User:Egm6321.f10.team03.schulze


 * Now Eq.2.4 and Eq.2.5 can be put in the form of Eq.2.3 as:


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\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]</math== Solution ==

$$\begin{align} & \Rightarrow \overset{\centerdot }{\mathop{\omega }}\,=\frac{-1}{\tau }\omega +\frac{Q}{\tau }ut \\ & \Rightarrow \omega =\frac{-1}{\tau }\omega t+\frac{Q}{2\tau }u{{t}^{2}}\to eq.B \\ & \overset{\centerdot }{\mathop{\phi }}\,=\omega \to eq.2\Rightarrow \phi =\omega t \\ & \Rightarrow \omega =\frac{-1}{\tau }\phi +\frac{Q}{2\tau }u{{t}^{2}}\to eq.B \\ & \Rightarrow \overset{\centerdot }{\mathop{\phi }}\,=\underbrace{\frac{-1}{\tau }}_{\underset{-}{\mathop{A}}\,}\phi +\underbrace{\frac{Q{{t}^{2}}}{2\tau }}_{\underset{-}{\mathop{B}}\,}u \\ \end{align}$$

= Problem 7 - Finding Integrating Factor and solve for y =

Given

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y' + {x}y = 2x + 3 $$ $$
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
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 * }

Find
1. Find the integrating factor 2. Solve for $$y$$

Solution
1. With an integrating factor, $$h(x)$$, we should get the equation
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h = e^{ \int a0(x) dx} $$
 * $$ \displaystyle
 * $$ \displaystyle

where a0(x) = x
(hy)' = h x^2 $$ $$ Using $$h(x) = x$$ yields
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
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 * }
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xy' + y = x^3 $$ $$ Dividing the above equation with $$x$$ yields the original equation
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
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 * }
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y' + \frac{1}{x} y = x^2 $$ $$ Hence, $$h(x) = x$$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
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2. Substituting $$h(x) = x$$ into equation 2 yields
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(xy)' = x^3 $$ $$ Next, we integrate the above equation
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }
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xy = \frac{x^4}{4} + C $$ $$ where $$C$$ is a constant. The final form of this equation is
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 5)
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 * $$ \displaystyle
 * $$ \displaystyle

y = \frac{x^3 }{4} + \frac{C}{x} $$ $$
 * $$\displaystyle (Eq. 6)
 * }
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Solution
Integrate equation 6
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\int \frac{h_y}{h} dy = - \int g(y) dy $$ $$ This yields
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7)
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\ln h = - \int^y g(y) dy $$ $$ The final equation is
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 8)
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 * $$ \displaystyle
 * $$ \displaystyle

h = e^{- \int^y g(y) dy} $$ $$
 * $$\displaystyle (Eq. 9)
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