User:Egm6321.f12.team1.pb/report3

=R*3.5 =

Given


and

Prove that
Equation ($$) Satisfies the condition
 * $$\frac {h_x} {h} = - \frac {1} {N} (N_x - M_y) =: n(x)$$,

 That is $$\frac {1} {N} (N_x - M_y) $$ is only a function of $$x$$ and an integrating factor $$h(x)$$ can be found to render it exact only if $$k_1(y)=d_1$$ (Constant).  Further show that ($$) includes the following equation

 as a particular case.

Solution
Given $$ \begin{array}{lcl} & N(x,y)  & = \bar {b} (x,y)c(y) \\ &         & = \left ( \int^x b(s)\mathrm{d}s+k_1(y) \right )c(y)  \\ &         & \\ \Rightarrow  & N_x(x,y) & = b(x)c(y) \\ &         & \end{array} $$  and  $$ \begin{array}{lcl} & M(x,y)  & = a(x)\bar {c} (x,y) \\ &         & = a(x) \left ( \int^y c(s)\mathrm{d}s+k_2(x) \right )  \\ &         & \\ \Rightarrow  & M_y(x,y) & = a(x)c(y) \\ &         & \end{array} $$  $$ \Rightarrow  \frac {1} {N} (N_x - M_y)  = \displaystyle\frac {1} {\left ( \int^x b(s)\mathrm{d}s+k_1(y) \right ) \cancel{c(y)}} (b(x)\cancel{c(y)} -a(x)\cancel{c(y)}) $$

The only dependency on $$y$$ of the right hand side of ($$) is through the term $$k_1(y)$$. So the left hand side is independent of $$y$$ if and only if $$k_1(y)$$ is a constant. That is $$ \frac {1} {N} (N_x - M_y) $$ is independent of $$y$$ if and only if $$k_1(y) = d_1$$, a constant$$.\qquad\blacksquare$$  In the case that $$k_1(y) = d_1$$, a constant, then


 * $$\bar {c} (x,y) = \int^y c(s)\mathrm{d}s+k_2(x) = \int^y \mathrm{d}s+k_2(x) $$

Substituting ($$), ($$), ($$) into ($$)


 * $$\bar b(x) y' + a(x)[y+k_2(x)]=0$$

$$\Rightarrow [a(x)y+a(x)k_2(x)]+\bar b(x) y'=0$$ $$\Rightarrow [a(x)y+k_3(x)]+\bar b(x) y'=0$$ <br\>Which has exactly the same form as ($$)$$.\qquad\blacksquare$$

=R*3.10=

Show That
The solution of ($$) is<br\>
 * $$x(t) = [\exp\{a(t-t_0)\}]x(t_0) + \int_{t_0}^t [\exp\{a(t-\tau)\}]b u(\tau) \mathrm{d} \tau$$

Solution
We can re-write ($$) as

$$\Rightarrow -\frac {1} {N} (N_t-M_x) = - a$$<br\> So the Integrating Factor is given as
 * $$h(t) = \exp \left[ \int^t -a \; \mathrm{d}s + k_1 \right] = \exp \left[ -at + k_1 \right] $$

Multiplying both sides of ($$) with the Integration Factor, we get<br\><br\>
 * $$-e^{\displaystyle \left\{ -at + k_1 \right\}} a\,x(t)+e^{\displaystyle \left\{ -at + k_1 \right\}} \dot x(t) = e^{\displaystyle \left\{ -at + k_1 \right\}} b u(t)$$

Which is in exact form<br\> $$\Rightarrow \frac {\mathrm{d}} {\mathrm{d}t} \left (e^{\displaystyle \left\{ -at + k_1 \right\}} x(t) \right ) = e^{\displaystyle \left\{ -at + k_1 \right\}} b u(t)$$<br\><br\>

Integrating with respect to $$t$$<br\>

Using the boundary condition at $$t=t_0, x(t)=x(t_0)$$<br\>

Subtracting ($$) from ($$)
 * $$e^{\displaystyle \left\{ -at + k_1 \right\}} x(t) - e^{\displaystyle \left\{ -at_0 + k_1 \right\}} x(t_0)= \int^t e^{\displaystyle \left\{ -a\tau + k_1 \right\}} b u(\tau) \mathrm{d} \tau - \int^{t_0} e^{\displaystyle \left\{ -a\tau + k_1 \right\}} b u(\tau) \mathrm{d} \tau + k_2$$

$$\Rightarrow e^{\displaystyle \left\{ -at + k_1 \right\}} x(t) = e^{\displaystyle \left\{ -at_0 + k_1 \right\}} x(t_0) + \int_{t_0}^t e^{\displaystyle \left\{ -a\tau + k_1 \right\}} b u(\tau) \mathrm{d} \tau$$<br\> Dividing both sides by $$e^{\displaystyle \left\{ -at + k_1 \right\}}$$, we get
 * $$ x(t) = e^{\displaystyle \left\{ a(t-t_0) \right\}} x(t_0) + \int_{t_0}^t e^{\displaystyle \left\{ a(t-\tau)\right\}} b u(\tau) \mathrm{d} \tau \qquad\blacksquare$$

<br\><br\> The homogeneous solution to ($$) is found by solving the same equation, only setting the forcing function to $$0$$. So the homogeneous solution is solution to the following equation<br\>
 * $$ -a x(t) + \dot x (t) = 0$$

And the homogeneous solution is
 * $$x(t) = e^{\displaystyle \left\{ a(t-t_0) \right\}} x(t_0)$$

The particular solution is the second term in $$x(t)$$
 * $$ x(t) = \int^t e^{\displaystyle \left\{ a(t-\tau)\right\}} b u(\tau) \mathrm{d} \tau$$

Show That
The solution of ($$) is<br\>
 * $$x(t) = \left [\exp\int_{t_0}^t a(\tau) \mathrm{d} \tau \right ]x(t_0) + \int_{t_0}^t \left [\exp\int_{\tau}^t a(s) \mathrm{d}s \right ]b u(\tau) \mathrm{d} \tau$$

Solution
We can re-write ($$) as

$$\Rightarrow -\frac {1} {N} (N_t-M_x) = - a(t)$$<br\> So the Integrating Factor is given as
 * $$h(t) = \exp \left[ \int^t -a(s) \; \mathrm{d}s + k_1 \right] $$

Multiplying both sides of ($$) with the Integration Factor, we get<br\><br\>
 * $$ -\exp \left[ \int^t -a(s) \; \mathrm{d}s + k_1 \right] a(t)\,x(t)+\exp \left[ \int^t -a(s) \; \mathrm{d}s + k_1 \right] \dot x(t) = \exp \left[ \int^t -a(s) \; \mathrm{d}s + k_1 \right] b(t) u(t)$$

Which is in exact form<br\> $$\Rightarrow \frac {\mathrm{d}} {\mathrm{d}t} \left (\exp \left[ \int^t -a(s) \; \mathrm{d}s + k_1 \right] x(t) \right ) = \exp \left[ \int^t -a(s) \; \mathrm{d}s + k_1 \right] b(t) u(t)$$<br\><br\>

Integrating with respect to $$t$$<br\>

Using the boundary condition at $$t=t_0, x(t)=x(t_0)$$<br\>

Subtracting ($$) from ($$)
 * $$ \exp \left[ \int^t -a(s) \; \mathrm{d}s + k_1 \right] x(t) - \exp \left[ \int^{t_0} -a(s) \; \mathrm{d}s + k_1 \right]  x(t_0)$$<br\>
 * $$ = \int^t \exp \left[ \int^\tau -a(s) \; \mathrm{d}s + k_1 \right] b(\tau) u(\tau) \mathrm{d} \tau -  \int^t \exp \left[ \int^\tau -a(s) \; \mathrm{d}s + k_1 \right]  b(\tau) u(\tau) \mathrm{d} \tau$$

$$\Rightarrow \exp \left[ \int^t -a(s) \; \mathrm{d}s + k_1 \right] x(t) =  \exp \left[ \int^{t_0} -a(s) \; \mathrm{d}s + k_1 \right]  x(t_0) + \int_{t_0}^t \exp \left[ \int^\tau -a(s) \; \mathrm{d}s + k_1 \right]  b(\tau) u(\tau) \mathrm{d} \tau$$<br\>

Dividing both sides by $$\exp \left[ \int^t -a(s) \; \mathrm{d}s + k_1 \right] $$, we get
 * $$ x(t) = exp \left[ \int^{t_0} -a(s) \; \mathrm{d}s - \int^{t} \{-a(s)\} \; \mathrm{d}s \right] x(t_0) + \int_{t_0}^t \exp \left[ \int^\tau -a(s) \; \mathrm{d}s -  \int^{t} \{-a(s)\} \; \mathrm{d}s  \right]  b(\tau) u(\tau) \mathrm{d} \tau $$

$$\Rightarrow x(t) = exp \left[ \int^t_{t_0} a(s) \; \mathrm{d}s\right] x(t_0) + \int_{t_0}^t \exp \left[ \int^t_\tau a(s) \; \mathrm{d}s   \right]  b(\tau) u(\tau) \mathrm{d} \tau \qquad\blacksquare$$