User:Egm6321.f10.team2.zou/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
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 only a function 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.

= 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


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$$ \displaystyle \exp (\int{-adt)=\exp (-at})$$ (Eq 3.16)
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Now we can multiply the entire equation 3.16 with the integrating factor


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$$ \displaystyle \exp (-at)\dot{x}(t)-\exp (-at)ax(t)=\exp (-at)bu(t)$$ (Eq 3.17)
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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:


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$$ \displaystyle \int\limits_^{t}{[\exp (-a\tau )x(\tau )]'=}\int\limits_^{t}{\exp (-a\tau )bu(\tau )dt}$$ (Eq 3.18)
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$$ \displaystyle \exp (-at)x(t)-\exp (-a{{t}_{0}})x({{t}_{0}})=\int\limits_^{t}{\exp (-a\tau )bu(\tau )dt}$$ (Eq 3.19)
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Solving the equation for $$x(t)$$ will yield us to the proof:


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$$ \displaystyle x(t)=[\exp a(t-{{t}_{0}})]x({{t}_{0}})+\int\limits_^{t}{[\exp a(t-\tau )]bu(\tau )d\tau }$$ (Eq 3.20)
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= Problem 4 =

From the lecture slide Mtg 14-2

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


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$$ \displaystyle \underline{x}(t)=[\exp (t-{{t}_{0}})\underline{A}]\underline{x}({{t}_{0}})+\int_^{t}{\left[ \exp (t-\tau )\underline{A} \right]\underline{B}\underline{u}(\tau )}d\tau $$ (Eq 4.1)
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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,


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$$ \displaystyle \underline{x}(t)=[\exp \int_^{t}{\underline{A}(s)ds}]\underline{x}({{t}_{0}})+\int_^{t}{\left[ \exp \int_{\tau }^{t}{\underline{A}(s)ds} \right]\underline{B}(\tau )\underline{u}(\tau )}d\tau $$ (Eq 4.2)
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= Problem 5 =

From the lecture slide Mtg 15-1

Given
System of Linear first order ODE:


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$$ \displaystyle \underline(t)=\underline{A}\underline{x}(t)+\underline{B}\underline{u}(t)$$ (Eq 5.1)
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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,


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$$ \displaystyle \underline{h}(t)=\exp \int_ – ^{t}{-\underline{A}ds}=\exp (\underline{M}-t\underline{A})$$ (Eq 5.2)
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where $$\displaystyle \underline{M}$$ is a matrix consists of constants.

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


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$$ \displaystyle \underline{x}(t)=\exp (t\underline{A}-\underline{M})*\left\{ \int_^{t}{\left[ \exp (M-\tau \underline{A}) \right]\underline{B}\underline{u}(\tau )d\tau +\underline{N}} \right\}$$ (Eq 5.3)
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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,


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$$ \displaystyle \underline{x}({{t}_{0}})=\exp (t\underline{A}-\underline{M})*\left\{ \int_^{\left[ \exp (M-\tau \underline{A}) \right]\underline{B}\underline{u}(\tau )d\tau +\underline{N}} \right\}$$ (Eq 5.4)
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Since,


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$$ \displaystyle \exp (t\underline{A}-\underline{M})*\exp (M-t\underline{A})=1$$
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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,


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$$ \displaystyle \underline{N}=\exp (M-{{t}_{0}}\underline{A})\underline{x}({{t}_{0}})-\int_^{\left[ \exp (M-\tau \underline{A}) \right]\underline{B}\underline{u}(\tau )}d\tau $$ (Eq 5.5)
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Substituting back to Eq 5.4, we have


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$$ \displaystyle \begin{align} \underline{x}(t) & =\exp (t\underline{A}-\underline{M})*\left\{ \int_^{t}{\left[ \exp (M-\tau \underline{A}) \right]\underline{B}\underline{u}(\tau )d\tau +\exp (M-{{t}_{0}}\underline{A})\underline{x}({{t}_{0}})-\int_^{\left[ \exp (M-\tau \underline{A}) \right]\underline{B}\underline{u}(\tau )}d\tau } \right\} \\ & =\exp (t\underline{A}-\underline{M}+M-{{t}_{0}}\underline{A})\underline{x}({{t}_{0}})+\exp (t\underline{A}-\underline{M})*\int_^{t}{\left[ \exp (M-\tau \underline{A}) \right]\underline{B}\underline{u}(\tau )}d\tau \\ & =[\exp (t-{{t}_{0}})\underline{A}]\underline{x}({{t}_{0}})+\int_^{t}{\left[ \exp (t-\tau )\underline{A} \right]\underline{B}\underline{u}(\tau )}d\tau \end{align}$$ (Eq 5.6)
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Thus we have derive $$\displaystyle \underline{x}(t)$$:


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$$ \displaystyle \underline{x}(t)=[\exp (t-{{t}_{0}})\underline{A}]\underline{x}({{t}_{0}})+\int_^{t}{\left[ \exp (t-\tau )\underline{A} \right]\underline{B}\underline{u}(\tau )}d\tau $$ (Eq 5.7)
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= References =