omega}=-\frac{1}{\tau }\omega+\frac{Q}{\tau }\delta $$ (Eq 7.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle \dot{\delta}=u $$ (Eq 7.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle \dot{x}(t)=\underline{A} \underline{x}(t)+\underline{B} \underline{u}(t) $$ (Eq 7.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Find
Make 7.1-7.3 in the form of 7.4

Solve

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

$$ \displaystyle \begin{bmatrix} \dot{\phi }\\ \dot{\omega }\\ \dot{\delta } \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0\\ 0 & -\frac{1}{\tau } & \frac{Q}{\tau }\\ 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} \phi \\ \omega \\ \delta \end{bmatrix}
 * style="width:95%" |
 * style="width:95%" |

+ \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix} u(t) $$ (Eq 7.5)
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle \underline{A}= \begin{bmatrix} 0 & 1 & 0\\ 0 & -\frac{1}{\tau } & \frac{Q}{\tau }\\ 0 & 0 & 0 \end{bmatrix}$$ (Eq 7.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle \underline{B}= \begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}$$ (Eq 7.7)
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 * style="width:95%" |
 * <p style="text-align:right">
 * }

=Problem 8:Exactness Condition for Second Order=

Given
Second order non-linear function can be shown as follows which was given in MT15-2 ;


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

$$  \displaystyle F\left ( x,y,{y}',{y}'' \right )=0 $$     (Eq 8.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

F is a function that derivative of function that is non-linear first order equation equals constant.


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

$$  \displaystyle F\left ( x,y,{y}',{y}'' \right )=\frac{\mathrm{d} \phi \left ( x,y,{y}' \right )}{\mathrm{d} x} $$ (Eq 8.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

F is exact if there is exist a fuunction $$ \phi (x,y,{y}')$$

If we want to show that an arbitrary second order function to be exact, this arbitrary function has to provide two exactness conditions.

First condition:An arbitrary second order function has to provide this form:


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

$$  \displaystyle F\left ( x,y,{y}',{y} \right )=g\left ( x,y,p \right )+f\left ( x,y,p \right ){y} $$     (Eq 8.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

where $$p={y}' $$

Second Condition:has two parts


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

$$  \displaystyle f_{xx}+2pf_{xy}+p^{2}f_{yy}=g_{xp}+pg_{yp}-g_{y} $$     (Eq 8.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle f_{xp}+pf_{yp}+2f_{y}=g_{pp} $$     (Eq 8.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

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

$$  \displaystyle g\left ( x,y,p \right )=\phi _{x}+\phi _{y}{y}' $$     (Eq 8.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle f\left ( x,y,p \right )=\phi _{p}^{} $$     (Eq 8.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Find
Given function in lecture mtg 17-2 as follows ;


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

$$  \displaystyle \left ( 15p^{4}cosx^{2} \right ){y}''+6xy^{2}{y}'+\left [ -6xp^{5}sinx^{2}+2y^{3} \right ]=0 $$     (Eq 8.8)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

For this equation check exactness.

Solution

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

$$  \displaystyle f= 15p^{4}cosx^{2} $$     (Eq 8.9)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle g=6xy^{2}{y}'+\left [ -6xp^{5}sinx^{2}+2y^{3} \right ] $$     (Eq 8.10)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle f_{xx}=-30p^{4}sinx^{2}-60p^{4}x^{2}cosx^{2} $$     (Eq 8.11)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle f_{xy}=0 $$     (Eq 8.12)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle f_{yy}=0 $$     (Eq 8.13)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle g_{xp}=6y^{2}-30p^{4}sinx^{2}-60p^{4}x^{2}cosx^{2} $$     (Eq 8.14)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle g_{yp}=12xy $$     (Eq 8.15)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle g_{y}=12xyp+6y^{2} $$     (Eq 8.16)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle f_{xp}=-120p^{3}xsinx^{2} $$     (Eq 8.17)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle f_{yp}=0 $$     (Eq 8.18)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle f_{y}=0 $$     (Eq 8.19)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle g_{pp}=-120p^{3}xsinx^{2} $$     (Eq 8.20)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

By substituting this tools into two equations (8.4 and 8.5) we observe that ;


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

$$  \displaystyle -30p^{4}sinx^{2}-60p^{4}x^{2}cosx^{2}+2p.0+p^{2}.0=6y^{2}-30p^{4}sinx^{2}-60p^{4}x^{2}cosx^{2}+p12xy-12xyp-6y^{2} $$     (Eq 8.21)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle -30p^{4}sinx^{2}-60p^{4}x^{2}cosx^{2}=-30p^{4}sinx^{2}-60p^{4}x^{2}cosx^{2} $$     (Eq 8.22)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle -120p^{3}xsinx^{2}+p.0+0=-120p^{3}xsinx^{2} $$     (Eq 8.23)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle -120p^{3}xsinx^{2}=-120p^{3}xsinx^{2} $$     (Eq 8.24)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

=Problem 9=

Given
For this problem, the equation is the same as which was given in problem 8. We can find first integral $$ \phi \left ( x,y,{y}' \right ) $$ by taking first derivative of f.


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

$$  \displaystyle \phi \left ( x,y,{y}' \right )=h\left ( x,y \right )+\int f\left ( x,y,{y}' \right )dp $$     (Eq 9.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle \phi \left ( x,y,{y}' \right )=h\left ( x,y \right )+3p^{5}cosx^{2} $$     (Eq 9.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$  \displaystyle g=\phi _{y}{y}'+\phi _{x} $$     (Eq 9.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

By substituting $$\phi _{x}$$ and$$ \phi _{y}$$ in the definition of g we concluded ;


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

$$ \displaystyle (6xy^2)y^{'}+2y^3=h_yy^{'}+h_x $$ (Eq 9.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

The following was also given:
 * {| style="width:100%" border="0"

$$ \displaystyle h_yy{'}=2y^3 $$ (Eq 9.5)
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 * style="width:95%" |
 * <p style="text-align:right">
 * }

Find
If  $$ \displaystyle h_{x}=6xy^{2}{y}'$$ then what is $$ \displaystyle \phi (x,y,p)$$ (Meeting 18-3)

Solution
For the second order ODEs we assume that :


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

$$  \displaystyle \phi =h(x,y)+\int f(x,y,p)dp $$     (Eq 9.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

So h(x,y) is only function of x and y. For the second option


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

$$  \displaystyle h_{x}=6xy^{2}{y}' $$     (Eq 9.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

tells us that after takind partial derivative of h over x we would find that another function depends on y prime.


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

$$  \displaystyle z(x,y,{y}')=6xy^{2}{y}' $$     (Eq 9.8)
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 * style="width:95%" |
 * <p style="text-align:right">
 * }

So the choice of derivation of h must be depend on x,y,and $${y}'$$ if we assume the choice is true.(?) However we can not expect that. Because h only depends on x and y.The derivation of h over x must yield a function depends on x and y.

=Contributions Of Team Members=


 * Problem 1:    Solved by:Oztekin,Sahin,Zou

Author:Oztekin,Zou Author:Oztekin
 * Problem 2:    Solved by:Oztekin,Sahin

Author:Zou
 * Problem 3-4-5: Solved by:Zou,Yaakow,Oztekin
 * Problem 6-7 : Solved by:Rasul,Patterson,Dube

Author:Dube,Oztekin
 * Problem 8-9 : Solved by:Dube,Oztekin

Author:Dube,Oztekin

= References =