User:Egm6321.f11.team2.kim/PR3

=Problem 3.2 - Integrating factor method =

Given
$$\displaystyle \left( \frac {1}{3}x^3+d_1\right)\left(y^4\right)y' + (5x^3+2)\left(\frac {1}{5}y^5+ sinx + d_2\right) = 0$$

Find
Show that the above equation is exact or can be made exact by IFM. Find intergrating factor h.

Solution
In order to exact condition First condition of exactness is that $$ \displaystyle M(x,y) + N(x,y)y' = 0$$

$$ \displaystyle (5x^3 +2)\left( \frac {1}{5}y^5 + sinx + d_2\right) = M(x,y) $$ $$ \displaystyle \left( \frac {1}{3}x^3+d_1\right)(y^4)y' = N(x,y)y' $$ $$\underbrace{ \left( \frac {1}{3}x^3+d_1\right)\left(y^4\right)}_{\displaystyle\color{blue}{N(x,y)}}y' + \underbrace {(5x^3+2)\left( \frac{1}{5}y^5+ sinx + d_2\right)}_{\displaystyle\color{blue}{M(x,y)}} = 0$$ So this equation satisfies the first condition. Second condition of exactness is that $$\displaystyle M_y(x,y) = N_x(x,y)$$ $$\displaystyle \frac {\partial M}{\partial y} = \frac {\partial N}{\partial x}$$ $$\displaystyle \frac {\partial M}{\partial y} = 5x^3y^4 + 2y^4 $$ $$\frac {\partial N}{\partial x} = x^2y^4 $$ It shows that this equation does not satisfy the second condition. Therefore, in order to make this equation to exact, integrating factor must be used. $$ \displaystyle h(x) = e^{-\int \frac{1}{N}(N_x -M_y)dx} $$ where d1 and d2 are integration constant at M(x,y) and N(x,y), so d1= d2= 0 at this point. Intergraing factor h is that $$\displaystyle h(x) = e^{-\int \frac{1}{N}(N_x -M_y)dx} = exp[-\int \frac{1}{\frac{1}{3}x^3y^4}(x^2y^4-5x^3y^4-2y^4)dx] =exp[-\int \frac {3}{x^3}(x^2-5x^3-2)dx]

$$ $$\displaystyle = exp\left[- \int \left(\frac{3}{x}-15-\frac{6}{x^3}\right)dx\right] = exp\left[-\left(3lnx - 15x + \frac{1}{2x^2}\right)\right] $$

=Problem 3.8 - =

Given
$$\displaystyle \dot x(t) = a(t)x(t) + b(t)u(t) $$

Find
Generalize $$\displaystyle \dot x(t) = ax(t) + bu(t) $$ to the case of linear time varient system.

Verify that your expression is indded the solution for

$$\displaystyle \dot \mathbf{X}(t)= \mathbf{A}(t)\mathbf{x}(t)+\mathbf{B}(t)\mathbf{u}(t) $$

Solution
From eq 3.2.2 $$\displaystyle \dot \mathbf{X}(t)= \mathbf{A}(t)\mathbf{x}(t)+\mathbf{B}(t)\mathbf{u}(t) $$ Let A(t)= a and B(t) = b the equation is following that $$\displaystyle \dot x(t) = ax(t) + bu(t)$$ rearranging the above equation. Thus, $$\displaystyle \dot x(t) - ax(t) = bt(u) $$ From the above equation, intergrating factor is that $$\displaystyle h(t) = e^{\int^t-ads}=e^{-at} $$ $$\displaystyle e^{-at}(\dot x(t)-ax(t)) = bu(t)e^{-at}$$ [LHS] : $$\displaystyle e^{-at}(\dot x(t)-ax(t)) = [e^{-at}x(t)]'$$ rearraing the eq $$\displaystyle [e^{-at}x(t)]' = bu(t)e^{-at} $$

interval is $$\displaystyle t_0< \tau < t$$ the equation is following that $$\displaystyle \int^t_{t_0} [e^{-a\tau}x(\tau)]'d\tau = \int^t_{t_0}bu(\tau)e^{-a\tau}d\tau $$ $$\displaystyle e^{-at}x(t)-e^{-at_0}x(t_0) = \int^t_{t_0}e^{-a\tau}bu(\tau)d\tau$$ dividing by $$\displaystyle e^{-at}$$ and rearraing for x(t) $$\displaystyle x(t) =\frac {\int^t_{t_0}e^{-a\tau}bu(\tau)d\tau}{e^{-at}} + e^{a(t-t_0)}x(t_0) = {\int^t_{t_0}e^{a(t-\tau)}bu(\tau)d\tau} + e^{a(t-t_0)}x(t_0) $$

=Problem 3.14 - =

Given
$$\displaystyle g=\cancel{xp^2}+\cancel{yp} = \phi_x + \phi_y p = \underbrace{(h_x + \cancel{yp})}_{\phi_x}+\underbrace{(h_y+\cancel{xp})}_{\phi_y}p $$ $$\displaystyle p(x)= y'(x) $$

Find
$$\displaystyle h_x + h_y p = 0$$

Find h(x,y).

Solution
$$\displaystyle h_x + h_y p = 0 $$ where

$$\displaystyle h_x = \frac {\partial h}{\partial x}$$ $$\displaystyle h_y = \frac {\partial h}{\partial y}$$ $$\displaystyle p = \frac {dy}{dx} $$ The eq 1.1.1 is following that

$$ \displaystyle \frac {\partial h(x,y)}{\partial x}+\frac {\partial h(x,y)}{\partial y} \frac {dy}{dx} = 0$$